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Masteriza [31]
3 years ago
9

If my weight on Earth is 140lbs, what is my mass?

Physics
1 answer:
SashulF [63]3 years ago
5 0

Answer:

63.57 kg

Explanation:

weight = 140 lbs

Let the mass  is m.

1 lbs = 4.45 N

The weight of an object is defined as the force with which our earth attracts the body towards its centre.

Weight is the product of mass of the body and the acceleration due to gravity of that planet.

W = m x g

On earth surface g = 9.8 m/s^2

Now convert lbs in newton

So, 140 lbs = 140 x 4.45 = 623 N

So, m x 9.8 = 623

m = 63.57 kg

Thus, the mass is 63.57 kg.

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Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter
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Wee can use here kinematics

as we know that

y = v*t + \frac{1}{2} at^2

for shorter tree we know that

y = 0 + \frac{1}{2}*9.8 * 2^2

y = 19.6 meter

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

y = v*t + \frac{1}{2} at^2

39.2 = 0 + \frac{1}{2}*9.8 * t^2

t = 2.83 s

so the time taken is 2.83 s

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To find the number of neutrons in an atom, what would you subtract?
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What part of the atmosphere contains dust, water vapor, and 75% of all gases
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Which property of a measurement is best estimated from the percent error? accuracy median precision range
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"Accuracy" would be the best option from the list regarding the property of a measurement that is best estimated from the percent error, since the higher the error is the lower the accuracy.


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A steel ball is dropped from a building's roof and passes a window, taking 0.115 s to fall from the top to the bottom of the win
marta [7]

Answer:

The building is 26.85m tall.

Explanation:

In order to solve this problem, we must start by drawing a diagram of the situation. (See attached picture).

We split the height of the building into three parts: y_{1}, the height of the window and y_2

In order to find each of those, we need to start by finding the velocities of the ball in points A and B. We will analyze the trajectory of the ball when bouncing back to the top of the building. Let's start by finding the velocity of the ball in A:

We can use the following formula to determine the velocity of the ball in part A:

\Delta y=V_{A}t+\frac{1}{2}at^{2}

which can be solved for V_{A}

so we get:

V_{A}=\frac{\Delta y-\frac{1}{2}at^{2}}{t}

and now we can substitute (remember the acceleration of gravity goes downward so we will consider it to be negative).

V_{A}=\frac{(1.30m)-\frac{1}{2}(-9.8m/s^{2})(0.115s)^{2}}{0.115s}

which yields:

V_{A}=11.87m/s

once we got the velocity at point A, we can now find the velocity at point B. We can do so by using the following formula:

a=\frac{V_{A}-V_{B}}{t}

which can be solved for V_{B} which yields:

V_{B}=V_{A}-at

so we can substitute values now:

V_{B}=11.87m/s-(-9.8)(0.115s)

Which yields:

V_{B}=13m/s

Now that we have the velocities at A and B, we can use them to find the values of y_{1} and y_{2}

Let's start with  y_{1}

We can use the following formula to find it:

y_{1}=\frac{V_{f}^{2}-V_{A}^{2}}{2a}

we know the final velocity of the rebound will be zero, so we can simplify our formula:

y_{1}=\frac{-V_{A}^{2}}{2a}

so we can substitute now:

y_{1}=\frac{-(11.87m/s)^{2}}{2(-9.8m/s^{2})}

which solves to:

y_{1}=7.19m

Now we can proceed and find the value of y_{2}

the value of y_{2} can be found by using the following formula:

y_{2}=V_{B}t-\frac{1}{2}at^{2}

in this case our t will be half of the tie spent below the bottom of the window, so:

t=\frac{2.04s}{2}=1.02s

so now we can substitute all the values in the given formula:

y_{2}=(13m/s)(1.02s)-\frac{1}{2}(-9.8m/s^{2})(1.02s)^{2}

which yields:

y_{2}=18.36m

so now that we have all the values we need, we can go ahead and calculate the height of the building:

h=y_{1}+window+y_{2}

when substituting we get:

h=7.19m+1.3m+18.36m

So the answer is:

h=26.85m

The building is 26.85m tall.

5 0
3 years ago
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