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murzikaleks [220]
3 years ago
15

A 3.0 kg object moving 8.0 m/s in the positive x direction has a one-dimensional elastic collision with an object (mass = M) ini

tially at rest. After the collision the object of unknown mass has a velocity of 6.0 m/s in the positive x direction. What is M?
1- 4.2 kg2- 5.0 kg3- 6.0 kg4- 7.5 kg
Physics
1 answer:
finlep [7]3 years ago
6 0
<h2>Option 2 is the correct answer.</h2>

Explanation:

Elastic collision means kinetic energy and momentum are conserved.

Let the mass of object be m and M.

Initial velocity object 1 be u₁,  object 2 be u₂

Final velocity object 1 be v₁,  object 2 be v₂

Initial momentum = m x u₁ + M x u₂ = 3 x 8 + M x 0 = 24 kgm/s

Final momentum = m x v₁ + M x v₂ = 3 x v₁ + M x 6 = 3v₁ + 6M

Initial kinetic energy = 0.5 m x u₁² + 0.5 M x u₂² = 0.5 x 3 x 8² + 0.5 x M x 0² = 96 J

Final kinetic energy = 0.5 m x v₁² + 0.5 M x v₂² = 0.5 x 3 x v₁² + 0.5 x M x 6² = 1.5 v₁² + 18 M

We have

            Initial momentum = Final momentum

            24 = 3v₁ + 6M

            v₁ + 2M = 8

             v₁ = 8 - 2M

            Initial kinetic energy = Final kinetic energy

            96 = 1.5 v₁² + 18 M

            v₁² + 12 M = 64

Substituting  v₁ = 8 - 2M

           (8 - 2M)² + 12 M = 64    

           64 - 32M + 4M² + 12 M = 64    

            4M² = 20 M

               M = 5 kg

Option 2 is the correct answer.  

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german

Answer:

Final velocity, V = 11.5m/s

Explanation:

Given the following data;

Initial velocity, U = 2.5m/s

Acceleration, a = 1.5m/s²

Time, t = 6secs

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

V = U + at

Substituting into the equation, we have

V = 2.5 + 1.5*6

V = 2.5 + 9

Final velocity, V = 11.5m/s

6 0
3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

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3 years ago
Which simple experiment could be conducted to show an example of how lightning is created? A) Rub a balloon so it picks up extra
alexandr1967 [171]

I think the answer is d

7 0
3 years ago
When you walk at an average speed (constant speed, no acceleration) of 20.7 m/s in 75.8 sec you will cover a distance of______?
nlexa [21]
1,569.06m
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The gauge pressure in the tires of your car is 210 kPa (30.5 psi) when the temperature is 25°C (77 °F). Several days later it is
DaniilM [7]

Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 210kPa + 101.325kPa

P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)

T = 25°C = 298.15K

Final P and T values:

P = ?, T = 0°C = 273.15K

Set the initial and final P/T values equal to each other and solve for the final P:

311.325/298.15 = P/273.15

P = 285.220kPa

Subtract 101.325kPa to find the final gauge pressure:

285.220kPa - 101.325kPa = 183.895271kPa

The final gauge pressure is 184kPa or 26.7psi.

8 0
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