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Sveta_85 [38]
1 year ago
15

An object that is dropped from a height H falls with a constant acceleration of g. The final

Physics
1 answer:
GuDViN [60]1 year ago
4 0

The height need to change by 4 to double the final velocity.

<h3>Final velocity of the object</h3>

The final velocity of an object during a free fall is related to maximum height of fall as given the equation below.

v = √2gh

v² = 2gh

v²/h = 2g

v₁²/h₁ = v₂²/h₂

when v₂ = 2v₁, change in height is calculated as;

h₂ = h₁v₂²/v₁²

h₂ = (h₁ (2v₁)²) / (v₁)²

h₂ = 4h₁v₁² / v₁²

h₂ = 4h₁

Thus, the height need to change by 4 to double the final velocity.

Learn more about final velocity here: brainly.com/question/25905661

#SPJ1

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Explain how the density of a gas is different from the density of a liquid.
zavuch27 [327]

Answer

The same number of particles in a gas spread further apart than in the liquid or solid states.

Explanation:

The same number of particles in a gas spread further apart than in the liquid or solid states. The same mass takes up a bigger volume. This means the gas is less dense. Density also depends on the material.

7 0
2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

3 0
3 years ago
Romeo (81.0 kg) entertains Juliet (53.0 kg) by playing his guitar from the rear of their boat at rest in still water, 2.70 m awa
den301095 [7]

Answer:

0.64 m

Explanation:

The first thing is calculate the center of mass of the system.

X_{cm}= \sum_{n=1}^{n}\frac{X_n\times M_n}{M_n}

now multiplying every coordinate x by the mass of each object (romeo, juliet and the boat) and dividing all by the total mass  taking by reference the position of juliet.

X_{cm}=\frac{53\times0 +81\times2.7+79\times1.35}{53+81+79}

X_cm = 1.4589 m

When the forces involved are internals, the center of mass don't change  

After the movement the center of mass remains in the same distance from the shore, but change relative to the rear of the boat.

X_{cm}=\frac{79\times1.35+(81+53)\times2.7}{53+81+79}

X_cm= 2.10 m

this displacement is how the boat move toward the shore.

2.10-1.46= 0.64 m

5 0
3 years ago
Describe three different ways to change your velocity when youre riding a bike?
netineya [11]

Answer:

describe three different ways to change your velocity when youre riding a bike?

6 0
2 years ago
Read 2 more answers
CHEGG You stretch a spring with spring constant k = 1.2x104 N/m to extend 6.0 cm away from its equilibrium position. How much do
lubasha [3.4K]

Answer:

The elastic potential energy of the spring change during this process is 21.6 J.    

Explanation:

Given that,

Spring constant of the spring, k=1.2\times 10^4\ N/m

It extends 6 cm away from its equilibrium position.

We need to find the elastic potential energy of the spring change during this process. The elastic potential energy of the spring is given by the formula as follows :

E=\dfrac{1}{2}kx^2\\\\E=\dfrac{1}{2}\times 1.2\times 10^4\times (0.06)^2\\\\E=21.6\ J

So, the elastic potential energy of the spring change during this process is 21.6 J.

4 0
3 years ago
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