What is the net force acting on the piano

You do a certain amount of work on an object initially at rest, and all the work goes into increasing the object’s speed. If you

EleoNora [17]

Answer:

$\sqrt{2}v$

Explanation:

The work done on the object at rest is all converted into kinetic energy, so we can write

$W=\frac{1}{2}mv^2$

Or, re-arranging for v,

$v=\sqrt{\frac{2W}{m}}$

where

v is the final speed of the object

W is the work done

m is the object's mass

If the work done on the object is doubled, we have W' = 2W. Substituting into the previous formula, we can find the new final speed of the object:

$v'=\sqrt{\frac{2W'}{m}}=\sqrt{\frac{2(2W)}{m}}=\sqrt{2}\sqrt{\frac{2W}{m}}=\sqrt{2}v$

So, the new speed of the object is $\sqrt{2}v$ .

3 0

4 years ago

Two trains, each having a speed of 33 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h fli

Alex Ar [27]

Answer:

66 km

Explanation:

Given that:

The speed of the two trains = 33 km/h

The speed of the bird = 60 km/h

The distance apart between the two trains = 60 km

From the given information, we are being told that the two trains are going at the same speed. Therefore, they will definitely collide at 30 km

We know that:

speed of the train = distance traveled × time

Making the time t the subject of the formula:

time = speed of the train / distance traveled

time = 30 km / 33 km/h

time = 0.909 / hr

Thus, the bird flying at a given speed of 60 km/h in a time of 0.909 / hr will cover a total distance of :

distance (d) = speed of the bird/ time

distance (d) = $\dfrac{60 \ km/hr}{0.909 \ /hr}$

distance (d) = 66 km

3 0

4 years ago

An object in motion will remain in motion and an object at rest will remain at rest until a greater force interrupts it. Explain

Fittoniya [83]

Answer:

A object, lets say a cup. This cup will never, ever move unless something or someone disturbs it. If something touches or hits this cup the cup will move. But, until the cup gets touched, nothing will EVER make it move.

Explanation:

<em>I hope this helps!!</em>

4 0

3 years ago

Please help me with my question

Rufina [12.5K]

Number 1!!!!!!!!!!!!!

6 0

3 years ago

What is the maximum acceleration of a platform that oscillates at amplitude 4.70 cm and frequency 6.50 Hz?

mylen [45]

Answer:

78.315 m/s²

Explanation:

Amplitude, A = 4.70 cm = 0.047 m

Frequency, f = 6.50 Hz

Angular frequency, ω = 2 π f = 2 x 3.14 x 6.50 = 40.82 rad/s

Maximum acceleration, a = ω² A

a = 40.82 x 40.82 x 0.047

a = 78.315 m/s²

8 0

4 years ago