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3 years ago

batman recharges his bat taser with 2.5mj of energy in 15 minutes how much power does it draw from the mains

Physics

1 answer:

aalyn [17]3 years ago

3 0

Power = (energy) / (time)

15 minutes = 900 seconds

Power = (2.5mJ) / (900 sec)

<em>Power = 0.00278 mWatts</em>

If your 'm' means 'Mega...', then power = 2,777.8 watts ... about the same as a hefty restaurant toaster, or home microwave oven when it's cooking.

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Answer is a. 180 degree

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B. the astronaut will never catch the first bounce.

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3 years ago

A 3.7kg block on a horizontal frictionless surface is attached to a spring whose force constant is 450 N/m. The block is pulled

earnstyle [38]

Answer:

U=1.44 J

Explanation:

Given that

m = 3.7 kg

K = 450 N/m

X= 0.08 m

We know that spring stored potential energy(U) and this potential energy given as

$U=\dfrac{1}{2}KX^2$

By putting the values

$U=\dfrac{1}{2}KX^2$

$U=\dfrac{1}{2}\times 450\times 0.08^2$

U=1.44 J

5 0

3 years ago

A car with a mass of 1.50 X 10^3 kg starts from rest and accelerates to a speed of 18.0 m/s in 12.0 s. Assume that the force of

Serga [27]

The average power is 20.3 kW

Explanation:

First of all, we calculate the work done on the car: the work-energy theorem states that the work done on the car is equal to the change in kinetic energy of the car, so we have

$W=K_f - K_i = \frac{1}{2}mv^2-\frac{1}{2}mu^2$

where

W is the work done

$K_i, K_f$ are the initial and final kinetic energy of the car

$m=1.50\cdot 10^3 kg = 1500 kg$ is the mass of the car

u = 0 is the initial velocity

v = 18.0 m/s is the final velocity

Substituting,

$W=\frac{1}{2}(1500)(18)^2=2.43\cdot 10^5 J$

Now we can find the average power developed by the car's engine, which is given by

$P=\frac{W}{t}$

where

$W=2.43\cdot 10^5 J$ is the work done

t = 12.0 s is the time taken

Substituting,

$P=\frac{2.43\cdot 10^5 J}{12.0}=20,250 W = 20.3 kW$

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

4 0

4 years ago

The volume of 1 kg of helium in a piston-cylinder device is initially 5 m3. Now helium is compressed to 2 m3 while its pressure

yuradex [85]

Answer:

-540 kJ

Explanation:

P = Pressure which is applied to the piston = 180 kPa

$V_2$ = Final volume = 2 m³

$V_1$ = Initial volume = 5 m³

Work done is given by

$W=P\Delta V\\\Rightarrow W=P(V_2-V_1)\\\Rightarrow W=180000(2-5)\\\Rightarrow W=180000\times -3\\\Rightarrow W=-540000\ J\\\Rightarrow W=-540\ kJ$

The work required to compress helium is -540 kJ

4 0

3 years ago