Answer:
This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.
U = (1/2)kx^2
U = (1/2)(5.3)(3.62-2.60)^2
U = 2.75706 J
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Explanation:
Answer:
(E) μs(mA +mB)g
Explanation:
We can apply for mB:
∑ Fx = mB*a (→)
⇒ Ffriction = mB*a ⇒ a = Ffriction / mB = μs*N / mB
⇒ a = μs*(mB*g) / mB ⇒ a = μs*g (acceleration of the system)
Now, for mA we have
∑ Fx = mA*a (→)
F - Ffriction = mA*a ⇒ F = mA*a + Ffriction
⇒ F = mA*(μs*g) + μs*(mB*g) ⇒ F = μs*g*(mA + mB)
We must know that the friction acts only between the two blocks
Answer:
<h3>The answer is 3.81 kg</h3>
Explanation:
The mass of the dog can be found by using the formula
f is the force
a is the acceleration
From the question we have
We have the final answer as
<h3>3.81 kg</h3>
Hope this helps you
Answer:
Image result for what is the force the when does when Gravity pushes you
The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.
Explanation:
Image result for what is the force the when does when Gravity pushes you
The important thing to remember is that gravity is neither a push nor a pull; what we interpret as a “force” or the acceleration due to gravity is actually the curvature of space and time — the path itself stoops downward.
Answer:
a) velocity v = 322.5m/s
b) time t = 19.27s
Explanation:
Note that;
ads = vdv
where
a is acceleration
s is distance
v is velocity
Given;
a = 6 + 0.02s
so,
Remember that
![v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7Bds%7D%7Bdt%7D%20%5C%5C%5Cfrac%7Bds%7D%7Bv%7D%20%3D%20dt%5C%5C%5Cint%5Climits%5Es_0%20%7B%5Cfrac%7Bds%7D%7B%5Csqrt%7B12s%2B0.02s%5E%7B2%7D%20%7D%20%7D%20%7D%20%5C%2C%20ds%20%3D%20%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt%20%5C%5Ct%3D%20%20%285%5Csqrt%7B2%7D%20%29%20ln%20%20%5Cfrac%7B%7C%20%5Bs%20%2B%20300%20%2B%20%5Csqrt%7B%28s%5E%7B2%7D%20%20%2B%20600s%29%7D%20%5D%20%7C%7D%7B300%7D%20.......2)
substituting s = 2km =2000m, into equation 1
v = 322.5m/s
substituting s = 2000m into equation 2
t = 19.27s
