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nalin [4]
2 years ago
8

A 70-Ω electrical appliance and a 100-Ω electrical appliance are plugged into the outlets of a house. By what factor will the po

wer used change if an additional 130-Ω electrical device is plugged in?
Physics
1 answer:
lukranit [14]2 years ago
8 0

Answer:

For an additional 130-Ω electrical device, the power used will change by a factor of 0.77.

Explanation:

Power is given as;

P = I²R

Where;

P is the power used due to resistance of the electrical appliance

I is the current passing through the appliance

R is the resistance offered by the appliance

For 70-Ω electrical appliance and  100-Ω electrical appliance, the total power used is;

P = 70I² + 100 I² =  (170I²) W

for an additional 130-Ω electrical device, the new the total power used is;

P = 130I² +  (170I²)  = (300 I²)W

Change \ in \ power \ used = \frac{I^2(300-170)}{I^2(170)} =0.77

Therefore, for an additional 130-Ω electrical device, the power used will change by a factor of 0.77.

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Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = 85\frac{1000}{3600}=23.61\ m/s

Initial velocity = u = 0

Equation of motion

v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2

Emergency acceleration is 2.99 m/s² (magnitude)

6 0
2 years ago
You are traveling in a spaceship at a speed of 0.70c away from Earth. You send a laser beam toward the Earth traveling at veloci
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2 years ago
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A box is moving along the x-axis and its position varies in time according to the expression:
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Answer:

38.4 m/s

Explanation:

a) at t = 3.2s. x = 6 * 3.2^2 = 61.44 m

b) at t = 3.2 + Δt. x = 6*(3.2 + \Delta t)^2

c) As Δt approaches 0. We can find the velocity by the ratio of Δx/Δt

v = \frac{\Delta x}{\Delta t} = \frac{x_2 - x_1}{\Delta t}

v = \frac{6*(3.2 + \Delta t)^2 - 61.44}{\Delta t}

v = \frac{6(3.2^2 + 6.4\Delta t + \Delta t^2) - 61.44}{\Delta t}

v = \frac{61.44 + 38.4\Delta t + \Delta t^2 - 61.44}{\Delta t}

v = \frac{\Delta t(38.4 + \Delta t)}{\Delta t}

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As Δt approaches 0, v = 38.4 + 0 = 38.4 m/s

3 0
3 years ago
I need help!! i’ll give 20 points
larisa86 [58]

Answer:

780 m to travel north

Explanation:

6 m over = 750

53 degree so it will take about 2 min to reach the destination

8 0
3 years ago
If an object is thrown in an upward direction from the top of a building 1.6 x 10^2 ft. high at an initial velocity of 21.82 mi/
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If an object is thrown in an upward direction from the top of a building 1.60 x 102 ft. high at an initial velocity of 21.82 mi/h, what is its final velocity when it hits the ground? (Disregard wind resistance. Round answer to nearest whole number and do not reflect negative direction in your answer.)


this question is troubling me i guessed 96 ft/s

can someone help me out and explain it thanks so much!!!!!!



7 0
3 years ago
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