371983
If the coefficient of friction between the rubber tires and the road way is 0.25 . Find the maximum speed with which a car can be driven round a curve of radius \(20 \mathrm{~m}\) without skidding. Given, \(g=9.8 \mathrm{~m} . \mathrm{s}^{-2}\) :
1 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(14 \mathrm{~m} . \mathrm{s}^{-1}\)
Explanation:
B Given, \(\mu=0.25\) \(\mathrm{r}=20 \mathrm{~m}\) \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{sec}^{2}\) To find velocity in the equilibrium \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}=\sqrt{0.25 \times 20 \times 9.8}\) \(\mathrm{v}=\sqrt{49}=7 \mathrm{~m} . \mathrm{s}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371984
Vehicles are streamlined to reduce
1 Static friction
2 Kinetic friction
3 Sliding friction
4 Fluid friction
Explanation:
D A streamlined shape is a shape that reduces friction drag between fluid such as air or water and an object moving through it. So, vehicles have streamlined shape to reduce fluid friction.
AP EAMCET-24.09.2020
LAWS OF MOTION (ADDITIONAL)
371985
Friction can perform
1 Positive work only
2 Zero work only
3 Negative work only
4 Positive, negative and zero work
Explanation:
D Friction can perform is work done by friction may be positive, negative and zero work. When frictional force is opposite to direction the work is negative. When friction force is in same direction of displacement work is positive and if there is no displacement then work is zero.
AP EAMCET (Medical)-07.10.2020
LAWS OF MOTION (ADDITIONAL)
371986
A cylinder rolls down on inclined plane of inclination \(30^{\circ}\), the acceleration of the cylinder is
1 \(\frac{g}{3}\)
2 \(g\)
3 \(\frac{g}{2}\)
4 \(\frac{2 g}{3}\)
Explanation:
A As we know that, acceleration of a cylinder rolls down on inclined plane. \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}\right)}\) Where, \(\mathrm{I}=\) moment of Inertia For solid cylinder, \(\mathrm{I}=\frac{\mathrm{mR}^{2}}{2}\) Then, \(\quad a=\frac{g \sin 30^{\circ}}{\left(1+\frac{m^{2}}{2} \times \frac{1}{m R^{2}}\right)}\) \(a=\frac{g \times 1 / 2}{1+1 / 2}=\frac{g}{3}\)
371983
If the coefficient of friction between the rubber tires and the road way is 0.25 . Find the maximum speed with which a car can be driven round a curve of radius \(20 \mathrm{~m}\) without skidding. Given, \(g=9.8 \mathrm{~m} . \mathrm{s}^{-2}\) :
1 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(14 \mathrm{~m} . \mathrm{s}^{-1}\)
Explanation:
B Given, \(\mu=0.25\) \(\mathrm{r}=20 \mathrm{~m}\) \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{sec}^{2}\) To find velocity in the equilibrium \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}=\sqrt{0.25 \times 20 \times 9.8}\) \(\mathrm{v}=\sqrt{49}=7 \mathrm{~m} . \mathrm{s}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371984
Vehicles are streamlined to reduce
1 Static friction
2 Kinetic friction
3 Sliding friction
4 Fluid friction
Explanation:
D A streamlined shape is a shape that reduces friction drag between fluid such as air or water and an object moving through it. So, vehicles have streamlined shape to reduce fluid friction.
AP EAMCET-24.09.2020
LAWS OF MOTION (ADDITIONAL)
371985
Friction can perform
1 Positive work only
2 Zero work only
3 Negative work only
4 Positive, negative and zero work
Explanation:
D Friction can perform is work done by friction may be positive, negative and zero work. When frictional force is opposite to direction the work is negative. When friction force is in same direction of displacement work is positive and if there is no displacement then work is zero.
AP EAMCET (Medical)-07.10.2020
LAWS OF MOTION (ADDITIONAL)
371986
A cylinder rolls down on inclined plane of inclination \(30^{\circ}\), the acceleration of the cylinder is
1 \(\frac{g}{3}\)
2 \(g\)
3 \(\frac{g}{2}\)
4 \(\frac{2 g}{3}\)
Explanation:
A As we know that, acceleration of a cylinder rolls down on inclined plane. \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}\right)}\) Where, \(\mathrm{I}=\) moment of Inertia For solid cylinder, \(\mathrm{I}=\frac{\mathrm{mR}^{2}}{2}\) Then, \(\quad a=\frac{g \sin 30^{\circ}}{\left(1+\frac{m^{2}}{2} \times \frac{1}{m R^{2}}\right)}\) \(a=\frac{g \times 1 / 2}{1+1 / 2}=\frac{g}{3}\)
371983
If the coefficient of friction between the rubber tires and the road way is 0.25 . Find the maximum speed with which a car can be driven round a curve of radius \(20 \mathrm{~m}\) without skidding. Given, \(g=9.8 \mathrm{~m} . \mathrm{s}^{-2}\) :
1 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(14 \mathrm{~m} . \mathrm{s}^{-1}\)
Explanation:
B Given, \(\mu=0.25\) \(\mathrm{r}=20 \mathrm{~m}\) \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{sec}^{2}\) To find velocity in the equilibrium \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}=\sqrt{0.25 \times 20 \times 9.8}\) \(\mathrm{v}=\sqrt{49}=7 \mathrm{~m} . \mathrm{s}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371984
Vehicles are streamlined to reduce
1 Static friction
2 Kinetic friction
3 Sliding friction
4 Fluid friction
Explanation:
D A streamlined shape is a shape that reduces friction drag between fluid such as air or water and an object moving through it. So, vehicles have streamlined shape to reduce fluid friction.
AP EAMCET-24.09.2020
LAWS OF MOTION (ADDITIONAL)
371985
Friction can perform
1 Positive work only
2 Zero work only
3 Negative work only
4 Positive, negative and zero work
Explanation:
D Friction can perform is work done by friction may be positive, negative and zero work. When frictional force is opposite to direction the work is negative. When friction force is in same direction of displacement work is positive and if there is no displacement then work is zero.
AP EAMCET (Medical)-07.10.2020
LAWS OF MOTION (ADDITIONAL)
371986
A cylinder rolls down on inclined plane of inclination \(30^{\circ}\), the acceleration of the cylinder is
1 \(\frac{g}{3}\)
2 \(g\)
3 \(\frac{g}{2}\)
4 \(\frac{2 g}{3}\)
Explanation:
A As we know that, acceleration of a cylinder rolls down on inclined plane. \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}\right)}\) Where, \(\mathrm{I}=\) moment of Inertia For solid cylinder, \(\mathrm{I}=\frac{\mathrm{mR}^{2}}{2}\) Then, \(\quad a=\frac{g \sin 30^{\circ}}{\left(1+\frac{m^{2}}{2} \times \frac{1}{m R^{2}}\right)}\) \(a=\frac{g \times 1 / 2}{1+1 / 2}=\frac{g}{3}\)
371983
If the coefficient of friction between the rubber tires and the road way is 0.25 . Find the maximum speed with which a car can be driven round a curve of radius \(20 \mathrm{~m}\) without skidding. Given, \(g=9.8 \mathrm{~m} . \mathrm{s}^{-2}\) :
1 \(5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
2 \(7 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
3 \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)
4 \(14 \mathrm{~m} . \mathrm{s}^{-1}\)
Explanation:
B Given, \(\mu=0.25\) \(\mathrm{r}=20 \mathrm{~m}\) \(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{sec}^{2}\) To find velocity in the equilibrium \(\mathrm{v}=\sqrt{\mu \mathrm{rg}}=\sqrt{0.25 \times 20 \times 9.8}\) \(\mathrm{v}=\sqrt{49}=7 \mathrm{~m} . \mathrm{s}^{-1}\)
AP EAMCET-25.09.2020
LAWS OF MOTION (ADDITIONAL)
371984
Vehicles are streamlined to reduce
1 Static friction
2 Kinetic friction
3 Sliding friction
4 Fluid friction
Explanation:
D A streamlined shape is a shape that reduces friction drag between fluid such as air or water and an object moving through it. So, vehicles have streamlined shape to reduce fluid friction.
AP EAMCET-24.09.2020
LAWS OF MOTION (ADDITIONAL)
371985
Friction can perform
1 Positive work only
2 Zero work only
3 Negative work only
4 Positive, negative and zero work
Explanation:
D Friction can perform is work done by friction may be positive, negative and zero work. When frictional force is opposite to direction the work is negative. When friction force is in same direction of displacement work is positive and if there is no displacement then work is zero.
AP EAMCET (Medical)-07.10.2020
LAWS OF MOTION (ADDITIONAL)
371986
A cylinder rolls down on inclined plane of inclination \(30^{\circ}\), the acceleration of the cylinder is
1 \(\frac{g}{3}\)
2 \(g\)
3 \(\frac{g}{2}\)
4 \(\frac{2 g}{3}\)
Explanation:
A As we know that, acceleration of a cylinder rolls down on inclined plane. \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{\left(1+\frac{\mathrm{I}}{\mathrm{mR}^{2}}\right)}\) Where, \(\mathrm{I}=\) moment of Inertia For solid cylinder, \(\mathrm{I}=\frac{\mathrm{mR}^{2}}{2}\) Then, \(\quad a=\frac{g \sin 30^{\circ}}{\left(1+\frac{m^{2}}{2} \times \frac{1}{m R^{2}}\right)}\) \(a=\frac{g \times 1 / 2}{1+1 / 2}=\frac{g}{3}\)
