All categories

3 years ago

5.) A 2000 N bear slides down a tree at a constant velocity, what is the

Physics

1 answer:

PilotLPTM [1.2K]3 years ago

4 0

Answer:

2000 N

1600 N

Explanation:

Step 1:

It is given that the bear's weight is 2000 N and it slides down the tree with constant velocity. Since it is sliding with constant velocity the overall force on the bear is zero. The weight of the bear acting downward balances the upward force caused due to friction. Hence the upward force equals the weight 2000 N.

Step 2:

It is given that the bear slides down with an acceleration 2 m/sec sq

Weight of the bear = 2000 N

Mass of the bear = 2000/10 (taking g = 10 m/sec sq)=200 kg

Force = Mass * Acceleration

Hence net force acting downward = 200*2=400 N.

Net force = Weight of bear - Force acting upward on the bear

400 = 2000 - Force acting upward on the bear

Force on the bear acting upward = 2000 - 400 = 1600 N

You might be interested in

A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is

Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

$m_1$ = Mass of block = 0.072 kg

$m_2$ = Mass of bullet = 4.67 g

$u_1$ = Initial Velocity of block = 0

$u_2$ = Initial Velocity of bullet = 629 m/s

$v_1$ = Final Velocity of block = 17 m/s

$v_2$ = Final Velocity of bullet

In this system the linear momentum is conserved

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s$

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

$K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J$

The final kinetic energy

$K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J$

Initial Kinetic energy > Final kinetic energy

3 0

3 years ago

Charge is placed on two conducting spheres that are very far apart and connected by a long thin wire. The radius of the smaller

mrs_skeptik [129]

Answer:

surface charge density on each sphere is $440 \times 10^{-9}$ C

Explanation:

given data

radius of smaller sphere = 5 cm

radius of larger sphere is 12 cm

electric field at surface of larger sphere = 660 kV/m = 660 × 1000 v/m

solution

we apply here electric field formula that is express as

E = $(\frac{1}{4\pi\epsilon })\times (\frac{Q_{1} }{R^{2} } )$ .................1

put here value

660000 = $9 \times 10^9 \times \frac{Q1}{0.12^2}$

Q1 = 1056 × $10^{-9}$

and

here field inside a conductor is zero so that electric potential ( V ) is constant

$\frac{Q{1} }{R} = \frac{Q{2} }{r}$ ..................2

so Q2 will be

Q2 = $\frac{5}{12} \times 1056 \times 10^{-9}$

Q2 = $440 \times 10^{-9}$ C

6 0

3 years ago

Which answer correctly describes the grey lines in this Hering illusion? (Optical Illusions Lesson)

Ratling [72]

The answer that correctly describes the grey lines in this Hering illusion is "The grey lines are bent." Option A. This is further explained below

<h3>What is the Hering illusion?</h3>

The Hering Illusion is one of several illusions in which a major component of a basic line picture is obscured.

In conclusion, The statement for the grey lines in this Hering illusion is "The grey lines are twisted.".