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Nikitich [7]
3 years ago
6

As an object falls:

Physics
1 answer:
alex41 [277]3 years ago
5 0

Answer:

c it is not accelerating on it's on but gravity pulls it there for velocity increases.

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A 1 200-kg car traveling initially at vCi 5 25.0 m/s in an easterly direction crashes into the back of a 9 000-kg truck moving i
sukhopar [10]

Answer:

The velocity of the truck after the collision is 20.93 m/s

Explanation:

It is given that,

Mass of car, m₁ = 1200 kg

Initial velocity of the car, v_{Ci}=25\ m/s

Mass of truck, m₂ = 9000 kg

Initial velocity of the truck, v_{Ti}=20\ m/s

After the collision, velocity of the car, v_{Cf}=18\ m/s

Let v is the velocity of the truck immediately after the collision. The momentum of the system remains conversed.

initial\ momentum=final\ momentum

1200\ kg\times 25\ m/s+9000\ kg\times 20\ m/s=1200\ kg\times 18+9000\ kg\times v

210000-21600=9000\ kg\times v

v=20.93\ m/s

So, the velocity of the truck after the collision is 20.93 m/s. Hence, this is the required solution.

8 0
3 years ago
Write any two different between work and power?​
attashe74 [19]

Answer:

1. Work Is a type of Physical Activity  

2. Power is Basically Having control of Society

Explanation:

3 0
3 years ago
Read 2 more answers
Can anyone help me plz I m confused with this last question?
Gemiola [76]

Answer:

764728497693575177015915715716245378tr7138

Explanation:

8 0
3 years ago
A student at another university repeats the experiment you did in lab. Her target ball is 0.860 m above the floor when it is in
dexar [7]

Answer:

K = 0.076 J

Explanation:

The height of the target, h = 0.860  m

The mass of the steel ball, m = 0.0120 kg

Distance moved, d = 1.50 m

We need to find the kinetic energy (in joules) of the target ball just after it is struck. Let t is the time taken by the ball to reach the ground.

h=ut+\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}}

Put all the values,

t=\sqrt{\dfrac{2\times 0.860 }{9.8}} \\\\=0.418\ s

The velocity of the ball is :

v=\dfrac{1.5}{0.418}\\\\= $$3.58\ m/s

The kinetic energy of the ball is :

K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 0.0120\times 3.58^2\\\\=0.076\ J

So, the required kinetic energy is 0.076 J.

6 0
3 years ago
FIND MEFIND V AT THE FIRST HILLFIND HEIGHT OF THE SECOND HILLFIND V AT POINT A
vitfil [10]

1)

At the starting point, the spring releases potential energy which is converted to kinetic energy of the truck. The formula fr calculating the elastic potential energy of the spring is expressed as

PE = 1/2kx^2

where

x is the extension of the spring

k is the spring constant

From the information given,

k = 8500

x = 7

Thus,

PE = 1/2 x 8500 x 7^2 = 208250 J

Since elastic potential energy of spring = kinetic energy of the truck, it means that

Kinetic energy = 208250

The formula for calculating kinetic energy is expressed as

KE = 1/2mv^2

where

m = mass of truck

v = velocity of truck

From the diagram,

m = 600

Thus,

208250 = 1/2 x 600 x v^2

208250 = 300v^2

v^2 = 208250/300 = 694.17

v = square root of 694.17

v = 26.35 m/s

The velocity at which the truck is moving is 26.35 m/s

The potential energy of the truck at that point is calculated by apply the formula,

Potential energy = mgh

where

g = acceleration due to gravity and its value is 9.81 m/s

h is the height of the truck and it is 20

m is the mass of the truck and it is 600

Thus,

Potential energy = 600 x 9.81 x 20 = 117720

Mechanical energy = potential energy + kinetic energy

Mechanical energy = 208250 + 117720 = 325970 J

7 0
1 year ago
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