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2 years ago

If I = 2.0 A in the circuit segment shown below, what is the potential difference VB - VA?

Physics

1 answer:

Tom [10]2 years ago

7 0

Answer:

10 V

Explanation:

The potential difference between two points is the amount of work required to carry a unit charge from one point to the other point. This would result in a potential difference between this two points.

The difference between the potential across two points B and A is $V_{BA}=V_B-V_A$

From the image attached:

$V_B-V_A=10-10I+20\\\\But\ I = 2A,hence:\\\\V_B-V_A=10-10(2)+20\\\\V_B-V_A=10-20+20\\\\V_B-V_A=10\ V$

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Suppose the ski patrol lowers a rescue sled carrying an injured skier, with a combined mass of 97.5 kg, down a 60.0-degree slope

Kitty [74]

a. 1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b.21,835 J work, in joules, is done by the rope on the sled this distance.

c. 23,170 J the work, in joules done by the gravitational force on the sled d. The net work done on the sled, in joules is 43,670 J.

<h3>What is friction work?</h3>

The work done by friction is the force of friction times the distance traveled times the cosine of the angle between the friction force and displacement

a. How much work is done by friction as the sled moves 28m along the hill?

ans. We use the formula:

friction work = -µ.mg.dcosθ

= -0.100 * 97.5 kg * 9.8 m/s² * 28 m * cos 60

= -1337.3 J

-1337.3 J work, in joules, is done by friction as the sled moved 28 m along the hill.

b. How much work is done by the rope on the sled in this distance?

We use the formula:

Rope work = -m.g.d(sinθ - µcosθ)

rope work = - 97.5 kg * 9.8 m/s² * 28 m (sin 60 – 0.100 * cos 60)

= 26,754 (0.816)

= 21,835 J

21,835 J work, in joules, is done by the rope on the sled this distance.

c. What is the work done by the gravitational force on the sled?

By using the formula:

Gravity work = mgdsinθ

= 97.5 kg * 9.8 m/s² * 28 m * sin 60

= 23,170 J

23,170 J the work, in joules done by the gravitational force on the sled .

D. What is the total work done?

By adding all the values

work done = -1337.3 + 21,835 + 23,170

= 43,670 J

The net work done on the sled, in joules is 43,670 J.

Learn more about friction work here:

brainly.com/question/14619763

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4 0

1 year ago

If the body temperature of a person is recorded as 37°C, what is the person’s body temperature in K? Round your answer to the ne

melisa1 [442]

K= 37°C+273.15
K= 310.15
Round to the nearest whole number
310K

6 0

2 years ago

Which is an example of a wave?

Travka [436]

Answer:

The answer is A

Explanation:

Someone shouting to you from across a room

4 0

1 year ago

Read 2 more answers

Suppose the gravitational force between two spheres is 100 N. If the distance between the spheres is halved, what is the new gra

natima [27]

Answer:

The gravitational force becomes 4 times i.e 400N

Explanation:

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3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

natali 33 [55]

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

where

G is the gravitational constant

M is the Earth's mass

R is the Earth's radius

If the projectile has an initial speed of 0.421 escape speed,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius: $0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

So, the projectile reaches a radial distance of 5.65 times the Earth's radius.

b) 2.36 times the Earth's radius

The kinetic energy needed to escape is:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:

$E=U=\frac{GMm}{R}$

Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

6 0

2 years ago