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deff fn [24]
3 years ago
6

a 1.34kg block sliding on a horizontal surface collides with a spring of force constant 1.93 N/cm. The block compresses the spri

ng 4.16cm from the extended position. Friction between the block and the surface dissipates 117mJ of mechanical energy as the block is brought to rest. Find the speed of the block at the instant of collision with the spring
Physics
1 answer:
sdas [7]3 years ago
7 0

Answer:

v=65m/s

Explanation:

Friction force to find the speed of the block at the instant of collision so

E_1=\frac{1}{2}*K*x^2

x=4.16cm

E_1=\frac{1}{2}*1.93N/cm*(4.16cm)^2

E_1=0.167 J

E_2=117mJ=0.117J

K_E=E_1+E_2=284J

K_E=\frac{1}{2}*m*v^2

Solve to v' to find the velocity

v=\sqrt{\frac{2*K_E}{m}}

v=\sqrt{\frac{2*0.283J}{1.34kg}}=\sqrt{0.424 m^2/s^2}

The speed of the block at the instant of collision with the spring is

v=65m/s

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A 0.50-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.9 m on a frictionless horizontal su
Sedaia [141]

Answer:

\boxed {\boxed {\sf 18 \ m/s}}

Explanation:

The ball is moving in a circle, so the force is centripetal.

One formula for calculating centripetal force is:

F_c= \frac{mv^2}r}

The mass of the ball is 0.5 kilograms. The radius is 1.9 meters. The centripetal force is 85 Newtons or 85 kg*m/s².

  • F_c= 85 kg*m/s²
  • m= 0.5 kg
  • r= 1.9 m

Substitute the values into the formula.

85 \ kg*m/s^2 = \frac{0.5 \ kg *v^2}{1.9 \ m}

Isolate the variable v. First, multiply both sides by 1.9 meters.

(1.9 \ m)(85 \ kg*m/s^2) = \frac{0.5 \ kg *v^2}{1.9 \ m}*1.9 \ m

(1.9 \ m)(85 \ kg*m/s^2) = {0.5 \ kg *v^2}

161.5 \ kg*m^2/s^2 = 0.5 \ kg*v^2

Divide both sides by 0.5 kilograms.

\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} = \frac{0.5 \ kg*v^2}{0.5 \ kg}

\frac {161.5 \ kg*m^2/s^2}{0.5 \ kg} =v^2

323 \ m^2/s^2 = v^2

Take the square root of both sides of the equation.

\sqrt {323 \ m^2/s^2} =\sqrt{ v^2

\sqrt {323 \ m^2/s^2} =v

17.9722007556 \ m/s =v

The original measurements have 2 significant figures, so our answer must have the same.

For the number we found, 2 sig fig is the ones place. The 9 in the tenth place tells us to round the 7 to an 8.

18 \ m/s =v

The maximum speed is approximately <u>18 meters per second.</u>

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3 years ago
A car moves at a constant velocity of 30 m/s and has 3.6 × 105 J of kinetic energy. The driver applies the brakes and the car st
LekaFEV [45]

The force needed to the stop the car is -3.79 N.

Explanation:

The force required to stop the car should have equal magnitude as the force required to move the car but in opposite direction. This is in accordance with the Newton's third law of motion. Since, in the present problem, we know the kinetic energy and velocity of the moving car, we can determine the mass of the car from these two parameters.

So, here v = 30 m/s and k.E. = 3.6 × 10⁵ J, then mass will be

K.E = \frac{1}{2} * m*v^{2}  \\\\m = \frac{2*KE}{v^{2} } = \frac{2*3.6*10^{5} }{30*30}=800 kg

Now, we know that the work done by the brake to stop the car will be equal to the product of force to stop the car with the distance travelled by the car on applying the brake.Here it is said that the car travels 95 m after the brake has been applied. So with the help of work energy theorem,

Work done = Final kinetic energy - Initial kinetic energy

Work done = Force × Displacement

So, Force × Displacement = Final kinetic energy - Initial Kinetic energy.

Force * 95 = 0-3.6*10^{5}\\ \\Force =\frac{-3.6*10^{5} }{95}=-3.79 N

Thus, the force needed to the stop the car is -3.79 N.

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3 years ago
Thunder is heard 7.2 seconds after a bolt of lightning is observed. if the speed of sound is 349 m/s, how far away did the light
Natasha2012 [34]

Answer:

2512.8 meters

Explanation:

s=d/t

by using speed formula

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Answer:

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A box with a mass of 18 kg is pushed across the floor. It has coefficient of friction of 0.39. Calculate the force of friction i
Taya2010 [7]

Answer:

68.8 N

Explanation:

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

Mass (m) of box = 18 Kg

Coefficient of friction (μ) = 0.39

Force of friction (F) =?

Next, we shall determine the normal force of the box. This is illustrated below:

Mass (m) of object = 18 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Normal force (N) =?

N = mg

N = 18 × 9.8

N = 176.4 N

Finally, we shall determine the force of friction experienced by the object. This is illustrated below:

Coefficient of friction (μ) = 0.39

Normal force (N) = 176.4 N

Force of friction (F) =?

F = μN

F = 0.39 × 176.4

F = 68.796 ≈ 68.8 N

Thus, the box experience a frictional force of 68.8 N.

3 0
3 years ago
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