Plz help

When the speed of your car is doubled, by what factor does its kinetic energy increase?

Tju [1.3M]

Answer:

a. 4

Explanation:

Hi there!

The equation of kinetic energy (KE) is the following:

KE = 1/2 · m · v²

Where:

m = mass of the car.

v = speed of the car.

Let´s see how would be the equation if the velocity is doubled (2 · v)

KE2 = 1/2 · m · (2 · v)²

Distributing the exponent:

KE2 = 1/2 · m · 2² · v²

KE2 = 1/2 · m · 4 · v²

KE2 = 4 (1/2 · m · v²)

KE2 = 4KE

Doubling the velocity increased the kinetic energy by 4.

7 0

3 years ago

Is the wavelength comparable to the size of atoms?

Helen [10]

It totally depends on what kind of wave you're talking about.

-- a sound wave from a trumpet or clarinet playing a concert-A pitch is about 78 centimeters long ... about 2 and 1/2 feet. This is bigger than atoms.

-- a radio wave from an AM station broadcasting on 550 KHz, at the bottom of your radio dial, is about 166 feet long ... maybe comparable to the height of a 10-to-15-story building. This is bigger than atoms.

-- a radio wave heating the leftover meatloaf inside your "microwave" oven is about 4.8 inches long ... maybe comparable to the length of your middle finger. this is bigger than atoms.

-- a deep rich cherry red light wave ... the longest one your eye can see ... is around 750 nanometers long. About 34,000 of them all lined up will cover an inch. These are pretty small, but still bigger than atoms.

-- the shortest wave that would be called an "X-ray" is 0.01 nanometer long. You'd have to line up 2.5 billion of <u>those</u> babies to cover an inch. Hold on to these for a second ... there's one more kind of wave to mention.

-- This brings us to "gamma rays" ... our name for the shortest of all electromagnetic waves. To be a gamma ray, it has to be shorter than 0.01 nanometer.

Talking very very very very roughly, atoms range in size from about 0.025 nanometers to about 0.26 nanometers.

The short end of the X-rays, and on down through the gamma rays, are in this neighborhood.

5 0

4 years ago

Two particles P and Q each moves towards ther along Straight Line (M)(N), 51m long. P starts from M with velocity 5m/s and const

bagirrra123 [75]

The time required by the particles are as follows:

a. t = 1.5 seconds

b. t = 3 seconds

c. t = 0.4 seconds

<h3>What is the time required?</h3>

The time required for the particles to be at several distances apart is calculated using the equation of motion given below:

$S = ut + \frac{1}{2}at^{2}$

a) Time required to be 30 m apart:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51 - 30

S1 + S2 = 21

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 21

2t^2 + 11t - 21 = 0

Solving for time, t by factorization, t = 1.5 seconds

b) Time required to meet:

Assuming the distance covered by P is S1 and distance covered by Q is S2.

S1 + S2 = 51

Substituting the values of velocity and acceleration in the equation of motion above:

5t + 1/2t^2 + 6t + 3t^2 = 51

2t^2 + 11t - 51 = 0

Solving for time, t by factorization, t = 3 seconds

c) Time required for velocity of P is ¾ of the velocity of Q:

Using the equation of motion: V = u + at

Vp = 3/4 Vq

4Vp= 3Vq

Substituting the values:

4(5 + t) = 3(6 + 3t)

5t = 2

t = 0.4 seconds

Learn more about distance, velocity and acceleration at: brainly.com/question/14344386

#SPJ1

7 0

2 years ago

What is helpful for motivating a person to achieve a fitness or health goal?

tresset_1 [31]

Answer:

I think it's b

Explanation:

seems about right

6 0

2 years ago

Read 2 more answers

Uma barra de ferro homogênea possui, a 20 °C, o comprimento de 6 m. Se o coeficiente de dilatação linear do ferro é 1,2 ∙ 10⁻⁵ °

Natali [406]

Answer:

25°C

Explanation:

Using the linear expansivity formula expressed as;

∝ = ΔL/lΔθ

∝ is coefficient of lineat expansion = 1.2 ∙ 10⁻⁵ °C⁻¹

ΔL is the change in length = 6.00036-6

ΔL = 0.00036m

l is the original length = 6m

Δθ is the change in temperature =θ₂-20

Substituting into the formula;

1.2 ∙ 10⁻⁵ °C⁻¹ = 0.00036/6(θ₂-20)

cross multiply

1.2 ∙ 10⁻⁵ * 6 = 0.00036/(θ₂-20)

7.2 ∙ 10⁻⁵= 0.00036/(θ₂-20)

0.00036 = 7.2 ∙ 10⁻⁵(θ₂-20)

0.00036 = 7.2 ∙ 10⁻⁵θ₂-144∙ 10⁻⁵

7.2 ∙ 10⁻⁵θ₂ = 0.00036+0.00144

7.2 ∙ 10⁻⁵θ₂ = 0.0018

θ₂ = 0.0018/0.000072

θ₂ = 25°C

Hence the temperature at which this bar must be acidic for its compression is 6,00036 m is 25°C

4 0

3 years ago