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kumpel [21]
3 years ago
14

Two charged objects have a repulsive force of .80 N. If the charge of one of the objects is doubled, then what is the new force?

Physics
1 answer:
sveticcg [70]3 years ago
6 0
The new force would be 1.6 N.
Since the charges multiply as variables (q1)x(q2), then it would simply be double (q1)x(q2), or 1.6.
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the answer is letter; C

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Do neither of these things ( c )

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If a bus travels 6 km in 10 minutes, what distance does it travel in 1 second ​
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A 2.0 kg block is pulled across a horizontal surface by a 15 N force at a constant velocity. What is the force of friction actin
lianna [129]

Answer:

<em>The force of friction acting on the block has a magnitude of 15 N and acts opposite to the applied force.</em>

Explanation:

<u>Net Force </u>

The Second Newton's law states that an object acquires acceleration when an unbalanced net force is applied to it.

The acceleration is proportional to the net force and inversely proportional to the mass of the object.

If the object has zero net force, it won't get accelerated and its velocity will remain constant.

The m=2 kg block is being pulled across a horizontal surface by a force of F=15 N and we are told the block moves at a constant velocity. This means the acceleration is zero and therefore the net force is also zero.

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2 years ago
A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block
MariettaO [177]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

7 0
3 years ago
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