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yarga [219]
4 years ago
8

Which descriptions best fit the labels? X: kinetic energy Y: potential energy X: potential energy Y: kinetic energy X: mechanica

l energy Y: electrical energy X: electrical energy Y: mechanical energy
Physics
1 answer:
Pani-rosa [81]4 years ago
3 0
W-APE. For example, work W done to accelerate a positive charge from rest is positive and results from a loss in PE, or a negative APE. There must be a minus sign in front of APE to make W positive. PE can be found at any point by taking one point as a reference and calculating the work needed to move a charge to the other point.

( The capital A’s in the words are supposed to be triangles ! I also hoped this helped ! Please mark me as brainliest !! )
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The earth is composed of four layers three of these layers are solid and only one is liquid which layer exits in the liquid stat
Vladimir [108]

Answer:

The outer core is a liquid

Explanation:

The outer core is a fluid composed of iron and nickel

7 0
4 years ago
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The current in a resistor is 3.0 A, and its power is 60 W. What is the voltage?
Novosadov [1.4K]

Answer:

20 volts

Explanation:

Use the equation P=VI

60=V(3)

V=20

6 0
2 years ago
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When water changes into vapor this is called?
Scorpion4ik [409]

When water changes into vapor, it is called evaporation.  BONUS:  This is formed by the boiling point of water, which is 230°F (Fahrenheit) or 110°C (Celsius).

4 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
Question 8
FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

8 0
3 years ago
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