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polet [3.4K]
4 years ago
12

(a) A 15.0 kg block is released from rest at point A in the figure below. The track is frictionless except for the portion betwe

en points B and C, which has a length of 6.00 m. The block travels down the track, hits a spring of force constant 2,150 N/m, and compresses the spring 0.200 m from its equilibrium position before coming to rest momentarily. Determine the coefficient of kinetic friction between the block and the rough surface between points B and C.
(b) What If? The spring now expands, forcing the block back to the left. Does the block reach point B?

If the block does reach point B, how far up the curved portion of the track does it reach, and if it does not, how far short of point B does the block come to a stop? (Enter your answer in m.)

Physics
1 answer:
castortr0y [4]4 years ago
4 0

Answer:

(a) coefficient of friction = 0.451

This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)

(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.

Explanation:

The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.

Thank you for reading and I hope this is helpful to you.

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Estimate how many raindrops would fall on 1 acre plot during a 1 inch rainfall​
Tom [10]

Answer:

1,580,088,782,700

Explanation:

There are 90,021 drops of water in one gallon, so multiplying all together. Roughly 1,580,088,782,700 drops of water will fall. Or about 1 1/2 trillion drops.

5 0
3 years ago
285 in scientific notation would be
polet [3.4K]
2.85 x 10^2 (ten to the second power)
3 0
3 years ago
A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a
trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden d_1=1.1cm

Speed of water in the hose is v_1=0.95m/sec

Number of holes n = 22

Diameter of each holes d_2=15cm

According to continuity equation A_1v_1=A_2v_2

d_1^2\times v_1=22\times d_2^2v_2

1.1^2\times 0.95=22\times 0.15^2\times v_2

v_2=2.322m/sec

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0
3 years ago
A sports car starts from rest it covers a distance of 900 m to attain a speed of 80m s determine the acceleration of the car and
Lady bird [3.3K]

The acceleration of the car and the time required to reach this speed will be 8 m/s² and 10 sec.

<h3>What is acceleration?</h3>

The rate of velocity change concerning time is known as acceleration.

Given data;

Initial velocity, u=0  m/s

Final velocity, v= 80 m/sec

Distance travelled,s =900m

From Newton's third equation of motion;

v²=u²+2as

a =(v²-u²)/2s

Substitute the given values;

a = (80²-0)/2 ×900

a = 6400/1800

a=8 m/s²

The time required to reach this speed is found in Newton's first equation of motion as;

v = u+at

Substitute the given values;

80 = 0 + 8t

t=80/8

t = 10 sec

Hence, the acceleration of the car and the time required to reach this speed will be 8 m/s² and 10 sec.

To learn more about acceleration, refer to the link brainly.com/question/2437624

#SPJ1

5 0
2 years ago
If atmospheric pressure suddenly changes from 1.00 atm to 0.896 atm at 298 k, how much oxygen will be released from 3.30 l of wa
xxMikexx [17]
At  a temperature of 298 K, the Henry's law constant is 0.00130 M/atm for oxygen. The solubility of oxygen in water 1.00 atm would be calculated as follows:

<span>S = (H) (Pgas) = 0.00130 M / atm x 0.21 atm = 0.000273 M
</span>
At 0.890 atm,
<span>S = (H)(Pgas) = 0.00130 M / atm x 0.1869 atm = 0.00024297 M</span>
<span>
If atmospheric pressure would suddenly change from 1.00 atm to 0.890 atm at the same temperature, the amount of oxygen that will be released from 3.30 L of water in an unsealed container would be as follows</span>
<span>
3.30 L x (0.000273 mol / L) = 0.0012012 mol</span>

3.30 L x (0.00024297 mol / L) = 0.001069068 mol

0.0012012 mol - 0.001069068 mol = 0.000132 mol
6 0
3 years ago
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