Answer:
Following are the solution to this question:
Explanation:
Law:




To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

Where,
= Mass of each object
= Initial Velocity of Each object
= Final Velocity
Our values are given as




Replacing we have that



Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s
Answer:


Explanation:
<u>Displacement
</u>
It's a vector magnitude that measures the space traveled by a particle between an initial and a final position. The total displacement can be obtained by adding the vectors of each individual displacement. In the case of two displacements:

Given a vector as its polar coordinates (r,\theta), the corresponding rectangular coordinates are computed with


And the vector is expressed as

The monkey first makes a displacement given by (0.198 km,0°). The angle is 0 because it goes to the East, the zero-reference for angles. Thus the first displacement is

The second move is (145 m , -15.8°). The angle is negative because it points South of East. The second displacement is

The total displacement is


In (magnitude,angle) form:




The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.