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2 years ago

8

A SMART car can accelerate from rest to a speed of 28 m/s in 20s. What distance does it travel in this time?

1 answer:

sashaice [31]2 years ago

6 0

Speed= distance/ time so distance = speed * time= 28 * 20= 560m .. so the answer is 560m.. l hope it helped :)

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A baton twirler is twirling her aluminum baton in a horizontal circle at a rate of 2.33 revolutions per second. A baton held hor

Nata [24]

Answer:

Explanation:

Given that;

horizontal circle at a rate of 2.33 revolutions per second

the magnetic field of the Earth is 0.500 gauss

the baton is 60.1 cm in length.

the magnetic field is oriented at 14.42°

we wil get the area due to rotation of radius of baton is

$\Delta A = \frac{1}{2} \Delta \theta R^2$

The formula for the induced emf is

$E = \frac{\Delta \phi}{\Delta t}$

$\phi = \texttt {magnetic flux}$

$E=\frac{\Delta (BA) }{\Delta t}$

$=B\frac{\Delta A}{\Delta t}$

B is the magnetic field strength

substitute

$\texttt {substitute}\ \frac{1}{2} \Delta \theta R^2 \ \ for \Delta A$

$E=B\frac{(\Delta \theta R^3/2)}{\Delta t} \\\\=\frac{1}{2} BR^2\omega$

The magnetic field of the earth is oriented at 14.42

$\omega =2.33\\\\L=60.1c,\\\\\theta=14.42\\\\B=0.5$

we plug in the values in the equation above

so, the induce EMF will be

$E=\frac{1}{2} \times (B\sin \theta)R^2\omega\\\\E=\frac{1}{2} \times (B\sin \theta)(\frac{L}{2} )\omega$

$=\frac{1}{2} \times0.5gauss\times\frac{0.0001T}{1gauss} \times\sin 14.42\times(\frac{60.1\times10^-^2m}{2} )^2(2.33rev/s)(\frac{2\pi rad}{1rev} )\\\\=2.5\times10^-^5\times0.2490\times0.0903\times14.63982\\\\=2.5\times10^-^5\times0.32917\\\\=8.229\times10^-^6V$

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2 years ago

Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

worty [1.4K]

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2 years ago

Read 2 more answers

A. In one short sentence, explain why we call the force of gravity an attractive force.

kondor19780726 [428]

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G $\frac{m1m2}{r}$

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

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3 years ago

A luge and its rider, with a total mass of 85 kg, emerge from adownhill track onto a horizontal straight track with an initial s

Verdich [7]

<span>(1) </span>Through the Second Law of motion, the equation for Force is:

F = m x a

Where m is mass and a is acceleration (deceleration)

<span>(2) </span>Distance is calculated through the equation,

D = Vi^2 / 2a

Where Vi is initial velocity

<span>(3) </span>Work is calculated through the equation,

W = F x D

Substituting the known values,

Part A:

<span>(1) </span> F = (85 kg)(2 m/s^2) = 170 N

<span>(2) </span> D = (37 m/s)^2 / (2)(2 m/s^2) = 9.25 m

<span>(3) </span> W = (170 N)(9.25 m) = 1572.5 J

Part B:

<span>(1) </span> F = (85 kg)(4 m/s^2) = 340 N

<span>(2) </span>D = (37 m/s)^2 / (2)(4 m/s^2) = 4.625 m

<span>(3) </span><span> W = (340 N)(4.625 m) = 1572.5 J</span>

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3 years ago

A migrating robin flies due north with a speed of 12 m/s relative to the air. The air moves due east with a speed of 6.7 m/s rel

mafiozo [28]

Here it is given that speed of migrating Robin is 12 m/s relative to air

so we can say that

$\vec v_{ra} = 12 m/s$ North

so it will be

Let North direction is along Y axis and East direction is along X axis

$\vec v_{ra} = 12\hat j$

also it is given that speed of air is 6.7 m/s relative to ground

$\vec v_a = 6.7 \hat i$

now as we know by the concept of relative motion

$\vec v_{ab} = \vec v_a - \vec v_b$

$\vec v_{ra} = \vec v_r - \vec v_a$

now by rearranging the terms

$\vec v_r = \vec v_{ra} + \vec v_a$

$\vec v_r = 12 \hat j + 6.7 \hat i$

now we need to find the speed of Robin which means we need to find the magnitude of its velocity which we found above

So here we will say

$v_r = \sqrt{12^2 + 6.7^2}$

$v_r = 13.7 m/s$

so the net speed of Robin with respect to ground will be 13.7 m/s

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3 years ago