Answer:
k = 17043.5 N/m = 17.04 KN/m
Explanation:
First we need to find the force applied by safe pn the spring:
F = Weight of Safe
F = mg
where,
F = Force Applied by the safe on the spring = ?
m = mass of safe = 800 kg
g = 9.8 m/s²
Therefore,
F = (800 kg)(9.8 m/s²)
F = 7840 N
Now, using Hooke's Law:
F = kΔx
where,
K = Spring Constant = ?
Δx = compression = 46 cm = 0.46 m
Therefore,
7840 N = k (0.46 m)
k = 7840 N/0.46 m
<u>k = 17043.5 N/m = 17.04 KN/m</u>
Answer:
Acceleration = 192.3 m/s² (Approx.)
Explanation:
Given:
Force = 125 N
Mass of ball = 0.65 kg
Find:
Acceleration
Computation:
We know that;
Acceleration = Force / Mas
So,
Acceleration = 125 / 0.65
Acceleration = 192.3 m/s² (Approx.)
935,500 joules because when we use the KE formula KE=1/2mv^2;
KE=1/2(750)(50)^2
KE=375(2500)
KE=935,500 Joules
Hope it helps
Answer:
The horizontal component of the velocity is 21.9 m/s.
Explanation:
Please see the attached figure for a better understanding of the problem.
Notice that the vector v and its x and y-components (vx and vy) form a right triangle. Then, we can use trigonometry to find the magnitude of vx, the horizontal component of the velocity.
To find vx, let´s use the following trigonometric rule of right triangles:
cos α = adjacent / hypotenuse
cos 5.7° = vx / 22 m/s
22 m/s · cos 5.7° = vx
vx = 21.9 m/s
The horizontal component of the velocity is 21.9 m/s.
Answer:
Explanation:
Given:
U1 = 1.6 m/s
U2 = -1.1 m/s
M1 = 1850 kg
M2 = 1400 kg
V1 = 0.27 m/s
Using momentum- collision equation,
M1U1 + M2U2 = M1V1 + M2V2
1850 × 1.6 - 1400 × 1.1 = 1850 × 0.27 + 1400 × V2
1420 = 499.5 + 1400V2
V2 = 0.6575 m/s
B.
KE = 1/2 × MV^2
KEa1 + KEa2 = KEb1 + KEb2
Delta KE = KE2 - KE1
KEa1 = 2368 J
KEb1 = 847 J
KEa2 = 67.433 J
KEb2 = 302.6 J
KE1 = KEa1 + KEb1
= 3215 J
KE2 = 370.033 J
Delta KE = -2845 J.
