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2 years ago

How are energy, work ang power related?

Physics

1 answer:

jolli1 [7]2 years ago

5 0

Answer:

Work is the change in energy and power is the rate of doing work.

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babymother [125]

Answer:

energy? or power? I'm not really sure but using the quizzes app really help with science problems

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2 years ago

Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe

Mila [183]

Your answer is 8. You add 2 + 1 + 5.3 to get 8.3. You round down to 8 because of the sig fig rules.

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2 years ago

A charge of 1.5 µC is placed on the plates of a parallel plate capacitor. The change in voltage across the plates is 36 V. How m

belka [17]

Answer:

Energy stored in the capacitor is $U=2.7\times 10^{-5}\ J$

Explanation:

It is given that,

Charge, $q=1.5\ \mu C=1.5\times 10^{-6}\ C$

Potential difference, V = 36 V

We need to find the potential energy is stored in the capacitor. The stored potential energy is given by :

$U=\dfrac{1}{2}q\times V$

$U=\dfrac{1}{2}\times 1.5\times 10^{-6}\times 36$

U = 0.000027 J

$U=2.7\times 10^{-5}\ J$

So, the potential energy is stored in the capacitor is $U=2.7\times 10^{-5}\ J$ . Hence, this is the required solution.

4 0

2 years ago

Read 2 more answers

Which of the following is NOT an indicator of a physical change?

S_A_V [24]

Answer:

1. C. The change is easily reversible

2. A. a physical change

Explanation:

Happy Holidays

8 0

2 years ago

Sapting Leng You have a string with a mass of 13.7 q. You stretch the string with a force of 8.39 N, giving it a length of 1.87

marshall27 [118]

Answer:

a) $\lambda=0.935\ \textup{m}$

b) $f=36.19\approx 36\ \textup{Hz}$

Explanation:

Given:

String vibrates transversely fourth dynamic, thus n = 4

mass of the string, m = 13.7 g = 13.7 × 10⁻¹³ kg

Tension in the string, T = 8.39 N

Length of the string, L = 1.87 m

a) we know

$L= n\frac{\lambda}{2}$

where,

$\lambda$ = wavelength

on substituting the values, we get

$1.87= 4\times \frac{\lambda}{2}$

$\lambda=0.935\ \textup{m}$

b) Speed of the wave (v) in the string is given as:

$v =f\lambda$

also,

$v=\sqrt\frac{T}{(\frac{m}{L})}$

equating both the formula for 'v' we get,

$f\lambda=\sqrt\frac{T}{(\frac{m}{L})}$

on substituting the values, we get

$f\times 0.935=\sqrt\frac{8.39}{(\frac{13.7\times 10^{3}}{1.87})}$

$f=\frac{33.84}{0.935}$

$f=36.19\approx 36\ \textup{Hz}$

5 0

2 years ago