Answer:
W = 34.64 ft-lbs
Explanation:
given,
Horizontal force = 4 lb
distance of push, d = 10 ft
angle of ramp, θ = 30°
Work done on the box = ?
We know,
W = F.d cos θ
W = 4 x 10 x cos 30°
W = 40 x 0.8660
W = 34.64 ft-lbs
Hence, work done on the box is equal to W = 34.64 ft-lbs
Answer:
22.505 seconds
Explanation:
V =19.8m/s
V = a*to
t1 = 19.8/3.3
= 6seconds
Distance travelled during acceleration
= 1/2 x 3.3 x 6²
= 59.4m
X_total = x1 + x2
X2 = 373-59.4
X2 = 313.6m
t2 = x2/v
= 313.6/19.8
= 16.505
Total = 16.505 + 6
= 22.505 seconds
the minimum time in which an elevator can travel the 373 m from the ground floor is 22.505 seconds.
I only know P and V and P is pressure and V is volume
Answer:
a) It takes her 1.43 s to reach a speed of 2.00 m/s.
b) Her deceleration is - 2.50 m/s²
Explanation:
The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:
v = v0 + a · t
Where:
v = velocty.
v0 = initial velocity.
a = acceleration.
t = time.
a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:
v = v0 + a · t
2.00 m/s = 0 m/s + 1.40 m/s² · t
t = 2.00 m/s / 1.40 m/s²
t = 1.43 s
It takes her 1.43 s to reach a speed of 2.00 m/s
b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:
v = v0 + a · t
0 = 2.00 m/s + a · 0.800 s
-2.00 m/s / 0.800 s = a
a = -2.50 m/s²
Her deceleration is - 2.50 m/s²
