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2 years ago

What are the types of waves

1 answer:

8 0

Waves come in two kinds, longitudinal and transverse. Transverse waves are like those on water, with the surface going up and down, and longitudinal waves are like of those of sound, consisting of alternating compressions and rarefactions in a medium

i hope this helps :)

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A 495-kg dragster accelerates from rest to a final speed of 105 m/s in 395 m, during which it encounters an average frictional f

drek231 [11]

Answer:

Explanation:

According to energy conservation which states that the workdone is equal to change in the system

Workdone = change in kinetic energy + (frictional force * distance)

Workdone = ΔK + fd

Workdone = kf-Ki + fd

Workdone = = 1/2(m(v-u)^2) + fd

Given

Mass m = 495kg

final velocity v = 105m/s

initial velocity = 0m/s

Force f= 1400N

distance d = 395m

Substitute

Workdone = 1/2(495(105-0)^2) + 1400(395)

Workdone = 2,728,687.5+553000

Workdone = 3,281,687.5 Joules

Time = 8.2secs

Power output = Workdone/Time

Power output = 3,281,687.5/8.2

Power output = 885,766.768

Power output = 8.858 * 10^5 watts

3 0

4 years ago

Which statement about force is incorrect

garik1379 [7]

Answer:

What are the options?

Explanation:

4 0

3 years ago

A Plane has a takeoff speed of 150 m/s and requires 1500m to reach that speed. Determine the acceleration of the plane and the t

mars1129 [50]

<u>Answer:</u>

The acceleration of the plane and the time required to reach this speed is (a)= 7.5 $m/sec^2$ and time(t) = 20 seconds

<u>Explanation: </u>

Given data Initial velocity $(V_i)$ = 0

Final velocity ( $V_f$ ) = 150 m/second

Distance (d) = 1500 m

We have the formula, $\mathrm{V}_{\mathrm{f}}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 \mathrm{ad}$

which gives $150^2$ = 0+2a(1500)

22500 = 3000 a

acceleration (a) = 7.5 $m/s^2$

$\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at}$

150 = 7.5 t

t= 150/7.5 = 20

t = 20 seconds.

5 0

3 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0

4 years ago

2. Vehicle-braking distance is the distance your vehicle travels after you see a problem

babunello [35]

The answer is B. False

8 0

3 years ago

Read 2 more answers