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3 years ago

If 270 watts of power is used in 42 seconds, how much work was done

Physics

1 answer:

navik [9.2K]3 years ago

4 0

Answer: W = 11340J

Explanation:

Hey there! I will give the following steps, if you have any questions feel free to ask me in the comments below.

So this is the Formula: Power = Work / Time.

Step 1: Find the Formula

P = W / T

Step 2: Make W the subject of the equation.

W = PT

Step 3: Given.

P = 270 Watts, T = 42 seconds

Step 4: Substitute these values into equation 2 .

W = 270(42)

Step 5: Simplify.

W = 11340J

The amount of work done was 11340.

~I hope I helped you! :)~

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

8 0

3 years ago

An object moves from the position +16m to the position +43m in 12s. What us the total displacement

NeX [460]

First method

initial distance = 16m

final distance= 43 m

total distance covered= final -initial

=43m -16m

=27m

Second method

Si= 16m

Sf =43 m

t= 12 s

first we will find V

V = (Sf-Si)/ t

V =( 43- 16)/ 12

V = 27/12 ⇒ V= 9/4

V= distance / time

distance= V×time

distance = (9/4) ×12

distance =27

3 0

3 years ago

If 270 watts of power is used in 42 seconds, how much work was done​

If 270 watts of power is used in 42 seconds, how much work was done