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vlabodo [156]
3 years ago
7

A 230.0 kg bear grasping a vertical tree slides down at constant velocity. What is the friction force between the tree and the b

ear?
Physics
2 answers:
vovangra [49]3 years ago
8 0
Well, it's equal to the weight of the bear, which is 230*10=2300 N
dezoksy [38]3 years ago
8 0
Hi there! Thanks for asking here on Brainly. <span>✉
</span>
Solving the dynamic equilibrium: 

<span>Friction = Gravity = ma
(230.0 kg)*(10 m/s2) =
23000 N

So the friction force between the tree and bear is 2300 N.

</span><span>Hope that helps! ★ <span>If you have further questions about this question or need more help, feel free to comment below or leave me a PM. -UnicornFudge aka Nadia </span></span>
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Ivan

Complete Question

The complete question is shown on the first uploaded image  

Answer:

a   it is always zero

b  0

c  \eta  =  -\epsilon _o E

Explanation:ss

Here the  net charge is  on the outer surface of the conductor thus this means that the net charge inside the conductor is zero

Generally the charge density of a conductor is dependent on the charge per unit area  which implies that the charge density is dependent on the net charge  so this  means that the charge density inside the conductor is zero

 

Generally the direction of electric field this from the  positive charge to the negative charge  so from the question we can deduce  that the negative charge is located on the surface of the conductor

    So We can mathematically define the charge density on the surface of the electric field as

             ∮E \cdot dA =  \frac{-Q}{\epsilon _o}

Where E is the electric field

          dA change in unit area

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          \epsilon _o  is the permittivity of free space

So

          EA  =  \frac{-Q}{\epsilon _o }

           \frac{Q}{A}  =  -\epsilon _o E

          \eta  =  -\epsilon _o E

Where \eta is the charge density

   

8 0
3 years ago
1. Determine the image distance in each of the following.
miskamm [114]

It is given that for the convex lens,

Case 1.

u=−40cm

f=+15cm

Using lens formula

v

1

−

u

1

=

f

1

v

1

−

40

1

=

15

1

v

1

=

15

1

−

40

1

v=+24.3cm

The image in formed in this case at a distance of 24.3cm in left of lens.

Case 2.

A point source is placed in between the lens and the mirror at a distance of 40 cm from the lens i.e. the source is placed at the focus of mirror, then the rays after reflection becomes parallel for the lens such that

u=∞

f=15cm

Now, using mirror’s formula

v

1

+

u

1

=

f

1

v

1

+

∞

1

=

15

1

v=+15cm

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A van is traveling on a road at a speed of 55 km/h. A girl sitting near the driver of the van throws a paper airplane to a boy a
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<span>A van is traveling on a road at a speed of 55 km/h relative to a
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3 years ago
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