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3 years ago

What happened to the combined energy of the two sleds when they collided?

1 answer:

8 0

Hi!
Here’s your answer:

What happened to the combined energy of the two sleds when they collided? It changed forms into another energy C. Because energy is conserved, the “lost” energy has actually been changed into other forms.

Hope this helps have a good day!!<3

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The magnetic field at the center of a wire loop of radius , which carries current , is 1 mT in the direction (arrows along the w

Citrus2011 [14]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The magnetic field is $B_{net} = \frac{1}{4} * mT$

And the direction is $-\r k$

Explanation:

From the question we are told that

The magnetic field at the center is $B = 1mT$

Generally magnetic field is mathematically represented as

$B = \frac{\mu_o I}{2R}$

We are told that it is equal to 1mT

$B = \frac{\mu_o I}{2R} = 1mT$

From the first diagram we see that the effect of the current flowing in the circular loop is (i.e the magnetic field generated)

$\frac{\mu_o I}{2R} = 1mT$

This implies that the effect of a current flowing in the smaller semi-circular loop is (i.e the magnetic field generated)

$B_1 = \frac{1}{2} \frac{\mu_o I}{2R}$

and for the larger semi-circular loop is

$B_2 = \frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

Now a closer look at the second diagram will show us that the current in the semi-circular loop are moving in the opposite direction

So the net magnetic field would be

$B_{net} = B_1 - B_2$

$= \frac{1}{2} \frac{\mu_o I}{2R} -\frac{1}{2} \frac{\mu_o I}{2 * (2R)}$

$=\frac{\mu_o I}{4R} -\frac{\mu_o I }{8R}$

$=\frac{\mu_o I}{8R}$

$= \frac{1}{4} \frac{\mu_o I}{2R}$

Recall $\frac{\mu_o I}{2R} = 1mT$

$B_{net} = \frac{1}{4} * mT$

Using the Right-hand rule we see that the direction is into the page which is $-k$

3 0

3 years ago

Part 1: A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the en

Blizzard [7]

Answer:

c. 2 m/s

Explanation:

The speed of a wave is given by:

$v=f \lambda$

where

v is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the wave in this problem, we have

f = 4 Hz is the frequency

$\lambda = 0.5 m$ is the wavelength

So, the speed of the wave is

$v=(4 Hz)(0.5 m)=2 m/s$

5 0

4 years ago

A rope is vibrated so that transverse waves propagate down it. if the distance between crests is 0.5 m and a new crest reaches t

xxMikexx [17]

<span>The speed of a wave, V, is f *lambda. Where f is the frequency and lambda is the distance. If a new crest reaches the end every 4 secs; it takes 8s to cover the distance. Hence, f, which is the number of oscillations covered is 8s. So we have V = 8 * 5 = 40 ms^1.</span>

7 0

3 years ago

If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the

PolarNik [594]

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁ , F₂ = m * a₂

fk = F₁ - F₂ ⇒ μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂ ⇒ μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

3 0

3 years ago

Brainly malfunction?|<br><br> I'm not sure why but...

Brut [27]

Answer:

its probably still trying to load ur next rank or whatever it did it to me too

Explanation:

6 0

3 years ago