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kupik [55]
3 years ago
8

A mass, M, is at rest on a frictionless surface, connected to an ideal horizontal spring that is unstretched. A person extends t

he spring 30 cm from equilibrium and holds it at this location by applying a 10 N force. The spring is brought back to equilibrium and the mass connected to it is now doubled to 2M. If the spring is extended back 30 cm from equilibrium, what is the necessary force applied by the person to hold the mass stationary there
Physics
1 answer:
nignag [31]3 years ago
7 0

Answer:

The necessary force applied by the person to hold the mass stationary there is 10 N

Explanation:

We are told that this person extends the spring 30 cm from equilibrium and holds it at this location by applying a 10 N force.

Thus, based on Hooke's law formula which is F = k Δx, we can say that the mass attached to the spring does not change the spring constant. Thus, the

same resistive spring force will still be in place and in turn, the same stretching force of 10N would still be required.

Thus;

The necessary force applied by the person to hold the mass stationary there is 10 N

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Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?
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Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

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2 years ago
Multiple-Concept Example 9 reviews the concepts that are important in this problem. A drag racer, starting from rest, speeds up
Mademuasel [1]

Answer:

V = 90.51 m/s

Explanation:

From the given information:

Initial speed (u) = 0

Distance (S) = 391 m

Acceleration (a) = 18.9 m/s²

Using the relation for the equation of motion:

v² - u² = 2as

v² - 0² = 2as

v² = 2as

v = \sqrt{2as}

v = \sqrt{2*18.9*391}

v = 121.57 m/s

After the parachute opens:

The initial velocity = 121.57 m/ss

Distance S' = 332 m

Acceleration = -9.92 m/s²

How fast is the racer can be determined by using the relation:

V=  \sqrt{v^2 + 2aS'}

V = \sqrt{121.57^2+ 2 (-9.92)(332)}

V = 90.51 m/s

6 0
3 years ago
The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth
FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × 10^{8} Pa

Explanation:

The pressure at the depth is given by,

P = h \rho g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

\rho = density of water = 1000 \frac{kg}{m^{3} }

g = acceleration due to gravity ≈ 9.8 \frac{m}{s^{2} }

P = 11033.15 × 9.8 × 1000

P = 1.08 × 10^{8} Pa

Thus, the pressure at the depth is 1.08 × 10^{8} Pa

4 0
3 years ago
A -0.06 charge that moves downward is in a uniform electric field with a strengthened of 200 N/C. What is the magnitude and dire
Nookie1986 [14]

Answer:

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Explanation:

The force on a positive charge will be in the same direction as the field, but the force on a negative charge will be in the opposite direction to the field. Thus the direction of the force is upward.

Given;

magnitude of charge, q = 0.06 C

magnitude of electric field, E = 200 N/C

The magnitude of the force is given by;

F = qE

F = 0.06 x 200 N/C

F = 12 N Upwards

Therefore, the magnitude of the force is 12 N Upwards

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A young hockey player stands at rest on the ice holding a 1.3-kg helmet. The player tosses the helmet directly in front of him w
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Answer

given,

initial speed of hockey player= 0 m/s

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final speed of the helmet, v = 6 m/s

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we need to calculate the mass of the hockey player, M = ?

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1.3 x 0 + M x 0 =  M x (-0.25) + 1.3 x 6

negative sign is taken because recoil velocity is in opposite direction

0 = -0.25 M + 7.8

0.25 M = 7.8

M = 31.2 Kg

Hence, the mass of the young hockey player is equal to 31.2 Kg

5 0
2 years ago
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