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slamgirl [31]
3 years ago
13

A cleaver physics professor wants to create a situation where a block starts from rest at the top of a 31.0° inclined plane and

encounters a spring at the bottom of the incline. The spring has a constant 3.4 kN/m and the block's mass is 33.0 kg. How far does the block travel before hitting the spring, if the spring was compressed 37 cm in it's initial collision?
Physics
1 answer:
UNO [17]3 years ago
8 0

Answer:

Explanation:

Let the length of inclined plane be L .

work done by gravity on the block

= force x length of path

= mg sinθ x L , m is mass of the block , θ is inclination of path

This in converted into potential energy of compressed spring

1/2 k x² = mgL sin31  , k is force constant . x is compression

.5 x 3400 x .37² = 33 x9.8 x sin31 L

L = 1.4

Length of incline = 1.4 m .

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Igoryamba

Answer:

Increases, increases

Explanation:

The current is directly proportional to the voltage and inversely proportional to the resistance. The implication of this is that, whenever the voltage is increased, the current increases simultaneously. On the other hand, if the resistance is increased, the current will decrease accordingly and vice versa.

Recall that power is given by P= V^2/R where;

P= power, V= voltage and R= resistance

We can see that power and resistance are inversely related hence decreasing the resistance increases the power output of the lightbulb.

8 0
3 years ago
Balance the equation <br> CuCl2 + H2S → CuS + HCl
Leviafan [203]

Answer:

CuCl2 + H2S -> CuS + 2HCl

Explanation:

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5 0
2 years ago
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You drop a ball from a height of 32 meters how much time passes before the ball hits the ground
tankabanditka [31]

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31 seconds

Explanation:

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The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizont
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Answer:

1923 N

Explanation:

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

Mass (m) = 65 Kg

Radius (r) = 2.5 m

Velocity (v) = 8.6 m/s

Centripetal force (F) =?

The centripetal force, F, can be obtained by using the following formula:

F = mv²/r

F = 65 × 8.6² / 2.5

F = 65 × 73.96 / 2.5

F = 4807.4 / 2.5

F = 1922.96 ≈ 1923 N

Thus, the magnitude of the centripetal's force acting on the student is approximately 1923 N

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3 years ago
The gravitational attraction between two masses of 3kg that are separated by a distance of 1cm is
Luba_88 [7]

Answer:

6.003×10¯⁶ N

Explanation:

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100 cm = 1 m

Therefore,

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1 cm = 0.01 m

Finally, we shall determine the gravitational attraction. This can be obtained as follow:

Mass 1 (M₁) = 3 Kg

Mass 2 (M₂) = 3 Kg

Distance apart (r) = 0.01 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

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F = 6.003×10¯¹⁰ / 1×10¯⁴

F = 6.003×10¯⁶ N

Thus the gravitational attraction is 6.003×10¯⁶ N

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