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3 years ago

What type of relationship exists between acceleration and mass?

1 answer:

7 0

Here, you can derive that by numerical method, as follows:
F = m.a
m = F/a

So, here we can see when we decrease one, other increase by same effect; we can say they are "Indirectly Proportional" to each other!

Hope this helps!

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Melting, freezing, and boiling are______ changes

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Melting, boiling, and freezing are state changes!
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3 years ago

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At which point is cool air circulating beneath warm air?

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A train moves with a constant velocity for 15s for 155m. How fast is the train moving?

bonufazy [111]

Answer: 10.3m/s

Explanation:

In theory and for a constant velocity the physics expression states that:

Eq(1): distance = velocity times time <=> d = v*t for v=constant.

If we solve Eq (1) for the velocity (v) we obtain:

Eq(2): velocity = distance divided by time <=> v = d/t

Substituting the known values for t=15s and d=155m we get:

v = 155 / 15 <=> v = 10.3

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3 years ago

Gina is driving her car down the street. She has a teddy bear sitting on the back seat. A dog runs in front of Gina's car, so sh

ser-zykov [4K]

Teddyber continue to move forward because Newton law 1. moving object continue to move until something external make it to stop. no seat belt on teddy ber so only dashboard can make her stop. same if people in car and no seatbelt.

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3 years ago

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The y-position of a damped oscillator as a function of time is shown in the figure.

NISA [10]

(1) The period of the oscillator is 1 second.

(2) The damping coefficient is 0.93.

<h3>What is period of oscillation?</h3>

The period of oscillation is the time taken to make one complete cycle.

From the graph, the time taken to make one complete oscillation is 1 second.

<h3>Damping coefficient</h3>

equation of the wave is given as;

y(t) = Ae^(-btx) cos(ωt)

3.5 = Ae^(-0) cos(0)

3.5 = A x 1

A = 3.5 cm

-3 = 3.5e^(-bx) cos(ω)

-3/3.5 = e^(-bx) cos(ω)

-0.857 = e^(-bx) cos(ω)

-0.857 / cos(ω) = e^(-bx)

ln[-0.857 / cos(ω)] = -bx

ln[-0.857 / cos(ω)] / b = - x ---- (1)

-2 = 3.5e^(-2bx) cos(2ω)

-0.57 = e^(-2bx) cos(2ω)

ln[-0.57 / cos(2ω)] = -2bx

ln[-0.57 / cos(2ω)] /2b = - x ------(2)

solve (1) and (2)

ln[-0.57 / cos(2ω)]/2b = ln[-0.857 / cos(ω)] /b

-0.57 / cos(ω) = 2(-0.857 / cos(ω))

2(-0.857/cosω) = -0.57/cos2ω

-(2 x 0.857) / (-0.57) = cosω/cos 2ω

3 = cosω/cos 2ω

3(cos 2ω) = cosω

3(2cos²ω - 1) = cos ω

6cos²ω - 6 = cosω

6cos²ω - cosω - 6 = 0

let cosω = y

6y² - y - 6 = 0

solve the quadratic equation;

y = 1.1 or -0.92

cosω = -0.92

ω = arc cos(-0.92)

ω = 2.74 rad/s

From equation (1)

ln[-0.857 / cos(ω)] / x = -b ---- (1)

let x = 1

ln(-0.857/cos(2.74) = -b

-0.93 = -b

b = 0.93

Thus, the damping coefficient is 0.93.

Learn more about damping coefficient here: brainly.com/question/14058210

#SPJ1

4 0

2 years ago