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vitfil [10]
3 years ago
5

Gravitational potential energy (U) depends on three things: the object’s mass (m), its height (h), and gravitational acceleratio

n (g), which is 9.81 m/s2 on Earth’s surface: U = mgh Energy is measured in joules (J). One joule is equal to one 1 kg•m2/s2. When calculating the energy of an object, it is helpful to convert the mass and height to kilograms and meters.
A. What is the mass of the 50-gram car, in kilograms?

B. Set Hill 1 to 75 cm and the other hills to 0 cm. What is the height in meters?

C. What is the potential energy of the car, in joules?
Physics
2 answers:
Viefleur [7K]3 years ago
8 0

A. 500N

B. 0.75 m

C. 7865432J


adoni [48]3 years ago
7 0

Explanation:

A. Mass of the car ,m = 50 g

1 g = 0.001 kg

50 g = 0.001 × 50 kg = 0.05 kg

B. Height of hill ,h= 75 cm = 0.75 m

1 cm = 0.01 m

Height of other hill ,h' = 0 cm = 0 m

C. Gravitational potential energy (U):

U=mgh

Mass of car = m = 0.05 kg

Height at which car is = h = 0.75 m

U= 0.05 kg\times 0.75 m\times 9.8 m/s^2=0.3675 kg/m^2/s^2=0.3675 J

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Answer:

Option C - 39.2 J

Explanation:

We are given that;

Mass; m = 2 kg.

Distance moved off the floor;d = 10 m.

Acceleration due to gravity;g = 9.8 m/s².

We want to find the work done.

Now, the Formula for work done is given by;

Work = Force × displacement.

In this case, it's force of gravity to lift up the boots, thus;

Formula for this force is;

Force = mass x acceleration due to gravity

Force = 2 × 9.8 = 19.2 N

∴ Work done = 19.6 × 2

Work done = 39.2 J.

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A ball is dropped from rest at point O. After falling for some time, it passes by a window of height 3.3 m and it does so in 0.2
stiv31 [10]

Answer:

Speed at which the ball passes the window’s top = 10.89 m/s

Explanation:

Height of window = 3.3 m

Time took to cover window = 0.27 s

Initial velocity, u = 0m/s

We have equation of motion s = ut + 0.5at²

For the top of window (position A)

                     s_A=0\times t_A+0.5\times 9.81t_A^2\\\\s_A=4.905t_A^2

For the bottom of window (position B)

                     s_B=0\times t_B+0.5\times 9.81t_B^2\\\\s_A=4.905t_B^2

\texttt{Height of window=}s_B-s_A=3.3\\\\4.905t_B^2-4.905t_A^2=3.3\\\\t_B^2-t_A^2=0.673

We also have

                 t_B-t_A=0.27

Solving

         t_B=0.27+t_A\\\\(0.27+t_A)^2-t_A^2=0.673\\\\t_A^2+0.54t_A+0.0729-t_A^2=0.673\\\\t_A=1.11s\\\\t_B=0.27+1.11=1.38s

So after 1.11 seconds ball reaches at top of window,

       We have equation of motion v = u + at

                                     v_A=0+9.81\times 1.11=10.89m/s

Speed at which the ball passes the window’s top = 10.89 m/s                

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Complete Question:

A hollow cylinder with an inner radius of 4.0mm and an outer radius of 30mm conducts a 3.0-A current flowing parallel to the axis of the cylinder. If the current density is uniform throughout the wire, what is the magnitude of the magnetic field at a point 12mm from its center?

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Inner radius, a = 4.0 mm = 0.004 m

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Let d² = r² - a²

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The magnitude of the magnetic field is given by:

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