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2 years ago

3. Why is static electricity not useful as a power

1 answer:

3 0

The answer is B, because all energy is released at once in static electricity.

There’s a quizlet that mentions these questions, if you are having trouble. I’d suggest to give them a look.

You might be interested in

Why are the peaks opposite in direction?

blagie [28]

Answer:

Explanation:

Peaks are in opposite direction because change in magnetic field at one end of the coil is opposite to the change in magnetic field at other end of the coil. Faraday's law predict that the direction of induced voltage is dependent on the nature of change in magnetic field

4 0

2 years ago

A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin

blsea [12.9K]

Answer:

The stress is calculated as $1.003\times 10^{9}\ Pa$

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = $0.560\times 10^{- 3}\ m$

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, $\Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m$

Now,

The stress in the wire is given by:

$Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}$ (1)

Now,

Force is due to the weight of the attached weight:

F = mg = $25.2\times 9.8 = 246.96\ N$

Cross sectional Area, A = $\pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}$

Using these values in eqn (1):

$\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa$

8 0

2 years ago

We have a toy gun with a spring constant of 50 N/m. The spring is compressed by 0.2 m. If you neglect friction and the mass of t

Arisa [49]

Answer:

$31.6\:\mathrm{m/s}$

Explanation:

The elastic potential energy of a spring is given by $Us=\frac{1}{2}kx^2$ , where $k$ is the spring constant of the spring and $x$ is displacement from point of equilibrium.

When released, this potential energy will be converted into kinetic energy. Kinetic energy is given by $KE=\frac{1}{2}mv^2$ , where $m$ is the mass of the object and $v$ is the object's velocity.

Thus, we have:

$Us=KE,\\\frac{1}{2}kx^2=\frac{1}{2}mv^2$

Substituting given values, we get:

$\frac{1}{2}\cdot 50\cdot 0.2^2=\frac{1}{2}\cdot 0.002\cdot v^2,\\v^2=\frac{50\cdot 0.2^2}{0.002},\\v^2=1000,\\v\approx \boxed{31.6\:\mathrm{m/s}}$

4 0

2 years ago

You are pushing on a heavy desk with a force of 65 N the desk does not slide the force of friction between the desk and the floo

astra-53 [7]

When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force. hence the statement is true.

7 0

3 years ago

I think I know this one but I don’t at the same time

galben [10]

Answer: B.) 6

Explanation:

To answer this problem you get the number of students who attended Wednesday (18) and the number of students who attended Tuesday (12) and subtract 18 - 12 = 6

Answer = B.) 6

8 0

2 years ago