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The average force necessary to stop a hammer with 25 NS momentun in 0.04 sec is ………... N
$ F = { \triangle P \over \triangle t } = { 25 \over 0.04 } = 625 N $
Newton's third law of motion leads to the law of consrevation of
Newton's third law of motion states that for every action, there is an equal and opposite reaction. This principle is fundamental to the conservation of momentum. When two bodies interact, the forces they exert on each other are equal in magnitude and opposite in direction, leading to the total momentum of the system remaining constant if no external forces are acting on it.
A ball falls on surface from 10 m height and rebounds to 2.5 m. If duration of
contact with floor is 0.01 sec. then average aceleration during
contact is _ __ ___$ms^{-2} $
$ \nu_o = \sqrt { 2 g h_1 } and \nu = \sqrt {2 g h_1 }$ $ madt = m ( \nu_2 + \nu_1 ) $ $ \therefore a = { \nu_2 + \nu_1 \over dt }$
A vehicle of 100 kg is moving with a velocity of 5 m / s . To stop it in 1/10 sec, the required force in opposite direction is ____ ____ N
To solve this, we use the formula for force, which is given by Newton's second law: \( F = ma \). Here, \( m \) is the mass of the vehicle (100 kg) and \( a \) is the acceleration. First, we need to find the acceleration, which is given by: \( a = \frac{\Delta v}{\Delta t} \). The change in velocity (\( \Delta v \)) is \( -5 \ \text{m/s} \) (since the vehicle is stopping), and the time (\( \Delta t \)) is \( \frac{1}{10} \ \text{s} \). Therefore, \( a = \frac{-5}{1/10} = -50 \ \text{m/s}^2 \). Now, applying Newton's second law: \( F = 100 \ \text{kg} \times -50 \ \text{m/s}^2 = -5000 \ \text{N} \). The negative sign indicates direction, but the magnitude of the force is 5000 N.
The linear momentum P of a particle varies with the time as follows $ P = a + bt^2 $ . Where a and b a re constants. The net force acting on the particle is
$ F = { dp \over dt } = 0 + 2 bt $ $ \therefore F \alpha t $
which of the following statement is correct?
A body with constant speed and zero acceleration means it is moving in a straight line with uniform motion. In uniform motion, both speed and velocity are constant, and the acceleration is zero. Therefore, the correct statement is: “A body has a constant speed and zero acceleration.â€
A force of 8 N acts on an object of mass 5kg in X- direction and another force of 6 N acts on it in Y - direction. Hence, the magnitude of acceleration of object will be
$ F = \sqrt { F_1^2 + F_2^2 } $ $ and a = { F/M } $
On the horizontal surface of a truck ( = 0.6 ) a block of mass 1 kg is placed. If the truck is accelerating at the rale of $ 5 m / s^ 2 $ then frictional force on the block will be_________ N
The frictional force can be found using the formula: \( f = \mu N \), where \( \mu \) is the coefficient of friction and \( N \) is the normal force. For a horizontal surface, the normal force \( N \) is equal to the weight of the block, \( mg \). Here, \( \mu = 0.6 \), \( m = 1 \ \text{kg} \), and \( g = 9.8 \ \text{m/s}^2 \). Thus, \( N = 1 \ \text{kg} \times 9.8 \ \text{m/s}^2 = 9.8 \ \text{N} \). The frictional force is then \( f = 0.6 \times 9.8 \ \text{N} = 5.88 \ \text{N} \). However, the maximum frictional force is \( \mu mg \), and if the truck is accelerating at \( 5 \ \text{m/s}^2 \), the frictional force required to prevent slipping is \( f = ma = 1 \ \text{kg} \times 5 \ \text{m/s}^2 = 5 \ \text{N} \). Therefore, the correct answer is 5 N.
A car turns a corner on a slippery road at a constant speed of 10 m/s. If the coefficient of friction is 0.5, the minimum radius of the arc at which the car turns is ________ meter.
$ v = \sqrt { rg } $ $ \therefore r = { v^2 / g } $
A person standing on the floor of a lift drops a coin. The coin reaches the floor of the lift in time to if the lift is stationary and the time $t_2$ if it is accelerated in upward direction. Than
$ d = { 1 \over 2 } gt_1^2 and \, d = {1 \over 2 } (g + a ) t_2^2 $ By comparing both eqn $ {1 \over 2 } gt_1^2 = { 1 \over 2 } (g+ a)t_2^2 $ $ \therefore t_1 \gt t_2 $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.