3000 N of friction exists between the bear and the tree.
The bear goes down at a constant speed, thus there is no acceleration. The forces add up to zero. As a result, upward friction and downward weight result in zero
i.e.,
F-mg=0
F = mg
F = (300 kg)/10 
(gravitational acceleration)
3000 N frictional force equals F.
How does friction force work?
- Friction is the force that prevents two solid objects from rolling or sliding over one another. Even though frictional forces, such as the traction required to walk without slipping, may be advantageous, they can also be a significant hindrance to motion.
Between solid surfaces, there are three basic types of friction:
- Rolling
- Sliding.
- Static. They are graded from strongest to weakest. Fluid friction happens when liquids or gasses are mixed together.
To learn more about Friction force, visit:
brainly.com/question/13707283
#SPJ4
Answer: Force = Mass X Acceleration
F = 5 x 2
F = 10 N
Answer:
<u>The transformation of energy in a torch light is as follows:</u>
1) When the torch is turned ON, the chemical energy in the batteries is converted into electrical energy.
2) The electrical energy is converted into heat and light energy. (We feel the torch to be hot after some time and we can see the light energy)
Hope this helped!
<h2>~AnonymousHelper1807</h2>
Answer:
The focal length of the appropriate corrective lens is 35.71 cm.
The power of the appropriate corrective lens is 0.028 D.
Explanation:
The expression for the lens formula is as follows;
Here, f is the focal length, u is the object distance and v is the image distance.
It is given in the problem that the given lens is corrective lens. Then, it will form an upright and virtual image at the near point of person's eye. The near point of a person's eye is 71.4 cm. To see objects clearly at a distance of 24.0 cm, the corrective lens is used.
Put v= -71.4 cm and u= 24.0 cm in the above expression.
f= 35.71 cm
Therefore, the focal length of the corrective lens is 35.71 cm.
The expression for the power of the lens is as follows;
Here, p is the power of the lens.
Put f= 35.71 cm.
p=0.028 D
Therefore, the power of the corrective lens is 0.028 D.
Answer:
dsfsdfgfdsggfgsgsfsfgfsgffsfgfgfgf
Explanation:
sgdfggsfdsgfgsgsmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm
