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3 years ago

What is the average acceleration during the time interval 0 seconds to 10 seconds?

Physics

1 answer:

zhuklara [117]3 years ago

3 0

Answer:

A. $0.5 m/s^2$

Explanation:

The average acceleration is equal to the ratio between the change in velocity and the time elapsed:

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

The change in velocity between t=0 s and t=10 s is the change on the vertical axis:

$\Delta v = 5 -0 = 5 m/s$

Whule the time interval is

$\Delta t = 10 s$

So the average acceleration is

$a=\frac{5 m/s}{10 s}=0.5 m/s^2$

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A large meteoroid enters the Earth's atmosphere at a speed of 20.0 km/s and is not significantly slowed before entering the ocea

Shalnov [3]

The shock wave from the meteoroid in the lower atmosphere has a Mach angle of 0.948°.

(a) The meteoroid's speed $v_s=20 \mathrm{~km} / \mathrm{s}$

$=20 \times 10^3 \mathrm{~m} / \mathrm{s}$

Air sound wave speed $&v=331 \mathrm{~m} / \mathrm{s} \\$

Speed of the shock wave in Mach $&\qquad \theta=\sin ^{-1}\left(\frac{v}{v_s}\right)$

$\begin{aligned}&=\sin ^{-1}\left(\frac{331 \mathrm{~m} / \mathrm{s}}{20 \times 10^3 \mathrm{~m} / \mathrm{s}}\right) \\&=0.948^{\circ}\end{aligned}$

Hence, 0.948° is the Mach angle of the shock wave from the meteoroid in the lower atmosphere.

<h3>What is the speed of the meteoroid?</h3>

A meteoroid's speed can be loosely broken down into three categories: slow, medium, and fast.

Slow meteors move around the sun at a leisurely pace of about 32 kilometers per second (20 miles per second).
Medium-speed meteors travel around the sun at approximately 50 kilometers per second (30 miles per second),
while fast meteors zoom past at over 120 kilometers per second (75 miles per second)!

To learn more about meteoroid, visit:

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5 0

1 year ago

A particle is moving in a circle of diameter 5. calculate the distance covered and the displacement when it completes 3 revoluti

Mamont248 [21]

Total distance covered is 47.1 m whereas displacement is zero.

<h3>Calculation:</h3>

Given,

Diameter, d = 5 m

No. of revolutions = 3

Radius, r = 5/2 = 2.5 m

To find,

Distance =?

Displacement =?

Distance covered in one revolution = 2πr

Put the values in this,

Distance = 2 × 3.14 × 2.5

= 15.7 m

Total distance covered in 3 revolution = 3 × 31.4

= 47.1 m

Displacement is the change in the position of the object or the distance between the initial and final position.

After 3 revolutions the particle comes back to its initial position. Therefore, the displacement is zero.

Hence, the total distance covered in 3 resolutions is 47.1 m whereas displacement is zero.

Learn more about distance and displacement here:

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5 0

8 months ago

an aircraft is flying at height of 3400m above the ground.if the angle subtended at a ground observation point by the aircraft p

Kamila [148]

Answer: 588.9 m/s

Explanation:

Given that :

θ = 30°

Height, h = 3400m

Time, t = 10 seconds

From trigonometry ;

Tanθ = opposite / hypotenus

Tan 30 = 3400 / x

x tan 30 = 3400

0.5773502x = 3400

x = 3400 / 0.5773502

x = 5888.9727

Recall ;

Speed = Distance / time

Speed = 5888.9727 / 10

Speed = 588.897 m/s

Speed = 588.9 m/s

7 0

3 years ago

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.600 rev

natulia [17]

Answer:

The final velocity $\omega_f = 0.4235 \frac{rev}{s}$

Explanation:

Given data

Mass of merry go round $M_m$ = 120 kg

Radius = 1.8 m

Initial angular velocity $\omega_i$ = 0.6 $\frac{rev}{sec}$

Mass of boy $M_{boy}$ = 25 kg

We know that the final velocity is given by

$\omega_f = \frac{\frac{1}{2}M_m \omega_i }{M_{boy} + \frac{1}{2} M_m }$

Put all the values in above formula we get

$\omega_f = \frac{\frac{1}{2}(120) 0.6}{25 + \frac{1}{2} (120) }$

$\omega_f = 0.4235 \frac{rev}{s}$

This is the final velocity.

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2 years ago

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Dmitriy789 [7]

Answer:

its the ultraviolet lights

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2 years ago