Ask question

Ask question

All categories

2 years ago

13

In an inelastic collision, as compared to an elastic collision, what is to be expected?.

1 answer:

motikmotik2 years ago

3 0

The loss or conservation of kinetic energy is the difference between an elastic and an inelastic collision. Kinetic energy is not preserved in an inelastic collision, and it will change forms into sound, heat, radiation, or another form. The kinetic energy in an elastic collision is preserved and does not change forms.

You might be interested in

How much force is needed to accelerate a 1,500 kg car at a rate of 8 m/s2?

lesya [120]

Force = mass * acceleration = 1500kg * 8m/s²

3 0

3 years ago

these are the types of attractions found between molecules. their strength determines the state of the substance at room tempera

Elina [12.6K]

<h2>Answer</h2>

The physical state of the elements depends upon the <u>attraction forces </u>and their <u>kinetic energy</u>.

<h2>Explanation</h2>

The elements or substances are fixed with each other with the help of different chemical forces including ionic bonding, covalent bonding, H- bonding etc. The strength of these forces is also one of the factors that affect their physical natures. For example, covalent or ionic bonds are the strongest bonds than all other bonds and metals that contain these forces are mostly in solid form. The kinetic motion of electrons in the element also affects the physical state of the element and potential of bonding.

7 0

4 years ago

Read 2 more answers

A cardinal (Richmondena cardinalis) of mass 3.70×10−2 kg and a baseball of mass 0.144 kg have the same kinetic energy. What is t

Radda [10]

Answer:

$\frac{p_{c}}{p_{b}}\approx 0.507$

Explanation:

Since the cardinal and ball have the same kinetic energy, it is possible to determine the ratio between speeds. (c for cardinal, b for baseball)

$K_{c} = K_{b}$

$\frac{1}{2}\cdot m_{c}\cdot v_{c}^{2}= \frac{1}{2}\cdot m_{b}\cdot v_{b}^{2}$

$\frac{v_{c}}{v_{b}}=\sqrt{\frac{m_{b}}{m_{c}} }$

The ratio is obtained by multiplying each side by $\frac{m_{c}}{m_{b}}$ :

$\frac{p_{c}}{p_{b}}=\frac{m_{c}}{m_{b}}\cdot \sqrt{\frac{m_{b}}{m_{c}} }$

$\frac{p_{c}}{p_{b}}= \sqrt{\frac{m_{c}}{m_{b}} }$

The value of this ratio is:

$\frac{p_{c}}{p_{b}}\approx 0.507$

3 0

3 years ago

Read 2 more answers

11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0

Greeley [361]

Answer:

Explanation:

position of centre of mass of door from surface of water

= 10 + 1.1 / 2

= 10.55 m

Pressure on centre of mass

atmospheric pressure + pressure due to water column

10 ⁵ + hdg

= 10⁵ + 10.55 x 1000 x 9.8

= 2.0339 x 10⁵ Pa

the net force acting on the door (normal to its surface)

= pressure at the centre x area of the door

= .9 x 1.1 x 2.0339 x 10⁵

= 2.01356 x 10⁵ N

pressure centre will be at 10.55 m below the surface.

When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .

7 0

3 years ago

A penny is dropped from the Statue of Liberty

Brrunno [24]

Answer:

a) The velocity is 2.94m/s

b) 0.441

Explanation:

a) Assume gravity is 9.8m/s^2

Use the equation below to solve for the velocity at 0.30 seconds

$vf=vi+at$ ,

vf =unknown velocity vi= initial velocity vi=0m/s a= 9.8m/s^2 t=0.30seconds

Step 1: Substitute the variables with the knowns

$vf=0m/s+(9.8m/s^2)*0.30seconds$

Step 2: Solve

$vf=2.94m/s$

b)

Use the equation below to solve for the displacement at 0.30 seconds

$x=vit+\frac{1}{2} at^{2}$

Step 1: Substitute the same variables with the knowns

$x=\frac{1}{2}*(9.8m/s^2)*(0.30seconds)^{2}$

Note that vi*t=0 as vi=0m/s

Step 2: Solve

x=0.441m

5 0

3 years ago