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Levart [38]
3 years ago
6

You have just landed on Planet X. You release a 100-g ball from rest from a height of 10.0 m and measure that it takes 3.40 s to

reach the ground. Ignore any force on the ball from the atmosphere of the planet. How much does the 100-g ball weigh on the surface of Planet X?
Physics
1 answer:
Nana76 [90]3 years ago
6 0

Answer:

w = 0.173 N

Explanation:

The weigh of any object is computed by multiplying its mass to the acceleration of gravity, so we need to find the gravity on that planet in order to compute the weigh we want.

The ball has a mass of 0.1 kg and its released from a height of 10 m, therefore it is in a free fall motion with gravity acting as a constant acceleration on the body, we can use the equations for free fall movement in order to determine the value for this acceleration:

y(t) = v_0 * t + y_0 - 0.5 * g * t^2

y(t) is the position in the end of the movement, when t = 3.4 s, so y(t) = 0 m.

v_0 is the initial velocity, in this case v_0 = 0 m/s.

y_0 is the initial position of the ball, in this case it is 10 m.

g is the gravity that we want to know.

Applying these values in the equation we have:

0 = 0*(3.4) + 10 - 0.5*g*(3.4)^2

0 = 10 - 0.5*11.56*g

0 = 10 -5.78*g

5.78*g = 10

g = 1.73 m/s^2

Then we can use this value to find out the weigh of the ball in that planet:

w = g*m = 0.1*1.73 = 0.173 N

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Explanation:

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6 0
3 years ago
A hula hoop is rolling along the ground with a translational speed of 26 ft/s. It rolls up a hill that is 16 ft high. Determine
nikklg [1K]

Answer:12.8 ft/s

Explanation:

Given

Speed of hoop v=26\ ft/s

height of top h=16\ ft

Initial energy at bottom is

E_b=\frac{1}{2}mv^2+\frac{1}{2}I\omega ^2

Where m=mass of hoop

I=moment of inertia of hoop

\omega=angular velocity

for pure rolling v=\omega R

I=mR^2

E_b=\frac{1}{2}mv^2+\frac{1}{2}mR^2\times (\frac{v}{R})^2

E_b=mv^2=m(26)^2=676m

Energy required to reach at top

E_T=mgh=m\times 32.2\times 16

E_T=512.2m

Thus 512.2 m is converted energy is spent to raise the potential energy of hoop and remaining is in the form of kinetic and rotational energy

\Delta E=676m-512.2m=163.8m

Therefore

163.8 m=mv^2

v=\sqrt{163.8}

v=12.798\approx 12.8\ ft/s

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3 years ago
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If a car changes ita velocity from 32km/hr to 54km/hr in 8.0 seconds, what is its acceleration
Andrews [41]
Firstly, let's convert the velocities in km/hr to m/s
32*1000/3600=8.89m/s
54*1000/3600=15m/s
From the formula, acceleration=V-U/t
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hope this helps.
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