The tendency for an object at rest to remain at rest

Small paragraph explaining how how potential and kinetic energy are related

weeeeeb [17]

Answer:

Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or state. While kinetic energy of an object is relative to the state of other objects in its environment, potential energy is completely independent of its environment.

Explanation:

4 0

2 years ago

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(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

2 years ago

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

Dmitrij [34]

Answer:

$\lambda_2 = 573.3 nm$

Explanation:

As we know that the position of maximum intensity on the screen is given as

$y = \frac{N\lambda L}{d}$

here we know that

$\lambda$ = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

$\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}$

so here we have

$8 (645 nm) = 9 \lambda_2$

$\lambda_2 = 573.3 nm$

4 0

3 years ago

Tina wants to install one unit that both heats and cools. Which should Tina install?

madreJ [45]

A heat pump? it might be furnace but i think its heat pump

8 0

2 years ago

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Assume that the function x(t) represents the length of tape that has unwound as a function of time. find θ(t), the angle through

bekas [8.4K]

We know that arc length (x(t)) is given with the following formula:
$x(t)=\theta(t) r$
Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:
$r(\theta)=r_0-\frac{D\cdot \theta}{2\pi}$
When we plug this back into the first equation we get:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}})\\ \frac{D\theta^2}{2\pi}-\theta r_0+x(t)=0\\$
We must solve this quadratic equation.
The final solution is:
$\theta=\frac{\pi r_0+\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D},\:\theta=\frac{\pi r_0-\sqrt{\pi \left(-2Dx(t)+\pi r_0^2\right)}}{D}$
It is rather complicated solution. If we asume that the tape has no thickness we get simply:
$x(t)=\theta(r_0-\frac{D\theta}{2\pi}});D=0\\
x(t)=\theta r_0\\
\theta(t)=\frac{x(t)}{r_0}$

8 0

3 years ago