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11111nata11111 [884]
3 years ago
8

What is the restoring force of a spring stretched 0.35 meters with a spring contact of 55 newtons?

Physics
2 answers:
Vesna [10]3 years ago
6 0
F=-kx = -19,25 answer D
Bogdan [553]3 years ago
6 0

Answer:

F=19.25 N

Explanation:

The restoring force for a spring is given by:

\vec{F}=-k \vec{x}

Where k is the spring constant and \vec{x} is the spring deformation.

The minus sign indicates that force is opposed to the deformation direction.

According to the problem:

k=55 \;N/m and  x=0.35 \;m

Thus, in magnitude, the restoring force is:

F=55\frac{N}{m} *0.35m=19.25\; N

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Lera25 [3.4K]

Answer:

<h2>14.66secs</h2>

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

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3 years ago
Which device uses a rotating magnetic field to produce an electric current?
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<span>The correct answer is C) a motor.
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7 0
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Explanation:

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u = 10 km/h \cdot \frac{1000 m/km}{3600 s/h}=2.78 m/s is the initial velocity

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