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natita [175]
3 years ago
9

A block of mass m is compressed against a spring (spring constant kk) on a horizontal frictionless surface. The block is then re

leased and subsequently falls from a high ground of height hh. The maximum compression of the spring before the block was released is \Delta xΔx. The velocity of the block after release is v_1v​1​​. The velocity of the block when it hits the ground is v_2v​2​​.
A.If the block started with a compression of \frac{1}{2}\Delta x​2​​1​​Δx, the block velocity after leaving would be \frac{1}{2}v​2​​1​​v.
B.The mechanical energy of this system is not conserved.
C.If height h is changed to h=2hh=2h, then the velocity of the block v_2v​2​​ will be changed into v_2=2vv​2​​=2v.
D.If the block started with a compression of 3\Delta x3Δx, the block’s velocity after leaving would be 6v6v.
E.None of the above statements are true.
Physics
1 answer:
Nimfa-mama [501]3 years ago
3 0

Answer:

A) True, B) False, C) False  and  D) false

Explanation:

Let's solve the problem using the law of conservation of energy to know if the statements are true or false

Let's look for mechanical energy

Initial

     Emo = Ke = ½ k Dx2

Final

     Em1= ½ m v12

     Emo = Em1

     ½ k Δx2 = ½ m v₁²

    v₁² = k / m Δx²

    v₁ = √ k/m   Δx

Now let's calculate the speed when it falls

   Vfy² = Voy² - 2gy

   Vfy² = - 2gy

   Vf² = v₁² + vfy²

A) True     v₁ = A Δx

.B) False. As there is no rubbing the mechanical energy conserves

.C) False the velocity is proportional to the square root of the height

     v2y = v2 √2

. D) false promotional compression speed

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Suppose a baseball pitcher throws the ball to his catcher.
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a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

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What are the wavelengths of electromagnetic wave in free space that have the following frequencies?.
irina [24]

The wavelengths of the light are 4.3 * 10^-12 m and 0.2 m respectively.

<h3>What is wavelength?</h3>

The term wavelength has to do with the horizontal distance that is covered by a wave. We know that a long wavelength implies that the wave is able to travel a long distance from one point to another.

Given that;

c = λf

c = speed of light

λ = wavelength of ight

f = frequency of light

Thus;

λ = 3 * 10^8/ 7.00 x 10^19

λ = 4.3 * 10^-12 m

λ = 3 * 10^8/1.50 x 10^9

λ = 2 * 10^-1 or 0.2 m

Learn more about wavelength:brainly.com/question/13533093

#SPJ1

Missing parts:

What are the wavelengths of electromagnetic wave in free space that have the following frequencies? (a) 7.00 x 10^19 Hz______ pm (b) 1.50 x 10^9 Hz__________ cm

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