363523
The frictional force due to air on a body of mass 25 \(kg\) falling with an acceleration of \(9.2\,m{s^{ - 2}}\) will be
1 \(0.15\,N\)
2 \(1.5\,N\)
3 \(15\,N\)
4 \({\rm{Zero}}\)
Explanation:
Let \(f\) be the frictional force exerted by air on the body \(mg - f = ma\) \(25 \times 9.8 - f = 25 \times 9.2\) \(f = 25 \times 0.6 = 1.5N\)
PHXI05:LAWS OF MOTION
363524
A body of mass 0.05 \(kg\) is observed to fall with an acceleration of \(9.5\,m{s^{ - 2}}\). The opposing force of air on the body is ______ (\(g = 9.8\,m{s^{ - 2}}\))
1 \(0.015\,N\)
2 \(0.15\,N\)
3 \(0.030\,N\)
4 \({\rm{zero}}\)
Explanation:
Here, mass of the body \(m = 0.05\,kg\) Acceleration \(a = 9.5\,m{s^{ - 2}},g = 9.8\,m{s^{ - 2}}\) \(\therefore \,\,mg - {f_{air}} = ma\) \({f_{air}} = m(g - a)\) \( = 0.05(9.8 - 9.5)\) \( = 0.015\,N\)
PHXI05:LAWS OF MOTION
363525
A player stops a football weighting \(0.5\;kg\) which comes flying towards him with a velocity of \(10\;m{\rm{/}}s\). If the impact lasts for \(\frac{1}{{50}}th\,\sec \) and the ball bounces back with a velocity of \(15\;m{\rm{/}}s\), then the average force involved is
1 \(250\;N\)
2 \(1250\;N\)
3 \(500\;N\)
4 \(625\;N\)
Explanation:
Here \(m=0.5 \mathrm{~kg} ; u=-10 \mathrm{~m} / \mathrm{s} ;\) \(t=1 / 50 s ; v=+15 m s^{-1}\) Force \( = m(v - u){\rm{/}}t = 0.5(10 + 15) \times 50 = 625\;N\)
MHTCET - 2021
PHXI05:LAWS OF MOTION
363526
A dynamometer \(D\) is attached to two bodies of masses \(M=6 \mathrm{~kg}\) and \(m=4 \mathrm{~kg}\). Forces \(F=20 N\) and \(f=10 N\) are applied to the masses as shown in the figure. The dynamometer reads
1 \(10\;N\)
2 \(20\;N\)
3 \(6\;N\)
4 \(14\;N\)
Explanation:
If \(a\) is the acceleration of the system, then \(a = \frac{{{\rm{ net force }}}}{{{\rm{ total mass }}}} = \frac{{20 - 10}}{{6 + 4}} = 1\;m{s^{ - 2}}\) It is along the direction of \(\vec{F}\). If \(R\) is the reading of the dynamometer, then \(R + Ma = F\,{\rm{or}}\,R = F - Ma = 20 - 6 \times 1 = 14{\mkern 1mu} \,N\)
363523
The frictional force due to air on a body of mass 25 \(kg\) falling with an acceleration of \(9.2\,m{s^{ - 2}}\) will be
1 \(0.15\,N\)
2 \(1.5\,N\)
3 \(15\,N\)
4 \({\rm{Zero}}\)
Explanation:
Let \(f\) be the frictional force exerted by air on the body \(mg - f = ma\) \(25 \times 9.8 - f = 25 \times 9.2\) \(f = 25 \times 0.6 = 1.5N\)
PHXI05:LAWS OF MOTION
363524
A body of mass 0.05 \(kg\) is observed to fall with an acceleration of \(9.5\,m{s^{ - 2}}\). The opposing force of air on the body is ______ (\(g = 9.8\,m{s^{ - 2}}\))
1 \(0.015\,N\)
2 \(0.15\,N\)
3 \(0.030\,N\)
4 \({\rm{zero}}\)
Explanation:
Here, mass of the body \(m = 0.05\,kg\) Acceleration \(a = 9.5\,m{s^{ - 2}},g = 9.8\,m{s^{ - 2}}\) \(\therefore \,\,mg - {f_{air}} = ma\) \({f_{air}} = m(g - a)\) \( = 0.05(9.8 - 9.5)\) \( = 0.015\,N\)
PHXI05:LAWS OF MOTION
363525
A player stops a football weighting \(0.5\;kg\) which comes flying towards him with a velocity of \(10\;m{\rm{/}}s\). If the impact lasts for \(\frac{1}{{50}}th\,\sec \) and the ball bounces back with a velocity of \(15\;m{\rm{/}}s\), then the average force involved is
1 \(250\;N\)
2 \(1250\;N\)
3 \(500\;N\)
4 \(625\;N\)
Explanation:
Here \(m=0.5 \mathrm{~kg} ; u=-10 \mathrm{~m} / \mathrm{s} ;\) \(t=1 / 50 s ; v=+15 m s^{-1}\) Force \( = m(v - u){\rm{/}}t = 0.5(10 + 15) \times 50 = 625\;N\)
MHTCET - 2021
PHXI05:LAWS OF MOTION
363526
A dynamometer \(D\) is attached to two bodies of masses \(M=6 \mathrm{~kg}\) and \(m=4 \mathrm{~kg}\). Forces \(F=20 N\) and \(f=10 N\) are applied to the masses as shown in the figure. The dynamometer reads
1 \(10\;N\)
2 \(20\;N\)
3 \(6\;N\)
4 \(14\;N\)
Explanation:
If \(a\) is the acceleration of the system, then \(a = \frac{{{\rm{ net force }}}}{{{\rm{ total mass }}}} = \frac{{20 - 10}}{{6 + 4}} = 1\;m{s^{ - 2}}\) It is along the direction of \(\vec{F}\). If \(R\) is the reading of the dynamometer, then \(R + Ma = F\,{\rm{or}}\,R = F - Ma = 20 - 6 \times 1 = 14{\mkern 1mu} \,N\)
363523
The frictional force due to air on a body of mass 25 \(kg\) falling with an acceleration of \(9.2\,m{s^{ - 2}}\) will be
1 \(0.15\,N\)
2 \(1.5\,N\)
3 \(15\,N\)
4 \({\rm{Zero}}\)
Explanation:
Let \(f\) be the frictional force exerted by air on the body \(mg - f = ma\) \(25 \times 9.8 - f = 25 \times 9.2\) \(f = 25 \times 0.6 = 1.5N\)
PHXI05:LAWS OF MOTION
363524
A body of mass 0.05 \(kg\) is observed to fall with an acceleration of \(9.5\,m{s^{ - 2}}\). The opposing force of air on the body is ______ (\(g = 9.8\,m{s^{ - 2}}\))
1 \(0.015\,N\)
2 \(0.15\,N\)
3 \(0.030\,N\)
4 \({\rm{zero}}\)
Explanation:
Here, mass of the body \(m = 0.05\,kg\) Acceleration \(a = 9.5\,m{s^{ - 2}},g = 9.8\,m{s^{ - 2}}\) \(\therefore \,\,mg - {f_{air}} = ma\) \({f_{air}} = m(g - a)\) \( = 0.05(9.8 - 9.5)\) \( = 0.015\,N\)
PHXI05:LAWS OF MOTION
363525
A player stops a football weighting \(0.5\;kg\) which comes flying towards him with a velocity of \(10\;m{\rm{/}}s\). If the impact lasts for \(\frac{1}{{50}}th\,\sec \) and the ball bounces back with a velocity of \(15\;m{\rm{/}}s\), then the average force involved is
1 \(250\;N\)
2 \(1250\;N\)
3 \(500\;N\)
4 \(625\;N\)
Explanation:
Here \(m=0.5 \mathrm{~kg} ; u=-10 \mathrm{~m} / \mathrm{s} ;\) \(t=1 / 50 s ; v=+15 m s^{-1}\) Force \( = m(v - u){\rm{/}}t = 0.5(10 + 15) \times 50 = 625\;N\)
MHTCET - 2021
PHXI05:LAWS OF MOTION
363526
A dynamometer \(D\) is attached to two bodies of masses \(M=6 \mathrm{~kg}\) and \(m=4 \mathrm{~kg}\). Forces \(F=20 N\) and \(f=10 N\) are applied to the masses as shown in the figure. The dynamometer reads
1 \(10\;N\)
2 \(20\;N\)
3 \(6\;N\)
4 \(14\;N\)
Explanation:
If \(a\) is the acceleration of the system, then \(a = \frac{{{\rm{ net force }}}}{{{\rm{ total mass }}}} = \frac{{20 - 10}}{{6 + 4}} = 1\;m{s^{ - 2}}\) It is along the direction of \(\vec{F}\). If \(R\) is the reading of the dynamometer, then \(R + Ma = F\,{\rm{or}}\,R = F - Ma = 20 - 6 \times 1 = 14{\mkern 1mu} \,N\)
363523
The frictional force due to air on a body of mass 25 \(kg\) falling with an acceleration of \(9.2\,m{s^{ - 2}}\) will be
1 \(0.15\,N\)
2 \(1.5\,N\)
3 \(15\,N\)
4 \({\rm{Zero}}\)
Explanation:
Let \(f\) be the frictional force exerted by air on the body \(mg - f = ma\) \(25 \times 9.8 - f = 25 \times 9.2\) \(f = 25 \times 0.6 = 1.5N\)
PHXI05:LAWS OF MOTION
363524
A body of mass 0.05 \(kg\) is observed to fall with an acceleration of \(9.5\,m{s^{ - 2}}\). The opposing force of air on the body is ______ (\(g = 9.8\,m{s^{ - 2}}\))
1 \(0.015\,N\)
2 \(0.15\,N\)
3 \(0.030\,N\)
4 \({\rm{zero}}\)
Explanation:
Here, mass of the body \(m = 0.05\,kg\) Acceleration \(a = 9.5\,m{s^{ - 2}},g = 9.8\,m{s^{ - 2}}\) \(\therefore \,\,mg - {f_{air}} = ma\) \({f_{air}} = m(g - a)\) \( = 0.05(9.8 - 9.5)\) \( = 0.015\,N\)
PHXI05:LAWS OF MOTION
363525
A player stops a football weighting \(0.5\;kg\) which comes flying towards him with a velocity of \(10\;m{\rm{/}}s\). If the impact lasts for \(\frac{1}{{50}}th\,\sec \) and the ball bounces back with a velocity of \(15\;m{\rm{/}}s\), then the average force involved is
1 \(250\;N\)
2 \(1250\;N\)
3 \(500\;N\)
4 \(625\;N\)
Explanation:
Here \(m=0.5 \mathrm{~kg} ; u=-10 \mathrm{~m} / \mathrm{s} ;\) \(t=1 / 50 s ; v=+15 m s^{-1}\) Force \( = m(v - u){\rm{/}}t = 0.5(10 + 15) \times 50 = 625\;N\)
MHTCET - 2021
PHXI05:LAWS OF MOTION
363526
A dynamometer \(D\) is attached to two bodies of masses \(M=6 \mathrm{~kg}\) and \(m=4 \mathrm{~kg}\). Forces \(F=20 N\) and \(f=10 N\) are applied to the masses as shown in the figure. The dynamometer reads
1 \(10\;N\)
2 \(20\;N\)
3 \(6\;N\)
4 \(14\;N\)
Explanation:
If \(a\) is the acceleration of the system, then \(a = \frac{{{\rm{ net force }}}}{{{\rm{ total mass }}}} = \frac{{20 - 10}}{{6 + 4}} = 1\;m{s^{ - 2}}\) It is along the direction of \(\vec{F}\). If \(R\) is the reading of the dynamometer, then \(R + Ma = F\,{\rm{or}}\,R = F - Ma = 20 - 6 \times 1 = 14{\mkern 1mu} \,N\)
