Answer:
2156J
Explanation:
Given parameters:
Height of lift = 10m
Mass = 22kg
Unknown:
Work done by the machine = ?
Solution:
Work done is the force applied to move a body through a certain distance.
So;
Work done = Force x distance
Here;
Work done = mass x acceleration due to gravity x height
Work done = 22 x 9.8 x 10 = 2156J
In this question a lot of information's are provided. Among the information's provided one information and that is the time of 4 seconds is not required for calculating the answer. Only the other information's are required.
Mass of the block that is sliding = 5.00 kg
Distance for which the block slides = 10 meters/second
Then we already know that
Momentum = Mass * Distance travelled
= (5 * 10) Kg m/s
= 50 kg m/s
So the magnitude of the blocks momentum is 50 kg m/s. The correct option among all the given options is option "b".
Answer:
P = 17.28*10⁶ N
Explanation:
Given
L = 250 mm = 0.25 m
a = 0.54 m
b = 0.40 m
E = 95 GPa = 95*10⁹ Pa
σmax = 80 MPa = 80*10⁶ Pa
ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m
We get A as follows:
A = a*b = (0.54 m)*(0.40 m) = 0.216 m²
then, we apply the formula
ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L
⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)
⇒ P = 24624000 N = 24.624*10⁶ N
Now we can use the equation
σ = P/A
⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa
So σ > σmax we use σmax
⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N
B.) Compressions
HOPPE IT HELPED
Answer:
0.231 N
Explanation:
To get from rest to angular speed of 6.37 rad/s within 9.87s, the angular acceleration of the rod must be
If the rod is rotating about a perpendicular axis at one of its end, then it's momentum inertia must be:
According to Newton 2nd law, the torque required to exert on this rod to achieve such angular acceleration is
So the force acting on the other end to generate this torque mush be:
