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Setler79 [48]
4 years ago
14

What's the momentum (in kg-m/s) of a cannonball with a mass of 21 moving with a velocity of 25 m/s?

Physics
1 answer:
lbvjy [14]4 years ago
8 0
P=mv
P=21*25=525 kg*m/s
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If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli
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Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

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N = normal force to cylinder

N==\frac{m(V)^{2}}{R}=m*(w)^2*R

Friction force\\Ff = k*N\\Ff= k*m*w^2*R

Gravitational force \\Fg = m*g

These must be balanced (the net force on the people will be 0) so set them equal to each other.

Fg = Ff

m*g = k*m*w^2*R

g=k*w^{2}*R

w^2 =\frac{g}{k*R}

w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}

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RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88

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If I bought 59 kids off the black market in a secret door in the stalls and I sell 21 kids then ask for 81 more then 12 off the
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You will be left with 106 kids

<h3>Meaning of word problem</h3>

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Vector vector a has a magnitude of 29 units and points in the positive y-direction. when vector vector b is added to vector a ,
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Good morning.

We have:

\mathsf{\overset{\to}{a} = 29\overset{\to}{j}}

Where j is the unitary vector in the direction of the y-axis.

We have that 

\mathsf{\overset{\to}{a}+\overset{\to}{b} = -18\overset{\to}{j}}

We add the vector -a to both sides:

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Therefore, the magnitude of b is 47 units.
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