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4 years ago

6

If the length of the ramp is 2m and the height of the ramp is 1m , what is the mechanical advantage?

1 answer:

slamgirl [31]4 years ago

5 0

Answer:

Mechanical Advantage is 2

Explanation:

M.A = length of ramp / height of ramp

= 2/1

=2

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A car is traveling at a speed of 20m/s and has a mass of 1200kg how much kinetic energy does it have

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Answer:

Explanation:

The formula for Kinetic Energy is

$KE=\frac{1}{2}mv^2$ . Filling in:

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3 years ago

Near the top of the Citigroup Center building in New York City, there is an object with mass of 4.8 x 105 kg on springs that hav

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Answer:

The force constant is $k =1.316 *10^{7} \ N/m$

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Explanation:

From the question we are told that

The mass of the object is $M = 4.8*10^{5} \ kg$

The period is $T = 1.2 \ s$

The period of the spring oscillation is mathematically represented as

$T =2 \pi \sqrt{ \frac{M}{k}}$

where k is the force constant

So making k the subject

$k = \frac{4 \pi ^2 M }{T^2}$

substituting values

$k = \frac{4 (3.142) ^2 (4.8 *10^{5}) }{(1.2)^2}$

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$E = \frac{1}{2} k x^2$

Where x is the spring displacement which is given as

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3 years ago

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

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$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

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3 years ago

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3 years ago

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