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2 years ago

9

An oven uses 4500 watts of power in 2 minutes, how much work was done?

1 answer:

jolli1 [7]2 years ago

4 0

First of all we state the formula
Power=work done/time
we can rearrange this formula as well
work done=power x time

Since the SI unit of time is in seconds we change the minutes to seconds
2mins= 60x2 = 120 seconds

Using our formula (work done=power x time) we simply put in the values
work done = 4500 x 120
work done = 540,000J

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What is the area of a rectangle whose sides are 3.2 m and 8.01 m? Answer in units of m²

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3 years ago

Read 2 more answers

In a college homecoming competition, eighteen students lift a sports car. While holding the car off the ground, each student exe

Nata [24]

Answer:

Explanation:

Given

Each student exert a force of $F=400 N$

Let mass of car be m

there are 18 students who lifts the car

Total force by 18 students $F=18\times 400=7200 N$

therefore weight of car $W=7200$

mass of car $m=\frac{W}{g}$

$m=\frac{7200}{9.8}=734.69 kg$

(b) $7200 N \approx 1618.624\ Pound-force$

$734.69 kg\approx 1619.71 Pounds$

6 0

2 years ago

A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers

Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0

2 years ago