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2 years ago

A sprinter accelerates from rest to a velocity of 12m/s in the first 6 seconds of the 100 meter dash .

Physics

1 answer:

GREYUIT [131]2 years ago

5 0

Answer:

a) 36 m

b) 64 m

Explanation:

Given:

v₀ = 0 m/2

v = 12 m/s

t = 6 s

Find: Δx

Δx = ½ (v + v₀) t

Δx = ½ (12 m/s + 0 m/s) (6 s)

Δx = 36 m

The track is 100 m, so the sprinter still has to run another 64 m.

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Answer:

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Explanation:

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1 year ago

Exercise 2.4.6: Suppose you wish to measure the friction a mass of 0.1 kg experiences as it slides along a floor (you wish to fi

JulijaS [17]

Answer:

b = 0.6487 kg / s

Explanation:

In an oscillatory motion, friction is proportional to speed,

fr = - b v

where b is the coefficient of friction

when solving the equation the angular velocity has the form

w² = k / m - (b / 2m)²

In this exercise we are given the angular velocity w = 1Hz, the mass of the body m = 0.1 kg, and the spring constant k = 5 N / m. Therefore we can disperse the coefficient of friction

let's call

w₀² = k / m

w² = w₀² - b² / 4m²

b² = (w₀² -w²) 4 m²

Let's find the angular velocities

w₀² = 5 / 0.1

w₀² = 50

w = 2π f

w = 2π 1

w = 6.2832 rad / s

we subtitute

b² = (50 - 6.2832²) 4 0.1²

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b = 0.6487 kg / s

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1 year ago

A red ball, initially at rest, is simultaneously hit by a blue ball traveling from west to east at 5 m/s and a green ball travel

andrezito [222]

Answer:

C. Between North and West

Explanation:

Since all have equal masses and the red ball and green ball are moving in south and east direction, the blue ball would most likely be moving between the north and West direction.

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2 years ago

A coaxial cable consists of a thin insulated straight wire carrying a current of 2.00 A surrounded by a cylindrical conductor ca

ella [17]

Answer:

B = 15μT

Explanation:

In order to calculate the magnitude of the magnetic field generated by the coaxial cable you use the Ampere's law, which is given by:

$B=\frac{\mu_oI}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

I: current

r: distance from the wire to the point in which B is calculated

In this case you have two currents with opposite directions, which also generates magnetic opposite magnetic fields. Then, you have (but only for r > radius of the cylindrical conductor) the following equation:

$B_T=B_1-B_2=\frac{\mu_o I_1}{2\pi r}-\frac{\mu_o I_2}{2\pi r}\\\\B_T=\frac{\mu_o}{2\pi r}(I_1-I_2)$ (2)

I1: current of the central wire = 2.00A

I2: current of the cylindrical conductor = 3.50A

r: distance = 2.00 cm = 0.02 m

You replace the values of all parameters in the equation (2), and you use the absolute value because you need the magnitude of B, not its direction.

$|B|=|\frac{4\pi*10^{-7}T/A}{2\pi (0.02m)}(2.00A-3.50A)|=1.5*10^{-5}T\\\\|B|=15*10^{-6}T=15\mu T$

The agnitude of the magnetic field outside the coaxial cable, at a distance of 2.00cm to the center of the cable is 15μT

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