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saveliy_v [14]
1 year ago
15

A 68 kg object, starting from rest, travels from point A to point B at a rate of 30 m/s in 2 hours. What is the applied force on

the object?
Physics
1 answer:
Natasha2012 [34]1 year ago
7 0

Answer:

\huge\boxed{\sf F = 0.28\ N}

Explanation:

<h3>Given Data:</h3>

Mass = m = 68 kg

Velocity = v = 30 m/s

Time = 2 hours = 2 × 60 × 60 = 7200 s

<h3>Required:</h3>

Force = F = ?

<h3>Formula to be used:</h3>

\displaystyle F = \frac{mv}{t}

<h3>Solution:</h3>

\displaystyle F = \frac{(68)(30)}{7200} \\\\F = \frac{2040}{7200} \\\\F = 0.28 N\\\\\rule[225]{225}{2}

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Explanation:

1 meter = 100 centimeters.

There are 275 centimeters.

275/100=2.75

So, each piece of wallpaper was 2.75 meters long.

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An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of
GaryK [48]

Explanation:

The given data is as follows.

     Velocity of bullet, c_{p} = 814.8 m/s

    Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

               t = \frac{24.7}{343}      (velocity in air = 343 m/s)

                 = 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

               d_{b} = c_{b} \times t

                          = 814.8 \times 0.072

                          = 58.66

                          = 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

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Total charge is uniformly distributed on a spherical surface of radius R. The sphere is centered at the origin and spins around
Flauer [41]

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3 years ago
An airplane pilot flies due west at a speed of 216 km/hr with respect to the air. After flying for a half an hour, the pilot fin
Yuki888 [10]

Answer:

speed wind  Vw = 54.04 km / h   θ = 87.9º

Explanation:

We have a speed vector composition exercise

In the half hour the airplane has traveled X = 108 km to the west, but is located at coordinated 119 km west and 27 km south

Let's add the vectors in each coordinate axis

   

X axis (East-West)

      -Xvion - Xw = -119

      Xw = -Xavion + 119

      Xw = 119 -108

      Xwi = 1 km

Calculate the speed for time of  t = 0.5 h

     Vwx = Xw / t

     Vwx= 1 /0.5

     Vwx = - 2 km / h

Y Axis (North-South)

    Y plane - Yi = -27

    Y plane = 0

    Yw = 27 km

    Vwy = 27 /0.5

    Vwy = 54 km / h

Let's use the Pythagorean theorem and trigonometry to compose the answer

 Vw = √ (Vwx² + Vwy²)

  Vw = R 2² + 54²

  Vw = 54.04 km / h

  tan θ = Vwy / Vwx

  tan θ = 54/2 = 27

  θ = Tan⁻¹ 1 27

  θ = 87.9º

The speed direction is 87. 9th measure In the third quadrant of the X axis in the direction 90-87.9 = 2.1º  west from the south

5 0
3 years ago
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