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Lady_Fox [76]
2 years ago
10

Young athlete has a mass of 42 kg one day there is no wind shear and hundred metre race in 14.2 second a sketch graph not in ske

tch showing her speed during the race is
icalculate the acceleration of the athlete during the first 3 seconds of the race
ii the escalating for of the athlete during the first 3 seconds of the race
iii the with whicj she crosses the finish line.​

Physics
2 answers:
Fudgin [204]2 years ago
8 0

Answer:

I don't get the question

Dmitrij [34]2 years ago
8 0

Answer:

ubc gu

Explanation:

b  ,yfhvc bnly

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The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61
nikitadnepr [17]

Answer:

(a) Angle of incidence for violet is more than the angle of incidence for red

(b) 2.4°

Explanation:

refractive index for violet , v = 1.66

refractive index for red, nR = 1.61

wavelength for violet, λv = 400 nm

wavelength for red, λR = 700 nm

Angle of refraction, r = 30°

(a) Let iv be the angle of incidence for violet.

Use Snell,s law

nv = Sin iv / Sin r

1.66 = Sin iv / Sin 30

Sin iv = 0.83

iv = 56°

Use Snell's law for red

nR  = Sin iR / Sin r  

where, iR be the angle of incidence for red

1.61 = Sin iR / Sin 30

Sin iR = 0.805

iR = 53.6°

So, the angle of incidence for violet is more than red.

(b) iv - iR = 56° - 53.6° = 2.4°

4 0
1 year ago
In the swing carousel amusement park ride, riders sit in chairs that are attached by a chain to a large rotating drum as shown i
irinina [24]

Answer:\theta =44.068^{\circ}

Explanation:

Given

time taken to complete the circle=7.9 s

radius of circle(r)=15 m

velocity of rider is given by =\frac{2\pi r}{t}

v=\frac{2\pi 15}{7.9}=11.93 m/s

Let us suppose T is the tension in the chain and \thetais the angle which chain makes with vertical

Therefore T\sin \theta =\frac{mv^2}{r}-1

T\cos \theta=mg --2

Divide 1 & 2 we get

tan\theta =\frac{v^2}{rg}

tan\theta =0.968

\theta =44.068^{\circ}

8 0
2 years ago
A solenoid that is 35 cm long and contains 450 circular coils 2.0 cm in diameter carries a 1.75-A current. (a) What is the magne
Taya2010 [7]

Answer:

Explanation:

a )  No of turns per metre

n = 450 / .35

= 1285.71

Magnetic field inside the solenoid

B = μ₀ n I

Where I is current

B = 4π x 10⁻⁷ x 1285.71 x 1.75

= 28.26 x 10⁻⁴ T

This is the uniform magnetic field inside the solenoid.

b )

Magnetic field around a very long wire at a distance d is given by the expression

B = ( μ₀ /4π ) X 2I / d

= 10⁻⁷ x 2 x ( 1.75 / .01 )

= .35 x 10⁻⁴ T

In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .

6 0
2 years ago
Read 2 more answers
An infinite conducting cylindrical shell of outer radius r1 = 0.10 m and inner radius r2 = 0.08 m initially carries a surface ch
gayaneshka [121]

Answer:

Explanation:

Solution is in the picture attached

8 0
2 years ago
A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
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