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3 years ago

10

Every year, new records in track and field events are recorded. Let's take an historic look back at some exciting races.

1 answer:

Hitman42 [59]3 years ago

4 0

First we need to turn Aouita's time for the race into seconds. There are 60 seconds in a minute, so 7 minutes and 29.45 seconds is (7 x 60) + 29.45 = 449.45. He ran 3000 meters in that time, so his average speed was 3000 meters divided by 449.45 seconds. 3000 / 449.45 = 6.67 m/s. So, on average, he covered 6.67 meters (more than 21 feet!) during each second of the race.

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liraira [26]

This affirmative is false

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4 years ago

Read 2 more answers

"A steel rotating-beam test specimen has an ultimate strength of 120 kpsi. Estimate the life of the specimen if it is tested at

MrRissso [65]

Answer:

life (N) of the specimen is 117000 cycles

Explanation:

given data

ultimate strength Su = 120 kpsi

stress amplitude σa = 70 kpsi

solution

we first calculate the endurance limit of specimen Se i.e

Se = 0.5× Su .............1

Se = 0.5 × 120

Se = 60 kpsi

and we know strength of friction f = 0.82

and we take endurance limit Se is = 60 kpsi

so here coefficient value (a) will be

a = $\frac{(f\times Su)^2}{Se}$ ......................1

put here value and we get

a = $\frac{(0.82\times 120)^2}{60}$

a = 161.4 kpsi

so coefficient value (b) will be

b = $-\frac{1}{3}log\frac{(f\times Su)}{Se}$

b = $-\frac{1}{3}log\frac{(0.82\times 120)}{60}$

b = −0.0716

so here number of cycle N will be

N = $(\frac{ \sigma a}{a})^{1/b}$

put here value and we get

N = $(\frac{ 70}{161.4})^{1/-0.0716}$

N = 117000

so life (N) of the specimen is 117000 cycles

7 0

3 years ago

A closely wound rectangular coil of 75.0 turns has dimensions of 22.0 cm by 45.0 cm . The plane of the coil is rotated from a po

pentagon [3]

Answer:51.52 V

Explanation:

Given

N=75 turns

$Area=22\times 45 cm^2$

$\theta =36^{\circ}$

$Emf induced =-N\frac{\mathrm{d}\phi }{\mathrm{d} t}$

$Emf=-75A\frac{B_2-B_1}{t}$

$Emf=-75\times 22\times 45\times 10^{-4}\frac{1(1-cos(90-36))}{6\times 10^{-2}}$

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8 0

3 years ago

Solve the problem. Unless stated otherwise, assume that the projectile flight is ideal, that the launch angle is measured from t

kirill115 [55]

Answer:

d = 185.26 meters

Explanation:

It is given that,

Launching angle of the projectile, $\theta=26^{\circ}$

Initial speed of the projectile, u = 48 m/s

Let at distance d the projectile hits the ground from the launch point. It is equal to range of the projectile. Its formula is given by :

$d=\dfrac{u^2\ sin2\theta}{g}$

Substituting all the values in above formula. So, we get :

$d=\dfrac{(48)^2\ sin2(26)}{9.8}$

d = 185.26 meters

So, the distance between the launch point and the point where it hits is 185.26 meters. Hence, this is the required solution.

3 0

3 years ago

an object with a mass of 3.2 kg has a force of 7.3 newtons applied to it. what is the resulting acceleration of the object?

ch4aika [34]

Answer:

2.28m/s^2

Explanation:

F=ma

a=F/m

a=7.3/3.2

a=2.28m/s^2

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3 years ago