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Yogita Ingle 6 years, 2 months ago
| Accuracy | Precision |
| It could be defined as the level of correctness of a measurement to its true value. | It could be defined as the sharp exactness of a measurement. |
| Has one factor used for measuring. | Has multiple factors for measurement. |
| Accurate items have to be precise in most cases. | Precise items may or may not be accurate. |
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Yogita Ingle 6 years, 2 months ago
Let the initial mass of the body be m and its velocity be v.
Initial momentum of the body = mv
Initial KE = mv^2/2
If the momentum increases by 10%, new momentum = mv + 10% of mv
= mv + 10/100mv
= mv + mv/10
=11mv/10
=m(11v/10)
New velocity = 11v/10
Increase in velocity = v/10
New KE = m(11v/10)^2/2
= 121/100mv^2/2
Increase in KE = 21%
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Yogita Ingle 6 years, 2 months ago
If the net external force acting on a system of bodies is zero, then the momentum of the system remains constant. This is the basic law of conservation of linear momentum.
1) When a bullet is fired from a gun, the recoil of the gun can be explained on the basis of the law of conservation of linear momentum.
2) When a heavy nucleus at rest disintegrates into two smaller nuclei, the products move in opposite directions, obeying the law of conservation of linear momentum.
3) The motion of rockets is obeyed by the law of conservation of linear momentum.
Muskann Tiwari 6 years, 2 months ago
Posted by Alok Maurya 6 years, 2 months ago
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Gaurav Seth 6 years, 2 months ago
acceleration = -av^2. (We are assuming that a is a constant and v represents velocity here)
Now, acceleration = dv/dt
Hence, dv/dt = -av^2
dv/v^2 = -adt
Integrating from t = 0, v = u to t = t and v =v
[1/u-1/v] = -at
1/v = 1/u+ at
1/v = (1+uat)/u
v = u/(1+uat)
Now v = dx/dt
dx/dt = u/(1+uat)
dx = u/(1+uat) dt
Integrating from x = 0, t = 0 to x = x, t = t
x = u/ua * ln|1+uat|
x = ln|1+uat|/a
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Ayush Vishwakarma 6 years, 2 months ago
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