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sertanlavr [38]
3 years ago
16

Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca

rs collide? kg • What must the total momentum of the system be after the train cars collide? kg •
Physics
2 answers:
Rudiy273 years ago
8 0

What is the total momentum of the system before the train cars collide?

 (1600) kg • ms

What must the total momentum of the system be after the train cars collide?

 (1600) kg •  ms

Brut [27]3 years ago
5 0

Let say the two train cars are of masses m_1 and m_2

now if the speed of two cars are v_1 and v_2

then we can say that the momentum of two cars before they collide is given by

P = m_1v_1 - m_2v_2

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

P = m_1v_1 - m_2v_2

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There must be a conducting wire and electromotive force or free electrons

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4 years ago
A girl weighing 45kg is standing on the floor, exerting a downward force of 200N on the floor. The force exerted on her by the f
sukhopar [10]

Answer:

c.

Equal to 200 N..........

7 0
3 years ago
A positive point charge Q is located at x=a and a negative point charge −Q is at x=−a. A positive charge q can be placed anywher
Scilla [17]

Answer:

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

Explanation:

The Coulomb's Law gives the force by the charges:

\vec{F} = K\frac{q_1q_2}{r^2}\^r

Let us denote the positon of the charge q on the y-axis as 'y'.

The force between 'Q' and'q' is

F_1 = K\frac{Qq}{x^2 + y^2}\\F_1_x = F_1\cos(\theta)

where Θ is the angle between F_1 and x-axis.

F_1_x = K\frac{Qq}{x^2 + y^2}(\frac{x}{\sqrt{x^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

whereas

F_2_x = K\frac{-Qq}{a^2 + y^2}(-\frac{a}{\sqrt{a^2 + y^2}}) = \frac{KQqa}{(a^2 + y^2)^{3/2}}

Finally, the x-component of the net force is

\vec{F}_x = \frac{2KQqa}{(a^2 + y^2)^{3/2}} \^x

8 0
4 years ago
A sinusoidal electromagnetic wave from a radio station passes perpendicularly through an open window that has area of 0.700 m2 .
Darina [25.2K]

Answer:

The  value is  E =  8.9 *10^{-5} \  J

Explanation:

From the question we are told that

   The  area is  A =  0.700 \  m^2

   The  root mean square value  is E_{rms} =  0.0400 \  V/m

   The time taken is t =  30.0 \  s

Generally the energy is mathematically represented as

     E =  c *  \sepsilon_o *  A  *  t  * E_{rms}^2

=>    E =  3.0*10^{8} *  8.85*10^{-12} *  0.700 *  30 * (0.04)^2

=>     E =  8.9 *10^{-5} \  J

     

3 0
3 years ago
A body initially at 100°C cools to 60°C in minutes and to 40°C. The temperature of body at the end of 15 minutes will be​
zhuklara [117]

The question is incomplete, the complete question is;

A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?

Answer:

See explanation

Explanation:

From Newton's law of cooling;

θ1 - θ2/t = K(θ1 + θ2/2 - θo]

Where;

θ1 and θ2 are initial and final temperatures

θo is the temperature of the surroundings

K is the constant

t is the time taken

Hence;

100 - 60/5 = K(100 + 60/2 - θo)

100 - 40/10 = K(100 + 40/2 - θo)

8= (80 - θo)K -----(1)

6= (70 - θo)K -----(2)

Diving (1) by (2)

8/6 = (80 - θo)/(70 - θo)

8(70 - θo) = 6(80 - θo)

560 - 8θo = 480 - θo

560 - 480 = -θo + 8θo

80 = 7θo

θo = 11.4°

Again from Newton's law of cooling;

θ = θo + Ce^-kt

Where;

t= 0, θ = 60° and θo = 11.4°

60 = 11.4 + C e^-K(0)

60 - 11.4 = C

C= 48.6°

To obtain K

40 = 11.4 + 48.6e^-10k

40 -11.4 = 48.6e^-10k

28.6/48.6 = e^-10k

0.5585 = e^-10k

-10k = ln0.5585

k= ln0.5585/-10

K= 0.0583

Hence, the temperature in 15 minutes;

θ= 11.4 + 48.6e^(-0.0583 × 15)

θ= 31.7°

4 0
3 years ago
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