Ask question

Ask question

All categories

2 years ago

8

A 360.0 $g$ block is dropped onto a vertical spring with a spring constant k = 254.0 $N/m$. The block becomes attached to the sp

ring, and the spring compresses 0.26 $m$ before momentarily stopping. While the spring is being compressed, what work is done by the block's weight?

1 answer:

tino4ka555 [31]2 years ago

7 0

Answer:

8.6 J

Explanation:

Work done by the block = change in energy in the spring

W = ½ kx²

W = ½ (254.0 N/m) (0.26 m)²

W = 8.6 J

You might be interested in

If we assume that, in the African savanna, the predator population remains constant, what other, nonmobile, factor would influen

RSB [31]

The food chain system would also affect it. If the antelopes don't have the right food or a low supply of it they might not stay in that area.

7 0

2 years ago

What change happens when matter<br> changes states?

Snowcat [4.5K]

Matter either loses or absorbs energy when it changes from one state to another. For example, when matter changes from a liquid to a solid, it loses energy. The opposite happens when matter changes from a solid to a liquid. For a solid to change to a liquid, matter must absorb energy from its surroundings.

4 0

2 years ago

What the kinetic energy quantities in calculation pls help me

Rashid [163]

Answer:

KE = 0.5 * m * v², where: m - mass, v - velocity.

Explanation:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s 2.

3 0

2 years ago

Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

kkurt [141]

Answer:

Approximately 1.62 × 10⁻⁴ V.

Explanation:

The average EMF in the coil is equal to

$\displaystyle \frac{\text{Final Magnetic Flux} - \text{Initial Magnetic Flux}}{2}$ ,

Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

8 0

3 years ago

Which tool would you use to measure the amount of rainfall?

zheka24 [161]

Answer:

a

Explanation:

5 0

2 years ago

Read 2 more answers