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Yuki888 [10]
3 years ago
14

7. A neutral aluminum rod is at rest on a foam insulating base. A negatively charged balloon is brought near one end of the rod

but not in direct contact with it. In what way, if any, will the charges in the rod be affected?
Physics
1 answer:
WITCHER [35]3 years ago
6 0

Answer:

If a negatively charged balloon is brought near one end of the rod but not in direct contact, then <u>the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon</u>, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.

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Part of the solar radiation entering the atmosphere about ___ actually reaches the earth's surface
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Answer:

48%

Explanation:

The earth's temperature has to be stable and this is the reason for the presence of incoming and outgoing heat. Of the solar energy reaching the atmosphere, about 29% is bounced back as reflected radiation by clouds, snow and particles in the atmosphere, about 23% is absorbed by the atmosphere and about 48% actually gets to the earth surface.

4 0
3 years ago
Which interactions can contribute to the intrinsic binding energy during enzymatic catalysis?
Pie

Answer:electrostatic interactions, hydrogen bonding, van der waals interactions.

Explanation:the above listed are the interaction that that contribute

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Sunny watched a video of her professor's lecture on genes and environment the night before class. In class the next day, her pro
IgorLugansk [536]

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7 0
4 years ago
Read 2 more answers
Which of the following would be the magnitude of the vector given a horizontal component of 30 and a
Scorpion4ik [409]

Answer:

50

Explanation:

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Brainliest, please :)

7 0
2 years ago
A 3.0-kg block starts at rest at the top of a 37° incline, which is 5.0 m long. Its speed when it reaches the bottom is 2.0 m/s.
Mama L [17]

Answer: f_{r} = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

W_{x} = horizontal component of the weight = mgsinФ

W_{y} = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - f_{r} = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8m/s^{2}

f_{r} = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

v^{2} = u^{2} + 2aS

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}

we slot in a into the equation below to get frictional force

mgsinФ - f_{r} = ma

3 * 9.8 * sin 37 - f_{r} = 3* 0.4

17.9633 - f_{r} =  1.2

f_{r} = 17.9633 - 1.2

f_{r} = 16.49N

4 0
4 years ago
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