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Cerrena [4.2K]
3 years ago
7

An airplane is at rest on a runway. It accelerates at 10 m/s2 for 15 seconds. How fast is it now traveling?​

Physics
1 answer:
KATRIN_1 [288]3 years ago
6 0

Answer:

150

Explanation:

v = at

v = 10(15)

v = 150 m/s

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If a car has a velocity of 85 km/hr, how long will it take to accelerate to 45 km/hr if the acceleration is -3 km/hr/sec?
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Answer:

Time, t = 13.34 seconds.

Explanation:

Given the following data;

Initial velocity, u = 85km/hr to meters per seconds = 85*1000/3600 = 23.61 m/s

Final velocity, v = 45km/hr to meters per seconds = 45*1000/3600 = 12.5 m/s

Acceleration, a = -3 km/hr/sec to meters per seconds square = -3*1000/3600 = -0.833m/s²

To find the time;

Acceleration = (v - u)/t

-0.833 = (12.5 - 23.61)/t

-0.833t = -11.11

t = 11.11/0.833

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A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

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3 years ago
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