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Rasek [7]
1 year ago
5

A metal sphere with a mass of 0.026 kg rolls along a friction-less surface at 3.7 m/s and strikes a stationary sphefe having a m

ass of 0.050 kg. The first sphere stops completely. The spheres are in contact for 0.007s before the second sphere is shot off down the track. What is the force applied to the second ball?
Physics
1 answer:
zubka84 [21]1 year ago
7 0

The force applied to the second ball by the first ball is 6.734 × 10^-4 N.

<h3>What is impulse of force?</h3>

The impulse of force is defined as the sum of the average force and the duration it is applied.

If the mass of the item remains constant, the impulse of force equals the change in momentum of the object.

Given that:  mass of a metal sphere: m = 0.026 kg.

Initial speed of the sphere: u = 3.7 m/s.

When the sphere stops completely, its change in momentum = mu - 0

= 0.026×3.7 N-s.

= 0.0962 N-s.

As the spheres are in contact for 0.007s before the second sphere is shot off down the track, the force applied  to the second ball =

change in momentum of 1st ball × time of contact

= 0.0962 × 0.007 N

= 0.0006734 N

= 6.734 × 10^-4 N.

Hence,  the force applied  to the second ball is  6.734 × 10^-4 N.

Learn more about impulse force here:

brainly.com/question/29787329

#SPJ1

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A 30.0 kg dolphin decelerates from 12.0 to 7.00 m/s in 2.60 s to join another dolphin in play. What average force (in N) was exe
Crazy boy [7]

Answer:

Force is 57.69 N to the opposite direction of motion of dolphin.

Explanation:

We have force is the product of mass and acceleration.

That is    

                 Force = Mass x Acceleration

                         F = ma

Mass of dolphin, m = 30 kg

We have equation of motion, v = u + at

Final velocity, v = 7 m/s

Initial velocity, u = 12 m/s

Time, t = 2.60 s

Substituting

                   7 = 12 + a x 2.6

                    a = -1.92 m/s²

Force,   F = 30 x -1.92 = -57.69 N

So the force is 57.69 N to the opposite direction of motion of dolphin.

7 0
3 years ago
Which of the following could be a step in a plan designed to mitigate the negative impacts of a natural hazard? I. design new te
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Designing warning and evacuation systems could be a step in a plan designed to mitigate the negative impacts of a natural hazard.
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3 years ago
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What would happen if you place 2 Plane Mirrors parallel to each other?
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Answer:

No image will be observed.

Explanation:

Images that are created by mirrors are virtual images. This virtual image can only be seen by an observer. In this case, an infinite number of images or no image will be created here as both will be reflecting their own images. Light will continuously bounce back and forth reflecting the same image.

6 0
3 years ago
The length of your eye decreases slightly as you age, making the lens a bit closer to the retina. Suppose a man had his vision s
julia-pushkina [17]

Answer:

<h2>A. Nearsightedness</h2>

Explanation:

A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.

At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.

4 0
3 years ago
A thermos contains m1 = 0.89 kg of tea at T1 = 31° C. Ice (m2 = 0.075 kg, T2 = 0° C) is added to it. The heat capacity of both w
lorasvet [3.4K]

Answer:

a) T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

b) T = 295.37 K

Explanation:

Given;

Initial temperature of tea T1 = 31 C

Initial temperature of ice T2 = 0 C

Mass of tea m1 = 0.89 kg

Mass of ice m2 = 0.075kg

The heat capacity of both water and tea c = 4186 J/(kg⋅K)

the latent heat of fusion for water is Lf = 33.5 × 10^4 J/kg

And T = the final temperature of the mixture

Heat loss by tea = heat gained by ice

m1c∆T1 = m2c∆T2 + m2Lf

m1c(T1-T) = m2c(T-T2) + m2Lf

m1cT1 - m1cT = m2cT - m2cT2 + m2Lf

m1cT + m2cT = m1cT1 + m2cT2 - m2Lf

T(m1c + m2c) = m1cT1 + m2cT2 - m2Lf

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

Substituting the values;

T = (m1cT1 + m2cT2 - m2Lf)/(m1c + m2c)

T = (0.89×4186×31 + 0.075×4186×0 - 0.075×33.5 × 10^4)/(0.89×4186 + 0.075×4186)

T = 22.37 °C

T = 273 + 22.37 K

T = 295.37 K

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3 years ago
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