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2 years ago

What is the velocity of an object that has been in free fall for 2.5 s?

Physics

1 answer:

Vaselesa [24]2 years ago

6 0

After 2.5 seconds of free fall, an item will be moving downhill at a speed of 25 m/s.Option D is correct.

<h3>What is velocity?</h3>

The change of displacement with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is;

t(time)= 2.5 s

u(falling velocity)=0 m/sec

v(striking velocity)=? (m/sec)

Newton's first equation of motion:

v=u+at

v=u+gt

v=10 × 2.5

v= 25 m/s downward(+ ve taken as downward)

The velocity of an object that has been in free fall for 2.5 s will be 25 m/s downward.

Hence, option D is correct.

To learn more about the velocity refer to the link;

brainly.com/question/862972

#SPJ1

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a small asteroid of mass 125 kg is orbiting a planet that has a mass of 3.52x 10^13 what is the radial distance between the aste

asambeis [7]

Answer:

r = 2.031 x 10⁶ m = 2031 km

Explanation:

In order for the asteroid to orbit the planet, the centripetal force must be equal to the gravitational force between asteroid and planet:

Centripetal Force = Gravitational Force

mv²/r = GmM/r²

v² = GM/r

r = GM/v²

where,

r = radial distance = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

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Therefore,

r = (6.67 x 10⁻¹¹ N.m²/kg²)(3.52 x 10¹³ kg)/(0.034 m/s)²

7 0

3 years ago

HELP ASAP TIMED TEST

balu736 [363]

Answer:

<em>Correct choice: b 4H</em>

Explanation:

<u>Conservation of the mechanical energy</u>

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

$U_1 = mgH$

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

$\displaystyle U_2=\frac{1}{2}mv^2$

Since the energy is conserved, U1=U2

$\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]$

For the speed to be double, we need to drop the snowball from a height H', and:

$\displaystyle mgH'=\frac{1}{2}m(2v)^2$

Operating:

$\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]$

Dividing [2] by [1]

$\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}$

Simplifying:

$\displaystyle \frac{H'}{H}=4$

Thus:

H' = 4H

Correct choice: b 4H

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