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beks73 [17]
2 years ago
13

A 5000-kg freight car runs into a 10,000-kg freight car at rest. They couple upon collision and move with a speed of 2 m/s. What

was the initial speed of the 5000-kg car?A) 4 m/sB) 5 m/sC) 6 m/sD) 8 m/sE) none of these
Physics
1 answer:
Vitek1552 [10]2 years ago
4 0

Answer:

C) 6 m/s

Explanation:

Given that

m₁=5000 kg

The initial velocity of 5000 kg car =u₁

m₂=10,000 kg

The initial velocity of 10000 kg car =u₂ = 0 m/s

After collision the final speed of the both car,v = 2 m/s

There is no any external force on the system that is why linear momentum will be conserved.

Linear momentum P = m v

m₁u₁ + m₂u₂ = (m₂ + m₁) v

5000 x u₁ + 10000 x 0 = (5000 + 10000) x 2

5000 x u₁ = 15000 x 2

5 x u₁ = 15 x 2

u₁ = 6 m/s

Therefore the answer is C.

C) 6 m/s

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This process is called polarization.
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2 years ago
How much heat is released when 35kg of water freezes?
inn [45]
You need an additional point of data here: the enthalpy of fusion, or conversely the enthalpy of melting (they differ only by their sign). For water (or ice) that value is gotten from sources such as the internet

<span>ΔH°(fus) = 6.01 kJ/mole </span>

<span>Since you have 35 000g, how many moles do you have? </span>
<span>Moles H2O = 35000 g/(18.015 g/mole) = 1942.8 moles</span>

<span>So, take that ΔH°(fus) in kJ/mole, multiply by the number of moles, and there ya go! 
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6.01 x 1942.8 = 11,676 kJ of energy is released

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3 0
2 years ago
When you bent the plastic ruler, what did you observe in its size? ​
WARRIOR [948]

Answer:

The size will increase.

Explanation:

When you bend a plastic ruler, it's size will increase because it is elastic and will exhibit elastic deformation. When it is been bent, it will continue to stretch until it get to a point where it will not be able to regain it formal shape, it size wound of increase. Therefore when the ruler get to elastic limit and you have bend it to the point it cannot regain it's formal shape back, it will remain bent and if further force is apply on it,it will break.

3 0
3 years ago
Calculate the kinetic energy in joules of a 1200 kg automobile moving at 18 m/s .
vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

Ke=\frac{1}{2}Mass*Velocity^2

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.

7 0
3 years ago
Read 2 more answers
Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
djverab [1.8K]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

As we know from the conservation of mass, the rate at which any amount of fluid mass (m_{1}) is entering in a system is equal to the rate at which the same amount of fluid mass (m_{2}) is leaving the system.

Rate of mass flow can be written as,

m = \rho A v

where \rho is the density of the fluid, A is the area through which the fluid is flowing and v is the velocity of the fluid.

Now, according to the problem, as the density of the fluid does not change, we can write

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas through which the fluid is passing and v_{1} and v_{2} are the velocities of the fluid through the respective cross-sectional areas.

As according to the problem, A_{2} > A_{1}, so from the above formula v_{2} < v_{1}.

Also we know that fluid pressure is created by the motion of the fluid through any area. When the fluid gains speed, some of its energy is used to move faster in the fluid’s direction of motion. It causes in a lower pressure.

So, as in this case v_{2} < v_{1} the pressure in the large cross-sectional area P_{2} will be greater than the pressure  P_{1} in the small cross sectional area, i.e.,

P_{2} > P_{1}.

6 0
3 years ago
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