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Alenkasestr [34]
3 years ago
14

A rocket with a mass of 1,000 kilograms accelerates at 3 meters per second per second. Use the equation F = m × a to find the fo

rce exerted by the rocket’s engine.
A.
3,000 N


B.
333.3 N


C.
30,000 N


D.
3,333 N
Physics
1 answer:
bija089 [108]3 years ago
5 0
C. 30,000 N is my answer
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What morbid structure traditionally has thirteen steps?
tatiyna

Answer:

A gallow

Explanation:

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What do we call air in the motion​
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6. several light bulbs are connected in series across a 115 v source
Natalka [10]

Explanation:

a) I = V / R

1.70 = 115 / R

R = 115 / 1.70

R = 67.647

R = 67.65 ohms

Therefore, equivalent resistance is 67.65 ohms

b) Equivalent resistance of circuit from above sum is 67.65 ohms

Given resistance of each bulb is 1.50 ohms

Number bulbs = Equivalent resistance / Resistance of each bulb

= 67.65 / 1.50

= 45

8 0
2 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
INT Raindrops acquire an electric charge as they fall. Suppose a 2.0-mm-diameter drop has a charge of 12 pC; these are both very
horsena [70]

Answer:

W = 2.3 10² F_{e}

Explanation:

The force of the weight is

         W = m g

         

let's use the concept of density

         ρ= m / v

the volume of a sphere is

         V = \frac{4}{3} π r³

         V = \frac{4}{3} π (1.0 10⁻³)³

         V = 4.1887 10⁻⁹ m³

the density of water ρ = 1000 kg / m³

          m = ρ V

          m = 1000 4.1887 10⁻⁹

          m = 4.1887 10⁻⁶ kg

therefore the out of gravity is

          W = 4.1887 10⁻⁶ 9.8

          W = 41.05 10⁻⁶ N

now let's look for the electric force

           F_e = q E

           F_e = 12 10⁻¹² 15000

           F_e = 1.8 10⁻⁷ N

         

the relationship between these two quantities is

          \frac{W}{F_e} = 41.05 10⁻⁶ / 1.8 10⁻⁷

           \frac{W}{F_e} = 2,281 10²

             

             W = 2.3 10² F_{e}

therefore the weight of the drop is much greater than the electric force

3 0
3 years ago
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