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Vlad [161]
3 years ago
9

A car travels 40 kilometers at an average speed of 80km/h and then travels 40 kilometers at an average speed of 40km/h. The aver

age speed of the car for this 80-km trip is
Physics
1 answer:
iris [78.8K]3 years ago
8 0

Speed = distance / time

Part 1: 40 km at 80 km per goir = 0.50 hours

Part 2: 40 km at 40 km per hour = 1 hour

The time of the trip was 1.5 hours

Total distance was 40 + 40 = 80 ins.

Speed = distance / time = 80/1.5 = 53.33 km/he ( round answer as needed)

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Which of the following is an example of an exerting force?
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4 0
3 years ago
A fuzzy bunny sees a butterfly and chases it right off of a 20 m high cliff at 7m/s
bogdanovich [222]

A) 2.02 s

To find the time it takes for the bunny to reach the rocks below the cliff, we just need to analyze its vertical motion. This motion is a free fall motion, which is a uniformly accelerated motion. So we can use the suvat equation:

s=u_y t+\frac{1}{2}at^2

where

s is the vertical displacement

u is the initial vertical velocity

a is the acceleration

t is the time

For the bunny here, choosing downward as positive direction,

u_y = 0 (initial vertical velocity is zero)

s = 20 m

a=g=9.8 m/s^2 (acceleration of gravity)

And solving for t, we find the time of flight:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s

B) 14.1 m

For this part, we need to consider the horizontal motion of the bunny.

The horizontal motion of the bunny is a uniform motion with constant velocity, which is the initial velocity of the bunny:

v_x = 7 m/s

Therefore the distance covered after time t is given by

d=v_x t

And substituting the time at which the bunny hits the ground,

t = 2.02 s

We find how far the bunny went from the cliff:

d=(7)(2.02)=14.1 m

C) 21.0 m/s at 70.5^{\circ} below the horizontal

The horizontal component of the velocity is constant during the entire motion, so it will still be the same when the bunny hits the ground:

v_x = 7 m/s

Instead, the vertical velocity is given by

v_y = u_y +at

And substituting t = 2.02 s, we find the vertical velocity at the moment of impact:

v_y = 0+(9.8)(2.02)=19.8 m/s

So, the magnitude of the final velocity is

v=\sqrt{v_x^2+v_y^2}=\sqrt{7^2+(19.8)^2}=21.0 m/s

And the angle is given by

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{19.8}{7})=70.5^{\circ}

below the horizontal

5 0
3 years ago
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