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3 years ago

11

Describe the relationship between kinetic energy and speed and give an example of how changing and object speed would affect its

kinetic energy

1 answer:

Andreas93 [3]3 years ago

6 0

Answer:

I'm not sure

Explanation:

I have had that question to Uchida c r go crew in to go be

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The intensity level of a "Super-Silent" power lawn mower at a distance of 1.0 m is 100 dB. You wake up one morning to find that

OLEGan [10]

Answer: The intensity level of sound in the bedroom is 80dB

Explanation:

Intensity of lawn mower at r=1m is 100dB

Beta1= 10dBlog(I1/Io)

100dB= 10dB log(I1/Io)

10^10= I1/Io

I1= Io(10^10)

10^12)×(10^10)= I1

I1=10^-2w/m^2

Intensity of lawn mower at r=20m

I2/I1=(r1/r2)^2 =(1/20)^2

I2= I1(1/400)

I2=2.5×10^-3W_m^2

Intensity of 4 lown mowers at 20m fro. Window

= 10dBlog(4I2/Io)

= 10^-4/10^-12

=80dB

6 0

3 years ago

What is the change in thermal energy E if the coefficent of kinetic friction between the box and floor is .4 , the distance the

Jet001 [13]

This question can be solved using the concept of friction energy.

The thermal energy change is b "258.4 J".

The change in thermal energy will be equal to the friction energy produced during the motion of the box.

$Change\ In\ Thermal\ Energy = E = Friction\ Energy\\\\E = \mu fd$

where,

μ = coefficient of kinetic friction = 0.4

f = force applied = 38 N

d = distance traveled by the box = 17 m

Therefore,

$E = (0.4)(38\ N)(17\ m)$

<u>E = 258.4 J</u>

Learn more about friction energy here:

brainly.com/question/1343045?referrer=searchResults

7 0

2 years ago

A Bullet Off mass 100 gm is fired From A Gun Off mass 5 Kg. If the backward velocity of the gun's 5 m / s, what is forward veloc

Elena L [17]

Answer:

250 m/s

Explanation:

The mass of the bullet, m₁ = 100 g = 0.1 kg

The mass of the gun, m₂ = 5 kg

The backward velocity of the gun, v₂ = -5 m/s

Given that the momentum is conserved, we have;

The total initial momentum = The total final momentum

The gun and the bullet are at rest, therefore, we have;

The initial momentum = 0

The total final momentum = m₁·v₁ + m₂·v₂

Where;

v₁ = The forward velocity of the bullet

Therefore, we get;

m₁·v₁ + m₂·v₂ = 0

0.1 kg × v₁ + 5 kg × (-5 m/s) = 0

0.1 kg × v₁ = 5 kg × 5 m/s

v₁ = (5 kg × 5 m/s)/(0.1 kg) = 250 m/s

The forward velocity of the bullet, v₁ = 250 m/s

6 0

3 years ago

An athlete is working out in the weight room. he steadily holds 50 kg above his head for ten seconds which statement is true abo

Andrews [41]

Hi. The answer to your question is the first option.

The athlete isn’t doing any work because he doesn’t move the weight.

Hope this helps :))

5 0

3 years ago

A proton of mass m is at rest when it is suddenly struck head-on by an alpha particle (which consistsof 2 protons and 2 neutrons

antiseptic1488 [7]

Answer:

3/5 v

Explanation:

The computation of speed will the alpha particle have after the collision is shown below:-

In a perfectly elastic the kinetic energy and collision the momentum are considered.

The velocity of the particles defines the below equation:

$VA_f=(\frac{m_A-m_B}{m_A+m_B})VA_i+(\frac{2m_B}{m_A+m_B})VB_i$

As we know that

$VA_i=v$

$\\VB_i=0$

Here, we consider A is the alpha particle and B is the proton and now by the above values we can solve the equation which is below:-

$VA_f=(\frac{4m-m}{4m+m})v$

$\\VA_f=\frac{3m}{5m}v$

$\\VA_f=\frac{3}{5}v$

Therefore the correct answer is $\frac{3}{5}v$

7 0

3 years ago