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dlinn [17]
3 years ago
13

The initial velocity of a truck is 10 m/s. How long must it accelerate at a constant acceleration of 2m/s2 before its average ve

locity is equal to three times its initial velocity
Physics
1 answer:
Yuliya22 [10]3 years ago
7 0

Answer:

10s

Explanation:

vf = vi+at

30=10+2t

20= 2t

t = 10 s

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An electric fan is running on HIGH. After fan has been running for 1.3 minutes, the LOW button is pushed. The fan slows down to
mafiozo [28]

Answer:

    wo = 18.75 rev / s

Explanation:

This is an exercise in endowment kinematics, it indicates that the final angular velocity is w_f = 109 rad / s, the time to reach this velocity is t = 1.87 s and the deceleration a = 4.7 rad / s²

         w_f = w₀ - a t

         w₀ = w_f + a t

         w₀ = 109 + 4.7 1.87

         w₀ = 117.8 rad / s

let's reduce to revolutions / s

         w₀ = 117.8 rad / s (1 rev / 2pi rad)

         w₀ = 18.75 rev / s

8 0
3 years ago
Which of the following describes a referee's job?
Serhud [2]

Answer:

C. Supervising the game to make sure teams are playing fairly

5 0
4 years ago
A 64.8 kg astronaut is on a space walk when the tether line to the shuttle breaks. The astronaut is able to throw a 11.0 kg oxyg
Umnica [9.8K]

dfdfdfdddfdddddddddddddddd

6 0
3 years ago
A sky diver, with parachute unopened, falls 625 m in 15.0 s. Then she opens her parachute and falls another 362 m in 139 s. What
Jobisdone [24]

Answer:

v_{avg} = 6.41 m/s

Explanation:

Average velocity is defined as the ratio of total displacement of the motion and total time taken in that motion

here we know that initially the sky diver drops without opening parachute by total displacement 625 m

then she open her parachute and drop another 362 m

so first it took time t = 15 s to drop without open parachute

then it took t = 139 s to drop next displacement

so here total displacement is given as

d = 625 m + 362 m

total time is given as

t = 15 s + 139 s

so average velocity is given as

v_{avg} = \frac{625 + 362}{15 + 139}

v_{avg} = 6.41 m/s

4 0
3 years ago
A car is traveling north at 17.7 m/s . After 6 it’s velocity is 141 in the same direction. Find the magnitude and direction of t
Furkat [3]

By equation of motion we have   v = u + at

Where u = Initial velocity, v = final velocity, t = time taken and a = acceleration

Here v = 141 m/s, u = 17.7 m/s and t = 6 s

On substitution we will get

        141 = 17.7+ 6a

       So, a = (141-17.7)/6 = 20. 55 m/s^{2}

       Aceeleration = 20. 55 m/s^{2} along north direction.


3 0
3 years ago
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