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ser-zykov [4K]
3 years ago
13

To complete a project, 200,000 Joules of work is needed. The time taken to complete the project is 20 seconds. How much power is

needed?
Physics
2 answers:
Leno4ka [110]3 years ago
5 0

Answer:

10,000

Explanation:

200,000/20 = power needed

200,000/20 = 10,000

Hope this Helps!

MArishka [77]3 years ago
3 0

Answer:

10,000 Watts

Explanation:

Power is the ratio of work to time, or work over time. It can be found using the following formula.

p=w/t

where p is the power, w is the work, and t is the time.

The project needed 200,000 Joules, and it took 20 seconds to complete the project. Therefore, w=200,000 and t=20. Substitute these values into the formula.

p=200,000/20

Divide

p=10,000

Add appropriate units. The units for power is Watts.

p=10,000 Watts

10,000 Watts of power is needed.

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Can someone see if it's right? thank you. ​
bearhunter [10]

Answer:

it's right you did a great job

8 0
2 years ago
A runner completes the 200-meter dash with a time of 19.80 seconds. What was the runner's average speed in miles per hour?
andriy [413]

Answer:

v = 22.54 mph.

Explanation:

Given that,

Distance moved, d = 200 m

Time, t = 19.8 s

We need to find the runner's average speed.

We know that,

1 mile = 1609.34 m

200 m = 0.124 miles

19.8 seconds = 0.0055 h

So,

Speed = distance/time

v=\dfrac{0.124}{0.0055}\\\\v=22.54\ mph

So, the runner's average speed is 22.54 mph.

4 0
3 years ago
A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate
maksim [4K]

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

        \texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

a=\frac{11.7-26.5}{5.2}=-2.85m/s^2

Angular acceleration

        \alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2

Angular acceleration = -3.95 rad/s²  

Magnitude =  3.95 rad/s²     

4 0
3 years ago
A charge of 5.0 coulombs moves through a circuit in 0.50 second. What is the current in the circuit
user100 [1]

Answer:

10

Explanation:

i = 5/.5 = 10 Amps.  Hope this helps :)

6 0
3 years ago
Consider two points in an electric field. The potential at point 1, V1, is 33 V. The potential at point 2, V2, is 175 V. An elec
Mnenie [13.5K]

Answer:

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

Explanation:

Since the electric potential at point 1 is V₁ = 33 V and the electric potential at point 2 is V₂ = 175 V, when the electron is accelerated from point 1 to point 2, there is a change in electric potential ΔV which is given by ΔV = V₂ - V₁.

Substituting the values of the variables into the equation, we have

ΔV = V₂ - V₁.

ΔV = 175 V - 33 V.

ΔV = 142 V

The change in electric potential energy ΔU = eΔV = e(V₂ - V₁) where e = electron charge = -1.602 × 10⁻¹⁹ C and ΔV = electric potential change from point 1 to point 2 = 142 V.

So, substituting the values of the variables into the equation, we have

ΔU = eΔV

ΔU = eΔV

ΔU = -1.602 × 10⁻¹⁹ C × 142 V

ΔU = -227.484 × 10⁻¹⁹ J

ΔU = -2.27484 × 10⁻²¹ J

ΔU ≅ -2.275 × 10⁻²¹ J

So, the required equation for the electric potential energy change is

ΔU  = e(V₂ - V₁) and its value ΔU = -2.275 × 10⁻²¹ J

5 0
3 years ago
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