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4 years ago

Despite a very strong wind, a tennis player

Physics

1 answer:

Gnoma [55]4 years ago

6 0

Answer:

Option 5. 1 and 3

Solution:

The only forces acting on the tennis ball after it has left contact with the racquet and the instant before it touches the ground are the force of gravity in the downward direction and the force by the air exerted on the ball.

The ball after it left follows the path of trajectory and as it moves forward in the horizontal direction the force of the air acts on it.

In the whole projectile motion of the ball, the acceleration due to gravity acts on the ball thus the force of gravity acts on the ball in the downward direction before it hits the ground.

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What will happen to the velocity and momentum of each ball when the small ball hits the heavier large ball?

kap26 [50]

Most of the momentum is transferred to the ball on top. Since the collision in this situation is elastic, momentum is conserved, meaning the momentum of both balls before hitting the floor is equal to the momentum of both balls right after the collision.

7 0

3 years ago

A baseball player hits a baseball with a bat. The mass of the ball is 0.25 kg. The ball accelerates at 200 m/s2. What is the for

Brums [2.3K]

Answer:

50N

Explanation:

Force (N) = mass (kg) × acceleration (m/s²)

0.25kg times 200m/s² = 50N

6 0

3 years ago

a person runs 27.0km west then turns around and runs 13.0km east what's the distance and displacement ?

Alexxandr [17]

Answer:

See below

Explanation:

Distance = 27 + 13 = 40 km

Displacement = 27 - 13 = 14 km

7 0

3 years ago

The differential equation below models the temperature of an 88°C cup of coffee in a 24°C room, where it is known that the coffe

densk [106]

Answer:

$y = 24+64\cdot e^{-150\cdot t}$

Explanation:

Let solve the differential equation by separating corresponding variables:

$\int\limits^t_0\, dt = -\frac{1}{150} \int\limits^y_{y_{o}} \frac{dy}{y-24}$

The solution of this equation is:

$t = -\frac{1}{150}\cdot (\ln|y-24|-\ln |y_{o}-24|)$

The explicit form of the temperature as a function of time is:

$\ln |y-24|=-150\cdot t + \ln |y_{o}-24|$

$y-24 = C\cdot e^{-150\cdot t}$

The value of the integration constant is:

$C = 64$

The complete expression is:

$y = 24+64\cdot e^{-150\cdot t}$

3 0

3 years ago

If a person weighs 1000 N on the surface of the Earth, how much will they weigh on the surface of the Moon? What “g” value shoul

Dafna1 [17]

Answer:

Hence the weight of the person on the moon is 162.4, and the value of g used is 1.624 m/s²

Explanation:

from the question,

W = mg........................ Equation 1

Where W = weight of the man on Earth, m = mass of the man, g = acceleartion due to gravity of the man

make m the subject of the equation

m = W/g.............. Equation 2

Given: W = 1000 N,

Constant: g = 10 m/s²

Therefore,

m = 1000/10

m = 100 kg

Weight on the moon

W' = mg'

W' = 100(1.624)

W' = 162.4 N.

Hence the weight of tthe person on the moon is 162.4, and the value of g used is 1.624 m/s²

5 0

3 years ago