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bogdanovich [222]
1 year ago
5

A 40,000 kg subway train is brought to a stop from a speed of 0.700 m/s in 0.250 m by a large spring bumper at the end of its tr

ack. What is the force constant k of the spring?

Physics
1 answer:
lozanna [386]1 year ago
5 0

Given,

The mass of the train, m=40000 kg

The initial velocity of the train, u=0.700 m/s

The compression in the spring bumper that stopped the train, x=0.250 m

The final velocity of the train, v=0 m/s

From the equation of motion,

v^2-u^2=2ax

Where a is the acceleration of the train.

On substituting the known values,

\begin{gathered} 0-0.700^2=2a\times0.250 \\ \Rightarrow a=\frac{-0.700^2}{2\times0.25} \\ =-0.98\text{ m/s}^2 \end{gathered}

The magnitude of the force applied by the train will be equal to the magnitude of the restoring force of the spring.

Therefore,

\begin{gathered} m|a|=kx \\ \Rightarrow k=\frac{m|a|}{x} \end{gathered}

Where k is the spring constant of the spring.

On substituting the known values,

\begin{gathered} k=\frac{40000\times0.98}{0.250} \\ =156800\text{ N/m} \end{gathered}

Therefore the spring constant of the spring is 156800 N/m

Thus the correct answer is option C.

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A jet airliner moving initially at 406 mph (with respect to the ground) to the east moves into a region where the wind is blowin
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Answer:

966 mph

Explanation:

Using as convention:

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While the components of the velocity of the wind are

v_{2x} = (568)(cos 15^{\circ})=548.6 mph\\v_{2y} = (568)(sin 15^{\circ})=147.0 mph

So the components of the resultant velocity of the jet are

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6 0
3 years ago
If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?
leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water                 ---> 1</span>

 

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

 

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m        (ANSWER)</span>

8 0
3 years ago
15 points for 2 questions
Elden [556K]
I believe the answer would be c because i think that you multiply the 2
8 0
3 years ago
Read 2 more answers
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