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Yogita Ingle 6 years, 3 months ago
1. It is becomes difficult to walk on a slippery road due to low friction. When we move on ice, it becomes difficult to walk due to low friction of ice.
2. We can not fix nail in the wood or wall if there is no friction. It is friction which holds the nail.
3. A horse can not pull a cart unless friction furnishes him a secure Foothold.
4. We can write on a paper or on a board.
5. Friction helps in applying the brakes.
6. It helps in walking on floor.
7. Coffee mug stays on the dashboard
8. Shuffling across a carpet to shock someone
9. Draging of atmosphire with earth is possible
10. Helps to pevents the life on earth by burning astroids.
Posted by #Aditi~ Angel???? 6 years, 3 months ago
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Posted by Suplex City ?? 6 years, 3 months ago
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Posted by #Aditi~ Angel???? 6 years, 3 months ago
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Gaurav Seth 6 years, 3 months ago
Given:
Mass of gun=M=100kg
mass of shell =m=0.020kg
Velocity of shell = v=80m/s
Recoil velocity of Gun=V=?
initial velocities of both gun and shell will be zero
so, initial momentum of system will be zero.
According to the law of conservation of momentum,
Initial momentum=Final momentum
so,
0=mv-MV
∴v=mv/M
=0.020x80/100
=0.016m/s
∴recoil velocity of Gun=0.016m/s
Posted by Jagabandhu Mishra 6 years, 3 months ago
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#Aditi~ Angel???? 6 years, 3 months ago
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Posted by Drishti Arora 6 years, 3 months ago
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#Aditi~ Angel???? 6 years, 3 months ago
Drishti Arora 6 years, 3 months ago
#Aditi~ Angel???? 6 years, 3 months ago
#Aditi~ Angel???? 6 years, 3 months ago
#Aditi~ Angel???? 6 years, 3 months ago
#Aditi~ Angel???? 6 years, 3 months ago
Gaurav Seth 6 years, 3 months ago
<i>For train A:</i>
Initial velocity, <i>u</i> = 72 km/h = 20 m/s
Time, <i>t</i> = 50 s
Acceleration, <i>a</i>I = 0 (Since it is moving with a uniform velocity)
From second equation of motion, distance (<i>s</i>I)covered by train A can be obtained as:
s = ut + (1/2)a1t2
= 20 × 50 + 0 = 1000 m
<i>For train B:</i>
Initial velocity, <i>u</i> = 72 km/h = 20 m/s
Acceleration, <i>a</i> = 1 m/s2
Time,<i> t</i> = 50 s
From second equation of motion, distance (<i>s</i>II) covered by train A can be obtained as:
<i>s</i>II = ut + (1/2)at2
= 20 X 50 + (1/2) × 1 × (50)2 = 2250 m
Length of both trains = 2 × 400 m = 800 m
Hence, the original distance between the driver of train A and the guard of train B is 2250 – 1000 – 800 = 450m.
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#Aditi~ Angel???? 6 years, 3 months ago
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Robinpreet Kaur 6 years, 3 months ago
1Thank You