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3 years ago

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A runner taking part in the 200-m dash must run around the end of a track that has a circular arc with a radius of curvature of

29.5 m. The runner starts the race at a constant speed. If she completes the 200-m dash in 24.4 s and runs at constant speed throughout the race, what is her centripetal acceleration as she runs the curved portion of the track

1 answer:

Leto [7]3 years ago

5 0

Answer: $2.27\ m/s^2$

Explanation:

Given

Length of the race track $L=200\ m$

the radius of curvature of the track $r=29.5\ m$

time taken to run on track is $t=24.4\ s$

Speed of runner is

$\Rightarrow v=\dfrac{L}{t}=\dfrac{200}{24.4}\\\\\Rightarrow v=8.196\ m/s$

Centripetal acceleration is

$\Rightarrow a_c=\dfrac{v^2}{r}=\dfrac{8.196^2}{29.5}\\\\\Rightarrow a_c=2.27\ m/s^2$

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The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

$f = \dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{1}{1.55}$

$f=0.645\ m$

$f=64.5\ cm$

We need to calculate the image distance

Using lens formula

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}$

Put the value into the formula

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2 years ago

When a golf ball is hit, it travels at 41 meters per second. The mass of a golf ball is 0.045kg. Calculate the kinetic energy of

Fittoniya [83]

Answer:

75.645 J

Explanation:

The kinetic energy is related to the mass and velocity by the formula ...

KE = 1/2mv²

For the given mass of 0.045 kg, and velocity of 41 m/s, the kinetic energy is ...

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The unit of energy, joule, is a derived unit equal to 1 kg·m²/s².

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2 years ago

A 13-kg sled is moving at a speed of 3.0 m/s. At which of the following speeds will the sled have twice as much kinetic energy?

AysviL [449]

K.E. = 1/2 mv²
K.E. is directly proportional to v^2
So, when K.E. increase by 2, K.E. increase by root. 2
v' = 1.41v
original v value was 3 so, final would be:
v' = 1.41*3 = 4.23
After round-off to it's tenth value, it will be:
v' = 4.2

So, option B is your answer!

Hope this helps!

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3 years ago

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

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$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

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$v_{2}^2=\dfrac{c^2}{8.93}$

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3 years ago

the Earth exerts a gravitational force of 500 newtons on an object what is the mass of the object in kilograms

MAXImum [283]

Use Newton's second law F = mass * acceleration
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2 years ago