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Advocard [28]
3 years ago
5

A sprinter accelerates from rest to a top speed of 24 km/h in 2 seconds and then runs at a constant velocity for the rest of the

100 meter dash,
a- what was the distance he covered in the first 2 seconds

b- what was his time for the whole race
Physics
1 answer:
Tems11 [23]3 years ago
5 0

Answer:

a. 6.67 meters

b. 16 seconds

Explanation:

a. v=(24×1000)/3600=6.67 m/s

v=u-at

a=(v-u)/t=(6.67-0)/2=3.335 m/s^2

s=ut+(1/2) at^2 = 0×2+.5×3.335×(2)²=6.67 meters

b. After 2seconds, he ran at a constant velocity. From a, in 2 seconds, he crossed 6.67 meters.

So, distance left =(100-6.67)=93.33 meters

to cross 93.33 meters,he needed,

s=vt

93.33=6.67t

t=13.99≡14 seconds

So, Total (93.33+6.67)=100 meters were covered in (14+2)=16 seconds

Hope, this helps you.

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Answer:

It makes sense because on that the day  the sun stops moving northward and starts moving southward

Explanation:

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3 years ago
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amm1812
O.99 m long .simple pendulum time period is 2s for second formula then use formula T=2pi.rt(lenght/gravity)
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Describe each class of lever and explain to characteristics of each
Nataly [62]

-- Class I lever

The fulcrum is between the effort and the load.

The Mechanical Advantage can be anything, more or less than 1 .

Example:  a see-saw

-- Class II lever

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The Mechanical Advantage is always greater than 1 .

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-- Class III lever

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3 years ago
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A record player turntable initially rotating at 3313 rev/min is braked to a stop at a constant rotational acceleration. The turn
Rus_ich [418]

Answer:

(A) It will take 22 sec to come in rest

(b) Work done for coming in rest will be 0.2131 J              

Explanation:

We have given the player turntable initially rotating at speed of 33\frac{1}{3}rpm=33.333rpm=\frac{2\times 3.14\times 33.333}{60}=3.49rad/sec

Now speed is reduced by 75 %

So final speed \frac{3.49\times 75}{100}=2.6175rad/sec

Time t = 5.5 sec

From first equation of motion we know that '

\alpha =\frac{\omega -\omega _0}{t}=\frac{2.6175-3.49}{4}=-0.158rad/sec^2

(a) Now final velocity \omega =0rad/sec

So time t to come in rest  t=\frac{0-3.49}{-0.158}=22sec

(b) The work done in coming rest is given by

\frac{1}{2}I\left ( \omega ^2-\omega _0^2 \right )=\frac{1}{2}\times 0.035\times (0^2-3.49^2)=0.2131J

4 0
3 years ago
in an effort to reach the store before it closed the motorist increase the speed of his car from 20 m per second to 60 meters pe
Ber [7]

Answer:

10

Explanation:

Givens

vi = 20 m/s

vf = 60 m/s

t = 4 second.

Formula

a = (vf - vi) / t

a = (60 - 20)/4

a = 40 / 4

a = 10 m/s^2

7 0
2 years ago
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