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MakcuM [25]
3 years ago
10

A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere

Physics
1 answer:
dmitriy555 [2]3 years ago
5 0

Question: A 2.0 kg sphere with a velocity of 6.0 m/s collides head-on and elastically with a stationary 10 kg sphere, What is thier velocities after collision.

Answer:

v = 6 m/s, v' = 0 m/s

Explanation:

From the question,

For Elastic collision,

mu+m'u' = mv+m'v'......................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, v = final veolocity of the first sphere, v' = final velocity of the second sphere.

Also,

The relative velocity before collision = relative velocity after collision

u-u' = v-v'............................ Equation 2

Given:  m = 2 kg, m' = 10 kg, u = 6 m/s, u' = 0 m/s

Substitute into equation 1 and 2

2(6)+10(0) = 2v+10v'

2v+10v' = 12.............. Equation 3

6-0 = v-v'

v-v' = 6 ................... Equation 4

Solve equation 3 and 4 simultaneously.

v = 6+v'............. Equation 5

Substitute equation 5 into equation 3

2(6+v')+10v' = 12

12+2v'+10v' = 12

12v' = 12-12

v' = 0/12

v' = 0 m/s.

Also substitute the value of v' into equation 5

v = 6+0

v = 6 m/s

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The temperature of air changes from 0 to 10°C while its velocity changes from zero to a final velocity, and its elevation change
aliya0001 [1]

Answer:

Final velocity = 119.83 m/s

Final elevation = 731.9 m

Explanation:

We are told that temperature of air changes from 0 to 10°C

Thus;

Change in temperature; ΔT = 10 - 0 = 10°C

Also, its velocity changes from zero to a final velocity. Thus;

v1 = 0 m/s

v2 is unknown

Also, its elevation changes from zero to a final elevation.

So, z1 = 0 and z2 is unknown

Now, we want to find v2 and z2 when the internal, kinetic and potential energy are equal.

Thus Equating the formula for both kinetic and internal energy gives;

½m(v2² - v1²) = mc_v•ΔT

m will cancel out and v1 is zero to give;

v2² = 2c_v•ΔT

v2 = √(2c_v•ΔT)

Where c_v is specific heat of constant volume of air with a constant value of 718 J/Kg.K

Thus;

v2 = √(2 × 718 × 10)

v2 = √14360

v2 = 119.83 m/s

To find z2, we will equate potential energy formula to that of the internal energy.

Thus;

mg(z2 - z1) = mc_v•ΔT

m will cancel out and since z1 is zero, then we have;

z2 = (c_v•ΔT)/g

z2 = 718 × 10/9.81

z2 = 731.9 m

4 0
3 years ago
If a rock is dropped from the top of a tower at the front of it and takes 3.6 seconds to hit the ground. Calculate the final vel
expeople1 [14]

Answer:

35.28m/s; 63.50m

Explanation:

<u>Given the following data;</u>

Time, t = 3.6 secs

Since it's a free fall, acceleration due to gravity = 9.8m/s²

Initial velocity, u = 0

To find the final velocity, we would use the first equation of motion;

V = u + at

Substituting into the equation, we have;

V = 0 + 9.8 * 3.6

V = 35.28m/s

Therefore, the final velocity of the penny is 35.28m/s.

To find the height, we would use the second equation of motion;

S = ut + \frac {1}{2}at^{2}

Substituting the values into the equation;

S = 0(3.6) + \frac {1}{2}*9.8*(3.6)^{2}

S = 0 + 4.9*12.86

S = 0.5 *36

S = 63.50m

Therefore, the height of the tower is 63.50m.

6 0
3 years ago
How much total energy is dissipated in 10. seconds
noname [10]

Answer : Total energy dissipated is 10 J

Explanation :

It is given that,

Time. t = 10 s

Resistance of the resistors, R = 4-ohm

Current, I = 0.5 A

Power used is given by :

P=\dfrac{E}{t}

Where

E is the energy dissipated.

So, E = P t.............(1)

Since, P=I^2R

So equation (1) becomes :

E=I^2Rt

E=(0.5\ A)^2\times 4\Omega \times 10\ s

E=10\ J

So, the correct option is (3)

Hence, this is the required solution.

7 0
3 years ago
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tiny-mole [99]
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What type of friction is using chalk in the summer to draw on the ground in Copley square?
Lera25 [3.4K]
The second option rolling friction
7 0
3 years ago
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