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Keith_Richards [23]
3 years ago
13

A train travels 99 kilometers in 2 hours, and then 90 kilometers in 3 hours. What is its average speed?

Physics
1 answer:
Agata [3.3K]3 years ago
3 0

Average speed =

(total distance)/(total time)

Average speed = (99+90)/(2+3)

That's (189 km) / (5 hr)

Average speed = 37.8 km/hr

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A baby is dropped from a 30 m cliff on Earth. Acceleration due to gravity on Earth is about 9.8 m/s^2. Calculate the following q
Katarina [22]

Answer:Dropping is not a good predictor of when labor will begin. In first-time mothers, dropping usually occurs 2 to 4 weeks before delivery, but it can happen earlier. In women who have already had children, the baby may not drop until labor begins.

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3 years ago
After crossing the finish line, a race car slows down from 47 m/s to 32m/s in 3seconds. What is the car’s acceleration?
Lisa [10]

Answer: -5 m/s^2

Explanation: a = v - u/t

                         = 32 - 47/3

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                         = -5 m/s^2

7 0
3 years ago
A projectile is launched horizontally from the top a 35.2m high cliff and lands a distance of 107.6m from the base of the cliff.
tankabanditka [31]

Answer:

v_o=40.14\ m/s

Explanation:

<u>Horizontal Launch </u>

It happens when an object is launched with an angle of zero respect to the horizontal reference. It's characteristics are:

  • The horizontal speed is constant and equal to the initial speed v_o
  • The vertical speed is zero at launch time, but increases as the object starts to fall
  • The height of the object gradually decreases until it hits the ground
  • The horizontal distance where the object lands is called the range

We have the following formulas

\displaystyle v_x=v_o

\displaystyle x=v_o.t

\displaystyle v_y=g.t

\displaystyle y=\frac{gt^2}{2}

Where v_o is the initial horizontal speed, v_y is the vertical speed, t is the time, g is the acceleration of gravity, x is the horizontal distance, and y is the height.

If we know the initial height of the object, we can compute the time it takes to hit the ground by using

\displaystyle y=\frac{gt^2}{2}

Rearranging and solving for t

\displaystyle 2y=gt^2

\displaystyle t^2=\frac{2\ y}{g}

\displaystyle t=\sqrt{\frac{2\ y}{g}}

We then replace this value in

\displaystyle x=v_o.t

To get

\displaystyle v_o=\frac{x}{t}

\displaystyle v_o=\frac{x}{\sqrt{\frac{2y}{g}}}

\displaystyle v_o=\sqrt{\frac{g}{2y}}.x

The initial speed depends on the initial height y=32.5 m, the range x=107.6 m and g=9.8 m/s^2. Computing v_o

\displaystyle v_o=\sqrt{\frac{9.8}{2(35.2)}}\ 107.6

The launch velocity is  

\boxed{v_o=40.14\ m/s}

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