Ask question

Ask question

All categories

2 years ago

8

Does an object in inelastic collisions exert less force than elastic collisions?

1 answer:

Montano1993 [528]2 years ago

7 0

Answer:

The object exerts <em>same</em> amount of force in elastic and inelastic collisions.

Explanation:

The force an object exerts is not different between the two types of collisions. What changes from elastic to inelastic is the amount of energy transformed from kinetic to other type during an inelastic collision.

You might be interested in

1pt A cannon fires a 5-kg ball horizontally from a

Klio2033 [76]

Answer: Both cannonballs will hit the ground at the same time.

Explanation:

Suppose that a given object is on the air. The only force acting on the object (if we ignore air friction and such) will be the gravitational force.

then the acceleration equation is only on the vertical axis, and can be written as:

a(t) = -(9.8 m/s^2)

Now, to get the vertical velocity equation, we need to integrate over time.

v(t) = -(9.8 m/s^2)*t + v0

Where v0 is the initial velocity of the object in the vertical axis.

if the object is dropped (or it only has initial velocity on the horizontal axis) then v0 = 0m/s

and:

v(t) = -(9.8 m/s^2)*t

Now, if two objects are initially at the same height (both cannonballs start 1 m above the ground)

And both objects have the same vertical velocity, we can conclude that both objects will hit the ground at the same time.

You can notice that the fact that one ball is fired horizontally and the other is only dropped does not affect this, because we only analyze the vertical problem, not the horizontal one. (This is something useful to remember, we can separate the vertical and horizontal movement in these type of problems)

7 0

2 years ago

Which of the following cars have the most kinetic energy

faltersainse [42]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2>

$\huge\boxed{OptionA}$

<h2>_____________________________________</h2><h2>DATA:</h2><h3>Blue Car: </h3>

mass = 4 kg

velocity = 5 m/s^2

<h3 /><h3>Orange truck:</h3>

Mass= 2kg

Velocity = 7m/s^2

<h3 /><h3>Grey Car:</h3>

mass = 6 kg

velocity = 4m/s^2

<h3 /><h3>Green Car:</h3>

Mass = 8 kg

Velocity = 3 m/s^2

<h2>_____________________________________</h2><h2>SOLUTION:</h2>

By the equation of kinetic energy,

K.E = $\frac{1}{2} mv^2$

Where,

K.E is kinetic energy

m is mass

v is velocity

<h2>_____________________________________</h2><h3>Kinetic energy of Blue car:</h3>

Directly substitute the variables in the equation,

K.E = $\frac{1}{2}x4x5^2$

Simplify the equation,

K.E = 50 J

<h2>_____________________________________</h2><h3>Kinetic Energy of Silver Car:</h3>

Directly substitude the variable in the equation,

K.E = $\frac{1}{2}x6x4^2$

Simplify the equation,

K.E = 48J

<h2>_____________________________________</h2><h3>Kinetic Energy of Green Car:</h3><h3 />

Substitute the variables in the equation,

K.E = $\frac{1}{2}x8x3^2$

Simplify the Equation,

K.E = 36J

<h2>_____________________________________</h2><h3>Kinetic Energy of Orange Truck:</h3><h3 />

Substitute the variable,

K.E = $\frac{1}{2}x 2x7^2$

Simplify the equation,

K.E = 49J

<h2>_____________________________________</h2>

As you can see that the highest value of kinetic energy is of Blue SUV thus it will be out answer.

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2><h3 /><h3 /><h3 /><h3 /><h2 /><h2 />

6 0

2 years ago

How does the law of conservation of energy apply to machines?

trasher [3.6K]

The answer is letter C

7 0

3 years ago

Read 2 more answers

Problem PageQuestion A cylinder measuring 2.3cm wide and 2.7cm high is filled with gas. The piston is pushed down with a steady

Alexandra [31]

Jehedhdhsh dbdbdbdbdbdbdbebddnsbdb
Dhdhdbdbdbdbdhfhbdbddbdbdbdbdbdbsbsbsbdbdbdb d s DVD’s

8 0

3 years ago

On the Moon the acceleration due to gravity is about one sixth that on Earth. If a golfer on the Moon imparted the same initial

swat32

Explanation:

We know that that the range of the ball on the earth

$R_{earth}=\frac{v_o^2sin2\theta}{g_{earth}}$

therefore, range of the ball on moon

$R_{moon}=\frac{v_o^2sin2\theta}{g_{moon}}$

$R_{moon}=\frac{v_o^2sin2\theta}{g_{earth}/12}$

therefore,

$R_{moon}=6R_{earth}$

Therefore, the range of ball will be 6 times on the moon than that on earth

6 0

2 years ago