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blsea [12.9K]
3 years ago
13

A 80.0-kg airplane pilot pulls out of a dive by following, at a constant speed of 180km/hr, the arc of a circle whose radius is

300m. A) At the bottom of the circle, what are the direction and magnitude of his acceleration
Physics
1 answer:
wolverine [178]3 years ago
6 0

Answer: 8.33m/s²

Explanation:

Mass of the airplane pilot = 80kg

The speed of plane v = 180km/h = 50m/s

The radius of circle r = 300 m

The acceleration of the plane will be calculated as:

a = v²/r

a = 50²/300

a = 2500/300

a = 8.33m/s²

Note that 180km/h was converted to m/s by

= (180 × 1000)/(60 × 60)

= 180000/3600

= 50m/s

1000 meters = 1 kilometer

60 minutes = 1 hour

60 seconds = 1 minute

3600 seconds = 1 hour

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Which of the following is distance divided by the change in time?
vovikov84 [41]

(distance covered) divided by (time to cover the distance) is SPEED.

(straight-line distance and direction between start-point and end-point) is DISPLACEMENT.

(amount and direction of change in speed) divided by (time for the change) is ACCELERATION.

(amount and direction of change in momentum) divided by (time for the change)  OR  (mass) times (acceleration) is FORCE.

4 0
4 years ago
A car accelerates from rest to 27 m/s in 8 seconds. What is the acceleration of the car?
Vlad1618 [11]

Answer:

<h3>The answer is 3.38 m/s²</h3>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

a =  \frac{v}{t}  \\

where

a is the acceleration

v is the velocity

t is the time

From the question

v = 27 m/s

t = 8 s

We have

a =  \frac{27}{8}  \\  = 3.375

We have the final answer as

<h3>3.38 m/s²</h3>

Hope this helps you

3 0
3 years ago
A student pushes a 12.0 kg box with a horizontal force of 20 N. The friction force on the box is 9.0 N. Which of the following i
antiseptic1488 [7]
Fnet=m*a
a= Fnet/m

a=(20N[E]+9.0N[E])/ 12.0kg

a=2.4167m/s^2

a ≈ 2.4m/s^2
8 0
4 years ago
Read 2 more answers
A ship leaves a port at noon and travels due west at 20 knots. At 6 PM, a second ship leaves the same port and travels northwest
Amanda [17]

Answer:

v = 12.44 Knots

Explanation:

First ship starts at Noon with speed 20 Knots towards West

now we know that 2nd ship starts at 6 PM with speed 15 Knots towards North West

so after time "t" of 2nd ship motion the two ships positions are given as

r_1 = 20(t + 6)\hat i

r_2 = 15(t)(cos45\hat i + sin45\hat j)

now we can find the distance between two ships as

x = \sqrt{(20(t + 6) - 10.6 t)^2 + (10.6t)^2}

now we have

x^2 = (120 + 9.4 t)^2 + (10.6 t)^2

x^2 = 200.72 t^2 + 14400 + 2256 t

now we will differentiate it with respect to time

2x\frac{dx}{dt} = 401.44 t + 2256

here we know that

t = \frac{90}{15} = 6 hours

so we have

x = 187.5

now we have

2(187.5) v = 401.44(6) + 2256

v = 12.44 Knots

5 0
3 years ago
What is the distance write the equation
Sedbober [7]

Answer:

135 m

Explanation:

Given:

vi = 12 m/s

vf = 18 m/s

t = 9s

Find: x

x = ½ (vf + vi) t

x = ½ (18 m/s + 12 m/s) (9 s)

x = 135 m

5 0
3 years ago
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