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3 years ago

An airplane went from 120 m/s to 180 m/s in 4.0 seconds. What was its acceleration?

1 answer:

3 0

15 m/s^2 The first thing to calculate is the difference between the final and initial velocities. So 180 m/s - 120 m/s = 60 m/s So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving 60 m/s / 4.0 s = 15.0 m/s^2 Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2

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Which of these most closely describes the shape of a ellipse

erica [24]

Oval —————————-ignore the line

4 0

4 years ago

How does the force required to lift an object in water compare to the force needed to lift the same object on land?it is the exa

padilas [110]

It takes less force to lift the object in the water than on land, since the water is helping with the weight of the object.

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3 years ago

While running a 100m race, a runner runs from the 20m to 30m mark in 77 frames of a video record. If the video camera recorded d

Marrrta [24]

Answer:

Speed of the runner during video interval is 6.49 m/s

Explanation:

According to the problem,

Number of frames recorded by camera in 1 second = 50

Time takes by camera to record 1 frame = (1/50) s

Time taken by camera to record 77 frames, t = $\frac{1}{50}\times 77$ s

Distance covered by the runner during the video recording, d = 10 m

Speed, v = $\frac{Distance}{time}=\frac{d}{t}$

Substitute the values of d and t in the above equation.

v = $\frac{10}{\frac{77}{50} }$

v = 6.49 m/s

3 0

4 years ago

In the figure, a proton is projected horizontally midway between two parallel plates that are separated by 0.50 cm, and are 5.60

noname [10]

a) Minimum speed of the proton: $6.05\cdot 10^6 m/s$

b) Angle of the velocity: $\theta=-5.1^{\circ}$

Explanation:

The proton experiences a vertical force due to the electric field, given by:

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the proton charge

$E=610,000 N/C$ is the magnitude of the electric field

The vertical acceleration of the proton is therefore

$a=\frac{qE}{m}$

where

$m=1.67\cdot 10^{-27}kg$ is its mass

Therefore, the vertical position of the proton at time t is

$y(t)=\frac{1}{2}at^2=\frac{1}{2}\frac{qE}{m}t^2$

where we assumed that the initial vertical velocity is zero (because the proton is fired horizontally) and the initial vertical position, halfway between the two plates, is the origin.

The horizontal motion of the proton instead is uniform, so the horizontal position is given by

$x(t)=v_0 t$