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serg [7]
2 years ago
6

A cannonball and a marble roll smoothly from rest down an incline. Is the cannonball’s(a) time to the bottom and(b) translationa

l kinetic energy at the bottom more than, less than, or the same as the marble’s?
Physics
1 answer:
Salsk061 [2.6K]2 years ago
5 0

Answer:

Explained

Explanation:

a) Cannonball and marble will reach the bottom at the same time, as acceleration due to gravity is independent of the mass of the body.

b)Translational Kinetic energy of the cannonball will be more than that of the marble as the cannonball has more mass than the marble.

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An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
solniwko [45]

Answer:

A) The new amplitude = 0.048 m

B) Period T = 0.6 seconds

Explanation: Please find the attached files for the solution

4 0
2 years ago
In a distant solar system, a giant planet has
sergeinik [125]

Answer:

mass of the planet: 5.9\,10^{26}\,kg

Explanation:

When a moon keeps a circular orbit around a planet, it is the force of gravity the one that provides the centripetal force to keep it in its circular trajectory of radius R. So if we can write that in such cases (being the mass of the planet "M" and the mass of the moon "m"), we can form an equation by making the centripetal force on the moon equal the force of gravity (using the Newton's Universal Law of Gravity):

m\frac{v^2}{R}=G\frac{M\,m}{R^2}

where we used here the tangential velocity (v) of the moon around the planet. This equation can be further simplified by dividing both sides by "m" and multiplying both sides by the orbital radius R:

m\frac{v^2}{R}=G\frac{M\,m}{R^2}\\v^2=G\frac{M}{R}

Notice that the mass of the moon has actually disappeared from the equation, which tells us that the orbiting velocity and period do not depend on the mass of the moon, but on the mass of the actual planet.

We know the orbital radius R (5.32\,10^5\,km=5.32\,10^8\,m, the value of the Universal Gravitational constant G, and we can estimate the value of the tangential velocity of the moon since we know it period: 36.3 hrs = 388800 seconds.

We know that the moon makes a full circumference (2\,\pi\,R) in 388800 seconds, therefore its tangential velocity is:

v=\frac{2\,\pi\,5.32\,10^8}{388800} \frac{m}{s} \\v=8.6\,10^3\,\frac{m}{s}

where we rounded the velocity to one decimal.

Notice that we have converted all units to the SI system, so when using the formula to solve for the mass of the planet, the answer comes directly in kg.

Now we use this value for the tangential velocity to estimate the mass of the planet in the first equation we made and simplified:

v^2=G\frac{M}{R}\\M=\frac{v^2\,R}{G} \\M=\frac{(8.6\,10^3)^2\,5.32\,10^8}{6.67\,10^{-11}}kg\\M=5.9\,10^{26}\,kg

8 0
2 years ago
In the system shown above, the pulley is a uniform disk with a mass of .75 kg and a radius of 6.5 cm. The coefficient of frictio
lord [1]

Answer:

i am answering the same question 3rd time

please find the answer in the images attached.

5 0
1 year ago
In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis
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M1 descending
−m1g + T = m1a 

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

a = (m2 − m1)g/m1 + m2
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5 0
2 years ago
Every winter I fly home to Chicago. It takes 3 hours. What is my average speed?
Tanya [424]

It depends on where you live when you're not visiting Chicago. We need to know the distance of the trip.

Your average speed on the trip is . . .

(total distance in miles) / (3 hours)

miles per hour

5 0
2 years ago
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