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igomit [66]
3 years ago
14

What is the difference in electrical potential energy between two places in an electric field?

Physics
1 answer:
Ierofanga [76]3 years ago
4 0
The question is poorly worded, and makes it harder for the student to learn the basics of the subject. "Potential Difference" is the difference in electrical potential between two places in an electric field ... not in electrical potential energy.
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Which of the following situations has more kinetic energy than potential energy?
MrMuchimi

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C. a rolling bowling ball

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The diagram shows a person using a piece of gym equipment to lift weights.
ki77a [65]

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C. The lower legs are levers, and the knees are fulcrums. The ankles hold the loads.

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If a plane has an airspeed of 40 m/s and is experiencing a crosswind of 30 m/s, what is its ground speed in m/s?
Anon25 [30]
<h2>Answer:50ms^{-1}</h2>

Explanation:

Let v_{a} be the airspeed.

Let v_{w} be the cross wind speed.

We know that,ground speed is the vector sum of airspeed and cross wind speed and airspeed is perpendicular to cross wind speed.

If v_{1} and v_{2} are two perpendicular vectors,the resultant vector has the magnitude \sqrt{|v|_{1}^{2}+|v|_{2}^{2}}

Given,

v_{a}=40ms^{-1}\\v_{c}=30ms^{-1}

So,the ground speed is \sqrt{40^{2}+30^{2}}=\sqrt{2500}=50ms^{-1}

6 0
3 years ago
A weightlifter lifts a 1250-N barbell 2 m in 3 s.How much power was used to lift the barbell?
Vlad [161]

The power is 833.3 W

Explanation:

First of all, we need to calculate the work done in lifting the barbell, which is equal to the change in gravitational potential energy of the barbell:

W=(mg)h

where

mg = 1250 N is the weight of the barbell

h = 2 m is the change in height

Substituting,

W=(1250)(2)=2500 J

Now we can calculate the power, which is equal to the work done per unit time:

P=\frac{W}{t}

where

W = 2500 J is the work done

t = 3 s is the time taken

Substituting,

P=\frac{2500}{3}=833.3 W

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brainly.com/question/7956557

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5 0
2 years ago
Two light bulbs have resistances of 400 Ω and 800 ΩThe two light bulbs are connected in series across a 120- V line. Find the cu
Natasha2012 [34]

1) Current in each bulb: 0.1 A

The two light bulbs are connected in series, this means that their equivalent resistance is just the sum of the two resistances:

R_{eq}=R_1 + R_2 = 400 \Omega + 800 \Omega=1200 \Omega

And so, the current through the circuit is (using Ohm's law):

I=\frac{V}{R_{eq}}=\frac{120 V}{1200 \Omega}=0.1 A

And since the two bulbs are connected in series, the current through each bulb is the same.

2) 4 W and 8 W

The power dissipated by each bulb is given by the formula:

P=I^2 R

where I is the current and R is the resistance.

For the first bulb:

P_1 = (0.1 A)^2 (400 \Omega)=4 W

For the second bulb:

P_1 = (0.1 A)^2 (800 \Omega)=8 W

3) 12 W

The total power dissipated in both bulbs is simply the sum of the power dissipated by each bulb, so:

P_{tot} = P_1 + P_2 = 4 W + 8 W=12 W

3 0
2 years ago
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