Answer:
Fr = 20 (N)
Explanation: See atached file ( free body diagram)
As for Newton´s low
∑ Fy = 0
-mg + N = 0 ⇒ - 40 + N = 0 ⇒ N = 40 [Newtons]
by definition : Fr = μs * N ⇒ Fr = 0,5 * 40 ⇒ Fr = 20 (N)
∑ Fx = 0 body is at rest
Fe - Fr = 0
Fr > Fe
Fr > 12 (N) the body is at rest
<span>FIRST SECTION
You should use the formula for uniformly accelerated linear movement.
Initial speed is 0 because it starts from rest.
d=(1/2)*a*t^2+vo*t =(1/2)*(4.0 m/s^2)*(3s)^2+0*3s=(1/2)*(4.0 m/s^2)*3^2*s^2+0=2.0 m*9=18m
You can calculate the final speed with the other formula:
v=a*t+vo=(4.0 m/s^2)*(3s)+0=(4.0 m/s)*(3)=12m/s
SECOND SECTION
You should use the formula for uniform linear movement.
Velocity is a constant: it remains in 12m/s.
d=v*t=12m/s*2s=12m*2=24m
THIRD SECTION
We should use the same formulas as the first section, but with different numbers.
Initial velocity will be 12m/s, and then velocity will start to decrease until it gets to 0.
We don’t know what the time is for this section.
Acceleration is negative, because it’s slowing down.
v=a*t+vo
0=-3.0 m/s^2*t+12m/s
3.0 m/s^2*t=12m/s
t=(12m/s)/(3.0 m/s^2)=4(1/s)/(1/s^2)=4s^2/s=4s
Now let’s use that time in the other formula:
d=(1/2)*a*t^2+vo*t =(1/2)*(-3.0 m/s^2)*(4s)^2+(12m/s)*3s=(-1.5 m/s^2)*4^2*s^2+12*3m*s/s=-1.5 m*4^2+36m=-1.5*16m+36m=-24m+36m=12m
Now let’s add the 3 stages:
d=18m+24m+12m=54m
</span>
The skydiver has a bunch of gravitational potential energy. The best example of chemical potential energy is gasoline.
Answer:
1058.78 ft/sec
Explanation:
Horizontal Component of Velocity; This is the velocity of a body that act on the horizontal axis. I.e Velocity along x-axis
The horizontal velocity of a body can be calculated as shown below.\
Vh = Vcos∅.......................... Equation 1
Where Vh = horizontal component of the velocity, V = The velocity acting between the horizontal and the vertical axis, ∅ = Angle the velocity make with the horizontal.
Given: V = 1178 ft/sec, ∅ = 26°
Substitute into equation 1
Vh = 1178cos26
Vh = 1178(0.8988)
Vh = 1058.78 ft/sec
Hence the horizontal component of the velocity = 1058.78 ft/sec
Answer:
Total mass of combination = 2+3+5 = 10kg.
Acceleration produced = 2m/s^2
hence force =( total mass × acceleration)= (2×10)= 20 N.
Net force on 3kg block = acceleration × mass = (2 × 2 )= 4 N
applied force on 2 kg block = 20N
Force between 2 kg and 3 kg block = (20-4) = 16N. ans
Net force on 3 kg block = 3 × 2 =6N.
Applied force on 3 kg block due to 2 kg block = 16N.
hence, force between 3 kg and 5 kg block = (16-6) = 10N .
answers:-
(a) 20 N
(b) 16N
(c) 10 N
