A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the speaker? What is the wavelength of this sound wave as it travels through air with a speed of about 341 m/s? What is the wavelength of the same sound wave as it enters some water, where it has a speed of about 1480 m/s?

Answer:

Fc₂ = 4.675 N

Explanation:

r₁ = 12 cm = 0.12 m

Fc= 1.7 N

Fc = mV²/r = m w² r   ( ∴  V= r w )

Fc = m w² r   ----------------------- (1)

As angular speed (w) and mass are constant.

let    m * w² = k   ------------------(2)

Put equation (2) in equation (1).

⇒  Fc = m w² r = k * r         ( ∴ m w² = k)

⇒  Fc =k * r     ----------------------( 3)

⇒  Fc₁ =k * r₁ = 1.7 N

⇒ k = 1.7/ r  = 1.7 / 0.12  N    ( as r₁=0.12 m )

⇒ k = 14.17

Centripetal force hen the rod is fixed at 33 cm from the axis:

From equation ( 3 )

Fc₂ =  k * r₂

Fc₂ = 14.17 * 0.33 N    ( ∴ r₂ = 0.33 m)

Fc₂ = 4.675 N

The force needed to push a block up an incline at a constant velocity is the sum of the gravitational force parallel to the incline and the frictional force, which is mg sin(θ) + μmg cos(θ). So the correct option is e.

The force required to push a block of mass m up an inclined plane at an angle θ with the horizon at a constant velocity, given the coefficient of friction between the plane and the block is μ, can be determined by analyzing the forces acting on the block. We need to consider both the component of the block's weight parallel to the incline and the frictional force.

Since the block is moving at a constant velocity, the net force along the incline must be zero. This implies that the applied force must counteract the combined forces of gravity pulling the block down the incline and the frictional force opposing the motion. The gravitational component parallel to the incline is mg sin(θ) and the frictional force is μmg cos(θ). Therefore, the required force F is:

F = mg sin(θ) + μmg cos(θ)

This corresponds to answer choice e.

To solve this problem we must use the perioricity equations given as a function of the mass and spring constant. Mathematically this can be expressed as:

[tex]T = 2\pi \sqrt{\frac{m}{k}}[/tex]

m = mass

k = Spring constant

Re-arrange to find the mass we have

[tex]m = \frac{T^2k}{4\pi^2 }[/tex]

Replacing with our values we have that

[tex]m = \frac{2.4^2*10}{4\pi^2}[/tex]

[tex]m = 1.459kg[/tex]

D) Mass is independent of acceleration due to gravity (as you can see at the equation previously given) for this reason the mass suspended on mars is given as the same found. Therefore the mass would be

m = 1.459kg

Final answer:

The mass suspended from the spring that would result in a period of 2.4 s is 0.142 kg.

Explanation:

To find the mass suspended from the spring that would result in a period of 2.4 s, we can use the formula for the period of oscillation:

T = 2π √(m/k)

Where T is the period, m is the mass, and k is the force constant of the spring.

Let's rearrange the formula to solve for m:

m = (T^2 · k) / (4π^2)

Substituting the given values:

m = (2.4^2 · 10) / (4π^2)

m = 0.142 kg

Answer:

2.2 m/s to the left

Explanation:

Momentum is conserved, so:

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

0 = (44 kg) (2.5 m/s) + (49 kg) v

v = -2.2 m/s

The canoe will move in the opposite direction of the woman with a velocity of approximately 2.24 m/s.

The velocity of the canoe after the woman jumps off can be determined by using the principle of conservation of momentum, which states that the total momentum before an event must equal the total momentum after the event when no external forces act on the system. In this scenario, the system consists of the woman and the canoe.

Before the woman jumps, the system is at rest, so its initial momentum is zero. When the woman jumps off the canoe to the right with a velocity of 2.5 m/s, by conservation of momentum, the canoe must move in the opposite direction to maintain the total momentum at zero.

The momentum possessed by the woman is given by the product of her mass and velocity (mw×vw).

Similarly, the momentum of the canoe is the product of its mass and its velocity in the opposite direction (mc×vc).

Momenta are equal in magnitude and opposite in direction:
mw ×vw = mc ×vc
Therefore, vc = (mw ×vw) / mc
Substituting the given values:

vc = (44 kg ×2.5 m/s) / 49 kg

= 2.24 m/s

The canoe will move to the left (opposite to the woman's direction) with a velocity of approximately 2.24 m/s.

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

[tex]W = U_f - U_i[/tex]

Where,

W = Work

[tex]U_f =[/tex] Final gravitational Potential energy

[tex]U_i =[/tex] Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

[tex]U_i = \frac{-GMm}{R}[/tex]

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is [tex]R_{orbit} \approx R[/tex] then the energy possessed by the spacecraft is

[tex]U_f = \frac{-GMm}{2R}[/tex]

Work needed to put it in orbit is the difference between the above two

[tex]W = U_f - U_i[/tex]

[tex]W = -GMm (\frac{1}{2R}-\frac{1}{R})[/tex]

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

[tex]W = \frac{GMm}{2R}[/tex]

Answer:

1.32 m

Explanation:

from the given figure we can find the velocity at the hole in the container

let it be v and we know that

then [tex]v= \sqrt{2gh}[/tex]

h= 15-4= 11 cm , g=9.81

[tex]v= \sqrt{2(9.81)(11)}[/tex]

v=14.69 m/s

now using

s=ut+0.5at^2

t= is the time required by water to reach bottom of the container

u = velocity in vertical direction is zero.

therefore,

[tex]0.04= 0\times t+\frac{1}{2}\times9.81\times t^2[/tex]

t= 0.09030 sec

let x be the distance far from the container will the water land

x=vt

x=14.69×0.09030 = 1.32 m

Answer:

4.192 N

Explanation:

Step 1: Identify the given parameters

Velocity of airflow = 45m/s

air temperature = 20⁰

plate length and width = 1m and 0.5m respectively.

Step 2: calculate drag force due to shear stress, [tex]F_{s}[/tex]

[tex]F_{s} = C_{f} \frac{1}{2} (\rho{U_{o}}WL)[/tex]

Note: The density and kinematic viscosity of air at 20⁰ at 1 atm, is 1.2 kg/m³ and 1.5 X 10⁻⁵ N.s/m²

⇒The Reynolds number ([tex]R_{eL}[/tex]) based on the length of the plate is

[tex]R_{eL} =\frac{VXL}{U}[/tex]

[tex]R_{eL} =\frac{45X1}{1.5 X 10^{-5}}[/tex]

[tex]R_{eL}[/tex] = 3 X10⁶  (flow is turbulent, Re ≥ 500,000)

⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "tripped" side of the plate is

[tex]C_{f} = \frac{0.074}{(R_{e})^\frac{1}{5}}[/tex]

[tex]C_{f} = \frac{0.074}{(3 X10^6)^\frac{1}{5}}[/tex]

[tex]C_{f}[/tex] = 0.0038

⇒The average shear stress coefficient ([tex]C_{f}[/tex]) on the "untripped" side of the plate is

[tex]C_{f} = \frac{0.523}{in^2(0.06XR_{e})} -\frac{1520}{R_{e}}[/tex]

[tex]C_{f} = \frac{0.523}{in^2(0.06 X 3X10^6)} -\frac{1520}{3X10^6}[/tex]

[tex]C_{f}[/tex] = 0.0031

The total drag force = [tex]\frac{1}{2}(1.2 X 45^2 X 1 X 0.5 (0.0038 +0.0031)[/tex]

The total drag force is 4.192 N

Answer:

77.08 C

Explanation:

[tex]m[/tex] = mass of the water = 500 g = 0.5 kg

[tex]c[/tex] = specific heat of water = 4186 J/(kg °C)

[tex]\Delta T[/tex] = Rate of change of temperature = 3 °C /min = (3/60 ) °C /s = 0.05  °C /s

[tex]k[/tex] = thermal conductivity of glass = 0.84

[tex]A[/tex] = Area of the element = 0.090 m²

[tex]t[/tex] = thickness of the element = 1.5 mm = 0.0015 m

[tex]T_{i}[/tex] = Temperature inside = 75 °C

[tex]T_{o}[/tex] = Temperature outside = ?

Using conservation of energy

Heat gained by water = Heat transferred through glass

[tex]m c \Delta T = \frac{kA(T_{o} - T_{i})}{t} \\(0.5) (4186 (0.05) = \frac{(0.84)(0.090)(T_{o} - 75)}{0.0015} \\104.65 = (50.4)(T_{o} - 75)\\T_{o} = 77.08 C[/tex]

Answer:

0.1040512455 N

[tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]

0.05925 N

[tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]

Explanation:

I = Current

B = Magnetic field

Separation between end points is

[tex]l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm[/tex]

Effective force is given by

[tex]F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N[/tex]

The force is 0.1040512455 N

[tex]tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}[/tex]

The angle the force makes is given by

[tex]\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]

The direction is [tex]36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction[/tex]

[tex]F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N[/tex]

The force is 0.05925 N

[tex]tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}[/tex]

[tex]\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]

The direction is [tex]29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction[/tex]

A. Magnitude of the force: [tex]\( 0.104 \, \text{N} \)[/tex]

B. Direction of the force: [tex]\( 323.8^\circ \)[/tex]

C. Magnitude of the force: [tex]\( 0.0593 \, \text{N} \)[/tex]

D. Direction of the force: [tex]\( 330^\circ \)[/tex]

Part A: Determine the magnitude of the magnetic force exerted by the external field on the wire

Given: Length of wire segments: [tex]\( L_1 = 40.0 \, \text{mm} = 0.040 \, \text{m} \) and \( L_2 = 55.0 \, \text{mm} = 0.055 \, \text{m} \)[/tex]

- Magnetic field, [tex]\( B = 0.300 \, \text{T} \)[/tex]

- Current, [tex]\( I = 5.10 \, \text{A} \)[/tex]

The force on a current-carrying wire in a magnetic field is given by [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \).[/tex]

For the first segment along the x-axis:

- [tex]\( \mathbf{L}_1 = 0.040 \, \text{m} \, \hat{i} \)[/tex]

[tex]\( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]

[tex]\[ \mathbf{F}_1 = I (\mathbf{L}_1 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.040 \, \hat{i} \times 0.300 \, \hat{k}) \][/tex]

[tex]\[ \mathbf{F}_1 = 5.10 \times 0.040 \times 0.300 \, \hat{j} \][/tex]

[tex]\[ \mathbf{F}_1 = 0.0612 \, \text{N} \, \hat{j} \][/tex]

For the second segment along the y-axis:

[tex]- \( \mathbf{L}_2 = 0.055 \, \text{m} \, \hat{j} \)[/tex]

[tex]- \( \mathbf{B} = 0.300 \, \text{T} \, \hat{k} \)[/tex]

[tex]\[ \mathbf{F}_2 = I (\mathbf{L}_2 \times \mathbf{B}) = 5.10 \, \text{A} \times (0.055 \, \hat{j} \times 0.300 \, \hat{k}) \][/tex]

[tex]\[ \mathbf{F}_2 = 5.10 \times 0.055 \times (-0.300) \, \hat{i} \][/tex]

[tex]\[ \mathbf{F}_2 = -0.08415 \, \text{N} \, \hat{i} \][/tex]

The net force [tex]\(\mathbf{F} = \mathbf{F}_1 + \mathbf{F}_2\)[/tex]:

[tex]\[ \mathbf{F} = -0.08415 \, \hat{i} + 0.0612 \, \hat{j} \][/tex]

The magnitude of the force is:

[tex]\[ |\mathbf{F}| = \sqrt{(-0.08415)^2 + (0.0612)^2} \][/tex]

[tex]\[ |\mathbf{F}| = \sqrt{0.00708 + 0.00375} \][/tex]

[tex]\[ |\mathbf{F}| = \sqrt{0.01083} \][/tex]

[tex]\[ |\mathbf{F}| \approx 0.104 \, \text{N} \][/tex]

Part B: The direction of the force can be found using the angle [tex]\( \theta \)[/tex] with respect to the positive x-axis:

[tex]\[ \theta = \tan^{-1} \left( \frac{F_y}{F_x} \right) \][/tex]

[tex]\[ \theta = \tan^{-1} \left( \frac{0.0612}{-0.08415} \right) \][/tex]

[tex]\[ \theta = \tan^{-1} \left( -0.727 \right) \][/tex]

[tex]\[ \theta \approx -36.2^\circ \][/tex]

Since this angle is measured counterclockwise from the positive x-axis, the direction clockwise is:

[tex]\[ 360^\circ - 36.2^\circ = 323.8^\circ \][/tex]

Part C: Determine the magnitude of the magnetic force exerted by the external field on the wire

For the wire directly from the origin to [tex]\( x = 20.0 \, \text{mm}, y = 35.0 \, \text{mm} \):[/tex]

- Length [tex]\( L = \sqrt{(20.0 \, \text{mm})^2 + (35.0 \, \text{mm})^2} = \sqrt{(0.020 \, \text{m})^2 + (0.035 \, \text{m})^2} \)[/tex]

- Length [tex]\( L = \sqrt{0.0004 + 0.001225} = \sqrt{0.001625} \)[/tex]- Length [tex]\( L \approx 0.0403 \, \text{m} \)[/tex]

[tex]\[ \mathbf{L} = 0.0403 \, \text{m} \][/tex]

Given current [tex]\( I = 4.90 \, \text{A} \)[/tex]:

[tex]\[ \mathbf{F} = I L B \sin(\theta) \][/tex]

[tex]\[ \theta = \angle \text{between } \mathbf{L} \text{ and } \mathbf{B} = 90^\circ \][/tex]

[tex]\[ \mathbf{F} = 4.90 \times 0.0403 \times 0.300 \][/tex]

[tex]\[ \mathbf{F} \approx 0.0593 \, \text{N} \][/tex]

Part D: - Current: [tex]\( 4.90 \, \text{A} \)[/tex]

- Magnetic field: [tex]\( 0.300 \, \text{T} \, \hat{k} \)[/tex]

- Displacement vector: [tex]\( \mathbf{L} = 0.020 \, \hat{i} + 0.035 \, \hat{j} \)[/tex]

- Force vector: [tex]\( \mathbf{F} = I (\mathbf{L} \times \mathbf{B}) \)[/tex]

[tex]\[ \mathbf{L} \times \mathbf{B} = (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]

[tex]\[ \mathbf{F} = 4.90 (0.020 \, \hat{i} + 0.035 \, \hat{j}) \times 0.300 \, \hat{k} \][/tex]

[tex]\[ \mathbf{F} = 4.90 \times (0.020 \times 0.300 \, \hat{j} - 0.035 \times 0.300 \, \hat{i}) \][/tex]

[tex]\[ \mathbf{F} = 4.90 \times (0.006 \, \hat{j} - 0.0105 \, \hat{i}) \][/tex]

[tex]\[ \mathbf{F} = 0.0294 \, \hat{j} - 0.05145 \, \hat{i} \][/tex]

The magnitude of the force:

[tex]\[ |\mathbf{F}| = \sqrt{(0.0294)^2 + (-0.05145)^2} \approx 0.0593 \, \text{N} \][/tex]

The angle with respect to the x-axis:

[tex]\[ \theta = \tan^{-1} \left( \frac{0.0294}{-0.05145} \right) = \tan^{-1} \left( -0.571 \right) \approx -30^\circ \][/tex]

The direction clockwise is:

[tex]\[ 360^\circ - 30^\circ = 330^\circ \][/tex]

To solve this problem it is necessary to apply the concepts related to rate of thermal conduction

[tex]\frac{Q}{t} = \frac{kA\Delta T}{d}[/tex]

The letter Q represents the amount of heat transferred in a time t, k is the thermal conductivity constant for the material, A is the cross sectional area of the material transferring heat, [tex]\Delta T[/tex], T is the difference in temperature between one side of the material and the other, and d is the thickness of the material.

The change made between glass and air would be determined by:

[tex](\frac{Q}{t})_{glass} = (\frac{Q}{t})_{air}[/tex]

[tex] k_{glass}(\frac{A}{L})_{glass} \Delta T_{glass} = k_{air}(A/L)_{air} \Delta T_{air}[/tex]

[tex]\Delta T_{air} = (\frac{k_{glass}}{k_{air}})(\frac{L_{air}}{L_{glass}}) \Delta T_{glass}[/tex]

[tex]\Delta T_{air} = (\frac{0.84}{0.0234})(\frac{5}{4}) \Delta T_{glass}[/tex]

[tex]\Delta T_{air} = 44.9 \Delta T_{glass}[/tex]

There are two layers of Glass and one layer of Air so the total temperature would be given as,

[tex]\Delta T = \Delta T_{glass} +\Delta T_{air} +\Delta T_{glass}[/tex]

[tex]\Delta T = 2\Delta T_{glass} +\Delta T_{air}[/tex]

[tex]20\°C = 46.9\Delta T_{glass}[/tex]

[tex]\Delta T_{glass} = 0.426\°C[/tex]

Finally the rate of heat flow through this windows is given as,

[tex]\Delta {Q}{t} = k_{glass}\frac{A}{L_{glass}}\Delta T_{glass}[/tex]

[tex]\Delta {Q}{t} = 0.84*24*10 -3*0.426[/tex]

[tex]\Delta {Q}{t} = 179W[/tex]

Therefore the correct answer is D. 180W.

Answer:

       v₂ = 0.56 m / s

Explanation:

This exercise can be done using Bernoulli's equation

        P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

Where points 1 and 2 are on the surface of the glass and the top of the straw

The pressure at the two points is the same because they are open to the atmosphere, if we assume that the surface of the vessel is much sea that the area of ​​the straw the velocity of the surface of the vessel is almost zero v₁ = 0

The difference in height between the level of the glass and the straw is constant and equal to 1.6 cm = 1.6 10⁻² m

We substitute in the equation

         [tex]P_{atm}[/tex] + ρ g y₁ = [tex]P_{atm}[/tex] + ½ ρ v₂² + ρ g y₂

         ½ v₂² = g (y₂-y₁)

        v₂ = √ 2 g (y₂-y₁)

Let's calculate

        v₂ = √ (2 9.8 1.6 10⁻²)

       v₂ = 0.56 m / s

Answer:

c. The net force is vertically downward.

Explanation:

At the top of  the loop, the only external force that keeps the ball moving around the loop, is the centripetal force.

Now, this centripetal force, is not a " new" force, it's just the vector sum of the two external forces (neglecting friction) , that act simultaneously  on the marble, making it to change its speed, in magnitude and direction: the gravity force (which it is always downward), and the normal force(which is always perpendicular to the contact surface, preventing that the marble comes trough the surface), in this case between the marble and the track, which, at the top of the loop, points down too.

So, the net  force, exactly at the top of the loop, is vertically downward.

Answer:

12 cm

Explanation:

[tex]P_1[/tex] = Initial pressure = [tex]P_a=1\times 10^5\ Pa[/tex]

[tex]P_2[/tex] = Final pressure = [tex]P_a+\rho_w gh[/tex]

h = Depth of cylinder = 36 cm

g = Acceleration due to gravity = 10 m/s²

[tex]\rho_w[/tex] = Density of water = 1000 kg/m³

[tex]h_1[/tex] = Depth of lake = 20 m

From the ideal gas relation we have

[tex]P_1V_1=P_2V_2\\\Rightarrow P_a(\pi r^2h)=(P_a+\rho_w gh_1)\pi r^2h'\\\Rightarrow 1\times 10^5\times 36=(1\times 10^5+1000\times 10\times 20)h'\\\Rightarrow h'=\dfrac{1\times 10^5\times 36}{1\times 10^5+1000\times 10\times 20}\\\Rightarrow h'=12\ cm[/tex]

The height of the cylinder of air in the bucket when the bucket is at the given depth is 12 cm

One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.

From the temperature the value is given as

[tex]\eta = 1-\frac{T_L}{T_H}[/tex]

Where,

T_L = Cold focus temperature

T_H = Hot spot temperature

Our values are given as,

T_L = 20\° C = (20+273) K = 293 K

T_H = 440\° C = (440+273) K = 713 K

Replacing we have,

[tex]\eta = 1-\frac{T_L}{T_H}[/tex]

[tex]\eta = 1-\frac{293}{713}[/tex]

[tex]\eta = 0.589[/tex]

Therefore the maximum possible efficiency the car can have is 58.9%

Answer:

0.49806

Explanation:

v = Velocity of wave

L = Length of tube

p denotes piccolo

f denotes flute

The fundamental frequency with both ends open is given by

[tex]f=\dfrac{v}{2L}\\\Rightarrow L=\dfrac{v}{2f}[/tex]

It can be seen that

[tex]L\propto \dfrac{1}{f}[/tex]

So,

[tex]\dfrac{L_p}{L_f}=\dfrac{f_f}{f_p}\\\Rightarrow \dfrac{L_p}{L_f}=\dfrac{257}{516} \\\Rightarrow \dfrac{L_p}{L_f}=0.49806[/tex]

The ratio of the piccolo's length to the flute's length is 0.49806

Answer:True

Explanation:

The given statement is true because when we drive a car in a circle with constant speed , the car experiences the centripetal acceleration towards the center.

But acceleration is change in velocity of object in a given time, here direction of velocity is constantly changing to give rise to acceleration.

If the magnitude of velocity is changing with time then the car would have experiences the tangential acceleration .

The heat required to convert 1.0-g  of ice cube at 0°C to vapor is 720 cal.

We know that when heat is supplied to a substance, change of state occurs as the object moves from a particular state of matter to another. The heat required to convert  1.0-g  of ice cube at 0°C to vapor is obtained from;

H = mLfus + mcdT + mLvap

H = (1 g * 80 cal/g) + (1 g * 1.00 cal/g⋅°C * (100 - 0)) + (1 g * 540 cal/g)

H = 80 cal + 100 cal + 540 cal

H = 720 cal

The heat required to convert 1.0-g  of ice cube at 0°C to vapor is 720 cal.

Learn more about vaporization: https://brainly.com/question/953809

The total amount of heat required to vaporize a 1-g ice cube at 0°C considering the stages of heat of fusion, heating up water, and heat of vaporization is 720 cal.

To calculate the total amount of heat energy required to vaporize a 1.0-g ice cube at 0°C, we need to consider the heat added to turn the ice into water (heat of fusion), heat to raise the temperature of that water from 0°C to 100°C, and finally heat to turn the water into steam (heat of vaporization).

First, we use the heat of fusion of ice, which is 80 cal/g, thus the heat needed to turn 1g of ice into water at 0°C is 1g * 80 cal/g = 80 cal.

Secondly, to raise the water temperature from 0 to 100°C, we use cwater = 1.00 cal/g⋅°C, thus the required heat is 1g * 100°C * 1.00 cal/g⋅°C = 100 cal.

Finally, we use the heat of vaporization of water which is 540 cal/g, for 1g of water, the heat needed is 1g * 540 cal/g = 540 cal.

Adding all those up give us a total heat of 80 cal + 100 cal + 540 cal = 720 cal.

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Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

[tex]k=\frac{2.303}{t}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  = ?

t = time passed by the sample  = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

[tex]k=\frac{2.303}{90.3min}\log\frac{400}{100}[/tex]

[tex]k=1.54\times 10^{-2}\text{ min}^{-1}[/tex]

Now we have to calculate the half-life of substance, we use the formula :

[tex]k=\frac{0.693}{t_{1/2}}[/tex]

[tex]1.54\times 10^{-2}\text{ min}^{-1}=\frac{0.693}{t_{1/2}}[/tex]

[tex]t_{1/2}=45min[/tex]

Therefore, the half-life of this substance will be, 45 minutes.

To solve this problem it is necessary to apply the concepts related to the simple harmonic movement, to the speed in terms of displacement and the timpo, as well as the angular frequency and the period of frequency.

PART A) According to the description given, 5 revolutions are made in one minute (or 60 seconds) that is to say that the frequency would be given by

[tex]f = \frac{1}{12s^{-1}}[/tex]

Therefore the angular velocity can be found as

[tex]\omega = 2\pi f[/tex]

[tex]\omega = 0.52rad/s[/tex]

The displacement that determines the maximum displacement based on the angular velocity time and the time in the simple harmonic movement, is equal to the radius of the circle, in other words

[tex]x = Acos(\omega t)[/tex]

Where,

A = Amplitude

[tex]\omega =[/tex] Angular velocity

t = time

If for our given values the value of the amplitude is 2m and the value of the angular velocity is 0.52rad/s

[tex]x = 2cos(0.52t)[/tex]

So the equation for the position of the shadow is of

[tex]x(t) = 2cos(0.52t)[/tex]

PART B) The equation for the velocity of the shadow is calculated as a expression of the displacement against the time, if we differenciate the previous value found, we have that,

[tex]\frac{dx}{dt} = \frac{d(2cos(0.52t))}{dt}[/tex]

[tex]v(t) = 2(-0.52)sin(0.523t)[/tex]

[tex]v(t) = -1.05sin(0.52t)[/tex]

The position of the shadow can be described using the equation x=2cos(10πt+φ) and the velocity of the shadow can be described using the equation v=-2*10π*sin(10πt+φ), using principles of simple harmonic motion.

The subject of this question involves physics and in particular, kinematics and rotational motion, as it relates to the shadow produced by a student standing on a rotating merry-go-round.

The position and velocity of the shadow will vary as the merry-go-round rotates, and these can be represented using mathematical equations. Let's denote the angular speed of the merry-go-round as 'w', which is given by the formula w = 2πn, where n is the number of rotations per minute. For the given question, n = 5 rotations per minute. Therefore, w = 10π rad/min.

(a) Position of the shadow: We can describe the position of the shadow in terms of simple harmonic motion - as the merry-go-round rotates, the shadow produced on the nearby building would be oscillating back and forth. This kind of motion can be mathematically represented as x=Acos(wt+φ), where x is the position of the shadow, A is the amplitude (which is the radius of the merry-go-round, hence 2 meters), w is the angular frequency, t is time, and φ is the phase constant.

(b) Velocity of the shadow: The velocity of the shadow is the derivative of the position with respect to time, which can be represented as v=-Aw*sin(wt+φ). In our case, the angular speed w = 10π rad/min, A is the amplitude (radius of the merry-go-round, hence 2 meters), and 't' is the time.

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Answer:

a. K = 1080.61 N/m

b. mₐ = 96.28 kg

Explanation:

T = 2π * √ m / K

T₁ = 2π * √m₁ / K , T₂ = 2π * √m₂ / K

m₁ = 36.2 kg , m₂ = m₁ + mₐ

T₁ = 1.15 s , T₂ = 2.2 s

Equal period to determine the force constant of the spring

a.

K = 4π²* m₁ / T₁²

K = 4π² * 36.2 kg / 1.15² s

K = 1080.61 N/m

So replacing and solve can find m₂

b.

m₂ = T₂² * m₁ / T₁²   ⇒  m₂ = 2.20² s * 36.2 kg  / 1.15 ² s

m₂ = 132.48

mₐ = m₂ - m₁

mₐ = ( 132.48 - 36.2 ) kg = 96.28 kg

Answer:b

Explanation:

Given

mass of block [tex]m=4 kg[/tex]

coefficient of static friction [tex]\mu =0.25 [/tex]

height of triangle is [tex]h=3 m[/tex]

[tex]F_{net}=mg\sin \theta -\mu _kmg\cos \theta [/tex]

[tex]a_{net}=g\sin \theta -\mu _kg\cos \theta [/tex]

[tex]a_{net}=9.8\sin 37-0.25\times 9.8\times \cos 37[/tex]

[tex]a_{net}=5.897-1.956=3.94 m/s^2[/tex]

here [tex]s=5 m[/tex]

[tex]v^2-u^2=2 a_{net}s[/tex]

[tex]v=\sqrt{2\times 3.94\times 5}[/tex]

[tex]v=6.27 m/s[/tex]

time taken to reach bottom of plane

[tex]v=u+at[/tex]

[tex]6.27=0+3.94\times t[/tex]

[tex]t=1.59 s\approx 1.6 s[/tex]                

Answer:

[tex]B=3A-5[/tex]

Explanation:

Variable Isolation

It's a common practice when dealing with equations that we have to isolate one variable in terms of other variables and/or constants. The isolation of the variable usually implies adding, subtracting, multiplying or dividing by constants. The following example shows how to isolate the A:

[tex]2A+3B=-4\\\\2A=-4-3B\\\\\displaystyle A=\frac{-4-3B}{2}[/tex]

We are required to find the equation where the variable has a coefficient of 1 and isolate it. The following equation fits into the description:

[tex]3A-B=5[/tex]

Isolating B:

[tex]B=3A-5[/tex]

a) 12.95*10^20 molecules

b) 5.84*10^20 molecules

Explanation:

a)Tylenol

Mass of Tylenol = 325 mg

(m)=325*10^-3 g

Molecular weight of Tylenol(M) = 151.16256 g/mol

the number of moles present  

n = m/M

= 325*10^-3 g /151.1656 g/mol

= 2.15 *10^-3 mol

the number of molecules present

N= nNa

=(2.15*10^ -3 mol)(6.023*10^23 molecules/mol)

N=12.95*10 ^20 molecules

b)Advil

Mass of Advil = 200 mg

m=200*10^-3 g

Molecular weight of Advil(M) =206.28082 g/mol

the number of moles

n=m/M

=200*10^ -3 g/ 206.28082 g/mol

=0.96955*10^ -3 mol

the number of molecules resent

N=nNa

=(0.96955*10^ -3)(6.023*10^ 23 molecules/ mol)

N=5.84*10^20 molecules

Tylenol contains more molecules of active ingredient hence it is a more effective drug.

We have to find the number of molecules in each of the drugs. In the drug, tylenol, the molar mass of the compound C8H9NO2 is;

8(12) + 9(1) + 14 + 2(16) = 96 + 9 + 14 + 32 =151 g/mol

Number of moles = 325 × 10^-3 g/151 g/mol = 0.0022 moles

1 mole contains 6.02 × 10^23 molecules

0.0022 moles contains  0.0022 moles ×  6.02 × 10^23 molecules /1 mole = 1.32 × 10^21 molecules

For Advil;

Molecular mass= 13(12) + 18(1) + 2(16) = 156 + 18 + 32 = 206 g/mol

Number of moles = 0.2 g/ 206 g/mol = 0.00097 moles

1 mole contains 6.02 × 10^23 molecules of advil

0.00097 moles contains  0.00097 moles × 6.02 × 10^23 molecules /1 mole = 5.8  × 10^20

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Answer:

[tex]f_{e}[/tex]  = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

    M = M₀ [tex]m_{e}[/tex]

Where M₀ is the magnification of the objective and  [tex]m_{e}[/tex] is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

       [tex]m_{e}[/tex] = 25 /  [tex]f_{e}[/tex]

The objective is focused on the distances of the tube (L)

     M₀ = -L / f₀

Substituting

     M = - L/f₀    25/[tex]f_{e}[/tex]  

1) Let's look for the focal length of the eyepiece (faith)

     [tex]f_{e}[/tex]  = - L 25 / f₀ M

     M = 400X = -400

     [tex]f_{e}[/tex]  = - 12 25 /0.40 (-400)

     [tex]f_{e}[/tex]  = 1.875 cm

Let's approximate two significant figures

    [tex]f_{e}[/tex]  = 1.9 cm

Final answer:

The focal length of the eyepiece lens in the microscope is approximately -0.01 mm.

Explanation:

To find the focal length of the eyepiece lens, we can use the formula for the magnification of a compound microscope, which is given by:

M = -fobjective/feyepiece

Where M is the overall magnification, fobjective is the focal length of the objective lens, and feyepiece is the focal length of the eyepiece lens.

Given that the overall magnification is 400x and the focal length of the objective lens is 0.40 cm, we can rearrange the formula to solve for the focal length of the eyepiece lens:

feyepiece = -fobjective/M = -0.40 cm/400 = -0.001 cm = -0.01 mm.

Therefore, the focal length of the eyepiece lens is approximately -0.01 mm.

Answer:

a) [tex]\angle r_{ag}=38.79^{\circ}[/tex]

b) [tex]\angle r_{gw}=44.95^{\circ}[/tex]

c) not possible

Explanation:

Given:

angle of incidence on the air-glass interface, [tex]\angle i_{ag}=90-20=70^{\circ}[/tex]

refractive index of glass with respect to air, [tex]n_g=1.5[/tex]

refractive index of water with respect to air, [tex]n_a=1.33[/tex]

a)

For angle of refraction in glass we use Snell's law:

[tex]n_g=\frac{sin\ i_{ag}}{sin\ r{ag}}[/tex]

[tex]1.5=\frac{sin\ 70}{sin\ r_{ag}}[/tex]

[tex]\angle r_{ag}=38.79^{\circ}[/tex]

b)

Now we have angle of incident for glass-water interface, [tex]\angle i_{gw}=\angle r_{ag}=38.79^{\circ}[/tex]

And the refractive index of water with respect to glass:

[tex]n_{gw}=\frac{n_w}{n_g}[/tex]

[tex]n_{gw}=\frac{1.33}{1.5}[/tex]

[tex]n_{gw}=0.8867 [/tex]

Therefore, angle of refraction in the water:

[tex]n_{gw}=\frac{sin\ i_{gw}}{sin\ r_{gw}}[/tex]

[tex]0.8867 =\frac{sin\ 38.79}{sin\ r_{gw}}[/tex]

[tex]\angle r_{gw}=44.95^{\circ}[/tex]

c)

For total internal reflection through water the light must enter the glass-water interface at an angle greater than the critical angle.

So,

[tex]n_{gw}=\frac{sin\ i_{(gw)_c}}{sin\ 90}[/tex]

[tex]0.8867 =\frac{sin\ i_{(gw)_c}}{1}[/tex]

[tex]i_{(gw)_c}=62.46^{\circ}[/tex]

Now this angle will become angle of refraction for the air-glass interface.

Hence,

[tex]n_g=\frac{sin\ i_{ag}}{sin\ i_{(gw)_c}}[/tex]

[tex]1.5=\frac{sin\ i_{ag}}{0.8867 }[/tex]

[tex]sin\ i_{ag}=1.33005[/tex]

Since we do not have any angle for sine for which the value exceeds 1, therefore it is not possible that the light will reflect from the water.

Answer:

Explanation:

Given

Coefficient of drag [tex]C_d=1.1[/tex]

Reynolds number [tex]Re.no.=103 to 105[/tex]

velocity [tex]v=2 m/s [/tex]

[tex]F_d=3 N[/tex]

if velocity if 2v i.e. [tex]4 m/s [/tex]

[tex]F_d=\frac{1}{2}C_d\rho A v^2----1[/tex]

keeping other factors as constant

[tex]F'_d=\frac{1}{2}C_d\rho A (2v)^2----2[/tex]

dividing  1 and 2

[tex]\frac{F_d}{F'_d}=\frac{v^2}{2v^2}[/tex]

[tex]F'_d=4F_d[/tex]

[tex]F'_d=4\times 3=12 N[/tex]

Answer:

c. is more than that of the fluid.

Explanation:

This problem is based on the conservation of energy and the concept of thermal equilibrium

[tex]heat= m s \Delta T [/tex]

m= mass

s= specific heat

\DeltaT=change in temperature

let s1= specific heat of solid and s2= specific heat of liquid

then

Heat lost by solid= [tex]20(s_1)(70-30)=800s_1 [/tex]

Heat gained by fluid=[tex]100(s_2)(30-20)=1000s_2 [/tex]

Now heat gained = heat lost

therefore,

1000 S_2=800 S_1

S_1=1.25 S_2

so the specific heat of solid is more than that of the fluid.

The correct option is:

(c) is more than that of the fluid.

Conservation of energy:

The conservation of energy suggests that the heat energy lost by the solid must be equal to the heat energy gained by the liquid.

Let the specific heat of the solid be [tex]s_s[/tex] and the specific heat of the liquid be [tex]s_l[/tex].

Heat energy lost by the solid is given by:

[tex]\Delta Q_s=ms_s\Delta T\\\\ \Delta Q_s=20s_s(70-30)\\\\ \Delta Q_s=800s_s[/tex]

Heat energy gained by the liquid:

[tex]\Delta Q_l=ms_l\Delta T\\\\ \Delta Q_l=100s_l(30-20)\\\\ \Delta Q_l=1000s_l[/tex]

According to the conservation of energy:

[tex]\Delta Q_s=\Delta Q_l\\\\800s_s=1000s_l\\\\\frac{s_s}{s_l}=\frac{1000}{800}\\\\\frac{s_s}{s_l}=\frac{5}{4}[/tex]

Hence, the specific heat of the solid is more than that of the fluid

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Answer:

The energy of an electron in an isolated atom depends on b. n only.

Explanation:

The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.

The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.

The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates [tex]E_n[/tex], and for the case of an hydrogen atom we have:

[tex]E_n=-\cfrac{13.6}{n^2}\, eV[/tex]

Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.

Answer:

A. [tex]W=600\ J[/tex]

B. [tex]Q=2112\ J[/tex]

C. [tex]\Delta U=1512\ J[/tex]

D. [tex]W=0\ J[/tex]

Explanation:

Given:

According to the question the final volume becomes twice of the initial volume.

Using ideal gas law:

[tex]P.V=n.R.T[/tex]

[tex]2\times 10^5\times V_i=0.2\times 8.314\times 360[/tex]

[tex]V_i=0.003\ m^3[/tex]

A.

Work done by the gas during the initial isobaric expansion:

[tex]W=P.dV[/tex]

[tex]W=P_i\times (V_f-V_i)[/tex]

[tex]W=2\times 10^5\times (0.006-0.003)[/tex]

[tex]W=600\ J[/tex]

C.

we have the specific heat capacity of oxygen at constant pressure as:

[tex]c_v=21\ J.mol^{-1}.K^{-1}[/tex]

Now we apply Charles Law:

[tex]\frac{V_i}{T_i} =\frac{V_f}{T_f}[/tex]

[tex]\frac{0.003}{360} =\frac{0.006}{T_f}[/tex]

[tex]T_f=720\ K[/tex]

Now change in internal energy:

[tex]\Delta U=n.c_p.(T_f-T_i)[/tex]

[tex]\Delta U=0.2\times 21\times (720-360)[/tex]

[tex]\Delta U=1512\ J[/tex]

B.

Now heat added to the system:

[tex]Q=W+\Delta U[/tex]

[tex]Q=600+1512[/tex]

[tex]Q=2112\ J[/tex]

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

[tex]W=0\ J[/tex]

Answer:

1. F = 45,458.17 N

2. P = 12,800,000 W

Explanation:

Part 1. The thrust force is the sum of the forces on the air and on the fuel.

For the air, 107 kg of air is accelerated from 281 m/s to 679 m/s in 1 second.

F = ma

F = (107 kg) (679 m/s − 281 m/s) / (1 s)

F = 42,586 N

For the fuel, 4.23 kg of fuel is accelerated from 0 m/s to 679 m/s in 1 second.

F = ma

F = (4.23 kg) (679 m/s − 0 m/s) / (1 s)

F = 2,872.17 N

So the thrust on the jet is:

F = 42,586 N + 2,872.17 N

F = 45,458.17 N

Rounded to three significant figures, the force is 45,500 N.

Part 2. Power = work / time, and work = force × distance, so:

Power = force × distance / time

Power = force × velocity

P = (45,458.17 N) (281 m/s)

P = 12,773,745.77 W

Rounded to three significant figures, the power is 12,800,000 W.