4. The Coyote has an initial position vector of 
.
4a. The Coyote has an initial velocity vector of 
. His position at time 
is given by the vector
where 
is the Coyote's acceleration vector at time . He experiences acceleration only in the downward direction because of gravity, and in particular 
where 
. Splitting up the position vector into components, we have 
with
The Coyote hits the ground when 
:
4b. Here we evaluate 
at the time found in (4a).
5. The shell has initial position vector 
, and we're told that after some time the bullet (now separated from the shell) has a position of 
.
5a. The vertical component of the shell's position vector is
We find the shell hits the ground at
5b. The horizontal component of the bullet's position vector is
where 
is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for :
Since static friction is the minimum force required to just start the motion of a stationary object.
Here if we need to start an object from rest then we required F = 700 N
So for the first part of the above problem Force will be F = 700 N
Now if the box is already moving then we will have to use kinetic friction force between box and floor
now we can write the equation of net force as
here
now we will have
Answer:
b) N = 560 N, c) fr = 138.56 N, d) μ = 0.247
Explanation:
a) In the attachment we can see the free body diagram of the system
b) Let's write Newton's second law on the y-axis
N + T_y -W = 0
N = W -T_y
let's use trigonometry for tension
sin θ = T_y / T
cos θ = Tₓ / T
T_y = T sin θ
Tₓ = T cos θ
we substitute
N = W - T sin 30
we calculate
N = 640 - 160 sin 30
N = 560 N
c) as the system goes at constant speed the acceleration is zero
X axis
Tₓ - fr = 0
Tₓ = fr
we substitute and calculate
fr = 160 cos 30
fr = 138.56 N
d) the friction force has the formula
fr = μ N
μ = fr / N
we calculate
μ = 138.56 / 560
μ = 0.247
250 kg*m/s
Momentum is found by multiplying mass by velocity
25*10=250
Answer:
See explanation
Explanation:
1) There are two main types of waves;
Mechanical waves and electromagnetic waves.
Mechanical waves usually require a medium for propagation, e.g sound, waves on a spiral spring, etc
Electromagnetic waves do not require a medium for propagation e.g microwaves, x-rays etc
2) The phenomenon is known as refraction. Refraction occurs at the boundary between two media because of the change in the speed of a wave as it moves from one medium to another.
