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3 years ago

9

a person using a machine applies a force of 100 newton's over a distance of 10 Meters to raise a 500 n object 1.5 meters what is

the work input?

1 answer:

Hoochie [10]3 years ago

7 0

Answer:

Work done by the machine (W) = 500 × 1.5 = 750 J

Work supplied to the machine (W) = 100 × 10 = 1000 J

Here, work supplied to the machine is input work = 1000 J

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after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

3 years ago

A 40.0-mH inductor is connected to a North American electrical outlet (ΔVrms = 120 V, f = 60.0 Hz). Assuming the energy stored i

trasher [3.6K]

Explanation:

It is given that,

Inductance, $L=40\ mH=40\times 10^{-3}\ H$

RMS value of voltage, $v_{rms}=120\ V$

Frequency, f = 60 Hz

We need to find the energy stored at t = (1 /185) s. It is assumed that energy stored in the inductor is zero at t = 0. So,

The current flowing through the inductor is given by :

$I_t=\dfrac{V_o}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{\sqrt{2} V_{rms}}{X_L}\ (sin\ \omega t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi f L}\ sin(2\pi f t-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt{2}}{2\pi\times 60\times 40\times 10^{-3}}\ sin(2\pi \times 60\times \dfrac{1}{185})-\dfrac{\pi}{2})$

$I_t=\dfrac{120\sqrt2}{15.07}\ sin(2\pi \times 60\times \dfrac{1}{185}-\dfrac{\pi}{2})$

I = 0.091 A

Energy stored in the inductor is, $U=\dfrac{1}{2}LI^2$

$U=\dfrac{1}{2}\times 40\times 10^{-3}\times (0.091)^2$

U = 0.000165 Joules

Hence, this is the required solution.

6 0

3 years ago

Question 7 Points 2

uranmaximum [27]

Answer:

The dynamo has a wheel that touches the back tyre. As the bicycle moves, the wheel turns a magnet inside a coil. This induces enough electricity to run the bicycle's lights. The faster the bicycle moves, the greater the induced voltage - and the brighter the lights.

8 0

2 years ago

Someone help please i need to finish this

Tom [10]

Answer:

4th answer

Explanation:

The gradient of a distance-time graph gives the speed.

gradient = distance / time = speed

Here, the gradient is a constant till 30s. So it has travelled at a constant speed. It means it had not accelarated till 30s. and has stopped moving at 30s.

6 0

2 years ago

Can work done=mass*acceleration*displacement(work=m*a*s)

Airida [17]

no, work is = force * distance or displacement

5 0

3 years ago