Use of lubricant
Use of ball bearers
Use of streamlined body
Use of graphite
Heavy crate sits at rest on the floor of a warehouse. you push on the crate with a force of 400 N, and it doesn't budge. The magnitude of the friction force on the crate in Newton is 400N
This is due to Friction force, which is defined as the resisting force that acts on a body when it is at rest (Static friction) or when it is in motion (Kinetic friction).
When a force is applied on a stationary body, the force of static friction starts to act on the body which prevents any relative motion between the object and surface. The magnitude of friction increases up to μsN, where μs is the coefficient of static friction. As the crate didn't budge, it means the amount of force applied was less than μsN. Hence the force applied was canceled by an equal and opposite amount of frictional force which was equal to 400N.
Learn more about frictional force here
brainly.com/question/1714663
#SPJ4
Answer:
Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s
Explanation:
When rod is released from rest then due to unbalanced torque about the hinge the system will rotate
Now moment of inertia of the system is given as
![I = \frac{ML^2}{12} + \frac{m_1L^2}{4} + \frac{m_2L^2}{4}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7BML%5E2%7D%7B12%7D%20%2B%20%5Cfrac%7Bm_1L%5E2%7D%7B4%7D%20%2B%20%5Cfrac%7Bm_2L%5E2%7D%7B4%7D)
now we have
![M = 0.120 kg](https://tex.z-dn.net/?f=M%20%3D%200.120%20kg)
![m_1 = 0.02 kg](https://tex.z-dn.net/?f=m_1%20%3D%200.02%20kg)
![m_3 = 0.08 kg](https://tex.z-dn.net/?f=m_3%20%3D%200.08%20kg)
now we have
![I = \frac{0.120(0.90)^2}{12} + \frac{0.02(0.90)^2}{4} + \frac{0.08(0.90)^2}{4}](https://tex.z-dn.net/?f=I%20%3D%20%5Cfrac%7B0.120%280.90%29%5E2%7D%7B12%7D%20%2B%20%5Cfrac%7B0.02%280.90%29%5E2%7D%7B4%7D%20%2B%20%5Cfrac%7B0.08%280.90%29%5E2%7D%7B4%7D)
so we have
![I = 8.1 \times 10^[-3} + 4.05 \times 10^[-3} + 0.0162](https://tex.z-dn.net/?f=I%20%3D%208.1%20%5Ctimes%2010%5E%5B-3%7D%20%2B%204.05%20%5Ctimes%2010%5E%5B-3%7D%20%2B%200.0162)
![I = 0.02835](https://tex.z-dn.net/?f=I%20%3D%200.02835)
now by energy conservation we can say work done by gravity must be equal to change in kinetic energy
so we have
![\frac{1}{2}I\omega^2 = m_1g \frac{L}{2} - m_2 g\frac{L}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7DI%5Comega%5E2%20%3D%20m_1g%20%5Cfrac%7BL%7D%7B2%7D%20-%20m_2%20g%5Cfrac%7BL%7D%7B2%7D)
![\frac{1}{2}(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%280.02835%29%5Comega%5E2%20%3D%20%280.08%20-%200.02%29%289.81%29%280.45%29)
![\omega = 4.32 rad/s](https://tex.z-dn.net/?f=%5Comega%20%3D%204.32%20rad%2Fs)
Now speed of 0.08 kg mass when it reaches to bottom point is given as
![v = \omega \frac{L}{2}](https://tex.z-dn.net/?f=v%20%3D%20%5Comega%20%5Cfrac%7BL%7D%7B2%7D)
![v = 4.32 (0.45)](https://tex.z-dn.net/?f=v%20%3D%204.32%20%280.45%29)
![v = 1.94 m/s](https://tex.z-dn.net/?f=v%20%3D%201.94%20m%2Fs)
Answer:
Explanation:
Energy = volt x charge
= 12 x 750000 J
= 9000,000 J
Latent heat of vaporisation
= 2260000 J / kg
kg of water vaporised
= 9000,000 / 2260000
= 3.982 kg .
Answer:
b) The gravitational pull will be weak.
Explanation:
Less mass requires less force. Gravity or gravitational pull is a force. In order for you to stay in place and not go through the ground, however, you require something to pulling you away from the Earth. That's why we have air resistance.