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sergeinik [125]

3 years ago

14

An object spins in a horizontal circle with a radius of 15.0 cm. The rotations are timed and the amount of time it takes for it

to go around once is 0.56 s. The centripetal force is measured to be 6.1 N.According to the experiment, the speed of the object is closest to:'

1 answer:

bezimeni [28]3 years ago

6 0

Answer:

1.7 m/s

Explanation:

Relevant Data provided as per the question below:-

Radius = 15.0 cm

Time = 0.56 s.

Based on the above information

The computation of the speed of the object is shown below:-

$Velocity = \frac{2\times \pi \times Radius}{Time}$

$Velocity = \frac{2\times \frac{22}{7} \times 0.15}{0.56}$

$= \frac{0.942857}{0.56}$

= 1.683 m/s

or

= 1.7 m/s

Therefore for computing the speed of the object or velocity we simply applied the above formula by considering the pi and all other given data

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If the sound source is moving then the pitch of the sound will ___________________ to the observer.

Alika [10]

Answer:

Increase

Explanation:

The frequency of sound determines the sound. If the frequency is lower the pitch will be lower. If the frequency is higher the pitch will be higher. This is affected by the motion of the sound source because when a sound source is moving faster the frequency will become higher.

3 0

3 years ago

A marble slides without friction in a vertical loop around the inside of a smooth, 28.6 cm diameter horizontal pipe. The marble'

Leviafan [203]

Answer:3.49 m/s

Explanation:

Given

Speed of marble at Bottom $v=4.22 m/s$

Diameter of loop $d=28.6 cm$

As Energy is conserved therefore Energy at top is equal to energy at bottom

$E_T=E_B$

$\frac{mv^2}{2}+mgh=\frac{mv_0^2}{2}$ ,where $v_0$ is the velocity at bottom

$\frac{v^2}{2}+gh=\frac{v_0^2}{2}$

$v_0^2=v^2+2gh$

$v^2=v_0^2-2gh$

$v=\sqrt{v_0^2-2gh}$

$v=\sqrt{4.22^2-2\times 9.8\times 0.286}$

$v=\sqrt{17.8084-5.6056}$

$v=3.49 m/s$

7 0

3 years ago

The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far befor

Mekhanik [1.2K]

Answer:

The new height the ball will reach = (1/4) of the initial height it reached.

Explanation:

The energy stored in any spring material is given as (1/2)kx²

This energy is converted to potential energy, mgH, of the ball at its maximum height.

If the initial height reached is H

And the initial compression of the spring = x

So, mgH = (1/2)kx²

H = kx²/2mg

The new compression, x₁ = x/2

New energy of loaded spring = (1/2)kx₁²

And the new potential energy = mgH₁

mgH₁ = (1/2)kx₁²

But x₁ = x/2

mgH₁ = (1/2)k(x/2)² = kx²/8

H₁ = kx²/8mg = H/4 (provided all the other parameters stay constant)

6 0

3 years ago

For a vertical spring-mass oscillator that is moving up and down, which of the following statements are true? (more than one sta

omeli [17]

Answer:

At the lowest point in the oscillation, the momentum is zero.

At the lowest point in the oscillation, $mg < ks_s$

Explanation:

Since spring block system is performing to and fro motion along straight line

So here we can say at the lowest position of its path the velocity will become zero.

So we can say that momentum of the spring block system is given as

$P = mv$

$P = 0$

Also we know that after reaching the lowest point the block will again go up towards its mean position

So at the lowest point of the spring block system the block will move upwards again

So this will accelerate upwards hence

$F_{spring} > mg$

$Ks > mg$

6 0

3 years ago

A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i

attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>

As per Newton's equation of motion, V= U+at
U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>

As per Newton's equation of motion,

V²-U² = 2aS

S= distance covered by the car
So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

4 0

2 years ago