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Karo-lina-s [1.5K]
3 years ago
6

What are the characteristics of image when object is between f1 and 2f1 for concave lense?​

Physics
1 answer:
k0ka [10]3 years ago
5 0

Answer:

The required diagram is shown in the figure. When an object is placed in front of the convex lens, i.e., between 2F

1

and F

1

, its image is formed beyond 2F

2

on the other side of the lens. The image is real, inverted and enlarged.

solution

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co
labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m

a)

The final velocity of cart 1 after collision is given as:

v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

7 0
3 years ago
Write short letters.
Svetradugi [14.3K]

Answer:

No. 67

Peter Street

12th Road

Chennai

24th June 201_

Dear Amrish

I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.

It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?

I am anxiously waiting for your reply.

Yours affectionately

your name

4 0
2 years ago
Read 2 more answers
what velocity must a 1340kg car have in order to havw the same momentum as a 2680 kg truck traveling at a velocity of 15m/s to t
kykrilka [37]
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.

Hope i helped
Have a good day :)

 
6 0
3 years ago
A car is traveling with a velocity of 5.5 m/s and has a mass of 1200 kg. What is the kinetic energy?!
vesna_86 [32]

Answer:

<h2>18150 J</h2>

Explanation:

The kinetic energy of the car can be found by using the formula

k =  \frac{1}{2} m {v}^{2}  \\

m is the Mass

v is the velocity

From the question we have

k =  \frac{1}{2}  \times 1200 \times  {5.5}^{2}  \\  = 600 \times 30.25

We have the final answer as

<h3>18150 J</h3>

Hope this helps you

3 0
2 years ago
A system had 150 kj of work done on it and its internal energy increased by 60 kj. How much energy did the system gain or lose a
mina [271]

Answer:

The system loses 90 kJ of heat

Explanation:

We can answer the question by using the 1st law of thermodynamics, which states that:

\Delta U=Q-W

where

\Delta U is the change in internal energy of the system

Q is the heat absorbed by the system (positive if absorbed, negative if released by the system)

W is the work done by the system (positive if done by the system, negative if done by the surrounding on the system)

In this problem, we have:

W=-150 kJ is the work done (negative, because it is done by the surrounding on the system)

\Delta U=+60 kJ is the increase in internal energy

Using the equation above, we can find Q, the heat absorbed/released by the system:

Q=\Delta U+W=+60 kJ+(-150 kJ)=-90 kJ

And the negative sign means that the system has lost this heat.

8 0
3 years ago
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