371683
When a force of 1 Newton acts on a mass of \(1 \mathrm{~kg}\) which is able to move freely, the object moves in the direction of force with a/an
1 speed of \(1 \mathrm{~km} / \mathrm{s}\)
2 acceleration of \(1 \mathrm{~m} / \mathrm{s}^{2}\)
3 speed of \(1 \mathrm{~m} / \mathrm{s}\)
4 acceleration of \(1 \mathrm{~km} / \mathrm{s}^{2}\)
Explanation:
B Given, Force, \(\mathrm{F}=1\) Newton Mass, \(\mathrm{m}=1 \mathrm{~kg}\) Applying second law of motion, \(\mathrm{F}=\mathrm{ma}\) \(1=1 \times a\) \(\mathrm{a}=1 \mathrm{~m} / \mathrm{s}^{2}\)
NDA (II) 2016
LAWS OF MOTION (ADDITIONAL)
371684
A body of mass \(m=3.513 \mathrm{~kg}\) is moving along the \(x\)-axis with a speed of \(5.00 \mathrm{~ms}^{-1}\). The magnitude of its momentum is recorded as
1 \(17.6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
2 \(17.565 \mathrm{~kg} \mathrm{~ms}^{-1}\)
3 \(17.56 \mathrm{~kg} \mathrm{~ms}^{-1}\)
4 \(17.57 \mathrm{~kg} \mathrm{~ms}^{-1}\)
Explanation:
B Given, Mass, \(\mathrm{m}=3.513 \mathrm{~kg}\) Velocity, \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\) The magnitude of its momentum is \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{p}=3.513 \times 5\) \(\mathrm{p}=17.565 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
AIEEE 2008
LAWS OF MOTION (ADDITIONAL)
371685
A diwali rocket is ejecting \(0.05 \mathrm{~kg}\) of gases per second at a velocity of \(400 \mathrm{~m} / \mathrm{s}\). The accelerating force on the rocket is
1 \(20 \mathrm{~N}\)
2 \(2 \mathrm{~N}\)
3 \(100 \mathrm{~N}\)
4 \(200 \mathrm{~N}\)
Explanation:
A Force \(=\) Rate of change in momentum. \(\mathrm{F}=\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}} \cdot \mathrm{V}\) \(=0.05 \times 400\) \(=20 \mathrm{~N}\)
AMU-2004
LAWS OF MOTION (ADDITIONAL)
371686
The linear momentum of a particle varies with time \(t\) as \(P=a+b t+c t^{2}\). Then which of the following is correct?
1 Force is dependent linearly on time
2 Velocity of particle is inversely proportional to time
3 Displacement of the particle is independent of time
4 Force varies with time in a quadratic manner.
Explanation:
A Given, \(\mathrm{P}=\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\) The force acting on particle \(=\) Rate of change of momentum \(F=\frac{d P}{d t}\) \(F=\frac{d}{d t}(a+\) \(F=b+2 c t\) \(\therefore \quad \mathrm{F}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\right)\) So, the force acting on the particle is linearly dependent on time.
371683
When a force of 1 Newton acts on a mass of \(1 \mathrm{~kg}\) which is able to move freely, the object moves in the direction of force with a/an
1 speed of \(1 \mathrm{~km} / \mathrm{s}\)
2 acceleration of \(1 \mathrm{~m} / \mathrm{s}^{2}\)
3 speed of \(1 \mathrm{~m} / \mathrm{s}\)
4 acceleration of \(1 \mathrm{~km} / \mathrm{s}^{2}\)
Explanation:
B Given, Force, \(\mathrm{F}=1\) Newton Mass, \(\mathrm{m}=1 \mathrm{~kg}\) Applying second law of motion, \(\mathrm{F}=\mathrm{ma}\) \(1=1 \times a\) \(\mathrm{a}=1 \mathrm{~m} / \mathrm{s}^{2}\)
NDA (II) 2016
LAWS OF MOTION (ADDITIONAL)
371684
A body of mass \(m=3.513 \mathrm{~kg}\) is moving along the \(x\)-axis with a speed of \(5.00 \mathrm{~ms}^{-1}\). The magnitude of its momentum is recorded as
1 \(17.6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
2 \(17.565 \mathrm{~kg} \mathrm{~ms}^{-1}\)
3 \(17.56 \mathrm{~kg} \mathrm{~ms}^{-1}\)
4 \(17.57 \mathrm{~kg} \mathrm{~ms}^{-1}\)
Explanation:
B Given, Mass, \(\mathrm{m}=3.513 \mathrm{~kg}\) Velocity, \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\) The magnitude of its momentum is \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{p}=3.513 \times 5\) \(\mathrm{p}=17.565 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
AIEEE 2008
LAWS OF MOTION (ADDITIONAL)
371685
A diwali rocket is ejecting \(0.05 \mathrm{~kg}\) of gases per second at a velocity of \(400 \mathrm{~m} / \mathrm{s}\). The accelerating force on the rocket is
1 \(20 \mathrm{~N}\)
2 \(2 \mathrm{~N}\)
3 \(100 \mathrm{~N}\)
4 \(200 \mathrm{~N}\)
Explanation:
A Force \(=\) Rate of change in momentum. \(\mathrm{F}=\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}} \cdot \mathrm{V}\) \(=0.05 \times 400\) \(=20 \mathrm{~N}\)
AMU-2004
LAWS OF MOTION (ADDITIONAL)
371686
The linear momentum of a particle varies with time \(t\) as \(P=a+b t+c t^{2}\). Then which of the following is correct?
1 Force is dependent linearly on time
2 Velocity of particle is inversely proportional to time
3 Displacement of the particle is independent of time
4 Force varies with time in a quadratic manner.
Explanation:
A Given, \(\mathrm{P}=\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\) The force acting on particle \(=\) Rate of change of momentum \(F=\frac{d P}{d t}\) \(F=\frac{d}{d t}(a+\) \(F=b+2 c t\) \(\therefore \quad \mathrm{F}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\right)\) So, the force acting on the particle is linearly dependent on time.
NEET Test Series from KOTA - 10 Papers In MS WORD
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LAWS OF MOTION (ADDITIONAL)
371683
When a force of 1 Newton acts on a mass of \(1 \mathrm{~kg}\) which is able to move freely, the object moves in the direction of force with a/an
1 speed of \(1 \mathrm{~km} / \mathrm{s}\)
2 acceleration of \(1 \mathrm{~m} / \mathrm{s}^{2}\)
3 speed of \(1 \mathrm{~m} / \mathrm{s}\)
4 acceleration of \(1 \mathrm{~km} / \mathrm{s}^{2}\)
Explanation:
B Given, Force, \(\mathrm{F}=1\) Newton Mass, \(\mathrm{m}=1 \mathrm{~kg}\) Applying second law of motion, \(\mathrm{F}=\mathrm{ma}\) \(1=1 \times a\) \(\mathrm{a}=1 \mathrm{~m} / \mathrm{s}^{2}\)
NDA (II) 2016
LAWS OF MOTION (ADDITIONAL)
371684
A body of mass \(m=3.513 \mathrm{~kg}\) is moving along the \(x\)-axis with a speed of \(5.00 \mathrm{~ms}^{-1}\). The magnitude of its momentum is recorded as
1 \(17.6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
2 \(17.565 \mathrm{~kg} \mathrm{~ms}^{-1}\)
3 \(17.56 \mathrm{~kg} \mathrm{~ms}^{-1}\)
4 \(17.57 \mathrm{~kg} \mathrm{~ms}^{-1}\)
Explanation:
B Given, Mass, \(\mathrm{m}=3.513 \mathrm{~kg}\) Velocity, \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\) The magnitude of its momentum is \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{p}=3.513 \times 5\) \(\mathrm{p}=17.565 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
AIEEE 2008
LAWS OF MOTION (ADDITIONAL)
371685
A diwali rocket is ejecting \(0.05 \mathrm{~kg}\) of gases per second at a velocity of \(400 \mathrm{~m} / \mathrm{s}\). The accelerating force on the rocket is
1 \(20 \mathrm{~N}\)
2 \(2 \mathrm{~N}\)
3 \(100 \mathrm{~N}\)
4 \(200 \mathrm{~N}\)
Explanation:
A Force \(=\) Rate of change in momentum. \(\mathrm{F}=\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}} \cdot \mathrm{V}\) \(=0.05 \times 400\) \(=20 \mathrm{~N}\)
AMU-2004
LAWS OF MOTION (ADDITIONAL)
371686
The linear momentum of a particle varies with time \(t\) as \(P=a+b t+c t^{2}\). Then which of the following is correct?
1 Force is dependent linearly on time
2 Velocity of particle is inversely proportional to time
3 Displacement of the particle is independent of time
4 Force varies with time in a quadratic manner.
Explanation:
A Given, \(\mathrm{P}=\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\) The force acting on particle \(=\) Rate of change of momentum \(F=\frac{d P}{d t}\) \(F=\frac{d}{d t}(a+\) \(F=b+2 c t\) \(\therefore \quad \mathrm{F}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\right)\) So, the force acting on the particle is linearly dependent on time.
371683
When a force of 1 Newton acts on a mass of \(1 \mathrm{~kg}\) which is able to move freely, the object moves in the direction of force with a/an
1 speed of \(1 \mathrm{~km} / \mathrm{s}\)
2 acceleration of \(1 \mathrm{~m} / \mathrm{s}^{2}\)
3 speed of \(1 \mathrm{~m} / \mathrm{s}\)
4 acceleration of \(1 \mathrm{~km} / \mathrm{s}^{2}\)
Explanation:
B Given, Force, \(\mathrm{F}=1\) Newton Mass, \(\mathrm{m}=1 \mathrm{~kg}\) Applying second law of motion, \(\mathrm{F}=\mathrm{ma}\) \(1=1 \times a\) \(\mathrm{a}=1 \mathrm{~m} / \mathrm{s}^{2}\)
NDA (II) 2016
LAWS OF MOTION (ADDITIONAL)
371684
A body of mass \(m=3.513 \mathrm{~kg}\) is moving along the \(x\)-axis with a speed of \(5.00 \mathrm{~ms}^{-1}\). The magnitude of its momentum is recorded as
1 \(17.6 \mathrm{~kg} \mathrm{~ms}^{-1}\)
2 \(17.565 \mathrm{~kg} \mathrm{~ms}^{-1}\)
3 \(17.56 \mathrm{~kg} \mathrm{~ms}^{-1}\)
4 \(17.57 \mathrm{~kg} \mathrm{~ms}^{-1}\)
Explanation:
B Given, Mass, \(\mathrm{m}=3.513 \mathrm{~kg}\) Velocity, \(\mathrm{v}=5 \mathrm{~m} / \mathrm{s}\) The magnitude of its momentum is \(\mathrm{p}=\mathrm{mv}\) \(\mathrm{p}=3.513 \times 5\) \(\mathrm{p}=17.565 \mathrm{~kg} \mathrm{~m} / \mathrm{s}\)
AIEEE 2008
LAWS OF MOTION (ADDITIONAL)
371685
A diwali rocket is ejecting \(0.05 \mathrm{~kg}\) of gases per second at a velocity of \(400 \mathrm{~m} / \mathrm{s}\). The accelerating force on the rocket is
1 \(20 \mathrm{~N}\)
2 \(2 \mathrm{~N}\)
3 \(100 \mathrm{~N}\)
4 \(200 \mathrm{~N}\)
Explanation:
A Force \(=\) Rate of change in momentum. \(\mathrm{F}=\frac{\mathrm{dP}}{\mathrm{dt}}=\frac{\mathrm{d}(\mathrm{mv})}{\mathrm{dt}}=\frac{\mathrm{dm}}{\mathrm{dt}} \cdot \mathrm{V}\) \(=0.05 \times 400\) \(=20 \mathrm{~N}\)
AMU-2004
LAWS OF MOTION (ADDITIONAL)
371686
The linear momentum of a particle varies with time \(t\) as \(P=a+b t+c t^{2}\). Then which of the following is correct?
1 Force is dependent linearly on time
2 Velocity of particle is inversely proportional to time
3 Displacement of the particle is independent of time
4 Force varies with time in a quadratic manner.
Explanation:
A Given, \(\mathrm{P}=\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\) The force acting on particle \(=\) Rate of change of momentum \(F=\frac{d P}{d t}\) \(F=\frac{d}{d t}(a+\) \(F=b+2 c t\) \(\therefore \quad \mathrm{F}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{a}+\mathrm{bt}+\mathrm{ct}^{2}\right)\) So, the force acting on the particle is linearly dependent on time.