Ask question

Ask question

All categories

3 years ago

11

Recall that if the colliding objects have hard surfaces, the collision is elastic. Based on this description, which bumper desig

n is more likely to result in an elastic collision? Predict how this bumper design will change the force that car 2 experiences during the collision.

1 answer:

Andrews [41]3 years ago

3 0

Answer:

wheres the pic

Explanation:

:/

You might be interested in

A robin in flight has 20.8 J of PE when it is 27.6 m high. What is the mass of the robin? (Unit = kg)

schepotkina [342]

Answer:

<h2> $\boxed{ \bold{ \purple{0.0769 \: kg \: }}}$ </h2>

Explanation:

$\sf{Potential Energy ( P.E ) \: = \: 20.8 \: joule}$

$\sf{distance \: = 27.6 \: metre}$

$\sf{mass = }$ ?

$\sf{acceleration \: due \: to \: gravity = 9.68 \: {metre \: per \: second}^{2} }$

Now, let's find the mass:

$\sf{PE \: = mass \times gravity \: \times \: distance}$

plug the values

⇒ $\sf{20.8 = m \times 9.8 \times 27.6}$

Multiply the numbers

⇒ $\sf{20.8 = 270.48 \: m}$

Swap the sides of the equation

⇒ $\sf{270.48m = 20.8}$

Divide both sides of the equation by 270.48

⇒ $\sf{ \frac{270.48m}{270.48} = \frac{20.8}{270.48} }$

Calculate

⇒ $\sf{0.0769}$ kg

Hope I helped!

Best regards!!

7 0

3 years ago

Read 2 more answers

Deduced hydrochloric acid is a strong acid

Zina [86]

Answer:

for real? i never knew that!:)

Explanation:

7 0

2 years ago

Does the thickness of the wire affect the strength of an electroscope

TiliK225 [7]

It probably does. I'm not 100% sure about it, but a thicker wire would increase the number of positive and negative charges in it.

4 0

3 years ago

How wood helps in home insulation plz answer fast<br>plz answwweeeeeeerrrrrrr

Usimov [2.4K]

Wood is a poor conductor and therefore a good insulator keeping heat inside

3 0

3 years ago

Read 2 more answers

If a particle with a charge of +4.3 × 10−18 C is attracted to another particle by a force of 6.5 × 10−8 N, what is the magnitude

AveGali [126]

Answer: 1.5×10^10 N/C

Explanation:

E= F/q

Where E= magnitude of the electric field

F= force of attraction

q= charge of the given body

Given F= 6.5×10^-8 N

q= 4.3× 10^-18 C

Therefore, E = 6.5×10 ^-8/ 4.3×10^-18

E = 1.5×10^10 N/C

7 0

3 years ago