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2 years ago

If you run a 500 W computer for 4 hours, your electric meter will read ; this is a measure of ___ "used"

Physics

2 answers:

emmasim [6.3K]2 years ago

7 0

Answer:

If you run a 500 W computer for 4 hours, your electric meter will read 2kWh; this is a measure of energy "used". The correct answer is D.

Explanation: Energy is defined as the capacity of doing work and is occurs in different forms( electrical, kinetic, potential energy). Electrical energy consumption can be calculated or measure through the multiplication of the power( W or kW) used by the number of hours during which the power is consumed. Therefore the unit of energy is kWh. From the question,

500W was used for 4 H

Therefore energy consumed

500× 4= 2000Wh. To convert to kWh

Divide by 1000,

2000÷ 1000= 2.

There energy consumed in kWh is,

2kWh energy.

Katen [24]2 years ago

3 0

Answer:

D) 2 kWh, energy

Explanation:

Energy = power×time

For electricity,

E = Pt.................... Equation 1

Where E = Electric Energy, P = power, t = time.

Given: P = 500 W, t = 4 hours.

Substitute into equation 1

E = 500(4)

E = 2000 Wh

E = (2000/1000) kWh

E = 2 kWh.

Hence the right option is D) 2 kWh, energy

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2 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

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Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

2 years ago

If you run a 500 W computer for 4 hours, your electric meter will read ____; this is a measure of _______ "used"

If you run a 500 W computer for 4 hours, your electric meter will read ; this is a measure of ___ "used"