A greater initial speed of the heavy cart could lead to a higher initial kinetic energy, which could result in a greater fraction of the kinetic energy being lost during a non-perfectly elastic collision. This could support the student's hypothesis.
Explanation:The student's hypothesis could be correct because the fraction of kinetic energy lost from the two-cart system during the collision could indeed depend on the initial speed of the heavy cart. If the first, heavier cart is moving at a higher speed before the collision, it would have a higher initial kinetic energy (K init). When it collides with the second cart, more kinetic energy could be transformed into other forms of energy such as heat or sound, especially since the collision is not perfectly elastic. This means a greater fraction of kinetic energy could be lost (K lost) in the process, supporting the student's hypothesis.
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The higher initial speed of the first cart leads to greater deformation and energy conversion to non-mechanical forms, resulting in a larger fraction of kinetic energy loss may be the reason the hypothesis might be correct.
The hypothesis that a greater fraction of kinetic energy is lost from the system during the collision when the speed of the first cart is greater might be correct due to the nature of inelastic collisions and the role of deformation and heat generation during such collisions.
One reason for this is that the energy loss in a collision often includes contributions from factors like the deformation of the carts and the generation of heat, both of which can increase with the relative speed at the moment of collision. As the speed of the first cart increases, the impact forces during the collision are higher, leading to greater deformation and more energy being converted into internal energy (such as heat and sound) rather than being retained as kinetic energy in the system.
This means that when the speed of the first cart is higher, a larger portion of the initial kinetic energy is likely to be transformed into non-mechanical forms of energy, resulting in a greater fraction of kinetic energy being lost from the two-cart system.
A 5 kg spinning ring of inner radius 30cm and outer radius 40cm is dropped on a flat disk, where it interlocks and causes the disk to spin. Before the ring was dropped, it was spinning at 60 Hz. If the disk has a mass of 2kg and a radius of 1 meter, how much force needs to be applied to the rim of the disk (at a 90 degree angle) to bring it to a stop in 5 seconds
The specific heat of a liquid x is 2.09 cal/g°c. A sample amount of grams of this liquid at 101 k is heated to 225 k. the liquid absorbs 5.23 kcals. what is the sample of liquid in grams? (round off decimal in the answer to nearest tenths)
Answer: 20 grams
Explanation:
The quantity of heat required to raise the temperature of a substance by one degree Celsius is called the specific heat capacity.
[tex]Q=m\times c\times \Delta T[/tex]
Q = Heat absorbed by liquid = 5.23 kcal = 5230 cal (1kcal=1000cal)
C = heat capacity of liquid = [tex]2.09cal/g^0C[/tex]
Initial temperature of the liquid = [tex]T_i[/tex] = 101 K
Final temperature of the liquid = [tex]T_f[/tex] = 225 K
Change in temperature ,[tex]\Delta T=T_f-T_i=(225-101)K=124K[/tex]
Putting in the values, we get:
[tex]5230=m\times 2.09cal/g^0C\times 124K[/tex]
[tex]m=20g[/tex]
Thus the sample of liquid in grams is 20
The key discovery about Cepheid variable stars that led in the 1920s to the resolution of the question of whether spiral "nebulae" were separate and distant galaxies or part of the Milky Way Galaxy was the direct relationship between the pulsation period and the absolute brightness or luminosity of the Cepheid variables. A measurement of ____ brightness of a variable star could then be used to determine the distance to the "nebula" containing it.
Answer:
Apparent
The key discovery about Cepheid variable stars that led in the 1920s to the resolution of the question of whether spiral "nebulae" were separate and distant galaxies or part of the Milky Way Galaxy was the direct relationship between the pulsation period and the absolute brightness or luminosity of the Cepheid variables. A measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.
Explanation:
A variable star is a star with changing apparent brightness. The changes can occur over years or in a fraction of seconds. For example the sun whose energy output varies by approximately 0.1 percent of its magnitude, over an 11-year solar cycle. This variable(apparent brightness) can be used to determine how far a variable star is (distance). Therefore, a measurement of apparent brightness of a variable star could then be used to determine the distance to the "nebula" containing it.
g A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased from 250 µT to 700 µT in 0.300 s, an emf of magnitude 60.0 mV is induced in the coil. What is the total length of wire in the coil?
Answer:
L = 182.4 m
Explanation:
Given:-
- The number of turns of the coil, N = 50
- The shape of the coil = square
- The angle between the coil and magnetic field, θ = 30°
- The change in magnetic field, ΔB = ( 700 - 250 ) μT
- The time duration in which magnetic field changes, Δt = 0.3 s
- The induced emf, E = 60.0 mV
Solution:-
- The problem at hand is an application of Faraday's law. The law states that the induced emf ( E ) is proportional to the negative rate of change of magnetic flux ( ΔФ / Δt ) and number of turns of the coil ( N ).
- The Faraday's law is mathematically expressed as:
E = - N* ( ΔФ / Δt )
Where,
- The flux ( Ф ) through a current carrying with an cross-sectional area ( A ) at a normal angle ( θ ) to the direction of magnetic field ( B ) is given by the following relationship.
Ф = B*A*cos ( θ )
- We need the rate of change of magnetic flux ( ΔФ / Δt ) for the Faraday's law. I.e the induced emf ( E ) is proportional to rate of change in magnetic field ( ΔB / Δt ), rate of change of angle between the coil and magnetic field ( Δθ / Δt ) or rate of change of cross-sectional area of the coil under the influence of magnetic field.
- To determine the exact relationship. We will derive the multi-variable function of flux ( Ф ) with respect to time "t":
Ф ( B , A , θ ) = B*A*cos ( θ )
- The first derivative would be ( Use chain and product rules )
( ΔФ / Δt ) = ΔB / Δt*A*cos ( θ ) + B*ΔA/Δt*cos ( θ ) - B*A*sin ( θ )*Δθ/Δt
- For the given problem the only dependent parameter that is changing is magnetic field ( B ) with respect to time "t". Hence, ( ΔA/Δt = Δθ/Δt = 0 ):
ΔФ / Δt = (ΔB/Δt)*A*cos ( θ )
- Substitute the rate of change of magnetic flux ( ΔФ / Δt ) into the expression for Faraday's Law initially stated:
E = - N*(ΔB/Δt)*A*cos ( θ )
- Plug in the values and evaluate the Area of the square coil:
A = - E / ( N*(ΔB/Δt)*cos ( θ ) )
A = - 0.06 / ( 50*[ (250-700)*10^-6/0.3 ] *cos ( 30° ) )
A = - 0.06 / -0.07216
A = 0.8314 m^2
- The square coil has equal sides ( x ). The area of a square A is given by:
A = x^2
x = √0.8314
x = 0.912 m
- The perimeter length of a single coil in terms of side length "x" is given as:
P = 4x
Whereas for a coil of N turns the total length ( L ) would be:
L = N*P
L = 4Nx
L = 4 * 50 * 0.912
L = 182.4 m ... Answer
In your research lab, a very thin, flat piece of glass with refractive index 1.50 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength lambda 0 in vacuum at normal incidence onto the surface of the glass. When lambda 0 = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. Use these measurements to calculate the thickness of the glass. Express your answer with the appropriate units. What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?
Answer:
The longest wavelength in vacuum for which there is constructive interference for the reflected light, λ = 3472.
Explanation:
Refractive index of Glass (given) = 1.5
For the case of a constructive interference,
2nt = (m + 1/2) λ
For case 1,
2nt = (m + 1/2) 496 nm
For case 2,
2nt = (m +1+ 1/2) 386 nm
2nt = (m+3/2) * 386 nm
(m + 1/2) 496 nm = (m+3/2) * 386 nm
m = 3
Inserting the value of m in 1.
2nt = (m + 1/2) 496 nm
2*1.5t = (3 + 1/2) * 496 nm
t = ((3 + 1/2) * 496 nm)/ 3
t = 578.6 nm
The thickness of the glass, t = 578.6 nm
b)
It is generally known that for constructive interference,
2nt = (m + 1/2) λ
λ = 2nt / ((m + 1/2))
For Longest Wavelength, m = 0
λ = 2*1.5*578.6/ (1/2)
λ = 3472 nm
[03.02]
Which statement correctly describes the relationship between current, voltage, and resistance? If we (1 point)
Group of answer choices
decrease the resistance, and do not change the voltage, the current will decrease
decrease the voltage, and do not change the resistance, the current will also decrease
increase the resistance, and do not change the current, the voltage will remain the same
increase the current, and increase the resistance, the voltage will also decrease
Answer:
Decrease the voltage,and do not change the resistance,the current will also decrease
Explanation:
Decrease the voltage,and do not change the resistance, the current will also decrease, because voltage is directly proportional to current
Answer:
B.If we decrease the voltage, and do not change the resistance, then the current will also decrease.
Explanation:
i just take the (pre)test
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A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. The drawing shows a spherical reservoir that contains 7.41 x 105 kg of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter of the delivery pipes.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The gauge pressure that water has at the House A [tex]P_A = 257020.68 Pa[/tex]
The gauge pressure that water has at the House B [tex]P_B = 188454 \ Pa[/tex]
Explanation:
From the question we are told that
The mass of water when full is [tex]m_f = 7.41* 10^{5} kg[/tex]
Generally the volume of water in this tank is mathematically represented as
[tex]V = \frac{m }{\rho}[/tex]
Where [tex]\rho[/tex] is the density of water with a value of with a value of [tex]\rho = 1000 kg /m^3[/tex]
substituting values
[tex]V = \frac{7.41 *10^5}{10^3}[/tex]
[tex]V = 741 m^3[/tex]
This volume is the volume of a sphere since the tank is spherical so
[tex]V = \frac{4 \pi ^3}{3}[/tex]
making r the subject of the formula
[tex]r =\sqrt[3]{ \frac{741 *3 }{4\pi} }[/tex]
[tex]r = 5.6134 m[/tex]
Now we can use this parameter to obtain the diameter
So
[tex]d = 2 * r[/tex]
substituting values
[tex]d = 2 * 5.6134[/tex]
[tex]d = 11.23m[/tex]
The pressure the water has at faucet in House A is mathematically evaluated as
[tex]P_A = \rho g h_A[/tex]
This height is obtained as follows
[tex]h_A = d+ 15[/tex]
The value 15 is gotten from the diagram
so
[tex]h_A = 15 + 11.23[/tex]
[tex]h_A = 26.22 m[/tex]
Now substituting values
[tex]P_A = 26.23 * 9.8 * 1000[/tex]
[tex]P_A = 257020.68 Pa[/tex]
The pressure the water has at faucet in House B is mathematically evaluated as
[tex]P_B = \rho g h_B[/tex]
This height is obtained as follows
[tex]h_B = d+ 15[/tex]
The value 15 is gotten from the diagram
so
[tex]h_B = d + 15 -h[/tex]
substituting values
[tex]h_B =11.23 + 15 -7[/tex]
[tex]h_A = 19.23 m[/tex]
Now substituting values
[tex]P_B = 19.23 * 9.8 * 1000[/tex]
[tex]P_B = 188454 \ Pa[/tex]
A 60kg60 kg board that is 6 m6 m long is placed at the edge of a platform, with 4 m4 m of its length extending over the edge. The board is held in place by blocks of masses M1M1 and M2M2 placed with their centers of mass on either end. If M2=30kgM2=30kg , what is the minimum value of M1M1 needed to keep the board from falling off the platform? 30kg
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The minimum mass of [tex]M_1 = 90\ kg[/tex] correct option is E
Explanation:
Free body diagram of the set up in the question is shown on the third uploaded image
The mass of board is [tex]M = 60kg[/tex]
The length of the board is [tex]L = 6 \ m[/tex]
The length extending over the edge is [tex]L_e = 4 \ m[/tex]
The second mass is [tex]M_2 = 30kg[/tex]
Now to obtain [tex]M_1[/tex] we take moment about the edge of the platform
[tex]M_1 g L_1 = Mg \frac{L}{2} + M_2 g L_2[/tex]
[tex]M_1 L_1 = M \frac{L}{2} + M_2 L_2[/tex]
Substituting value
[tex]M_1 (2) = (60)(1) + (30)(4)[/tex]
[tex]M_1 = 90 \ kg[/tex]
The minimum value of M1 needed to keep the board from falling off the platform is 90 kg.
From the information given, we are to find:
the mass (M1) of placed on the left side edge of the boardGiven that:
the mass of the board = 60 kgthe length of the board = 6 mIf the mass on the right side = 30 kg, and the length of the board L1 = 2mThen, the length L2 which extend over the edge = 4mConsider the center of gravity in the board that lies at the length of the board midpoint.
Then, the distance (D) of the gravity center from the platform end = 3 - 2
= 1 m
∴
Considering the moment about the platform end, the mass (M1) placed on the left side edge of the board can be computed as:
[tex]\mathbf{M_1gL_1 = MgD + M_2gL_2} \\ \\ \mathbf{M_1L_1 = MD + M_2gL_2} \\ \\ \mathbf{M_1(2) = 60 \ kg \times 1 + 30 \ kg \times (4)} \\ \\ \mathbf{ M_1 =\dfrac{60 \ kg + 120 kg }{2} } \\ \\ \mathbf{ M_1 =\dfrac{180 \ kg}{2} } \\ \\ \mathbf{ M_1 =90 \ kg }[/tex]
Therefore, we can conclude that the minimum value M1 needed to keep the board from falling off the platform is 90 kg.
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horizontal circular platform rotates counterclockwise about its axis at the rate of 0.919 rad/s. You, with a mass of 73.5 kg, walk clockwise around the platform along its edge at the speed of "1.05" m/s with respect to the platform. Your 20.5 kg poodle also walks clockwise around the platform, but along a circle at half the platform's radius and at half your linear speed with respect to the platform. Your 18.5 kg mutt, on the other hand, sits still on the platform at a position that is 3/4 of the platform's radius from the center. Model the platform as a uniform disk with mass 90.7 kg and radius 1.91 m. Calculate the total angular momentum of the system
Answer:
The total angular momentum is 292.59 kg.m/s
Explanation:
Given that :
Rotation of the horizontal circular platform [tex]\omega[/tex] = 0.919 rad/s
mass of the platform (m) = 90.7 kg
radius (R) = 1.91 m
mass of the poodle [tex]m_p[/tex] = 20.5 kg
Your mass [tex]m'[/tex] = 73.5 kg
speed v = 1.05 m/s with respect to the platform
[tex]V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525 \ m /s \\ \\r = \frac{R}{2}[/tex]
r = 0.955
Mass of the mutt [tex]m_m[/tex] = 18.5 kg
[tex]r' = \frac{3}{4} \ R[/tex]
Your angular momentum is calculated as:
Your angular velocity relative to the platform is [tex]\omega' = \frac{v}{R} = \frac{1.05}{1.91 } = 0.5497 \ rad/s[/tex]
[tex]Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s[/tex]
[tex]I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2[/tex]
[tex]L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s[/tex]
For poodle :
[tex]Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \ rad/s[/tex]
Actual [tex]\omega_p = \omega - \omega' = 0.919 -0.550 = 0.369 \ rad/s[/tex]
[tex]I_p = m_p(\frac{R}{2} )^2 = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2[/tex]
[tex]L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \ kgm/s[/tex]
[tex]I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \ kgm^2[/tex]
[tex]L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s[/tex]
Disk [tex]I = \frac{mr^2}{2} = \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2[/tex]
[tex]L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s[/tex]
Total angular momentum of system is:
L = [tex]L_D +L_Y+L_P+L_M[/tex]
= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s
= 292.59 kg.m/s
A 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second. The kinetic energy is joules
Answer:2250J
Explanation:
mass(m)=20kg
velocity(v)=15m/s
Kinetic energy=(m x v^2)/2
Kinetic energy =(20 x 15^2)/2
Kinetic energy =(20x15x15)/2
Kinetic energy=4500/2
Kinetic energy=2250J
The kinetic energy of the ball will be 2250 joules.
We have a 20 kilogram ball rolls down a 10 meter ramp at the rate of 15 meters per second.
We have to determine its kinetic energy is joules.
What is the formula to calculate the kinetic energy of the body of mass 'm' moving with velocity 'v' ?The kinetic energy of the body is as follows -
K.E. = [tex]\frac{1}{2} mv^{2}[/tex]
According to the question -
Mass of ball = 20 kg
Velocity of ball = 15 m/s
Substituting the values in the above formula, we get -
K.E. = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2} \times 20\times 15\times 15[/tex] = 225 x 10 = 2250 joules.
Hence, the kinetic energy of the ball will be 2250 joules.
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A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the center of the point source the z axis. When a sheet of cardboard in which a square hole 0.020 m on a side has been cut is placed between the point source and the screen, 0.50 m from the point source with the hole centered on the z axis, a bright square shows up on the screen. If, instead, a second sheet of cardboard with a similar square hole is placed between the point source and screen, 0.25 m from the point source with the hole centered on the z axis, the bright square it casts on the screen is identical to the bright square from the first sheet. What is the length of the side of the hole in this sheet?
Answer:
The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]
Explanation:
From the question we are told that
The distance of the point source from the screen is [tex]d = 1.0 m[/tex]
The length of a side of the first square hole is [tex]L_1 = 0.020 \ m[/tex]
The distance of the cardboard from the point source is [tex]D_1 = 0.50\ m[/tex]
The distance of the second cardboard from the point source is [tex]D_2 = 0.25 \ m[/tex]
Let take the [tex]\alpha_{max }[/tex] as the angle at which the light is passing through the edges of the cardboards square hole
Since the bright square casted on the screen by both square holes on the individual cardboards are then it means that
[tex]\alpha_{max} __{1}} = \alpha_{max} __{2}}[/tex]
This implies that
[tex]tan (\alpha_{max} __{1}}) = tan (\alpha_{max} __{2}})[/tex]
Looking at this from the SOHCAHTOA concept
[tex]tan (\alpha_{max} __{1}}) = \frac{opposite}{Adjacent}[/tex]
Here opposite is the length of the side of the first cardboard square hole
and
Adjacent is the distance of the from the first cardboard square hole to the point source
And for
[tex]tan (\alpha_{max} __{2}}) = \frac{opposite}{Adjacent}[/tex]
Here opposite is the length of the side of the second cardboards square hole (let denote it with [tex]L_2[/tex])
and
Adjacent is the distance of the from the second cardboards square hole to the point source
So
[tex]tan (\alpha_{max} __{1}}) = \frac{0.020}{0.50}[/tex]
And
[tex]tan (\alpha_{max} __{2}}) = \frac{L_2}{0.25}[/tex]
Substituting this into the above equation
[tex]\frac{0.020}{0.50} = \frac{L_2}{0.25}[/tex]
Making [tex]L_2[/tex] the subject
[tex]L_2 = \frac{0.25 *0.020}{0.50}[/tex]
[tex]L_2 = 0.01m[/tex]
Since it is a square hole the sides are the same hence
The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]
A biker can ride at 12 m/s on a level road when there is no wind and at 7.5 m/s on a level road when there is a head wind of 5 m/s. Assume that fluid drag and resistance due to friction are the only forces acting on the biker. The friction force is given by the product of rolling resistance and bike velocity. The density of ambient air is 1.2 kg/m3 If the biker delivers 10.2500 W of power while riding the bike, determine the value of LaTeX: C_D A
Answer:
The answer is "[tex]1.94 \ m^2[/tex]".
Explanation:
Formula:
[tex]F= \frac{1}{2} \times C_D \times density \times area \times velocity^2\\\\Power (P) = F \times velocity \\\\P = \frac{C_D}{2} \times density \times area \times velocity^3\\[/tex]
Given value:
[tex]\ P = 10.25 \ W \\\\\ density = 1.2 \ kg/m^3 \\\\\ velocity= 2.5 \\\\[/tex]
[tex]10.25 = \frac{C_D A}{2} \times 1.2 \times 2.5^3\\\\C_D A= \frac{10.25 \times 2 }{1.2 \times 2.5^3}\\\\C_D A = 1.94 m^2\\[/tex]
In what direction must a force be applied so that the forces on the 1 kg object are balanced
Answer:
towards the object
Explanation:
Final answer:
To balance the forces on the 1 kg object, a force should be applied in the downward direction.
Explanation:
The forces on the 1 kg object can be balanced by applying a force in the opposite direction to the net force acting on the object. In this case, the net force is the sum of the weight of the object and the tension in the string. Since the weight acts downward and the tension in the string acts upward, the force should be applied in the downward direction to balance the forces on the 1 kg object.
A 16ft seesaw is pivoted in the center. At what distance from the center would a 200lb person sit to balance a 120lb person on the opposite end?
Answer:
9.6 ft
Explanation:
Distance is inversely proportional to weight
distance = k / (weight), where
k is a constant
or you could say,
distance * weight = k
In this scenario,
120 * 16 = 200 * distance
On rearranging, making, distance the subject of formula, we have
Distance = 120 * 16 / 200
Distance = 1920 / 200
Distance = 9.6 ft
So the 200 pounds person should sit 9.6 feet away from the centre to balance the see saw
Answer:
4.8 ft
Explanation:
torque = wt × distance
t1 = 120lb x 8 ft =960
t2 = 200lb x X ft
set them equal to each other.
120(8) = 200x
960 = 200x
x = 4.8 ft
A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10 s time interval, the magnitude of the field increases uniformly from 250 to 350 mT. (a) Determine the emf induced in the loop (in V). (Enter the magnitude.) V (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop
Answer:
Magnitude of induced emf is 0.00635 V
Explanation:
Radius of circular loop r = 45 mm = 0.045 m
Area of circular loop [tex]A=\pi r^2[/tex]
[tex]A=3.14\times 0.045^2=0.00635m^2[/tex]
Magnetic field is increases from 250 mT to 350 mT
Therefore change in magnetic field [tex]dB=250-350=100mT[/tex]
Emf induced is given by
[tex]e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}[/tex]
[tex]e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V[/tex]
Magnitude of induced emf is equal to 0.00635 V
2. In a series circuit, all resistors have identical currents.
a) What is the relationship between the power and resistance of these resistors?
b) In a parallel circuit, all resistors have identical voltages. What is the relationship
between the power and resistance of these resistors?
Answer:
Explanation:
In series connection of resistors, same current flows in the circuit.
Power dissipated by a resistor is
P = i²R
And since same current flows in them, it implies that the power is directly proportional to Resistance, so the higher the resistance of the resistor the higher the power dissipated in the series connection. Also, the lower the resistance, the lower the power dissipated.
2. In parallel connection, same voltage is applied across the resistor.
So, power dissipated by each resistor is
P = V² / R
So, since the same voltage is applied across parallel connection, then, power dissipated in each resistor is inversely proportional to the resistance.,
So, the higher the resistance, the lower the power and the lower the resistance, the higher the power.
Calcula el peso aparente de una bola de aluminio de 50 cm3, cuando se encuentra totalmente sumergida en alcohol. Datos: la densidad del aluminio es 2,7 g / cm3 y la densidad del alcohol es 0,8 g / cm3
Answer:
W_apparent = 93.1 kg
Explanation:
The apparent weight of a body is the weight due to the gravitational attraction minus the thrust due to the fluid where it will be found.
W_apparent = W - B
The push is given by the expression of Archimeas
B = ρ_fluide g V
ρ_al = m / V
m = ρ_al V
we substitute
W_apparent = ρ_al V g - ρ_fluide g V
W_apparent = g V (ρ_al - ρ_fluide)
we calculate
W_apparent = 980 50 (2.7 - 0.8)
W_apparent = 93100 g
W_apparent = 93.1 kg
A rectangular coil of wire with a dimension of 4 cm x 5 cm and 10 turns is located between the poles of a large magnet that produces a uniform magnetic field of 0.75 T. The surface of the coil which is originally parallel to the field is rotated in 0.10 s, so that its surface is perpendicular to the field. Calculate the average induced emf across the ends of coil as the coil rotates.
Answer:0.15 V
Explanation:
Given
Dimension of coil [tex]4cm\times 5cm[/tex]
Area of coil [tex]A=4\times 5=20\ cm^2[/tex]
Magnetic field [tex]B=0.75\ T[/tex]
Time of rotation [tex]t=0.1\ s[/tex]
No of turns [tex]N=10[/tex]
Initial flux associated with the coil
[tex]\phi_i=N(B\cdot A)[/tex]
[tex]\phi_i=N(BA\cos \theta )[/tex]
where [tex]\theta [/tex]=angle between magnetic field and area vector of coil
[tex]\phi_i=N(BA\cos 90 )[/tex]
Finally when coil is perpendicular to the field
[tex]\phi_f=N(B\cdot A)[/tex]
[tex]\phi_i=N(BA\cos 0 )[/tex]
and induced emf is given by
[tex]e=-\frac{d\phi }{dt}[/tex]
[tex]e=-\frac{\phi_1-\phi_2}{t-0}[/tex]
[tex]e=-\frac{(0-10\times 0.75\times 20\times 10^{-4})}{0.1}[/tex]
[tex]e=0.15\ V[/tex]
In part (a), suppose that the box weighs 128 pounds, that the angle of inclination of the plane is θ = 30°, that the coefficient of sliding friction is μ = 3 /4, and that the acceleration due to air resistance is numerically equal to k m = 1 3 . Solve the differential equation in each of the three cases, assuming that the box starts from rest from the highest point 50 ft above ground. (Assume g = 32 ft/s2 and that the downward velocity is positive.)
Answer:
v(t) = 21.3t
v(t) = 5.3t
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
Explanation:
When no sliding friction and no air resistance occurs:
[tex]m\frac{dv}{dt} = mgsin \theta[/tex]
where;
[tex]\frac{dv}{dt} = gsin \theta , 0 < \theta < \frac{ \pi}{2}[/tex]
Taking m = 3 ; the differential equation is:
[tex]3 \frac{dv}{dt}= 128*\frac{1}{2}[/tex]
[tex]3 \frac{dv}{dt}= 64[/tex]
[tex]\frac{dv}{dt}= 21.3[/tex]
By Integration;
[tex]v(t) = 21.3 t + C[/tex]
since v(0) = 0 ; Then C = 0
v(t) = 21.3t
ii)
When there is sliding friction but no air resistance ;
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta[/tex]
Taking m =3 ; the differential equation is;
[tex]3 \frac{dv}{dt}=128*\frac{1}{2} -\frac{\sqrt{3} }{4}*128*\frac{\sqrt{3} }{4}[/tex]
[tex]\frac{dv}{dt}= 5.3[/tex]
By integration; we have ;
v(t) = 5.3t
iii)
To find the differential equation for the velocity (t) of the box at time (t) with sliding friction and air resistance :
[tex]m \frac{dv}{dt}= mg sin \theta - \mu mg cos \theta - kv[/tex]
The differential equation is :
= [tex]3 \frac{dv}{dt}=128*\frac{1}{2} - \frac{ \sqrt{ 3}}{4}*128 *\frac{ \sqrt{ 3}}{2}-\frac{1}{3}v[/tex]
= [tex]3 \frac{dv}{dt}=16 -\frac{1}{3}v[/tex]
By integration
[tex]v(t) = 48 + Ce ^{\frac{t}{9}[/tex]
Since; V(0) = 0 ; Then C = -48
[tex]v(t) = 48 -48 e ^{ \frac{t}{9}}[/tex]
A solid, cylindrical wire conductor has radius R = 30 cm. The wire carries a current of 2.0 A which is uniformly distributed over the cross-section of the wire (current density is constant). What is the magnitude of the magnetic field due to the current in the wire at a radial distance of r = 200 cm from the center axis of the wire? HINT: Use Ampere’s law, noting that B is tangential.
Answer:
Explanation:
The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.
Applying Ampere's law :-
∫ B dl = μ₀ I ; I is current passing through ampere's loop
B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2
B = 2 x 10⁻⁷ T.
1. Electromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when they broadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to check your teeth. Even though they have different names and different applications, these types of radiation are really all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillating electric and magnetic fields. Which of the following statements correctly describe the various applications listed above?A. All these technologies use radio waves, including low-frequency microwaves. B. All these technologies use radio waves, including high-frequency microwaves. C. All these technologies use a combination of infrared waves and high-frequency microwaves. D. Microwave ovens emit in the same frequency band as some wireless Internet devices. E. The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies. F. All these technologies emit waves with a wavelength in the range 0.10 to 10.0 m. G. All the technologies emit waves with a wavelength in the range 0.01 to 10.0 km.2. Despite their extensive applications in communication systems, radio waves a not the only form of EM waves present in our atmosphere. Another form of EM radiation plays an even more important role in our life (and the life of our planet) sunlight. The sun emits over a wide range of frequencies; however, the fraction of its radiation that reaches earth's surface is mostly in the visible spectrum. (Note that about 35% of the radiation coming from the sun is absorbed directly by the atmosphere before even reaching the earth's surface). The earth, then, absorbs this radiation and reemits it as infrared waves. Based on this information, which of the following statements is correct? A. The earth absorbs visible light and emits radiation with a shorter wavelength. B. The earth absorbs visible light and emits radiation with a longer wavelength. C. The earth absorbs visible light and emits radiation with a lower frequency. D. The earth absorbs visible light and emits radiation with a higher frequency.
For the first question
Answers:
b.) All these technologies use radio waves, including high-frequency microwaves
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
For second question
Answer:
B. The earth absorbs visible light and emits radiation with a longer wavelength
C. The earth absorbs visible light and emits radiation with a lower frequency.
Both options are correct
Explanation:
In the EM spectrum, wavelenght reduces with increasing energy and frequency increases with increasing energy. The order of increasing energy is radio wave, microwave, infrared rays, visible rays, ultraviolet rays, xrays, gamma rays.
If the earth absorbs visible rays of higher energy, and radiates infrared rays of lower energy, then this means that the radiated radiation will have a longer wavenght, and a lower frequency.
1) The statements that correctly describe the various applications given in the question are;
B. All these technologies use radio waves, including high-frequency microwaves
D. Microwave ovens emit in the same frequency band as some wireless Internet devices.
E. The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies.
F. All these technologies emit waves with a wavelength in the range 0.10 to 10.0 m.
2) The correct statements on the EM spectrum in the question are;
B. The earth absorbs visible light and emits radiation with a longer wavelength.
C. The earth absorbs visible light and emits radiation with a lower frequency.
1) We are told about different waves being used by different technologies such as;
- Radio and TV emitting radio waves
- microwave oven using microwaves to cook food
- Dentists checking teeth with x-rays
All these waves are electromagnetic waves.
Looking at the options, options B, D, E and F are true because;
The given technologies all make use of radio waves as well as high frequency microwaves. Also, from research it is known that the frequency band emitted by microwaves is the same as that of some internet devices since some of the waves emitted by internet devices are the shortest wavelengths.The wavelength of the waves emitted by the given technologies are usually within the range of 0.1m to 10m2) The correct options are Options B and C because;
In the EM radiation spectrum, when the energy increases, the wavelength usually reduces and the frequency usually increases. The order of radio waves according to increase in energy is;
radio waves < infrared radiation < visible rays < ultraviolet rays < xrays < gamma rays.
Now, if we say that the earth absorbs visible rays of higher energy, and radiation of infrared rays with lower energy, then from our definition, we can conclude that radiated waves will have a longer wavelength, and a lower frequency.
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The drawing shows three identical rods (A, B, and C) moving in different planes. A constant magnetic field of magnitude 4.50 T is directed along the +y axis. The length of each rod is L = 1.3 m, and the speeds are the same, vA = vB = vC = 2.6 m/s. For each rod, find the magnitude of the motional emf, and indicate which end (1 or 2) of the rod is positive. rod A V ---Select--- End 1 is positive. End 2 is positive. No emf in rod. rod B V ---Select--- End 1 is positive. End 2 is positive. No emf in rod. rod C V ---Select--- End 1 is positive. End 2 is positive. No emf in rod.
Answer:
A)
The emf is zero because the velocity of the rod is parallel to the direction of the magnetic field, so the charges experience no force.
B)
The emf is vBL
= (2.6 m/s)(4.50 T)(1.3 m)
= 15.21 V.
The positive end is end 2.
C)
The emf is zero because the magnetic force on each charge is directed perpendicular to the length of the rod.
A particle of positive charge ???? is assumed to have a fixed position at P. A second particle of mass m and negative charge −q moves at constant speed in a circle of radius r1, centered at P. Derive an expression for the work W that must be done by an external agent in the second particle in order to increase the radius of the circle of motion, centered at P, to r2.
Answer:
[tex]W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]
Explanation:
To find the work W to put the negative charge in the new orbit you can use the following formula:
[tex]W=\Delta K\\\\K=\frac{1}{2}mv^2[/tex]
That is, the total work is equal to the change in the kinetic energy of the negative charge. Then you calculate the speed of the electron, by using the second Newton Law and the expression for the electrostatic energy:
[tex]F=ma_c\\\\-k\frac{(q_1)(q_2)}{r_1^2}=m\frac{v^2}{r_1}\\\\v^2=k\frac{q_1q_2}{mr_1}[/tex]
r1: radius of the first orbit
m: mass of the negative charge
v: velocity of the charge
k: Coulomb's constant
q1: charge of the fixed particle at point P
q2: charge of the negative charge
Hence, the velocity of the charge in a new orbit with radius r2 is:
[tex]v'^2=k\frac{q_1q_2}{mr_2}[/tex]
Finally the work required to put the charge in the new orbit is:
[tex]W=\Delta K =\frac{1}{2}m[v'^2-v^2]\\\\W=\frac{1}{2}m[k\frac{q_1q_2}{mr_2}-k\frac{q_1q_2}{mr_1}]\\\\W=\frac{1}{2}kq_1q_2[\frac{1}{r_2}-\frac{1}{r_1}][/tex]
A fellow astronaut passes by you in a spacecraft traveling at a high speed. The astronaut tells you that his craft is 21.1 m long and that the identical craft you are sitting in is 17.3 m long. (a) According to your observation, how long is your craft? m (b) According to your observation, how long is the astronaut's craft? m (c) According to your observation, what is the speed of the astronaut's craft relative to your craft?
Answer:
A) 21.1 m
B) 17.3 m
C) 3.267x10^7 m/s
Explanation:
This is a case of special relativity.
Let the relative speed of astronauts ship to my ship be v.
According to my observation,
My craft is 21.1 m long, according to my observation, astronauts craft is 17.3 m long.
If we fix the reference frame as my ship, then the rest lenght of our identical crafts is 21.1 m and the relativistic lenght is 17.3 m
l' = 21.1 m
l = 17.3 m
From l = l'(1 - p^2)^0.5
Where p is c/v, and c is the speed of light
17.3 = 21.1 x (1 - p^2)^0.5
0.82 = (1 - p^2)^0.5
Square both sides
0.67 = 1 - p^2
P^2 = 0.33
P = 0.1089
Revall p = v/c
v/c = 0.1089
But c = speed of light = 3x10^8 m/s
Therefore,
v = 3x10^8 x 0.1089 = 3.267x10^7 m/s
Following are the response to the given points:
a) Its ship travels to the distance of [tex]21.1\ m[/tex]
b) The astronaut's craft would be at a range of [tex]17.3\ m[/tex]
c) Relativity's use of length contraction:
[tex]\to L=L_0(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to \frac{L}{L_0}=(\sqrt{1-\frac{v^2}{c^2}})[/tex]
Here,
[tex]\to \frac{L}{L_0}=\frac{17.3}{21.1}=0.81[/tex]
Hence
[tex]\to 0.81=(\sqrt{1-\frac{v^2}{c^2}}) \\\\\to 0.6561=1-\frac{v^2}{c^2}\\\\\to \frac{v^2}{c^2} =1- 0.6561\\\\\to \frac{v^2}{c^2} =0.3439\\\\\to \frac{v}{c} =0.58\\\\\to v= 0.58 c[/tex]
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An engineer is designing a contact lens. The material has is an index of refraction of 1.55. In order to yield the prescribed focal length, the engineer specifies the following dimensions: inner radius of curvature = +2.42 cm outer radius of curvature = +1.98 cm where the inner radius of curvature describes the surface that touches the eye, and the outer radius of curvature describes the surface that first interacts with incoming light. What is the focal length of this contact lens (in cm)?
Answer:20 cm
Explanation:
Given
Refractive index of material [tex]n=1.55[/tex]
Outer radius [tex]R_1=1.98\ cm[/tex]
Inner radius [tex]R_2=2.42\ cm[/tex]
using lens maker formula
[tex]\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})[/tex]
[tex]\frac{1}{f}=(1.55-1)(\frac{1}{1.98}-\frac{1}{2.42})[/tex]
[tex]f=\frac{10.895}{0.55}[/tex]
[tex]f=19.81\approx 20\ cm[/tex]
2. Categorize each statement as true or false. A cylindrical capacitor is essentially a parallel-plate capacitor rolled into a tube.The dielectric constant indicates the distance by which the two plates of a capacitor are separated.The charge on a capacitor increases quickly at first, then much more slowly as the capacitor charges.The voltage across a capacitor in an RC circuit increases linearly during charging.One of the principle purposes of a capacitor is to store electric potential energy.A capacitor charges rapidly when connected to an RC circuit with a battery. True False
The true statement is A. A cylindrical capacitor is a parallel-plate capacitor rolled into a tube, C. The charge on a capacitor increases quickly at first, then much more slowly as the capacitor charges E. One of the principal purposes of a capacitor is to store electric potential energy. F. A capacitor charges rapidly when connected to an RC circuit with a battery and the false statement are B. The dielectric constant indicates that the distance by which the two plates of a capacitor are separated and D. The voltage across a capacitor in an RC circuit increases linearly during charging.
Let's look at each statement one by one to categorize them as true or false.
a. True. A cylindrical capacitor can be thought of as a parallel-plate capacitor with the plates rolled into cylindrical shapes.
b. False. The dielectric constant is a measure of a material's ability to increase the capacitance of a capacitor, not a measure of the distance between the plates.
c. True. The charge on a capacitor follows an exponential curve, increasing rapidly at first and then more slowly as it approaches its maximum charge.
d. False. The voltage increases exponentially, not linearly, when charging a capacitor in an RC circuit.
e. True. Capacitors store electric potential energy in the electric field between their plates.
f. True. Initially, the capacitor in an RC circuit charges quickly, but the rate of charging decreases over time as it gets closer to full charge.
In a Tesla coil, a long solenoid with length l and cross-sectional area A is closely wound with N 1 turns of wire. A coil with N 2 turns surrounds it at its center. You measure the mutual inductance. Then your friend gives you a new solenoid made out of twice as much wire, so that it has twice as many turns and is twice as long. How much larger is the mutual inductance M with the new solenoid instead of the old one
The mutual inductance of a Tesla coil system with two solenoids will double if the length and number of turns of the primary solenoid are both doubled, assuming all other factors remain constant.
When we are given a Tesla coil configuration with a solenoid of length l and cross-sectional area A, closely wound with N1 turns of wire, and another coil with N2 turns surrounding it at its center, we can calculate the mutual inductance based on the properties of the solenoids. If a new solenoid is introduced that has twice as many turns and is twice as long, the mutual inductance M of the system will be affected.
To understand how the mutual inductance changes, let's remember that for a closely wound solenoid, the mutual inductance can be calculated by a formula incorporating the number of turns, permeability of the core material, cross-sectional area, and the length of the solenoid. The mutual inductance is directly proportional to the product of the number of turns of each coil, the magnetic permeability of the core, and the area of the cross-section, and inversely proportional to the length of the solenoid. Therefore, if we double the length l and the number of turns N1 while keeping all other factors constant, the mutual inductance will also double.
A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
before it hits the ground. We can ignore air resistance.
Answer: Vertical displacement = -27.6m
And takes 2.375 s
A pumpkin thrown at a horizontal speed of 4.0 m/s and travels 9.5 m horizontally before hitting the ground will take 2.375 seconds to hit the ground.
How to find time?Given the following information:
Horizontal speed of the pumpkin = 4.0 m/s
Horizontal distance traveled before hitting the ground = 9.5 m
Ignoring air resistance
Use the following formula to calculate the time it takes the pumpkin to hit the ground:
time = horizontal distance / horizontal speed
time = 9.5 m / 4.0 m/s = 2.375 seconds
Therefore, it takes the pumpkin 2.375 seconds to hit the ground.
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A young child hold a string attached to a balloon. What is the reaction force to the balloon pulling up on the earth?
Answer:
As the Ballon pulls up, the distance between the Ballon and the center of the earth increases, the gravitational pull reduces and the gravity potential energy increases.
Consider a rough asphalt roadway with 17 cm curbs and one 3.75 m lane in each direction and a 1.45 m shoulder on each side. The road has a longitudinal slope of 0.75% and is crowned in the center with a transverse slope of 2.25%. The local ordinance requires no ponding in the roadway (only on the shoulders) for a storm with an intensity of 90 mm/hr. A) Determine the maximum distance between curb inlets.
Answer:
Explanation:
Half width - 3.75 m
Rainfall intensity - 90 mm/hr
Longitudinal slope - 0.75%
Cross slope - 2.25%
Allowable limit of gutter flow, Q road = (Design constant x Intensity of rain x Area) / 360
= (0.91×90×(9×L1×10∧-4)/360 = 0.000204 L1
Using the standard design chart, allowable limit of gutter flow =0.018 m3/s
Therefore,0.018 = 0.000204 L1
L1 = 0.018 / 0.000204 = 88.24
Inlet spacing to be adapted is 88m