Answer:
120s^-1
Explanation:
v=12v
I=10A
and since rate is with time, therefore rate=energy/time.
H=IV
10×12=120/s
therefore the rate is 120s^-1
Distance and force.
You need to know how much force was needed and how far it went to calculate how much work was done. The answer is C.
Answer:
A) 0.44 rad/s
B) 3.091 rad/s
Explanation:
Initial angular momentum must be equal to final angular momentum.
W1 = 0.80 rad/s
I = 1360 kgm^2
Initial moment Iw = 0.8 x 1360 = 1088 rad-kg-m^2/s
For second case
I = (65 x 4)4.2 + 1360 = 2452 kg-m2-rad/s
New momentum = 2452 x W2
Where w2 = final angular momentum
= 2452W2
Equating both moment,
1088 = 2452W2
W2 = 0.44 rad/s0
If they were on it before,
Initial momentum
I = 1360 - (60 x 4)4.2
= 352 rad-kg-m^2/s
352W1 = 1088
W1 = 3.091 rad/s
From Hooke's law, the length of the spring after extension and after the mass is attached is 1.788 meters. Option B is the answer.
<h3>
HOOKE'S LAW</h3>
Hooke's law state that in an elastic material, the force applied is directly proportional to the extension provided that the elastic limit is not exceeded.
Given that a 12.0 kg mass is attached to 1.2 m long spring with a spring constant of 200.0 N/m.
The given parameters are:
- Initial length
= 1.2 m
- Spring constant K = 200 N/m
According to Hooke's law
F = Ke
But F = mg
mg = Ke
Substitute all the parameters into the formula to get extension e
12 x 9.8 = 200e
e = 117.6 / 200
e = 0.588 m
The length of the spring after extension and after the mass is attached will be
=
+ e
= 1.2 + 0.588
= 1.788 m
Therefore, the correct answer is option B because the length of the spring after extension and after the mass is attached is 1.788 meters.
Learn more about Hooke's law here: brainly.com/question/12253978
Answer:
the instantaneous velocity is 51 m/s
Explanation:
Given;
acceleration, a = 2 + 5t²
Acceleration is the change in velocity with time.
![a = \frac{dv}{dt} \\\\a = 2 + 5t^2\\\\The \ acceleration \ (a) \ is \ given \ so \ we \ have \ to \ find \ the \ velocity \ (v)\\\\To \ find \ the \ velocity, \ integrate\ both \ sides \ of \ the \ equation\\\\2 + 5t^2 = \frac{dv}{dt} \\\\\int\limits^3_0 {(2 + 5t^2)} \, dt = dv\\\\v = [2t + \frac{5t^3}{3} ]^3_0\\\\v = 2(3) + \frac{5(3)^3}{3} \\\\v = 6 + 5(3)^2\\\\v = 6 + 45\\\\v = 51 \ m/s](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%5C%5C%5C%5Ca%20%3D%202%20%2B%205t%5E2%5C%5C%5C%5CThe%20%5C%20acceleration%20%5C%20%28a%29%20%5C%20is%20%5C%20given%20%5C%20so%20%5C%20we%20%5C%20have%20%5C%20to%20%5C%20find%20%5C%20the%20%5C%20velocity%20%5C%20%28v%29%5C%5C%5C%5CTo%20%5C%20find%20%5C%20the%20%5C%20velocity%2C%20%5C%20integrate%5C%20both%20%5C%20sides%20%5C%20of%20%5C%20the%20%5C%20equation%5C%5C%5C%5C2%20%2B%205t%5E2%20%3D%20%5Cfrac%7Bdv%7D%7Bdt%7D%20%5C%5C%5C%5C%5Cint%5Climits%5E3_0%20%7B%282%20%2B%205t%5E2%29%7D%20%5C%2C%20dt%20%3D%20dv%5C%5C%5C%5Cv%20%3D%20%5B2t%20%2B%20%5Cfrac%7B5t%5E3%7D%7B3%7D%20%5D%5E3_0%5C%5C%5C%5Cv%20%3D%202%283%29%20%2B%20%5Cfrac%7B5%283%29%5E3%7D%7B3%7D%20%5C%5C%5C%5Cv%20%3D%206%20%2B%205%283%29%5E2%5C%5C%5C%5Cv%20%3D%206%20%2B%2045%5C%5C%5C%5Cv%20%3D%2051%20%5C%20m%2Fs)
Therefore, the instantaneous velocity is 51 m/s