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3 years ago

A ball is thrown forward at 5 m/s how would the path of the ball differ on earth than on the moon

2 answers:

6 0

In this problem, we are given that a ball is thrown upward and that it has an initial velocity of 5 m/s. Its path both on Earth and on the moon will be trajectory in nature. This is because while its horizontal velocity is maintained, its vertical velocity increases.

Degger [83]3 years ago

6 0

Answer:

The ball on the moon would have a more straight path than on the Earth, and it would have a longer range

Explanation:

In both cases, the motion of the ball is the motion of a projectile, which has a parabolic trajectory. In fact, the path is the resultant of two indipendent motions:

- A horizontal uniform motion, with constant horizontal velocity 5 m/s

- An accelerated vertical motion, with constant acceleration downward equal to the gravitational acceleration of the planet

On the Moon, the value of the gravitational acceleration is about 1/6 that of the Earth. This means that the downward acceleration of the ball on the Moon is smaller than on the Earth: so, the ball will take more time to reach the ground on the Moon, and so it will have a longer range, and a more straight trajectory.

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4 years ago

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An airplane is flying in horizontal flight at a constant velocity. The weight of the airplane is 40,000 N. The wings produce a l

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Explanation:

the plane is not going up and the rest are true

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3 years ago

If a fully loaded shopping cart has a head-on collision with an empty cart both moving at the same speed toward each other, whic

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Answer:

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4 years ago

Light of wavelength 633 nm falls on a single slit 0.5 mm wide and forms a diffraction pattern on a screen 1.0 m away from the sl

andrey2020 [161]

Answer:

The position of the first dark spot on the positive side of the central maximum is 1.26 mm.

Explanation:

Given that,

Wavelength of light is 633 nm.

Slit width, d = 0.5 mm

The diffraction pattern forms on a screen 1 m away from the slit. We need to find the position of the first dark spot on the positive side of the central maximum.

For destructive interference :

$\dfrac{dY}{D}=n\lambda$

Y is the distance of the minima from central maximum

Here, n = 1

$Y=\dfrac{n\lambda D}{d}\\\\Y=\dfrac{1\times 633\times 10^{-9}\times 1}{0.5\times 10^{-3}}\\\\Y=0.00126\ m\\\\Y=1.26\times 10^{-3}\ m\\\\Y=1.26\ mm$

So, the position of the first dark spot on the positive side of the central maximum is 1.26 mm.

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3 years ago

A water tank is in the shape of an inverted cone with depth 10 meters, and top radius 8 meters. Water is flowing into the tank a

Elis [28]

Answer:

The value of leaking rate in the question is repeated. By searching on the web I could find the correct value wich is 0.002h^2 m^3 /min.

The depth of the water has to be equal to 7.07 m in order to have a stationary volume.

Explanation:

In order to have a stationary water level the flow of water that comes into the tank (0.1 m^3/min) must be equal to the flow of water that goes out of the tank (0.002*h^2 m^3/min), therefore:

0.002*h^2 = 0.1

h^2 = 0.1/0.002

h^2 = 50

h = sqrt(50) = 7.07 m

7 0

3 years ago