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navik [9.2K]

Answer:yayyyy

Explanation:

4 0

3 years ago

Read 2 more answers

Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that

Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

$g_e$ = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

$m=M-0.1M\\\Rightarrow m=0.9M$

$g_e=G\frac{M}{r^2}$

$g=G\frac{0.9M}{r^2}$

Dividing the equations we get

$\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2$

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0

3 years ago

You throw a ball upward with a speed of 14m/s. What is the acceleration of the ball after it leaves your hand? Ignore air resist

omeli [17]

The acceleration of the ball after leaving the hand is $9.8 m/s^2$ downward

Explanation:

In order to find the acceleration of the ball during its motion, we have to study which forces are acting on it.

After the ball leaves the hand, if we neglect air resistance, there is only one force acting on the ball: the force of gravity, whose magnitude is

$F=mg$

where m is the mass of the ball and g is the acceleration of gravity ( $g=9.8 m/s^2$ ), acting in the downward direction.

According to Newton's second law, the acceleration of the ball is given by

$a=\frac{\sum F}{m}$

where

$\sum F$ is the net force acting on the ball

After the ball leaves the hand, the only force acting on it is the force of gravity, so we can substitute (mg) into the previous equation:

$a=\frac{mg}{m}=g=9.8 m/s^2$

This means that the acceleration of the ball remains $9.8 m/s^2$ downward for the entire motion, after leaving the hand.

Learn more about Newton's second law:

brainly.com/question/3820012

#LearnwithBrainly

5 0

3 years ago

The momentum of an object is determined to be 7.2 x 10-3 cm kg x m/s. Express this as provided or use any equivalent unit. How i

Leno4ka [110]

Complete Question:

The momentum of an object is determined to be 7.2 × 10-3 kg⋅m/s. Express this quantity as provided or use any equivalent unit. (Note: 1 kg = 1000 g).

Answer:

7.2 gm/s.

Explanation:

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

Given the following data;

Momentum = 7.2 * 10^-3 kgm/s

1 kg = 1000 g

Substituting the unit in kilograms with grams, we have;

Momentum = 7.2 * 10^-3 * 1000 gm/s

<em>Momentum = 7.2 gm/s. </em>

7 0

3 years ago

A 5-kg ball collides inelastically head-on with a 10-kg ball, which is initially stationary. Which of the following statements i

NARA [144]

Answer:

The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball.

Explanation:

In inelastic collision, the total momentum is always conserved after collision but the kinetic energy is reduced.

Momentum is Mass X velocity.

5 kg ball is in motion, while 10 kg ball is stationary; that is zero velocity.

The momentum of 10 kg ball before collision is zero while the momentum of 5 kg ball before collision is more than zero. Therefore, the magnitude of change in momentum will not be equal.

Next possible options are in kinetic Energy

Initial Kinetic energy = $\frac{1}{2}mu^2$

Final kinetic energy = $\frac{1}{2}mv^2$

Change in kinetic energy = Final Kinetic Energy - Initial Kinetic Energy

Change in kinetic energy of 5kg ball = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2$

Since the 5-kg ball has initial velocity (u), the magnitude of the change in velocity will be reduced.

Change in kinetic energy of 10kg ball:

the ball is initially at rest, therefore the initial velocity (u) will be zero (0)

Δ K.E = $\frac{1}{2}mv^2 -\frac{1}{2}mu^2 = \frac{1}{2}m(v-u)^2 = \frac{1}{2}m(v-0)^2 = \frac{1}{2}mv^2$

From the solution above, the magnitude of the change in velocity experienced by 10 kg ball is higher than 5 kg ball.

Hence, The magnitude of the change of velocity the 5-kg ball experiences is less than that of the 10-kg ball

4 0

4 years ago