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3 years ago

7.

Physics

2 answers:

SSSSS [86.1K]3 years ago

7 0

Free fall is a special type of motion in which the gravity is the only force that acting upon an object.

<u>Explanation</u>:

In the free fall the object have gravity force only. when you dropped the ball from top of the building, and it will start falling. The balls falls due to gravity. Imagine when you have held the ball in your hand, gravity is acting, but the frictional force between your fingers and ball is acting on the opposite side. When the ball is kept in air, gravity is the only force that acts on the ball. Hence in free fall, the object has no another forces except gravity.

Igoryamba3 years ago

5 0

D. Free fall

Explanation:

An object is said to be in free fall when there is only one force acting on the body, which is the force of gravity.

Near the Earth's surface, the force of gravity acting on a body is given by

F = mg

where

m is the mass of the body

g is the acceleration of gravity (its value is $9.8 m/s^2$ )

The direction of this force is downward (towards the Earth's centre).

If we apply Newton's second law on an object in free-fall, we can find its acceleration. In fact, we have:

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

And substituting F,

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

So, every object in free-fall accelerates at $9.8 m/s^2$ towards the ground.

Learn more about free fall here:

brainly.com/question/1748290

brainly.com/question/11042118

brainly.com/question/2455974

brainly.com/question/2607086

#LearnwithBrainly

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(15pts) A hungry 12.0 kg fish is coasting from west to east at 75 cm/s when it suddenly swallows a 1 kg fish swimming towards it

faust18 [17]

Answer:

The speed of the big fish after swallowing the small fish is 0.38 m/s.

Explanation:

Consider west to east direction as positive and the opposite direction as negative.

Given:

Mass of big fish (m₁) = 12.0 kg

Initial velocity of big fish (u₁) = 75 cm/s = 0.75 m/s

Mass of small fish (m₂) = 1 kg

Initial velocity of small fish (u₂) = -4 m/s (Direction is opposite to u₁)

After swallowing the small fish, both the fishes move together with same velocity. Let the velocity be 'v'.

So, as there are no effects of drag or any other forces, the given scenario can be considered as a case of inelastic collision where the objects move together with same velocity after collision.

The momentum is conserved in inelastic collision. Therefore,

Initial momentum of the fishes = Final momentum of the fishes

$m_1u_1+m_2u_2=(m_1+m_2)v\\\\v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}$

Now, plug in the given values and solve for 'v'. This gives,

$v=\frac{12.0\times 0.75+1\times (-4)}{12.0+1}\\\\v=\frac{9-4}{13}\\\\v=\frac{5}{13}=0.38\ m/s$

Therefore, the speed of the big fish after swallowing the small fish is 0.38 m/s

3 0

3 years ago

What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?

Alexandra [31]

600/3 = 200
the slope is 200m/min

OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour

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3 years ago

A ____________ galaxy has arms of gas and dust; where as _____________ galaxy is a category of galaxy that does not have a disti

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Answer:

spiral, an irregular

Explanation:

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3 years ago

An object is at rest in front of a compressed spring. It travels over a surface that exerts a kinetic frictional force on it and

PolarNik [594]

Answer:

the object will travel 0.66 meters before to stop.

Explanation:

Using the energy conservation theorem:

$E_i+K_i+W_f=K_f+U_f$

The work done by the friction force is given by:

$W_f=F_f*d\\W_f=\µ*m*g*d\\W_f=0.35*4*9.81*d\\W_f=13.7d[J]$

so:

$\frac{1}{2}1800*(10*10^{-2})+0-13.7d=0+0\\d=0.66m$

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3 years ago

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

IceJOKER [234]

Answer:

Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

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3 years ago