Answer:
reactants are needed to start a chemical reaction
Explanation:
am I correct
Answer:
on moon he can jump 4.2 m high
Explanation:
given data
gravitational acceleration on moon a(m) = 1/6
jump = 0.7 m
to find out
how high could he jump on the moon
solution
we know gravitational acceleration on earth a = g = 9.8 m/s²
so on moon am =
= 1.633 m/s²
so if he jump on earth his speed will be for height 0.7 m i s
speed v = ![\sqrt{2gh}](https://tex.z-dn.net/?f=%5Csqrt%7B2gh%7D)
v = ![\sqrt{2(9.8)0.7}](https://tex.z-dn.net/?f=%5Csqrt%7B2%289.8%290.7%7D)
v = 3.7 m/s
so if he hump on moon
height will be
height =
put here value
height =
height = 4.2 m
so on moon he can jump 4.2 m high
Incomplete question as the mass of ball is missing.So I assume the mass of ball is 0.35 kg.The complete question is here
A baseball has mass 0.35 kg. (a) If the velocity of a pitched ball has a magnitude of 43.5 m/s and the batted ball’s velocity is 57.5 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.Express your answer to three significant figures and include the appropriate units.
Answer:
J=4.90 kg.m/s
Explanation:
First, we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction So we have:
![v_{i}=43.5m/s\\ v_{f}=57.5m/s](https://tex.z-dn.net/?f=v_%7Bi%7D%3D43.5m%2Fs%5C%5C%20v_%7Bf%7D%3D57.5m%2Fs)
We plug our values for m and vi.So we get the initial momentum of ball:
![p_{1}=(0.35kg)(43.5m/s)\\p_{1}=15.225kg.m/s\\](https://tex.z-dn.net/?f=p_%7B1%7D%3D%280.35kg%29%2843.5m%2Fs%29%5C%5Cp_%7B1%7D%3D15.225kg.m%2Fs%5C%5C)
We plug our values for m and vf.So we get the final momentum of ball:
![p_{2}=(0.35kg)(57.5m/s)\\p_{2}=20.125kg.m/s\\](https://tex.z-dn.net/?f=p_%7B2%7D%3D%280.35kg%29%2857.5m%2Fs%29%5C%5Cp_%7B2%7D%3D20.125kg.m%2Fs%5C%5C)
So the change in momentum is:
Δp=p₂ - p₁
Δp=(20.125 kg.m/s) - (15.225 kg.m/s)
Δp=4.90 kg.m/s
We get the impulse applied to the ball by bat as follow
J=Δp
J=4.90 kg.m/s
Answer:
62.64 RPM.
Explanation:
Given that
m= 4.6 g
r= 19 cm
μs = 0.820
μk = 0.440.
The angular speed of the turntable = ω rad/s
Condition just before the slipping starts
The maximum value of the static friction force =Centripetal force
![\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s](https://tex.z-dn.net/?f=%5Cmu_s%5C%20m%20g%3Dm%5C%20%5Comega%5E2%5C%20r%5C%5C%20%5Comega%5E2%3D%5Cdfrac%7B%5Cmu_s%5C%20m%20g%7D%7Bm%20r%7D%5C%5C%20%5Comega%3D%5Csqrt%7B%5Cdfrac%7B%5Cmu_s%5C%20%20g%7D%7B%20r%7D%7D%5C%5C%20%5Comega%3D%5Csqrt%7B%5Cdfrac%7B0.82%5Ctimes%2010%7D%7B%200.19%7D%7D%5C%20rad%2Fs%5C%5C%5Comega%3D%096.56%5C%20rad%2Fs)
![\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM](https://tex.z-dn.net/?f=%5Comega%3D%5Cdfrac%7B2%5Cpi%20N%7D%7B60%7D%5C%5CN%3D%5Cdfrac%7B60%5Ctimes%20%5Comega%7D%7B2%5Cpi%20%7D%5C%5CN%3D%5Cdfrac%7B60%5Ctimes%206.56%7D%7B2%5Cpi%20%7D%5C%20RPM%5C%5CN%3D62.64%5C%20RPM)
Therefore the speed in RPM will be 62.64 RPM.