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sergejj [24]
3 years ago
10

Two objects have the same momentum. Which of the following is true? I. The masses of the objects are equal. Ii. The speeds of th

e objects are equal. Iii. The velocities of the objects are equal. Iv. The direction of motion of each object is the same.
Physics
1 answer:
Vaselesa [24]3 years ago
3 0

Answer:

Iv. The direction of motion of each object is the same.

Explanation:

Momentum of an object is defined as:

p= mv

where

m is the mass of the object

v is its velocity

Since v is a vector, it follows that p is also a vector, and it has the same direction of the velocity. This means that if two objects have same momentum, then the direction of their momentum vector is also the same: therefore, the direction of motion of each object must be the same.

The other options are wrong, because they just state that only one of the two quantities involved in the equation (mass or velocity) is the same, but the two objects can actually have the same momentum even if they have different masses and velocities (in fact, the only thing that matters is that the product between mass and velocity is the same).

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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
A need the answers pleaseeeee
IRISSAK [1]

It's just asking you to sit down and COUNT the little squares in each sector.

It'll help you keep everything straight if you take a very sharp pencil and make a tiny dot in each square as you count it.  That way, you'll be able to see which ones you haven't counted yet, and also you won't count a square twice when you see that it already has a dot in it.

(If, by some chance, this is a picture of the orbit of a planet revolving around the sun ... as I think it might be ... then you should find that both sectors jhave the same number of squares.)  

7 0
3 years ago
How much energy (in kilojoules) is required to convert 200 mL of diethyl ether at its boiling point from liquid to vapor if its
LiRa [457]

Answer:

55.96kJ

Explanation:

Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether

Volume (v) = 200mL, density (d) = 0.7138g/mL

Mass = d × v = 0.7138 × 200 = 142.76g

Enthalpy of vaporization of diethyl ether = 29kJ/mol

MW of diethyl ether (C2H5)2O = 74g/mol

Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g

Energy = 142.76g × 0.392kJ/g = 55.96kJ

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3 years ago
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Anuta_ua [19.1K]

Sure , I'm subscribing!!!

8 0
2 years ago
Read 2 more answers
The charge that passes a cross-sectional area A=10-4 m2 varies with time according
Anarel [89]

Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

= 6A.

6 0
3 years ago
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