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d1i1m1o1n [39]
3 years ago
5

A race car, traveling at constant speed, makes one lap around a circular track of radius 200 m. When the car has traveled halfwa

y around the track, what is the magnitude of the displacement from the starting point?
Physics
1 answer:
mario62 [17]3 years ago
8 0

The magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

Answer:

Explanation:

Since the race track is a circular track, the distance for one lap will be equal to the circumference of the circular track. And the circumference will be equal to the circumference of the circle.

Since the radius of the track is given as 200 m, then the circumference of the circular track will be

Circumference = 2πr = 2 × 3.14 × 200

So the circumference of the circular track = 1256 m.

So the starting point or position of the track is considered as zero and if the car has traveled half way means, the car has covered half of the circumference of the track.

As the circumference = 1256 m, then half of the circumference of the circle = 1256/2 = 256 m.

So the displacement is the measure of difference between the final position and initial position. As here the initial position is zero and the final position is the halfway around the track which is equal to 256 m.

Then Displacement = Final-Initial = 256-0= 256 m.

So the magnitude of the displacement of the car from the starting point to halfway around the track is 256 m.

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3 years ago
Electromagnetic radiation of 5.16Ă—1016 Hz frequency is applied on a metal surface and caused electron emission. Determine the w
IgorC [24]

Answer:

Work function of the metal, W_o=3.38\times 10^{-17}\ J

Explanation:

We are given that  

Frequency of the electromagnetic radiation,  f=5.16\times 10^{16} Hz

The maximum kinetic energy of the emitted electron, K=4.04\times 10^{-19}\ J

We need to find the work function of the metal.

We know that the maximum kinetic energy of ejected electron

K=h\nu-w_o

Where h=Plank's constant=6.63\times 10^{-34} J.s

\nu =Frequency of light source

w_o=Work function

Substitute the values in the given formula  

Then, the work function of the metal is given by :

W_o=h\nu -K

W_o=6.63\times 10^{-34}\times 5.16\times 10^{16}-4.04\times 10^{-19}

W_o=3.38\times 10^{-17}\ J

So, the work function of the metal is 3.38\times 10^{-17}\ J. Hence, this is the required solution.

8 0
3 years ago
A
zhuklara [117]

Hello! For this excersice let's applicate the formula:

\boxed{\boxed{d=vt}}

Data:

d = Distance = 150 m

v = Velocity = ¿?

t = Time = 0,9 s

Now, let's replace according the formula:

\boxed{150 \ m = v * 0,9 \ s}

  • Clear v:

\boxed{\dfrac{150 \ m}{0,9 \ s} = v}

  • Resolving:

\boxed{166,67 \ m/s = v}

The velocity is <u>166,67 meters per second.</u>

7 0
3 years ago
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