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Ierofanga [76]
3 years ago
13

Sawyer launches his 180 kg raft on the Mississippi River by pushing on it with a force of 75N. How long must Sawyer push on the

raft on to accelerate it to a speed 2 m/s?
Physics
1 answer:
Daniel [21]3 years ago
6 0

Answer: 4.8 s

Explanation:

We have the following data:

m=180 kg the mass of the raft

F=75 N the force applied by Sawyer

V=2 m/s the raft's final speed

V_{o}=0 m/s the raft's initial speed (assuming it starts from rest)

We have to find the time t

Well, according to Newton's second law of motion we have:

F=m.a (1)

Where a is the acceleration, which can be expressed as:

a=\frac{\Delta V}{\Delta t}=\frac{V-V_{o}}{t-t_{o}} (2)

Substituting (2) in (1):

F=m\frac{V-V_{o}}{t-t_{o}} (3)

Where t_{o}=0

Isolating t from (3):

t=\frac{m(V-V_{o})}{F} (4)

t=\frac{180 kg(2 m/s-0 m/s)}{75 N}

Finally:

t=4.8 s

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A flywheel with radius of 0.400 mm starts from rest and accelerates with a constant angular acceleration of 0.600 rad/s2rad/s2.
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A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c
tensa zangetsu [6.8K]

Answer:

E = \frac{3kQ^2}{5R}

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

V = \frac{kq}{r}

now we can say that

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q = \frac{Qr^3}{R^3}

now electric potential is given as

V = \frac{k\frac{Qr^3}{R^3}}{r}

V = \frac{kQr^2}{R^3}

now work done to bring a small charge from infinite to the surface of this sphere is given as

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dW = \frac{kQr^2}{R^3} dq

here we know that

dq = \frac{3Qr^2dr}{R^3}

now the total energy of the sphere is given as

E = \int dW

E = \int_0^R  \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})

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E = \frac{3kQ^2}{5R}

7 0
3 years ago
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