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monitta
3 years ago
10

A 100 kilowatt bulb burns for 3 hours how much energy did it use

Physics
1 answer:
belka [17]3 years ago
4 0

Answer:

1.08×10⁹ J

Explanation:

From the question given above, the following data were obtained:

Power (P) = 100 kilowatt

Time (t) = 3 hours

Energy (E) =?

Next, we shall convert 100 kilowatt to Watts. This can be obtained as follow:

1 KW = 1000 W

Therefore,

100 KW = 100 KW × 1000 W / 1 KW

100 KW = 100000 W

Thus, 100 KW is equivalent to 100000 W.

Next, we shall convert 3 hrs to second (s). This can be obtained as follow:

1 h = 3600 s

Therefore,

3 h = 3 h × 3600 s / 1 h

3 h = 10800 s

Thus, 3 h is equivalent to 10800 s.

Finally, we shall determine the amount of energy consumed as follow:

Power (P) = 100000 W

Time (t) = 10800 s

Energy (E) =?

P = E/t

100000 = E / 10800

Cross multiply

E = 100000 × 10800

E = 1.08×10⁹ J

Therefore, 1.08×10⁹ J of energy was consumed.

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an ice skater starts with a velocity of 2.25 m/s in a 50.0 direction. after 8.33 m/s in a 120 direction. what is the y-component
Nata [24]

The y-component of the acceleration is 0.28 m/s^2

Explanation:

The y-component of the ice skater acceleration can be calculated with the equation

a_y = \frac{v_y-u_y}{t}

where

v_y is the y-component of the final velocity

u_y is the y-component of the initial velocity

t is the time elapsed

Here we have:

  • Initial velocity is u=2.25 m/s at \theta_1=50.0^{\circ}, so its y-component is u_y = u sin \theta_1 = (2.25)(sin 50.0^{\circ})=1.72 m/s
  • Final velocity is v=4.65 m/s at \theta_2=120.0^{\circ}, so its y-component is v_y = v sin \theta_2 = (4.65)(sin 120.0^{\circ})=4.03 m/s

The time elapsed is

t = 8.33 s

Therefore, the y-component of the acceleration is

a_y = \frac{4.03-1.72}{8.33}=0.28 m/s^2

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Answer:

5.2\ \text{m/s}

70^{\circ} south of east

Explanation:

v_a = 3 m/s

\theta_a = 20^{\circ} north of east

v_b = 6 m/s

\theta_b = 40^{\circ} south of east = 360-40=320^{\circ} north of east

x and y component of v_a

v_{ax}=v_a\cos \theta\\\Rightarrow v_{ax}=3\times \cos 20^{\circ}\\\Rightarrow v_{ax}=2.82\ \text{m/s}

v_{ay}=v_a\sin\theta\\\Rightarrow v_{ay}=3\times \sin20^{\circ}\\\Rightarrow v_{ay}=1.03\ \text{m/s}

x and y component of v_b

v_{bx}=v_b\cos \theta\\\Rightarrow v_{bx}=6\times \cos 320^{\circ}\\\Rightarrow v_{bx}=4.6\ \text{m/s}

v_{by}=v_b\sin\theta\\\Rightarrow v_{by}=6\times \sin320^{\circ}\\\Rightarrow v_{by}=-3.86\ \text{m/s}

\Delta v=v_b-v_a\\\Rightarrow \Delta v=(4.6-2.82)\hat{i}+(-3.86-1.03)\hat{j}\\\Rightarrow \Delta v=1.78\hat[i}-4.89\hat{j}

Magnitude

|\Delta v|=\sqrt{(-4.89)^2+1.78^2}\\\Rightarrow \Delta v=5.2\ \text{m/s}

Direction

\theta=\tan{-1}|\dfrac{-4.89}{1.78}|\\\Rightarrow \theta=70^{\circ}

The magnitude of the change in velocity vector is 5.2\ \text{m/s} and the direction is 70^{\circ} south of east.

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