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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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Can an object be accelerated while traveling at constant velocity? Why or why not?

AysviL [449]

Answer:

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes. If an object's velocity is constant, however, its acceleration will be zero.

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3 years ago

An infinite line of charge with charge density λ1 = -2 μc/cm is aligned with the y-axis as shown. 1) what is ex(p), the value of

Natasha2012 [34]

As we know by Gauss's law that

$\int E.dA = \frac{q}{\epsilon_0}$

so for line charge the gaussian surface is cylindrical in shape

so we will have

$E(2\pi RL) = \frac{\lambda L}{\epsilon_0}$

now by rearranging the terms

$E = \frac{\lambda}{2\pi \epsilon_0 R}$

so here we will have to find the x component of electric field so it is given by above equation

$E_x = \frac{\lambda}{2\pi \epsilon_0 x}$

here x = distance from the wire where we need to find electric field

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3 years ago

Calculate the average induced voltage between the tips of the wings of a Boeing 747 flying at 940 km/hr above East Lansing. The

ANTONII [103]

Answer:

option (b)

Explanation:

B = 0.7 x 10^-4 T

v = 940 km/h = 940 x 5 / 18 = 261.11 m /s

l = 60 m

Th emotional emf is given by

e = B v l

e = 0.7 x 10^-4 x 261.11 x 60 = 1.096 V

e = 1.1 V

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3 years ago

Give a example of frequency which an elephant can hear and a tiger cannot hear include a unit in ansawer

lisov135 [29]

P1.5.3 thats what i got........... hope its right not sure though

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3 years ago

E asteroid belt circles the sun between the orbits of mars and jupiter. one asteroid has a period of 6.0 earth years. part a wha

anzhelika [568]

To solve the problem, use Kepler's 3rd law :

T² = 4π²r³ / GM

Solved for r :

r = [GMT² / 4π²]⅓

but first covert 6.00 years to seconds :

6.00years = 6.00years(365days/year)(24.0hours/day)(6...
= 1.89 x 10^8s

The radius of the orbit then is :

r = [(6.67 x 10^-11N∙m²/kg²)(1.99 x 10^30kg)(1.89 x 10^8s)² / 4π²]⅓
= 6.23 x 10^11m

6 0

3 years ago