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kupik [55]
3 years ago
10

How do charges act on insulators and conductors?

Physics
1 answer:
umka2103 [35]3 years ago
5 0

Answer:

An insulator holds charge within its atomic structure. Objects with like charges repel each other, while those with unlike charges attract each other. A conducting object is said to be grounded if it is connected to the Earth through a conductor.

Explanation:

Source:https://courses.lumenlearning.com/physics/chapter/18-2-conductors-and-insulators/

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10 points
MariettaO [177]

Answer:

true

Explanation:

yes because I said so

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3 years ago
Light travels fastest when moving through
vitfil [10]
A vaccum unlike sound,light can travel through any matter including a great vacuum of nothing (space) 
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3 years ago
A 9.0 kg bowling ball races down the lane at 15 m/s before striking a bowling pin (at rest) with a mass of 0.85 kg. If the 0.85
artcher [175]

Answer:

v = 10.75\,\frac{m}{s}

Explanation:

The system ball-pin is modelled by the Principle of Moment Conservation:

(9\,kg)\cdot (15\,\frac{m}{s} ) + (0.85\,kg)\cdot (0\frac{m}{s} ) = (9\,kg)\cdot v + (0.85\,kg)\cdot (45\,\frac{m}{s} )

The velocity of the bowling ball after the collision is:

v = 10.75\,\frac{m}{s}

8 0
3 years ago
Read 2 more answers
If a spear is jabbeb exactly at that place where the fish appears in water, the fish is not killed
Korvikt [17]

You have learned your lesson well, Suhay. Your statement is correct.

The light rays from the fish BEND when they flow out of the water into the air. But our primitive brain still believes that the light rays flow STRAIGHT from the fish. The result is that the fish does not APPEAR to be at that place where it really is.

7 0
3 years ago
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
dexar [7]

Answer:

1.08 s

Explanation:

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

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

3 0
3 years ago
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