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snow_tiger [21]
3 years ago
10

An 80 kg man is standing in an elevator which is moving down and slowing down at a rate of 2.0 m/s2. how hard must the floor of

the elevator push up on the man?
Physics
1 answer:
V125BC [204]3 years ago
8 0
F = mass * acceleration
F = 80kg * 2m/s^2
F = 160N
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While running at a constant velocity, how should you throw a ball with respect to you so that you can catch it yourself?
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You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x). 
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Give a specific example of a machine,and describe how its mechanical efficiency might be calculated
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A spearfisher stands in shallow water and sees a fish a few feet in front of her. She throws her spear directly toward the posit
DENIUS [597]

Answer:

the spear will end up above the fish relative to the actual position of the fish.

Explanation:

due to refraction of light coming from the fish the fish will appear slightly above from its real position

So due to this refraction the spearfisher will throw the spear directly at the image of the fish due to which it will not reach the position of fish but it will reach the position above the fish.

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A person jogs eight complete laps around a quarter-mile track in a total time of 12.5 min. Calculate (a) the average speed and (
Margarita [4]

\large\displaystyle\text{$\begin{gathered}\sf \huge \bf{\underline{Data:}} \end{gathered}$}

  • \large\displaystyle\text{$\begin{gathered}\sf 1\ mile = 1609.34 \ m \end{gathered}$}
  • \large\displaystyle\text{$\begin{gathered}\sf  1/4 \ mile = 402.33 \ m \end{gathered}$}

                           \large\displaystyle\text{$\begin{gathered}\sf 12.5 \not{min}*\frac{60 \ s}{1\not{min}}=750 \ s \end{gathered}$}

                   \large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                         \large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \  V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}

                  \large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \  in \ m/s} \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}  

                          \large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}

                          \large\displaystyle\text{$\begin{gathered}\sf V=\frac{D}{T} \ \ \ \ \ V=\frac{402.33 \ m}{750 \ s}   \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V= 0.53 \ m/s \end{gathered}$}

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2 years ago
7. Answer each of the following questions if the student applied a 550N force to a 100kg box on the same surface. What would be
blagie [28]
F=ma
550=100a
a=550/100
a=5.5 m/s²
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