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svlad2 [7]
1 year ago
14

A 300-kg bear grasping a vertical tree slides down at constant velocity. the friction force between the tree and the bear is:___

______
a. 30 n.
b. 300 n.
c. 3000 n.
d. more than 3000 n.
Physics
1 answer:
Nady [450]1 year ago
3 0

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 m/s^{2} (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

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A cyclist intends to cycle up a 7.70o hill whose vertical height is 126m. Assuming the mass of bicycle plus person is 75.0kg, ca
Ksivusya [100]

Answer:

92704.5 J

596.44737 N

Explanation:

m = Mass of person + bicycle = 75 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Vertical height = 126 m

\theta = Angle = 7.7°

d = Diameter = 0.388 m

Work done against gravity is given by

P=mgh\\\Rightarrow P=75\times 9.81\times 126\\\Rightarrow P=92704.5\ J

Work done is 92704.5 J

Force required is given by

F=\dfrac{mgrsin\theta}{\pi d}\\\Rightarrow F=\dfrac{75\times 9.81\times sin7.7\times 5.12}{\pi\times 0.388}\\\Rightarrow F=596.44737\ N

The force is 596.44737 N

7 0
2 years ago
a sleepy atudent drops a calculator out of a window yhata 20.7 m off the geound. we can ignore air resistance. what js the veloc
frez [133]

The acceleration of gravity is

9.8 m/s^2 down.

When an object falls out of a hand, its speed after 1.8s is

(9.8)x(1.8) = 17.6 m/s down.

It doesn't matter what it is, how much it weighs, or how high it was dropped from.

If it's more than 17.6 m/s, then this happened on a different, bigger planet.

If it's less than 17.6 m/s, then it must have hit something on the way down, like some air or something.

6 0
2 years ago
As the temperature increases the number of effective collisions between reacting particles in a chemical reaction
Reil [10]
Answer: option A) initially increases, then decreases.

Justification:

The increase of the rate of effective collisions among particles as the temperature increases is explained by the collision theory in virtue of the increase of the kinetic energy.

This is, as the temperature increase so the kinetic energy increase and the higher the kinetic energy the greater the number of collisions and the greater the chances that this energy overcome the activation energy (the energy needed to start the reaction).

Now, as the reaction progress the number of reactants particles naturally decrease (some of them have been converted into product) so this lower number of particles means lower concentration which means lower collisions and, thereafter, a decrease in the reaction rate.

8 0
3 years ago
6. An earthquake releases two types of traveling seismic waves, called transverse and longitudinal waves. The average speed of t
zubka84 [21]

Answer:

The distance away the center of the earthquake is 1083.24 km.

Explanation:

Given that,

Speed of transverse wave = 9.1\ km/s

Speed of longitudinal wave = 5.7 km/s

Time = 71 sec

We need to calculate the distance of transverse wave

Using formula of distance

d=v\times t

d=9.1\times t....(I)

The distance of longitudinal wave

d=5.7\times (t+71)....(II)

From the first equation

t=\dfrac{d}{9.1}

Put the value of t in equation (II)

d =5.7\times(\dfrac{d}{9.1}+71)

\dfrac{9.1d-5.7d}{9.1}=71\times5.7

d0.3736=404.7

d =1083.24\ km

Hence, The distance away the center of the earthquake is 1083.24 km.

6 0
2 years ago
How are solution useful for society
Butoxors [25]
Solutions are basically a release from a problem. This is more than helpful. 
8 0
3 years ago
Read 2 more answers
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