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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:

r = 0.11 m

Explanation:

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<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

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