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pickupchik [31]
2 years ago
6

In 1993, Ileana Salvador of Italy walked 3.0 km in under 12.0 min. Suppose that during 115 m of her walk Salvador is observed to

steadily increase her speed from 4.20 m/s to 5.00 m/s. How long does this increase in speed take?
Physics
1 answer:
dexar [7]2 years ago
6 0

Answer:

t = 25 seconds

Explanation:

Given that,

Distance, d = 115 m

Initial speed, u = 4.2 m/s

Final speed, v = 5 m/s

We need to find the time taken in increasing the speed.

We know that,

Acceleration, a=\dfrac{v-u}{t} ....(1)

The third equation of kinematics is as follows :

v^2-u^2=2ad\\\\\text{Put the value of a in above equation}\\\\v^2-u^2=2\times \dfrac{v-u}{t}\times d\\\\\because (a^2-b^2)=(a-b)(a+b)\\\\(v-u)(v+u)=\dfrac{2\times (v-u)d}{t}\\\\t=\dfrac{2d}{v+u}\\\\\text{Putting all the values}\\\\t=\dfrac{2\times 115}{4.2+5}\\\\t=25\ s

Hence, it will take 25 seconds to increase the speed.

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2 years ago
Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
nalin [4]

Answer:

t = 1.02 s

Explanation:

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v = u - at

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so,

0 = 2 - 1.962 × t

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5 0
2 years ago
Sandy is on a road trip. She leaves at 8:00 AM. It takes her 2 hours to drive 200 kilometers. She stops at a rest stop for half
lutik1710 [3]
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v= \frac{S}{t}

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t=2h+0.5h+1.5 h=4 h
Therefore, the average velocity is 
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2 years ago
A mercury atom in the ground state absorbs 20.00 electronvolts of energy and is ionized by losing an electron. How much kinetic
sukhopar [10]
We are given a mercury atom in the ground state which absorbs 20 eV of energy. It is then ionized by losing an electron. We need to calculate the kinetic energy that the electron has after ionization.

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To determine the kinetic energy, we can use this equation:

KE = 20 Joules / Coulombs * </span>1.60217662 × 10-19<span> coulombs 
KE = 1.25x10^20 Joules 

Therefore, the amount of kinetic energy that the electron has after ionization is </span>1.25x10^20 Joules or 1.25x10^17 kJ. <span />
8 0
2 years ago
A source charge generates an electric field of 4286 N/C at a distance of 2.5 m. What is the magnitude of the source charge?
mrs_skeptik [129]
The Answer is 3.0uc. I took the quiz.
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