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adelina 88 [10]

3 years ago

11

A crate slides down a ramp that makes a 20∘ angle with the ground. To keep the crate moving at a steady speed, Paige pushes back

on it with a 68 N horizontal force.
Part A
How much work does Paige do on the crate as it slides 3.0 m down the ramp?

1 answer:

prisoha [69]3 years ago

6 0

Answer:

the answer is 69.7687j

Explanation:

W =F sin Φ

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Over the time interval after a difference in potential is applied between the ends of a wire, what would happen to the drift vel

Dmitrij [34]

The the drift velocity of the electrons is determined by atom vibrations in the crystal lattice.

<h3>How to explain the information?</h3>

Assume we could increase the average time between collisions in a typical metal to get to a limit of zero resistance. The free electrons would therefore be continuously accelerated by a constant applied voltage, according to the classical paradigm of conduction. Both the current and the drift speed would gradually pick up over time.

Although it is not the scenario implied by the question, it is possible to switch to zero resistance by using a superconducting wire instead of the usual metal. In this scenario, the maximum current is constrained, the drift velocity of the electrons is determined by atom vibrations in the crystal lattice, and it is difficult to produce a potential difference across the superconductor.

Learn more about electrons in:

brainly.com/question/860094

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5 0

2 years ago

a parent swings a 18.5 kg child in a circle of radius 1.05m, making 5 revolutions in 13.4s. what centripetal acceleration does t

Zolol [24]

Answer: $0.146 m/s^{2}$

Explanation:

The <u>centripetal acceleration</u> $a_{c}$ of an object moving in a uniform circular path is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$V$ is the tangential velocity

$r=1.05 m$ is the radius of the circle

On the other hand, the tangential velocity is expressed as:

$V=\omega r$ (2)

Where $\omega$ is the angular velocity, which can be found knowing the child makes 5 revolutions in 13.4s:

$\omega=\frac{5 rev}{13.4 s}=0.37 rev/s$ (3)

Substituting (3) in (2):

$V=(0.37 rev/s)(1.05 m)$ (4)

$V=0.39 m/s$ (5)

Substituting (5) in (1):

$a_{c}=\frac{(0.39 m/s)^{2}}{1.05 m}$ (6)

Finally:

$a_{c}=0.146 m/s^{2}$

6 0

3 years ago

Read 2 more answers

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

=> $0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t$

=> $v_t = 0.02175 \ m/s$

3 0

4 years ago

One positve and one negative charhe jave a force of -5. 00 N between them. If the force changed to -170 N , by what factor did t

stepladder [879]

Answer:

The distance is shortenend by factor .1715

Explanation:

5 n = 1/r^2

sqrt (1/5) = r

170 n = 1 / ( x sqrt(1/5))^2

(xsqrt 1/5)^2 = 1/170

x sqrt 1/5 = .076696

x = .1715

5 0

2 years ago

A spring with spring constant k is suspended vertically from a support and a mass m is attached. The mass is held at the point w

garik1379 [7]

Answer:

The oscillation frequency of the spring is 1.66 Hz.

Explanation:

Let m is the mass of the object that is suspended vertically from a support. The potential energy stored in the spring is given by :

$E_s=\dfrac{1}{2}kx^2$

k is the spring constant

x is the distance to the lowest point form the initial position.

When the object reaches the highest point, the stored potential energy stored in the spring gets converted to the potential energy.

$E_P=mgx$

Equating these two energies,

$\dfrac{1}{2}kx^2=mgx$

$\dfrac{k}{m}=\dfrac{2g}{x}$ .............(1)

The expression for the oscillation frequency is given by :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2g}{x}}$ (from equation (1))

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{2\times 9.8}{0.18}}$

f = 1.66 Hz

So, the oscillation frequency of the spring is 1.66 Hz. Hence, this is the required solution.

8 0

4 years ago