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3 years ago

What is produced by AC and DC generators?

Physics

2 answers:

Goryan [66]3 years ago

8 0

I think the answer is C.

lina2011 [118]3 years ago

6 0

Answer:

I'm positive the Answer to this would be C :)

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A vertical spring with a spring constant of 420 N/m is mounted on the floor. From directly above the spring, which is unstrained

mixas84 [53]

Answer:

19.53 cm

Explanation:

The computation of the height is as follows:

Here we applied the conservation of the energy formula

As we know that

P.E of the block = P.E of the spring

m g h = ( 1 ÷ 2) k x^2

where

m = 0.15

g = 9.81

k = 420

x = 0.037

So now put the values to the above formula

(0.15) (9.81) (h) = 1 ÷2 × 420 × (0.037)^2

1.4715 (h) = 0.28749

h = 0.19537 m

= 19.53 cm

3 0

3 years ago

A vertical scale on a spring balance reads from 0 to 155 N . The scale has a length of 10.0 cm from the 0 to 155 N reading. A fi

Harrizon [31]

Answer:

mass of the fish is 8.11 kg

Explanation:

As we know that the frequency of oscillation of spring block system is given as

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

here we know that the reading of scale varies from 0 to 155 N from length varies from x = 0 to x = 10 cm

Now we have

$k = \frac{155}{0.10} N/m$

$k = 1550 N/m$

so now we have

$2.20 = \frac{1}{2\pi}\sqrt{\frac{1550}{m}}$

$m = 8.11 kg$

so mass of the fish is 8.11 kg

4 0

3 years ago

What would spines be helpful for?

ANEK [815]

A and c...............................

3 0

3 years ago

Read 2 more answers

Which term describes the quantity of energy transferred from a warmer object to a cooler object?

Mars2501 [29]

Answer: B

Explanation:

6 0

3 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

3 years ago