All categories

3 years ago

A camcorder has a power rating of 15 watts. If the output voltage from its battery is 5 volts, what current does it use?

Physics

1 answer:

MrMuchimi3 years ago

5 0

Formula

W = E * I

Givens

E = 5 volts

W = 15 watts

I = ?

Solution

W = E * I

15 = 5 * I

15/5 = I

I = 3 amps. Answer

You might be interested in

If the load on Pulley E is 20N, how much effort is needed to life it?

Mariana [72]

It would need to be over 20 because if the load of the Pulley E is 20 and the effort is 20, then they will be equal and the Pulley would not move, so your answer is at least 20

7 0

4 years ago

The sphere that contains the Earth's crust is the?

Serggg [28]

The LITHOSPHERE contains the earth crust

8 0

4 years ago

A pendulum consisting of a sphere suspended from a light string is oscillating with a small angle with respect to the vertical.

anyanavicka [17]

Answer:

Period remains the same, Max KE increases, Max acceleration is the same

Explanation:

4 0

3 years ago

Grandfather clocks are designed so they can be adjusted by moving the weight at the bottom of the pendulum up or down. Suppose y

ella [17]

Answer:

3. Lower the weight.

Explanation:

The time period of a pendulum is given by

$T=2\pi\sqrt{\dfrac{L}{g}}$

where,

L = Length of pendulum

g = Acceleration due to gravity = 9.81 m/s²

It can be seen from the equation that the time period depends on the length of the pendulum and the acceleration due to gravity of the planet.

The acceleration due to gravity of the planet is constant so the time period is proportional to the length of the pendulum.

$T\propto \sqrt{L}$

Here, the time period has reduced as the clock runs fast.

So, in order to increase the time period we have to lower the weight.

8 0

3 years ago

Students use a simple pendulum with a length of 36.9cm to determine the value of "g". If it takes 14.2s to complete 10 oscillati

wel

Explanation:

It is given that,

Length of the simple pendulum, l = 36.9 cm = 0.369 m

If it takes 14.2 s to complete 10 oscillations, $T=\dfrac{14.2}{10}=1.42\ s$

(a) The time period of the simple pendulum is given by :

$T=2\pi\sqrt{\dfrac{l}{g}}$

$g=\dfrac{4\pi^2l}{T^2}$

$g=\dfrac{4\pi^2\times 0.369}{(1.42)^2}$

$g=7.22\ m/s^2$

(b) On the surface of moon, $g'=\dfrac{g}{6}$

At earth, $g=9.8\ m/s^2$

$g'=1.63\ m/s^2$

As the value of g is less on the moon, so the time period on the moon increases.

(c) The time period on the earth, T = 3 s

On earth, $T=2\pi\sqrt{\dfrac{l}{g}}$

$3=2\pi\sqrt{\dfrac{l}{9.8}}$ ..............(1)

On moon, $T'=2\pi\sqrt{\dfrac{l}{g'}}$

$T'=2\pi\sqrt{\dfrac{l}{1.63}}$ ..............(2)

On solving equation (1) and (2),

T' = 18.03 s

Hence, this is the required solution.

7 0

4 years ago