145787
A physical quantity \(A\) on multiplication with velocity results in another quantity \(B\). If the quantity \(B\) is energy, then the quantity \(A\) is
1 mass
2 momentum
3 force
4 acceleration
5 Power
Explanation:
B According to given question, \(\text { A } \times \text { Velocity }=\text { Energy }\) \(\mathrm{A} \times \mathrm{v}=\frac{1}{2} \mathrm{mv}^{2} \quad\left\{\text { Energy }=\frac{1}{2} \mathrm{mv}^{2}\right\}\) \(A=\frac{1}{2} m v\) \(A=\frac{p}{2}\) \(p=\text { momentum } \{p=m v\}\)
Kerala CEE 2021
Laws of Motion
145788
A body of mass \(10 \mathrm{~kg}\) has constant velocity of \(10 \mathrm{~ms}^{-1}\). A constant force is applied for a duration of \(4 \mathrm{~s}\) on the object and gives it a speed of \(2 \mathrm{~ms}^{-1}\) in opposite direction. Then, find acceleration produced in it.
145789
Five objects of different masses are simultaneously released vertically downwards from height ' \(h\) ' (in air). Which physical quantity associated with the objects will change at the instant they strike the ground? (neglect the air resistance.)
1 Time
2 Momentum
3 Velocity
4 Acceleration
Explanation:
B We know that, \(\begin{array}{rrr} \text { height }(\mathrm{h})=\frac{1}{2} \mathrm{gt}^{2} \\ \text { So, } \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\end{array}\) Time is same for all objects. \(\mathrm{v}=\mathrm{gt}\) Velocity is also same for all objects Mass of objects are different so momentum \((\mathrm{P}=\mathrm{mv})\) are different for objects. So, momentum \(\mathrm{p}\) will be different for all.
MHT-CET 2020
Laws of Motion
145790
A force of \(0.03 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\). which is initially at rest for a duration of 10 seconds. Then find the momentum acquired by the body.
145787
A physical quantity \(A\) on multiplication with velocity results in another quantity \(B\). If the quantity \(B\) is energy, then the quantity \(A\) is
1 mass
2 momentum
3 force
4 acceleration
5 Power
Explanation:
B According to given question, \(\text { A } \times \text { Velocity }=\text { Energy }\) \(\mathrm{A} \times \mathrm{v}=\frac{1}{2} \mathrm{mv}^{2} \quad\left\{\text { Energy }=\frac{1}{2} \mathrm{mv}^{2}\right\}\) \(A=\frac{1}{2} m v\) \(A=\frac{p}{2}\) \(p=\text { momentum } \{p=m v\}\)
Kerala CEE 2021
Laws of Motion
145788
A body of mass \(10 \mathrm{~kg}\) has constant velocity of \(10 \mathrm{~ms}^{-1}\). A constant force is applied for a duration of \(4 \mathrm{~s}\) on the object and gives it a speed of \(2 \mathrm{~ms}^{-1}\) in opposite direction. Then, find acceleration produced in it.
145789
Five objects of different masses are simultaneously released vertically downwards from height ' \(h\) ' (in air). Which physical quantity associated with the objects will change at the instant they strike the ground? (neglect the air resistance.)
1 Time
2 Momentum
3 Velocity
4 Acceleration
Explanation:
B We know that, \(\begin{array}{rrr} \text { height }(\mathrm{h})=\frac{1}{2} \mathrm{gt}^{2} \\ \text { So, } \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\end{array}\) Time is same for all objects. \(\mathrm{v}=\mathrm{gt}\) Velocity is also same for all objects Mass of objects are different so momentum \((\mathrm{P}=\mathrm{mv})\) are different for objects. So, momentum \(\mathrm{p}\) will be different for all.
MHT-CET 2020
Laws of Motion
145790
A force of \(0.03 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\). which is initially at rest for a duration of 10 seconds. Then find the momentum acquired by the body.
145787
A physical quantity \(A\) on multiplication with velocity results in another quantity \(B\). If the quantity \(B\) is energy, then the quantity \(A\) is
1 mass
2 momentum
3 force
4 acceleration
5 Power
Explanation:
B According to given question, \(\text { A } \times \text { Velocity }=\text { Energy }\) \(\mathrm{A} \times \mathrm{v}=\frac{1}{2} \mathrm{mv}^{2} \quad\left\{\text { Energy }=\frac{1}{2} \mathrm{mv}^{2}\right\}\) \(A=\frac{1}{2} m v\) \(A=\frac{p}{2}\) \(p=\text { momentum } \{p=m v\}\)
Kerala CEE 2021
Laws of Motion
145788
A body of mass \(10 \mathrm{~kg}\) has constant velocity of \(10 \mathrm{~ms}^{-1}\). A constant force is applied for a duration of \(4 \mathrm{~s}\) on the object and gives it a speed of \(2 \mathrm{~ms}^{-1}\) in opposite direction. Then, find acceleration produced in it.
145789
Five objects of different masses are simultaneously released vertically downwards from height ' \(h\) ' (in air). Which physical quantity associated with the objects will change at the instant they strike the ground? (neglect the air resistance.)
1 Time
2 Momentum
3 Velocity
4 Acceleration
Explanation:
B We know that, \(\begin{array}{rrr} \text { height }(\mathrm{h})=\frac{1}{2} \mathrm{gt}^{2} \\ \text { So, } \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\end{array}\) Time is same for all objects. \(\mathrm{v}=\mathrm{gt}\) Velocity is also same for all objects Mass of objects are different so momentum \((\mathrm{P}=\mathrm{mv})\) are different for objects. So, momentum \(\mathrm{p}\) will be different for all.
MHT-CET 2020
Laws of Motion
145790
A force of \(0.03 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\). which is initially at rest for a duration of 10 seconds. Then find the momentum acquired by the body.
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Laws of Motion
145787
A physical quantity \(A\) on multiplication with velocity results in another quantity \(B\). If the quantity \(B\) is energy, then the quantity \(A\) is
1 mass
2 momentum
3 force
4 acceleration
5 Power
Explanation:
B According to given question, \(\text { A } \times \text { Velocity }=\text { Energy }\) \(\mathrm{A} \times \mathrm{v}=\frac{1}{2} \mathrm{mv}^{2} \quad\left\{\text { Energy }=\frac{1}{2} \mathrm{mv}^{2}\right\}\) \(A=\frac{1}{2} m v\) \(A=\frac{p}{2}\) \(p=\text { momentum } \{p=m v\}\)
Kerala CEE 2021
Laws of Motion
145788
A body of mass \(10 \mathrm{~kg}\) has constant velocity of \(10 \mathrm{~ms}^{-1}\). A constant force is applied for a duration of \(4 \mathrm{~s}\) on the object and gives it a speed of \(2 \mathrm{~ms}^{-1}\) in opposite direction. Then, find acceleration produced in it.
145789
Five objects of different masses are simultaneously released vertically downwards from height ' \(h\) ' (in air). Which physical quantity associated with the objects will change at the instant they strike the ground? (neglect the air resistance.)
1 Time
2 Momentum
3 Velocity
4 Acceleration
Explanation:
B We know that, \(\begin{array}{rrr} \text { height }(\mathrm{h})=\frac{1}{2} \mathrm{gt}^{2} \\ \text { So, } \mathrm{t}=\sqrt{\frac{2 \mathrm{~h}}{\mathrm{~g}}}\end{array}\) Time is same for all objects. \(\mathrm{v}=\mathrm{gt}\) Velocity is also same for all objects Mass of objects are different so momentum \((\mathrm{P}=\mathrm{mv})\) are different for objects. So, momentum \(\mathrm{p}\) will be different for all.
MHT-CET 2020
Laws of Motion
145790
A force of \(0.03 \mathrm{~N}\) acts on a body of mass \(0.5 \mathrm{~kg}\). which is initially at rest for a duration of 10 seconds. Then find the momentum acquired by the body.
