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defon
3 years ago
11

MY NOTES A spring with a mass of 2 kg has a damping constant 14 kg/s. A force of 3.6 N is required to keep the spring stretched

0.3 m beyond its natural length. The spring is stretched 0.6 m beyond its natural length and then released. Find the position of the mass at any time t. (Assume that movement to the right is the positive x-direction and the spring is attached to a wall at the left end.)
Physics
1 answer:
il63 [147K]3 years ago
7 0
Answer: What’s the question?
Explanation: everything looks good for this!! I understand now :) I think ❤️ pls give me Brainly!
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A car travels 10 meters east in 4 seconds. What is the car's velocity? *
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your velocity is 2.5 m/sec

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In being served, a tennis ball is accelerated from rest to a speed of 31.6 m/s. The average power generated during the serve is
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Answer:

The force acting on the ball is 92.4 N.

Explanation:

Given that,

Initial speed of the ball, u = 0

Final speed of the ball, v = 31.6 m/s

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P=\dfrac{W}{t}

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P=\dfrac{Fd}{t}

Since, v=\dfrac{d}{t}

So,

P=F\times v

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F=\dfrac{P}{v}\\\\F=\dfrac{2920\ W}{31.6\ m/s}\\\\F = 92.4\ N

So, the force acting on the ball is 92.4 N. Hence, this is the required solution.

7 0
3 years ago
A. A land speed car can decelerate at 9.8m/s. How long does it take the car to come to a complete stop from a run of 885 km/hr (
Nimfa-mama [501]

Answer:

A. 25.08 s

B. 3082.53 m

C. 3×10⁵ m/s²

Explanation:

A. Determination of the time.

This can be obtained as illustrated below:

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Time (t) =.?

v = u + at

0 = 245.8 + (–9.8 × t)

0 = 245.8 – 9.8t

Collect like terms

0 – 245.8 = – 9.8t

– 245.8 = – 9.8t

Divide both side by –9.8

t = –245.8 / –9.8

t = 25.08 s

Therefore, it will take 25.08 s for the car to come to a complete stop.

B. Determination of the distance travelled by the car.

Acceleration (a) = –9.8 m/s²

Initial velocity (u) = 245.8 m/s

Final velocity (v) = 0 m/s

Distance (s) =?

v² = u² + 2as

0² = 245.8² + (2 × –9.8 × s)

0 = 60417.64 – 19.6s

Collect like terms

0 – 60417.64 = – 19.6s

– 60417.64 = – 19.6s

Divide both side by –19.6

s = –60417.64 / –19.6

s = 3082.53 m

Thus, the car travelled a distance of 3082.53 m before stopping completely.

C. Determination of the acceleration of the object.

Initial velocity (u) = 0 m/s

Final velocity (v) = 600 m/s

Distance (s) = 0.6 m

Acceleration (a) =?

v² = u² + 2as

600² = 0² + (2 × a × 0.6)

360000 = 0 + 1.2a

360000 = 1.2a

Divide both side by 1.2

a = 360000 / 1.2

a = 300000 = 3×10⁵ m/s²

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3 years ago
If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for
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Answer:

\ d_{out} = 100 \ m.

Explanation:

Given data:

F_{in} = 50 \ \rm N

F_{out} = 10 \ \rm N

d_{in} = 20 \ m

Let the distance traveled by the object in the second case be d_{out}.

In the given problem, work done by the forces are same in both the cases.

Thus,

W_{in} = W_{out}

F_{in}.d_{in} = F_{out}.d_{out}

\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}

\ d_{out} = \frac{50 \times 20}{10}

\ d_{out} = 100 \ m.

5 0
3 years ago
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