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Elis [28]
3 years ago
8

A 49 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 3.3 m/s just before hitting the ground.

(a) What change occurs in the gravitational potential energy of the bear-Earth system during the slide? (b) What is the kinetic energy of the bear just before hitting the ground? (c) What is the average frictional force that acts on the sliding bear?
Physics
1 answer:
hichkok12 [17]3 years ago
8 0

Answer:

a) \Delta U_g=-5.3kJ

b) K=0.27kJ

c) F_f=0.45kN

Explanation:

the gravitational potential energy is given by:

U_g=m.g.h\\

\Delta U_g=m.g.h_f-m.g.h_i\\\Delta U_g=49kg*9.8m/s^2*(0m-11m)\\\Delta U_g=-5.3kJ

The kinetic energy is given by:

K=\frac{1}{2}m.v^2\\

the initial kinetic energy is zero because the motion started from rest, so:

K=\frac{1}{2}*49kg*(3.3m/s^2)^2\\K=0.27kJ

applying the conservation of energy theorem:

U_g-W_f=K_f\\W_f=-(\Delta K+\Delta U)\\W_F=5.3kJ-0.27kJ\\W_F=-5.0kJ

The work done by the friction force is given by:

W_f=F_f.h.cos(\theta)\\

the angle of the force is 180 degrees because it's against the movement:

F_f=\frac{W_f}{h.cos(\theta)}\\\\F_f=\frac{-5.0kJ}{11m.cos(180^o)}\\\\F_f=0.45kN

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A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the
alex41 [277]

Answer:

(a)v_1 = a_1t_1 = 1.76 t_1

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1

(b) The velocity of the car before the driver begins braking is

v_1 = 1.76*20 = 35.2m/s

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2

We can use the following equation of motion to calculate how far the car has travel since braking to stop

s_2 = v_1t_2 + a_2t_2^2/2

s_2 = 35.2*5 - 7.04*5^2/2 = 88 m

Also the distance from start to where the driver starts braking is

s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

(c) The distance from the limb to where the car stops is 550 - 440 = 110 m

8 0
3 years ago
A force vector points due east and has a magnitude of 150 newtons. A second force is added to . The resultant of the two vectors
balandron [24]

Answer:

a. 240 N due east

b. 540 N due west

Explanation:

Let east be the reference direction

(a) if the resultant force has a magnitude of 390 N and points east, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of

390 - 150 = 240 N

(b) if the resultant force has a magnitude of 390 N and points west, it would be -390N is eastern reference, and the 1st force is 150N due East, then the additional force would also due east and has a magnitude of

-390 - 150 = -540 N

This force would point west

6 0
3 years ago
Name and describe the three types of binary systems.
Pachacha [2.7K]

Answer and Explanation:

The binary system is a star system about half of the star are the part of the binary system

The important binary systems are

  • Visual binary : in visual binary system the two stars resolved visually with nay sort of optical device the orbit of binary orbit is very large about center of mass.
  • Spectroscopic binary : in this binary system the stars are detected by close analysis of light and after detection it is found that there are two seller spectrum present instead of one.
  • Eclipsing binary : eclipsing binary system is a several types of variable stars this binary system is very important in astrophysics

5 0
3 years ago
What is F = m x a in Newton's laws of motion?
madam [21]

Answer:

This is Newton's second law.

<u>Newton's second law text:</u>

(If a resultant force acts on a body, then an acceleration will give it an acceleration, the magnitude of which is directly proportional to the amount of the net force, and a direction is in the direction of the net force itself)

F=ma

net force = mass x acceleration

I hope I helped you^_^

8 0
3 years ago
A 13 500 N car traveling at 50.0 km/h rounds a curve of radius 2.00 × 102 m. Find the following: a. the centripetal acceleration
Afina-wow [57]

Answer:

a. 0.947 m/s^2

b. 1304.54 N

c. 0.0966

Explanation:

mass of car = 13500 N = 13500/9.8 = 1377.55 kg

velocity = 50 km/h = 50,000 m/h = 13.9 m/s

raidus = 204 m

a. centripetal acceleartion = v^2/r = 13.9^2/204 = 0.947 m/s^2

b. centripetal force = m*centripetal acceleration = 1377.55 * 0.947 = 1304.54 N

c. In order for the car to round the curve safely, static friction = centripetal force

static friction = coefficient of friction (mu) * mg = mu* 1377.55*9.8 = 13500mu

13500mu = 1304.54

mu = 1304.54/13500 = 0.0966

5 0
2 years ago
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