Answer:
Speed of the soccer ball with respect to ground is ZERO
Explanation:
As we know that truck is moving with speed 20 mph
Now a soccer ball is shoot from truck with respect to the motion of truck in opposite direction
The relative speed of the soccer ball is
now we know that
so we have
so we have
Well, isn’t v=d/t, then by making subject d, d=vt, turn 3 hours into seconds, 1hr=3600sec, hence 3hr=x, cross multiply and you would have converted it into seconds, then put it into your equation of d=vt
Answer:
N = 1364 N
Explanation:
given data
accelerate upward = 5.70 m/s²
mass = 88.0 kg
solution
normal force is in upward direction so, weight of the student in downward direction and acceleration is in upward direction so formula is express as
N - mg = ma ...........................1
N = m × (g+a)
put here value
N = 88.0 × (9.8 + 5.70)
N = 1364 N
B strength training I think that’s the answer
Answer:
303.29N and 1.44m/s^2
Explanation:
Make sure to label each vector with none, mg, fk, a, FN or T
Given
Mass m = 68.0 kg
Angle θ = 15.0°
g = 9.8m/s^2
Coefficient of static friction μs = 0.50
Coefficient of kinetic friction μk =0.35
Solution
Vertically
N = mg - Fsinθ
Horizontally
Fs = F cos θ
μsN = Fcos θ
μs( mg- Fsinθ) = Fcos θ
μsmg - μsFsinθ = Fcos θ
μsmg = Fcos θ + μsFsinθ
F = μsmg/ cos θ + μs sinθ
F = 0.5×68×9.8/cos 15×0.5×sin15
F = 332.2/0.9659+0.5×0.2588
F =332.2/1.0953
F = 303.29N
Fnet = F - Fk
ma = F - μkN
a = F - μk( mg - Fsinθ)
a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0
303.29-0.35( 666.4 - 303.29*0.2588)/68.0
303.29-0.35(666.4-78.491)/68.0
303.29-0.35(587.90)/68.0
(303.29-205.45)/68.0
97.83/68.0
a = 1.438m/s^2
a = 1.44m/s^2
