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iren2701 [21]
3 years ago
11

5. Which pair of words, in order, correctly completes

Physics
1 answer:
sattari [20]3 years ago
3 0

Answer:

B is an answer.

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A force of 19 newtons is applied on a cart of 2 kilograms, and it experiences a frictional force of 1.7 newtons. What is the acc
amid [387]
A= f/m

a= 19/2

a= 9.5m/s^2
8 0
3 years ago
Railroad cars are loosely coupled so that there is a noticeable time delay from the time the first and last car is moved from re
olga nikolaevna [1]

Answer:

Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This loose coupling can be very important for braking as well).

Explanation:

8 0
3 years ago
using mass and distance, identify and compare the sun's and moon's contribution to the formation of tides on earth
vekshin1

Answer:

Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.

7 0
3 years ago
As a mercury atom absorbs a photon of
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The energy absorbed by photon is 1.24 eV.
This is the perfect answer.
8 0
2 years ago
A student pushes a 0.2 kg box against a spring causing the spring to compress 0.15 m. When the spring is released, it will launc
german

Answer:

The maximum height the box will reach is 1.72 m

Explanation:

F = k·x

Where

F = Force of the spring

k = The spring constant = 300 N/m

x  = Spring compression or stretch = 0.15 m

Therefore the force, F of the spring = 300 N/m×0.15 m = 45 N

Mass of box = 0.2 kg

Work, W, done by the spring = \frac{1}{2} kx^2 and the kinetic energy gained by the box is given by KE = \frac{1}{2} mv^2

Since work done by the spring = kinetic energy gained by the box we have

\frac{1}{2} mv^2 =  \frac{1}{2} kx^2  therefore we have v = \sqrt{\frac{kx^2}{m} } = x\sqrt{\frac{k}{m} } = 0.15\sqrt{\frac{300}{0.2} } = 5.81 m/s

Therefore the maximum height is given by

v² = 2·g·h or h = \frac{v^2}{2g} = \frac{5.81^{2} }{2*9.81} = 1.72 m

6 0
3 years ago
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