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Rina8888 [55]
3 years ago
11

Two identical 0.400 kg masses are pressed against opposite ends of a light spring of force constant 1.75 N/cm, compressing the s

pring by 39.0 cm from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless horizontal table.
Physics
1 answer:
nlexa [21]3 years ago
7 0

Answer:

0.853 m/s

Explanation:

Total energy stored in the spring = Total kinetic energy of the masses.

1/2ke² = 1/2m'v².................... Equation 1

Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.

make v the subject of the equation,

v = e[√(k/m')].................... Equation 2

Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.

Substitute into equation 2

v = 0.39[√(1.75/0.8)

v = 0.39[2.1875]

v = 0.853 m/s

Hence the speed of each mass = 0.853 m/s

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Two fish swimming in a river have the following equations of motion:
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Answer:

The second fish, X2, is moving faster than the first fish, X1

Explanation:

The given parameters for the equation of motion of the fishes are;

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X2 = 1.3 m + (-2.7 m/s)×t

The given equation are straight line equations in the slope and intercept form, where the slope is the speed and in m/s and the intercept is the starting point of swimming of the fishes

For the first fish, the intercept = -6.4 m, the slope = the  speed = -1.2 m/s

For the second fish, the intercept = 1.3 m, the slope = the  speed = -2.7 m/s

Whereby the fishes are swimming in the opposite direction of the measurement of length, we have;

The magnitude of the speed of the second fish \left | -2.7 \ m/s \right | = 2.7 \ m/s, is larger than the magnitude of the speed of the first fish \left | -1.2 \ m/s \right | = 1.2 \ m/s

Therefore, the second fish, X2, is moving faster than the first fish, X1.

8 0
3 years ago
Which form of Kepler’s third law can you use to relate the period T and radius r of a planet in our solar system as long as the
ser-zykov [4K]

T2=r In the form of Kepler's law that can use to relate the period T and radius of the planet in our solar systems

<u>Explanation:</u>

<u>Kepler's third law:</u>

  • Kepler's third law states that For all planets, the square of the orbital

     period (T) of a planet is proportional to the cube of the average orbital    radius (R).

  • In simple words T (square) is proportional to the R(cube) T²2 ∝1 R³3
  • T2 / R3 = constant = 4π ² /GM

      where G = 6.67 x 10-11 N-m2 /kg2

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Why does an object hover over a large magnet
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MatroZZZ [7]

Answer:

Explanation:

Given

Inclination \theta =32.9^{\circ}

Distance of landing point R=7.78\ m

Considering athlete to be an Projectile

range of projectile is given by

R=\frac{u^2\sin 2\theta }{g}

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R'=9.07\ m

 

3 0
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