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2 years ago

A child in a boat throws a 6-kg package horizontally to the right with a speed of 8 m/s.

Physics

1 answer:

victus00 [196]2 years ago

4 0

Answer:

0.5 m/s

Explanation:

Using conservation of momentum, P1=P2, the system starts off with zero momentum as nothing is moving. But in the second part of the equation(P2) the child throws the package to the right. By Newton's third law, the child and the boat should move to the left. Plugging in what we know, 0= -90v + 6kg*8m/s. Solving for v you will get 0.5 meters per second. The mass is 90kg as that the is mass of the child and boat combined. I also made it negative as the boat and child move left (I designated this as the negative direction) and the package's momentum is positive as it is moving to the right.

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balandron [24]

Answer:

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3 years ago

A car is traveling at 75 m/s. 50 seconds later it is traveling at 25 m/s. What is the car’s acceleration?

scoray [572]

Answer:

a = -1 m/s^2

Explanation:

Vi = 75 m/s

Vf = 25 m/s

t = 50 s

Plug those values into the following equation:

Vf = Vi + at

25 = 75 + 50a

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3 years ago

A ball is thrown straight up. At the top of its path its acceleration is

timurjin [86]

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3 years ago

A particle is moving with (SHM) of period 8.0s and amplitude5.0m

nadezda [96]

Answer:

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

Explanation:

Given the description of period and amplitude, the SHM could be described by:

$f(x)=5\,sin(\frac{\pi}{4}x)$

and its angular velocity can be calculated doing the derivative:

$f(x)=5\, \,sin(\frac{\pi}{4}x)\\f'(x)=5\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$

And therefore, the tangential velocity is calculated by multiplying this expression times the radius of the movement (3 m):

$velocity(x)=15\,\frac{\pi}{4}\,cos(\frac{\pi}{4}x)$ and is given in m/s.

Then the maximum speed is obtained when the cosine function becomes "1", and that gives:

Max speed = $\frac{15\, \pi}{4} \,\, \frac{m}{s}$

The acceleration is found from the derivative of the velocity expression, and therefore given by:

$acceleraton(x)=-15\,\frac{\pi^2}{16}\,sin(\frac{\pi}{4}x)$

and the maximum of the function will be obtained when the sine expression becomes "-1", which will render:

Max acceleration = $\frac{15\,\pi^2}{16} \,\,\frac{m}{s^2}$

6 0

3 years ago

Using the definition of moment of inertia, calculate icm, the moment of inertia about the center of mass, for this object.

Step2247 [10]

<span> The masses have no inertia about their own CM, and "the object" is the two masses. </span>
<span>1. Icm (at point A) = 2mr^2
hope this helps</span>

3 0

2 years ago