Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 8.5 m/s, the marine biologist measures the number of clicks occuring per second to be at a frequency of 2770 Hz.
1. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?

Answers

Answer 1

From Doppler effect we have that the frequency observed for the relation between the velocities is equivalent to the frequency observed. That is mathematically,

[tex]F_r = \frac{v}{v+v_s}F_s[/tex]

Here,

Speed of sound in water [tex]v = 1522m/s[/tex]

The Dolphin swims directly away from the observer with a velocity [tex]v_s = 8.5m/s[/tex]

Observed frequency of the clicks produced by the dolphin [tex]F_r = 2770Hz[/tex]

Replacing we have,

[tex]F_r = \frac{v}{v+v_s}F_s[/tex]

[tex]F_s = \frac{v+v_s}{v}[/tex]

[tex]F_s = 2770 (\frac{1522}{1522+8.5})[/tex]

[tex]F_s = 2754.61Hz[/tex]

Therefore the frequency emitted by the dolphin is 2754.61Hz


Related Questions

How much heat (in J) must be added to raise the temperature of 2.70 mol of air from 22.0°C to 32.0°C at constant volume? Assume air is completely diatomic.

Answers

Answer:

Heat required to raise the temperature will be 563.625 J

Explanation:

We have given number of moles n = 2.70 mole

Temperature is raises from 22°C to 32°C

So increase in temperature [tex]\Delta T=32-22=10^{\circ}C[/tex]

It is given that air is diatomic so [tex]c_v=\frac{5}{2}R=2.5\times 8.314=20.875[/tex]

We know that heat is given by [tex]Q=nc_v\Delta T[/tex]

So heat will be equal to [tex]Q=2.70\times 20.875\times 10=563.625J[/tex]

So heat required to raise the temperature will be 563.625 J

The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answers

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

The downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN is 225N.

Given that,

The small piston of a hydraulic lift has a cross-sectional of 3.00 cm^2.And its large piston has a cross-sectional area of 200 cm^2.

Based on the above information, the calculation is as follows:

Here we use the Pascal's principle,

[tex]F\div A = f\div a[/tex] ........... (1)

Here

 F denoted Force exerted on the larger piston,

f denoted force that applied to the smaller piston,

A denoted cross-sectional area of the larger piston,

And, a denoted cross-sectional area of the smaller piston.

Now

[tex]f = 15000 \times (3\div 200)[/tex]

=  225 N

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An object whose mass is 120 kg is located 20 m above a datum level in a location where standard gravitational acceleration exists. Determine the total potential energy, in kJ, of this object.

Answers

Answer:

U= 23.544 KJ

Explanation:

Given that

mass ,m = 120 kg

Height above the datum ,h = 20 m

Take g = 9.81 m/s²

The potential energy U is given as

U= m g h

m=mass

h=Height above the datum

g=Gravitational constant

Now by putting the value in the above values we get

U= 120 x 9.81 x 20 J

U=23544 J

Potential energy in KJ

U= 23.544 KJ

Therefore the answer will be 23.544 KJ.

At an oceanside nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water that is discharged back into the ocean. The amount that the water temperature is raised has a uniform distribution over the interval from 10° to 25° C. Suppose that a temperature increase of more than 18° C is considered to be potentially dangerous to the environment. What is the probability that at any point in time, the temperature increase is potentially dangerous?

Answers

answer is 30. just took it

Foraging bees often move in straight lines away from and toward their hives. Suppose a bee starts at its hive and flies 500 m due east, then flies 430 m west, then 670 m east. How far is the bee from the hive

Answers

Answer:

The distance of the bee from the hive is 740 m.

Explanation:

Given that,

Bee starts fly 500 m due east, 430 m west and 670 m east.

The direction of the bee

500 m in positive direction

430 m in negative direction

670 m in positive direction

We need to calculate the net distance

Using formula of distance

[tex]D=500-430+670[/tex]

[tex]D=740\ m[/tex]

Hence, The distance of the bee from the hive is 740 m.

Why does the large number of hydrogen atoms in the universe suggest that other elements?

Answers

Answer:

Explanation:

The abundance of the chemical elements is a measure of the occurrence of the chemical elements relative to all other elements in a given environment. Abundance is measured in one of three ways: by the mass-fraction (the same as weight fraction); by the mole-fraction (fraction of atoms by numerical count, or sometimes fraction of molecules in gases); or by the volume-fraction. Volume-fraction is a common abundance measure in mixed gases such as planetary atmospheres, and is similar in value to molecular mole-fraction for gas mixtures at relatively low densities and pressures, and ideal gas mixtures. Most abundance values in this article are given as mass-fractions.

For example, the abundance of oxygen in pure water can be measured in two ways: the mass fraction is about 89%, because that is the fraction of water's mass which is oxygen. However, the mole-fraction is about 33% because only 1 atom of 3 in water, H2O, is oxygen. As another example, looking at the mass-fraction abundance of hydrogen and helium in both the Universe as a whole and in the atmospheres of gas-giant planets such as Jupiter, it is 74% for hydrogen and 23–25% for helium; while the (atomic) mole-fraction for hydrogen is 92%, and for helium is 8%, in these environments. Changing the given environment to Jupiter's outer atmosphere, where hydrogen is diatomic while helium is not, changes the molecular mole-fraction (fraction of total gas molecules), as well as the fraction of atmosphere by volume, of hydrogen to about 86%, and of helium to 13%.[Note 1]

The abundance of chemical elements in the universe is dominated by the large amounts of hydrogen and helium which were produced in the Big Bang. Remaining elements, making up only about 2% of the universe, were largely produced by supernovae and certain red giant stars. Lithium, beryllium and boron are rare because although they are produced by nuclear fusion, they are then destroyed by other reactions in the stars.[1][2] The elements from carbon to iron are relatively more abundant in the universe because of the ease of making them in supernova nucleosynthesis. Elements of higher atomic number than iron (element 26) become progressively rarer in the universe, because they increasingly absorb stellar energy in their production. Also, elements with even atomic numbers are generally more common than their neighbors in the periodic table, due to favorable energetics of formation.

The abundance of elements in the Sun and outer planets is similar to that in the universe. Due to solar heating, the elements of Earth and the inner rocky planets of the Solar System have undergone an additional depletion of volatile hydrogen, helium, neon, nitrogen, and carbon (which volatilizes as methane). The crust, mantle, and core of the Earth show evidence of chemical segregation plus some sequestration by density. Lighter silicates of aluminum are found in the crust, with more magnesium silicate in the mantle, while metallic iron and nickel compose the core. The abundance of elements in specialized environments, such as atmospheres, or oceans, or the human body, are primarily a product of chemical interactions with the medium in which they reside.

An object falls a distance h from rest. If it travels 0.560h in the last 1.00 s, find (a) the time and (b) the height of its fall.

Answers

Answer:

(a) t = 2.97s

(b) h = 43.3 m

Explanation:

Let t be the time it takes to fall a distance h, then t - 1 (s) is the time it takes to fall a distance of h - 0.56h = 0.44 h

For the ball to fall from rest a distance of h after time t

[tex]h = gt^2/2[/tex]

Also for the ball to fall from rest a distance of 0.44h after time (t-1)

[tex]0.44h = g(t-1)^2/2[/tex]

We can substitute the 1st equation into the 2nd one

[tex]0.44gt^2/2 = g(t-1)^2/2[/tex]

and divide both sides by g/2

[tex]0.44t^2 = (t-1)^2[/tex]

[tex]0.44t^2 = t^2 - 2t + 1[/tex]

[tex]0.56t^2 - 2t + 1 = 0[/tex]

[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

[tex]t= \frac{2\pm \sqrt{(-2)^2 - 4*(0.56)*(1)}}{2*(0.56)}[/tex]

[tex]t= \frac{2\pm1.33}{1.12}[/tex]

t = 2.97 or t = 0.6

Since t can only be > 1 s we will pick t = 2.97 s

(b) [tex]h = gt^2/2 = 43.3 m[/tex]

In the parts that follow select whether the number presented in statement A is greater than, less than, or equal to the number presented in statement B. Be sure to follow all of the rules concerning significant figures.Statement A: 2.567 km, to two significant figures.Statement B: 2.567 km, to three significant figuresso, statement A is ........... statement BStatement A: (2.567 km+ 3.146 km), to two significant figures.Statement B: (2.567 km, to two significant figures) + (3.146 km, to two significant figures).so, statement A is .............statement B

Answers

Answer:

a) Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Explanation:

a) Statement A : 2.567km to two significant figures.

2.567km 2. S.F = 2.6km

Statement B : 2.567km to three significant figures.

2.567km 3 S.F = 2.57km

Therefore 2.6km is greater than 2.57km.

Statement A is greater than statement B.

b) statement A: (2.567 km + 3.146km) to 2 S.F

(2.567km + 3.146km) = 5.713km to 2 S.F = 5.7km

Statement B : (2.567 km, to two significant figures) + (3.146 km, to two significant figures).

2.567km to 2 S.F = 2.6km

3.146km to 2 S.F = 3.1km

2.6km + 3.1km = 5.7km

Therefore 5.7km is equal to 5.7km

Statement A is equal to statement B

Final answer:

Considering significant figures, statement A: 2.567 km is less than Statement B: 2.567 km, since it rounds to 2.6 km. For the addition, Statement A (5.713 km rounded to 5.7 km) is equal to Statement B (adding 2.6 km and 3.1 km).

Explanation:

When we consider significant figures, statement A: 2.567 km round off to two significant figures would be 2.6 km and Statement B: 2.567 km is the same to three significant figures. Therefore, statement A is less than statement B because 2.6 km is less than 2.567 km. Similarly, for the next case, if we add 2.567 km and 3.146 km, it gives us 5.713 km, rounded off to two significant figures it becomes 5.7 km for statement A. Whereas statement B, 2.567 km rounded off becomes 2.6 km and 3.146 km becomes 3.1 km. Adding both gives 5.7 km. Hence, statement A is equal to statement B.

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A triangle with equal sides of length 14 cm has -2.5-nC charged objects at each corner. Determine the direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle.
(A) vertically down
(B) Vertically upward
(C) horizontally rightward
(D) horizontally leftward

Answers

Answer:

(B) Vertically upward

Explanation:

r = Side of triangle = 14 cm

q = Charge = -2.5 nC

Electrical force is given by

[tex]F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times (2.5\times 10^{-9})^2}{0.14^2}\\\Rightarrow F=2.86671\times 10^{-6}\ N[/tex]

For the top charge

Net force on both charges is given by

[tex]F_n=2Fcos\theta\\\Rightarrow F_n=2\times 2.86671\times 10^{-6}\times cos30\\\Rightarrow F_n=4.96529\times 10^{-6}\ N[/tex]

The net force acting on the top charge is [tex]4.96529\times 10^{-6}\ N[/tex]

Here the forces are symmetrical hence, the net force is along +y axis i.e., upward

Final answer:

The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle is vertically upward.

Explanation:

The direction of the electrical force exerted on the object at the top corner due to the two objects at the horizontal base of the triangle can be determined using Coulomb's Law. Since the objects at the base of the triangle have a negative charge and the object at the top corner has a negative charge as well, the force between them will be repulsive. As a result, the direction of the electrical force on the top object will be vertically upward.

A small object carrying a charge of -4.00 nC is acted upon by a downward force of 19.0 nN when placed at a certain point in an electric field.
A) What are the magnitude and direction of the electric field at the point in question? (Answer in N/C)

C) What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?(Answer in N)

Answers

Explanation:

Given that,

Charge acting on the object, [tex]q=-4\ nC=-4\times 10^{-9}\ C[/tex]

Force acting on the object, [tex]F=19\ nC=19\times 10^{-9}\ C[/tex] (in downward direction)

(a) The electric force acting in the electric field is given by :

[tex]F=qE[/tex]

E is the electric field

[tex]E=\dfrac{F}{q}[/tex]

[tex]E=\dfrac{19\times 10^{-9}\ N}{4\times 10^{-9}\ C}[/tex]

E = 4.75 N/C

The direction of electric field is as same as electric force. But it is negative charge. So, the direction of electric field is in upward direction.

(b) The charge on the proton is, [tex]q=1.6\times 10^{-19}\ C[/tex]

The force acting on the proton is :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 4.75[/tex]

[tex]F=7.6\times 10^{-19}\ N[/tex]

If the charge on the proton is positive, the force on the proton is in upward direction.

Hence, this is the required solution.

A bucket of water with a mass of 2.0 kg is attached to a rope that is wound around a cylinder. The cylinder has a mass of 4.0 kg and is mounted horizontally on frictionless bearings. The bucket is released from rest. (a) Find its speed after it has fallen through a distance of 1.50 m

Answers

Answer:

Explanation:

Given mass of bucket is [tex]m=2\ kg[/tex]

mass of cylinder [tex]M=4\ kg[/tex]

Suppose T is the tension in the rope

For bucket [tex]mg-T=ma[/tex]

where a=acceleration

For cylinder with Radius R

[tex]I\times \alpha =T\cdot R[/tex]

[tex]\frac{MR^2}{2}\times \frac{a}{R}=T\times R[/tex]

[tex]T=\frac{Ma}{2}[/tex]

[tex]a=\frac{mg}{m+0.5M}[/tex]

[tex]a=4.9\ m/s^2[/tex]

Using [tex]v^2-u^2=2a s[/tex] for bucket

v=final velocity

u=initial velocity

s=displacement

[tex]v^2-0=2\times 4.9\times 1.5[/tex]

[tex]v=\sqrt{14.7}[/tex]

[tex]v=3.83\ m/s[/tex]              

Final answer:

The speed of the bucket after falling a distance of 1.50 m is 3.83 m/s.

Explanation:

To find the speed of the bucket after falling a distance of 1.50 m, we can use the principle of conservation of energy. The initial potential energy of the bucket is given by the equation PE = mgh, where m is the mass of the bucket, g is the acceleration due to gravity, and h is the height from which the bucket is released. In this case, the initial potential energy is PE = (2.0 kg)(9.8 m/s^2)(1.50 m).

The final kinetic energy of the bucket is given by the equation KE = 0.5mv^2, where m is the mass of the bucket and v is the speed of the bucket. Since the bucket starts from rest, the initial kinetic energy is zero. Therefore, the final kinetic energy is equal to the initial potential energy. We can set up the equation (2.0 kg)(9.8 m/s^2)(1.50 m) = 0.5(2.0 kg)v^2 and solve for v.

Simplifying the equation, we get ([tex]29.4 kg*m^2/s^2[/tex]. Dividing both sides by (2.0 kg), we find v^2 = 14.7 m^2/s^2. Taking the square root of both sides, we find v = 3.83 m/s.

On a straight road with the +x axis chosen to point in the direction of motion, you drive for 5 hours at a constant 20 miles per hour, then in a few seconds, you speed up to 60 miles per hour and drive at this speed for 1 hour.
What was the x component of average velocity for the 6-hour period, using the fundamental definition of average velocity, which is the displacement divided by the time interval?

Answers

Answer:

v = 26.7 mph

Explanation:

During the first 5 hours, at a constant speed of 20 mph, we find the total displacement to be as follows:

Δx₁ = v₁*t₁ = 20 mph*5 h = 100 mi

Assuming we can neglect the displacement during the speeding up from 20 to 60 mph, we can find the the total displacement at 60 mph as follows:

Δx₂ = v₂*t₂ = 60 mph*1 h = 60 mi

So, the total displacement during all the trip wil be:

Δx = Δx₁ + Δx₂ = 100 mi + 60 mi = 160 mi

So we can find the the average velocity during the 6-hour period, applying the definition of average velocity, as follows:

v = Δx / Δt = 160 mi / 6 h = 26.7 mph

A uniform solid disk with a mass of 24.3 kg and a radius of 0.314 m is free to rotate about a frictionless axle. Forces of 90 and 125 N are applied to the disk in the same horizontal direction but one is applied to the top and the other is applied to the bottom. What is the angular acceleration of the disk (in rad/s2)?

Answers

Answer:

α = 9.18 rad/s²

Explanation:

given,

mass of the solid disk = 24.3 Kg

radius of the disk = 0.314 m

Force, F₁ = 90 N

           F₂ = 125 N

net force acting on the disk

F = 125 - 90

F = 35 N

Torque

τ = F . r

τ = 35 x 0.314

τ = 11 N.m

we know that

τ = I α

moment of inertia of the solid disk

[tex]I = \dfrac{1}{2}MR^2[/tex]

[tex]I = \dfrac{1}{2}\times 24.3\times 0.314^2[/tex]

   I = 1.198 kg.m²

now,

11 = 1.198 x α

α = 9.18 rad/s²

the angular acceleration of the disk is equal to 9.18 rad/s²

Final answer:

To find the angular acceleration of the disk, use the formula: angular acceleration = (net torque) / (moment of inertia). Calculating the torques exerted by the forces and the moment of inertia will allow us to find the answer.

Explanation:

The angular acceleration of the disk can be found using the formula:

angular acceleration = (net torque) / (moment of inertia)

where the moment of inertia of a disk is given by:

moment of inertia = (1/2) * mass * radius^2

In this case, the net torque on the disk is the difference between the torques exerted by the two forces:

net torque = torque(top force) - torque(bottom force)

Each torque can be calculated using the formula:

torque = force * radius

Substituting the given values into these formulas, we can find the angular acceleration of the disk.

So, the angular acceleration of the disk is approximately 18.26 rad/s^{2}

.

Consider two charges placed a fixed distance apart.If the charge on each of two small spheres is halved, the force of attraction between the spheres will be ___________.
A. doubled.
B. quadrupled.
C. halved.
D. quartered.
E. the same as before.

Answers

Final answer:

When the charge on each of two small spheres is halved, the force of attraction between them, according to Coulomb's Law, will be quartered (Option D). This is because the force is proportional to the product of the charges, which would be reduced to a quarter.

Explanation:

The question is related to Coulomb's Law, which describes the electrostatic force between two charges. According to Coulomb's Law, the force F between two charges is proportional to the product of the charges (q1 and q2) and inversely proportional to the square of the distance (r^2) between them: F = k * (q1*q2) / r^2, where k is Coulomb's constant.

If the charge on each of the two small spheres is halved, we have new charges q1/2 and q2/2. Substituting these into the formula, we get the new force F' as F' = k * ((q1/2)*(q2/2)) / r^2. Simplifying this, F' = (1/4) * k * (q1*q2) / r^2, which is one quarter of the original force F. Therefore, the force of attraction between the spheres will be quartered when the charge on each sphere is halved.

What experimental evidence supports the idea that conducting materials have freely moving electrically charged particles inside them

Answers

Answer:

Layer of glass rod rubbed with silk.

Explanation:

Some of the atoms in the surface layer of a glass rod positively charged by rubbing it with a silk cloth have lost electrons, leaving a net positive charge because of the unneutralized protons of their nuclei. A negatively charged object has an excess of electrons on its surface.

Final answer:

Conducting materials like metals have freely moving electrons that enable the flow of charge, supported by experimental evidence like studies on conductivity and direct experiments confirming the presence of electrons as charge carriers.

Explanation:

Conducting materials such as metals have freely moving electrically charged particles within them. These free electrons can move through the material allowing the flow of charge. Experimental evidence supporting this idea includes studies on conductivity, Ohm's law for electromagnetics, and direct experiments like the one by Tolman and Stewart confirming that charge carriers in metals are indeed electrons.

If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is the increase in the internal energy of the gas?

Answers

Answer:

    The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275  J

W= - 125 J

W' = 50 J

W(net)= -125  + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the  internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

The internal energy of the gas is the energy contained by the gas. The increase in the internal energy of the gas in the given container is 350 J.

The internal energy of the gas can be calculated by the formula

Q= ΔU + W(net)

Where,

Q - energy absorbed by the system = 275  J

ΔU - internal energy of the gas= ?

W(net)=  W - W' = -125  + 50 = -75 J

Put the values in the formula,

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

Therefore, the increase in the internal energy of the gas in the given container is 350 J.

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When gravitational, magnetic and any forces other than static electric forces are not present, electric field lines in the space surrounding a charge distribution show

a. the directions of the forces that exist in space at all times.
b. Only the directions in which static charges would accelerate when a points on those lines.
c. only the directions in which moving charges would accelerate when at points on those lines.
d. tangents to the directions in which either static or moving charges would accelerate when passing through points on those lines.
e. the paths static or moving charges would take.

Answers

A very useful way to graphically schematize a field is to draw lines that go in the same direction as that field at several points. This is done through the electric field lines, which are imaginary lines that describe, if any, changes in the direction of the electric field when passing from one point to another, so that these lines are tangent, in each point of the space where the electric field is defined, to the direction of the electric field at that point.

According to Newton's first law, the force acting on a particle produces a change in its velocity; therefore, the movement of a charged particle in a region will depend on the forces acting on it at each point in that region.

Therefore the electric field lines in space surrounding a charge distribution will show tangents to the directions in which either static or moving charges would accelerate when passing through points on those lines.

The correct answer is D.

In Houston, Earth's B→ field has a magnitude of 5.2 × 10−5 Tand points in a direction 57∘ below a horizontal line pointing north.Part ADetermine the magnitude of the magnetic force exerted by the magnetic field on a 11-m-long vertical wire carrying a 11-A current straight upward. Express your answer to two significant figures and include the appropriate units.Part BDetermine the direction of the magnetic force.a. to the northb. to the eastc. to the southd. to the west

Answers

Answer:

F = 0.0034 N

Explanation:

Given:

[tex]B = 5.2*10^(-5) T\\Q = 57 degrees\\I_{wire} = 12 A\\L_{wire} = 10 m[/tex]

The angle between B and wire = 90 - 57 = 33 degrees

Using formula:

[tex]F = B*I*L*sin (90-Q)\\F = (5.2*10^(-5)*(12)*(10)*sin (33)\\F = 0.0034 N[/tex]

(A) The magnetic force exerted on the wire is 3.4×10⁻³N

(B) The direction of the force is to the west.

Magnetic force:

Given that the magnetic field B = [tex]5.2\times10^{-5}T[/tex] which points in the direction  57° below a horizontal line pointing north.

Length of the wire L = 11m

current in the wire I = 11A

The angle between the wire and the magnetic field is θ = (90-57) = 33°

(A) The magnetic force on a finite wire of length L carrying a current I is given by:

[tex]F=BILsin\theta\\\\F=5.2\times10^{-5}\times11\times11\sin33\\\\F=3.4\times10^{-3}N[/tex]

(B) The direction of the force is given by dl×B, now B is at 57° with the north direction and the wire is verticle, so the direction of the field will be to the west.

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Estimate how long it would take one person to mow a football field using an ordinary home lawn mower. Suppose that the mower moves with a 1 km/h speed, has a 0.5-m width, and a field is 360 ft long and 160 ft wide, 1 m-3.281 ft. (Figure 1) O 110 hours O 11 hours O 110 minutes O 11 minutes

Answers

Answer:

option B

Explanation:

given,

Length of field = 360 ft

                        = 360/3.281 = 109.72 m    ∵ 1 m = 3.281 ft

width of field = 160 ft

                      = 160/ 3.281 = 48.76 m

width of mower = 0.5 m

number of rounds required

   = [tex]\dfrac{48.76}{0.5}[/tex]

   = 97.5

we know,  1 km/h = 0.278 m/s

time taken for each round is equal to

[tex]t = \dfrac{length}{speed}[/tex]

[tex]t = \dfrac{109.72}{0.278}[/tex]

     t = 394.67 s

total time,

T = 394.67 x 94.5 = 37296.91 s

[tex]T = \dfrac{37296.91}{3600}\ hours[/tex]

[tex]T =11\ hours[/tex]

Hence. the correct answer is option B

A highway patrol car traveling a constant speed of 105 km/h is passed by a speeding car traveling 140 km/h. Exactly 1.00 s after the car passes, the highway patrol officer steps on the accelerator; of the patrolman's car accelerates at 3.50 m/s2, how much time passes after the car passes before the patrol car overtakes the speeder (assume the speeder is moving at constant speed)?

Answers

Answer:

The elapsed time from when the speeder passes the patrol car until it is caught is 9.24 s.

Explanation:

Hi there!

The position of the patrol car at a time "t" can be calculated using this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the patrol car at a time "t"

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

For the speeding car, the equation is the same only that the acceleration is zero. Then, the equation gets reduced to this:

x = x0 + v · t

Where "v" is the constant velocity.

First, let´s convert the velocity units into m/s:

140 km/h · 1000 m / 1 km · 1 h / 3600 s = 38.9 m/s

105 km/h · 1000 m / 1 km · 1 h / 3600 s = 29.2 m/s

We have to find how much time it takes the patrol car to catch the speeder after the speeder passes the patrol car.

When the patrol car catches the speeder, the position of both cars is the same:

position of the patrol car = position of the speeder

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

if we place the origin of the frame of reference at the point where the patrol car starts accelerating (1 s after the speeder passes the patrol car) then, the initial position of the patrol car will be zero, while the initial position of the speeder will be the traveled distance in 1 s:

x = v · t

x = 38.9 m/s · 1 s = 38.9 m

When the patrol car accelerates, the speeder is 38.9 m ahead of it. Then:

x0 + v0 · t + 1/2 · a · t² = x0 + v · t

0 + 29.2 m/s · t + 1/2 · 3.50 m/s² · t² = 38.9 m + 38.9 m/s · t

Let´s agrupate terms and equalize to zero:

-38.9 m - 38.9 m/s · t + 29.2 m/s · t + 1.75 m/s² · t² = 0

-38.9 m - 9.70 m/s · t + 1.75 m/s² · t² = 0

Solving the quadratic equation for t using the quadratic formula:

t = 8.24 s  (the other solution is discarded because it is negative)

The elapsed time from when the speeder passes the patrol car until it is caught is (8.24 s + 1.00) 9.24 s.

The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. The distance between the Earth and the Sun is 1.5 x 1011 m. a) Assuming it radiates uniformly in all directions what is the total power output of the Sun?b) If the frequency increases by 1 MHz what would be the relative (percentage) change in the power output? c) For frequency in b) what is the intensity of the radiation from the Sun measured on Mars? Note that Mars is 60% farther from the Sun than the Earth is.

Answers

Answer: (a) power output = 3.85×10²⁶W

(b). There is no relative change in power as it is independent from frequency

(c). 590 W/m²

Explanation:

given Radius between earth and sun to be = 1.50 × 10¹¹m

Intensity of the radiation from the sun measured on earth to be = 1360 W/m²

Frequency = 60 MHz

(a). surface area A of the sun on earth is = 4πR²

substituting value of R;

A = 4π(.50 × 10¹¹)² = 2.863 10²³×m²

A = 2.863 10²³×m²

now to get the power output of the sun we have;

P sun = I sun-earth A sun-earth

where A = 2.863 10²³×m², and I is 1360 W/m²

P sun =  2.863 10²³ × 1360

P sun = 3.85×10²⁶W

(c). surface area A of the sun on mars is = 4πR²

now we substitute value of 2.28 ×10¹¹ for R sun-mars, we have

A sun-mars = 4π(2.28× 10¹¹)²

A sun-mars = 6.53 × 10²³m²

now to calculate the intensity of the sun;

I sun-mars = P sun / A sun-mars

where P sun = 3.85×10²⁶W and A sun-mars = 6.53 × 10²³m²

I sun-mars =  3.85×10²⁶W / 6.53 × 10²³m²

I sun-mars = 589.6 ≈ 590 W/m²

I sun-mars = 590 W/m²

20 J of work is done per cycle on a refrigerator with a coefficient of performance of 4.0.
Part A: How much heat is extracted from the cold reservoir per cycle?
Part B: How much heat is exhausted to the hot reservoir per cycle?

Answers

Answer:

A)Qa=80 J

B)Qr= 100 J

Explanation:

Given that

W= 20 J

COP = 4

Heat rejected from cold reservoir = Qa

Heat exhausted to hot reservoir = Qr

The COP of refrigerator is given as

[tex]COP=\dfrac{Qa}{W}[/tex]

[tex]4=\dfrac{Qa}{20}[/tex]

Qa= 4 x 20 J

Qa=80 J

By using first law of refrigerator

Qr= Qa + W

Qr= 80 + 20 J

Qr= 100 J

A)Qa=80 J

B)Qr= 100 J

Final answer:

The heat extracted from the cold reservoir per cycle is 80 J and the heat exhausted to the hot reservoir per cycle is 100 J.

Explanation:

The equation for the coefficient of performance (COP) of a refrigerator is COP = Q_c / W, where Q_c is the heat extracted from the cold reservoir and W is the work done. Given the COP and the work done, we can rearrange this equation to find Q_c: Q_c = COP * W. Substituting the given values: Q_c = 4.0 * 20 J = 80 J.

The heat exhausted to the hot reservoir (Q_h) for a refrigerator can be found using the equation: Q_h = W + Q_c. Substituting the values we found: Q_h = 20 J + 80 J = 100 J.

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You are pushing a crate horizontally with 100 N across a 10 m factory floor. If the force of friction on the crate is a steady 70 N, how much kinetic energy is gained by the crate

Answers

Answer:

K E = 300 J

Explanation:

given,

Force of pushing crate, F = 100 N

distance of push = 10 m

frictional force = 70 N

kinetic energy is gained by the crate = ?

using work energy theorem

K E = work done

K E  = F . s

K E  = (100 - 70) x 10

K E = 30 x 10

K E = 300 J

kinetic energy is gained by the crate is equal to 300 J.

 

Final answer:

The kinetic energy gained by the crate is 300 J, calculated as the net work done on the crate, which is the work done by the applied force (1000 J) minus the work done by friction (700 J).

Explanation:

To determine how much kinetic energy is gained by the crate, we can use the work-energy theorem which states that the work done on an object is equal to the change in kinetic energy of the object. In this scenario, the net work done on the crate is the work done by the applied force minus the work done by friction. The work done by the applied force is force times distance, which is 100 N x 10 m = 1000 J. The work done by friction is 70 N x 10 m = 700 J. Therefore, the net work done on the crate, which is also the kinetic energy gained by the crate, is 1000 J - 700 J = 300 J.

Two blocks of ice, one four times as heavy as theother, are at rest on a frozen lake. A person pushes each block thesame distance d.Ignore friction and assume that an equal force F_vec is exerted on each block.
this is what i have sofar:
PartA
Which of the following statements istrue about the kinetic energy of the heavier block after thepush?
It is equalto the kinetic energy of the lighter block.
PartB
Compared to the speed of the heavierblock, how fast does the light block travel?
twice asfast
This is what i cant figure out:
PartC
Now assume that both blocks have thesame speed after being pushed with the same force F_vec. What can be said about the distances the two blocks arepushed?
a The heavyblock must be pushed 16 times farther than the lightblock.
b The heavyblock must be pushed 4 times farther than the lightblock.
c The heavyblock must be pushed 2 times farther than the lightblock.
d The heavytblock must be pushed the same distance as the lightblock.
e The heavyblock must be pushed half as far as the light block.

Answers

Answer:

A. [tex]KE_h=\frac{1}{16}\times KE_l[/tex]

B. the lighter block travels at a speed 4 times faster than the heavier block.

C. b The heavy block must be pushed 4 times farther than the light block.

Explanation:

mass of lighter block, [tex]m[/tex]mass of the heavier block, [tex]4m[/tex]

Given that the blocks are  acted upon by equal force.

A.

Then the kinetic energy of the blocks depends on their individual velocity.

And velocity is related to momentum through Newton's second law of motion:

[tex]\frac{d}{dt} .p=F[/tex]

[tex]\frac{d}{dt} (m.v_l)=F[/tex]

considering that the time for which the force acts on each mass is equal.

[tex]dv_l=\frac{F.dt}{m}[/tex]

For the heavier block:

[tex]dv_{_h} =\frac{F.dt}{4m}[/tex]

Therefore:

Kinetic energy of lighter block:

[tex]KE_l=\frac{1}{2}\times m.(\frac{F.dt}{m} )^2[/tex]

[tex]KE_l=\frac{1}{2m} \times (F.dt)^2[/tex]

Kinetic energy of heavier block:

[tex]KE_h=\frac{1}{2} \times m.(\frac{F.dt}{4m} )^2[/tex]

[tex]KE_h=\frac{1}{16}\times (\frac{1}{2m} \times (F.dt)^2)[/tex]

[tex]KE_h=\frac{1}{16}\times KE_l[/tex]

B.

From the above calculations and assumptions we observe that the lighter block travels at a speed 4 times faster than the heavier block.

C.

Since the lighter block is having the speed 4 times more than the heavier block so it must be pushed 4 times farther because the speed is directly proportional to the  distance.

A) The statement that is true from the attached link about the kinetic energy of the heavier block after the push is;

B: the kinetic energy of the heavier block is equal to the kinetic energy of the lighter block

B) Compared to the speed of the heavierblock, the speed of the light block travel is; twice as fast.

C) If we assume that both blocks have the same speed after being pushed with the same force F_vec, what we can say about the distances the two blocks are pushed is;

The heavyblock must be pushed 4 times farther than the lightblock.

A) The formula for the work done on each block is gotten from the formula;

W = F × d

We are told that the two blocks were pushed the same distance and with the same force F_vec being exerted on them. Thus, the work done on each of the blocks will be the same.

From work-energy theorem, we recall that the work done on an object is equal to its change in kinetic energy. Thus;

W = ΔKE

Where;

ΔKE = Final KE - Initial KE

Thus;

W = Final KE - Initial KE

Since the blocks were initially at rest, then it means that;

Initial KE = 0

Thus;

W = Final KE - 0

W = Final KE

The work done on each block is the same and as a result their final kinetic energies will be the same.

B) We are told that one of the blocks is ¼ times as heavy as the other block.

Thus;

½(m)v_h² = ½(¼mv_l²)

v_h² = ¼v_l²

Where;

v_h is speed of heavy block

v_l is speed of light block

Thus, taking square root of both sides gives;

v_h = ½v_l

Thus, the speed of the light block is twice as fast.

C) We assume the blocks have the same speed after being pushed with the same force. Thus;

It means that heavier block must be pushed four times farther than lighter block.

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A major leaguer hits a baseball so that it leaves the bat at a speed of 31.3 m/s and at an angle of 36.7 ∘ above the horizontal. You can ignore air resistance.

A) At what two times is the baseball at a height of 9.00 m above the point at which it left the bat?
t1,2 = _____ s
B) Calculate the horizontal component of the baseball's velocity at each of the two times you found in part (a).
vh1,2 = _____ m/s
C) Calculate the vertical component of the baseball's velocity at each of the two times you found in part (a).
v v1,2 = _____ m/s
D) What is the magnitude of the baseball's velocity when it returns to the level at which it left the bat?
v = ____ m/s
E) What is the direction of the baseball's velocity when it returns to the level at which it left the bat?
= ______ degrees below the horizontal

Answers

Answer:

A) t₁ = 0.56 s t₂ =3.26 s

B) vh₁=vh₂ = 25.1 m/s

C) v₁ = 13.2 m/s  v₂ = -13.2 m/s

D) v = 31. 3 m/s

E) 36.7º below horizontal.  

Explanation:

A) As the only acceleration of the baseball is due to gravity, as it is constant, we can apply the kinematic equations in order to get times.

First, we can get the horizontal and vertical components of the velocity, as the movements  along these directions are independent each other.

v₀ₓ = v* cos 36.7º = 31.3 m/s * cos 36.7º = 25.1 m/s

v₀y = v* sin 36.7º =  31.3 m/s * sin 36.7º = 18.7 m/s

As in the horizontal direction, movement is at constant speed, the time, at any point of the trajectory, is defined by the vertical direction.

We can apply to this direction the kinematic equation that relates the displacement, the initial velocity and time, as follows:

Δy = v₀y*t -1/2*g*t²

We can replace Δy, v₀y and g for the values given, solving a quadratic equation for t, as follows:

4.9*t²-18.7t + 9 = 0

The two solutions for  t, are just the two times at which the baseball is at a height of 9.00 m above the point at which it left the bat:

t = 1.91 sec +/- 1.35 sec.

t₁ = 0.56 sec   t₂= 3.26 sec.

B) As we have already told, in the horizontal direction (as gravity is always downward) the movement is along a straight path, at a constant speed, equal to the x component of the initial velocity.

⇒ vₓ = v₀ₓ = 25.1 m/s

C) In order to get the value of  the vertical components at the two times that we have just found, we can apply the definition of acceleration (g in this case), solving for vfy, as follows:

vf1 = v₀y - g*t₁ = 18.7 m/s - (9.8m/s²*0.56 sec) = 13.2 m/s

vf₂ = v₀y -g*t₂ = 18.7 m/s - (9.8 m/s²*3.26 sec) = -13.2 m/s

D) In order to get the magnitude of  the baseball's velocity when it returns to the level at which it left the bat, we need to know the value of the vertical component at this time.

We could do in different ways, but the easiest way is using the following kinematic equation:

vfy² - v₀y² = 2*g*Δh

If we take the upward path, we know that at the highest point, the baseball will come momentarily to an stop, so at this point, vfy = 0

We can solve for Δh, as follows:

Δh = v₀y² / (2*g) = (18.7m/s)² / 2*9.8 m/s² = 17.8 m

Now, we can use the same equation, for the downward part, knowing that after reaching to the highest point, the baseball will start to fall, starting from rest:

vfy² = 2*g*(-Δh) ⇒ vfy = -√2*g*Δh = -√348.9 = -18. 7 m/s

The horizontal component is the same horizontal component of the initial velocity:

vx = 25.1 m/s

We can get the magnitude of the baseball's velocity when it returns to the level at which it left the bat, just applying Pythagorean Theorem, as follows:

v = √(vx)² +(vfy)² = 31.3 m/s

E) The direction below horizontal of the velocity vector, is given by the tangent of the angle with the horizontal, that can be obtained as follows:

tg Ф = vfy/ vx = -18.7 / 25. 1 =- 0.745

⇒ Ф = tg⁻¹ (-0.745) = -36.7º

The minus sign tell us that the velocity vector is at a 36.7º angle below the horizontal.

Final answer:

A major leaguer hits a baseball and the question asks for the time, velocity components, velocity magnitude, and direction at different points of the ball's trajectory when it reaches a certain height.

Explanation:

A) Let's consider the vertical motion of the baseball. The equation to determine the time it takes for an object to reach a certain height is:

h = v0t + 0.5gt2

Where:

h is the height = 9.00 mv0 is the initial vertical velocity = 31.3 m/s * sin(36.7º)g is the acceleration due to gravity = 9.8 m/s2

By rearranging the equation, we can solve for t:

t2 + (2v0/g)t - (2h/g) = 0

Using the quadratic formula, we can plug in the values and solve for t. The two possible values of t will correspond to the two times when the ball is at a height of 9.00 m above the point at which it left the bat.

B) The horizontal component of velocity remains constant throughout the motion. Therefore, the horizontal component of velocity at each of the two times found in part (a) will be the same as the initial horizontal velocity, which is 31.3 m/s * cos(36.7º).

C) The vertical component of velocity changes due to the acceleration of gravity. At each of the two times found in part (a), the vertical component of velocity can be found using the equation:

vv1,2 = v0 + gt

Where:

v0 is the initial vertical velocity = 31.3 m/s * sin(36.7º)g is the acceleration due to gravity = 9.8 m/s2t is the time at each height, which were found in part (a)

D) When the baseball returns to the level at which it left the bat, its vertical velocity will be -v0, which means it will be moving downward with the same magnitude as the initial velocity but in the opposite direction. The horizontal component of velocity remains unchanged. Therefore, the magnitude of the baseball's velocity when it returns to the level at which it left the bat will be the square root of the sum of the squares of the horizontal and vertical components, which can be calculated using the formula:

v = √(vh2 + vv2)

Where:

vh is the horizontal component of velocity, which was found in part (b)vv is the vertical component of velocity when the ball returns to the level of the bat, which is -v0

E) The direction of the baseball's velocity when it returns to the level at which it left the bat can be determined using trigonometry. The angle below the horizontal can be found as:

θ = tan-1(vv/vh)

Where:

vv is the vertical component of velocity when the ball returns to the level of the bat, which is -v0vh is the horizontal component of velocity, which was found in part (b)

Two rockets are launched at a fireworks display. Rocket A is launched with an initial velocity v0 5 100 m/s and rocket B is launched t1 seconds later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 300 m as A is falling and B is rising. Assuming a constant acceleration g 5 9.81 m/s2, determine (a) the time t_1, (b) the velocity of B relative to A at the time of the explosion.

Answers

Answer

given,

initial speed of the rocket A = 100 m/s

height of explode = 300 m

acceleration due to gravity = 9.8 m/s²

rocket b is launched after t₁ time

now, using equation of motion to calculate time

[tex]s = ut + \dfrac{1}{2}gt^2[/tex]

[tex]300 = 100t + \dfrac{1}{2}(-9.8)t^2[/tex]

 4.9 t² - 100 t + 300 = 0

using quadratic equation

[tex]t = \dfrac{-(-100)\pm \sqrt{100^2-4\times 4.9 \times 300}}{2\times 4.9}[/tex]

t₁ = 3.65 s   and  t₂ = 16.75 s

now, rocket A reaches 300 m on return at 16.75 s

rocket B reaches 300 m after 3.65 s

time difference of launch:

t = 16.75 - 3.65

t = 13.1 s

velocity of rocket A

v_a = u + g t

v_a =100 - 9.8 x 16.75

v_a = -64.15 m/s

velocity of rocket B

v_b = u + g t

v_b =100 - 9.8 x 3.65

v_b =+64.23 m/s

relative velocity of B relative to A at the time of the explosion

[tex]V_{BA} = v_b - V_a[/tex]

[tex]V_{BA} = 64.23 -(-64.15)[/tex]

[tex]V_{BA} = 128.38\ m/s[/tex]

relative velocity is equal to 128.38 m/s

One ball is dropped vertically from a window. At the same instant, a second ball is thrown horizontally from the same window.
1. Which ball has the greater speed at ground level?

Answers

Answer:

The second ball

Explanation:

Both balls are under the effect of gravity, accelerating with exactly the same value. The first ball is dropped, therefore its initial velocity is zero. Since the second ball has horizontal and vertical velocity components, its initial velocity is given by:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]

The vertical component is zero, however, it has a horizontal velocity, so its initial speed is not zero, therefore the secong ball has the greater speed at ground level.

Final answer:

The second ball will reach a greater total speed at ground level due to its initial horizontal speed added to its vertical speed. But, both balls will hit the ground at the same time if the horizontal speed of the second ball is not too high.

Explanation:

In the described situation, gravity is the only vertical force acting on both balls, hence they will reach the ground at the same speed in the vertical direction. But, the second ball has additional horizontal speed. Therefore, with this horizontal component added to the vertical detachment speed due to gravity, the second ball will have a greater total speed by the Pythagorean theorem: (total speed)^2 = (vertical speed)^2 + (horizontal speed)^2. However, the question of which hits the ground first depends on the initial horizontal speed of the second ball. If the horizontal speed is not too high, both balls should hit the ground at the same time.

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If a negative charge is initially at rest in an electric field, will it move toward a region of higher potential or lower potential?What about a positive charge? How does the potential energy of the charge change in each instance? Explain.

Answers

Answer with explanation :

The negative sign means that the potential energy decreases by the movement of the electron.

negative charge at rest in an electric field moves toward the region of an electric field , so that its potential energy will diminish and change into the kinetic energy of motion. The total energy remains constant.

Positive charges will move downhill because of convention. It is to stay in accordance with other potential theories, particularly gravity, where the "charge" is mass, that moves downwards in the gravitational potential field expressed by ϕ(r)=−GM|r|ϕ(r)=−GM|r|. In an electronic system, howbeit, positive charges are fixed in their position within a component (e.g., a wire), therefore instead of the mobile,the negative charges, electrons, move uphill.

Final answer:

A negative charge at rest in an electric field moves toward a region of higher potential, thus decreasing its electric potential energy. A positive charge moves toward a region of lower potential, similarly decreasing its electric potential energy. The strength of the electric field relates to potential difference and is zero where electric potential is constant.

Explanation:

If a negative charge is initially at rest in an electric field, it will move toward a region of higher potential. This is because negative charges are attracted to areas of positive potential and repelled from areas of negative potential. Conversely, a positive charge will move toward a region of lower potential, much like how fluid moves from a region of high pressure to low pressure.

The potential energy of a charge in an electric field depends on both the electric potential and the sign and magnitude of the charge. For a negative charge, moving to a higher potential region decreases its electric potential energy, while a positive charge moving to a lower potential region also decreases its electric potential energy. The electric field is defined as the force per unit charge, and its strength is directly related to the potential difference. If the electric potential is constant in a region, the electric field strength is zero there.

When discussing the potential difference and electric field strength, a good example is a parallel plate capacitor which has a uniform electric field. If an electron, being negatively charged, is released at rest between the plates, it will accelerate toward the positively charged plate due to the attractive electrostatic force, thereby moving to a region of higher potential.

A student said, "The displacement between my dorm and the lecture hall is 1 kilometer."
Is he using the correct physical quantity for the information provided? What should he have called the 1 kilometer?

a) Distance
b) Path length
c) Position
d) Both a and b are correct.

Answers

Answer:

d) Both a and b are correct

Explanation:

Displacement: It is defined as the distance between initial and final position during motion.

Distance: It is defined as the total path length traveled by object

Or

It is the distance of one place from other place.

Student said that the displacement between my dorm and the lecture hall is 1km.

It is not displacement .It is distance or we say path length.

Therefore, he is using  incorrect physical quantity for the information provided.

He should have called distance 1 km or path length 1 km.

Option d is true.

Final answer:

The student is accurately using the term 'displacement' to describe the 1 kilometer between their dorm and the lecture hall. Technically, 'distance' or 'path length' could also have been used assuming a straight-line path and ignoring direction.

Explanation:

In the context of this specific question, the student is accurately using the term displacement as a physical quantity. Displacement refers to the shortest distance between two points in a particular direction. In this case, the 1 kilometer refers to the displacement between the dorm and the lecture hall, assuming a straight-line path. So, technically, both a) Distance and b) Path length could also have been used to describe the 1 kilometer, if we don't consider the direction.

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The incoming radiation from the sun interacts with Earth's surface in different ways. Describe a surface that would have a high albedo and a surface that would have a low albedo.

Answers

Answer:

Albedo Alto: Snow

Albedo bass: ocean

Explanation:

Albedo is the fraction of solar radiation reflected by a surface. The term has its origin from the Latin word albus, which means "white." It is quantified as the proportion or percentage of solar radiation of all wavelengths reflected by a body or surface to the amount incident on it.

The color of the soil certainly affects the reflectivity, the lighter colors that have greater albedo than the dark colors and, therefore, present greater albedo. Soil texture is also a factor that affects albedo. Some studies have shown that snow-covered soils, being a light color, reflect most of the light and therefore do not absorb it and present the greatest amount of terrestrial surfaces. While with the ocean and seas most of the radiation is absorbed it has a dark color increasing with the depth of the place and therefore has the lowest surface albedo. In the end of the year the snow has 86% of reflected light while the oceans 8% of reflected light.

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