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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 0
3 years ago
Read 2 more answers
A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe
ikadub [295]

Answer:

f_{fr}=1590.85 N

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r} (1)

  • f(fr) is the friction force
  • N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

Ncos(25)-mg-f_{fr}sin(25)=0 (2)          

Let's combine (1) and (2) to find f(fr)

f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}

f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}  

f_{fr}=1590.85 N

I hope it helps you!

5 0
3 years ago
Consider a rectangular ice floe 5.00 m high, 4.00 m long, and 3.00 m wide. a) What percentage of the ice floe is below the water
artcher [175]

Answer:

(a) 92 %

(b) 6.76 %

Explanation:

length, l = 4 m, height, h = 5 m, width, w = 3 m, density of water = 1000 kg/m^3

density of ice = 920 kg/m^3, density of mercury = 13600 kg/m^3

(a) Let v be the volume of ice below water surface.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of water x g = Total volume of ice block x density

                                                                      of ice x g

v x 1000 x g = V x 920 x g

v / V = 0.92

% of volume immersed in water = v/V x 100 = 0.92 x 100 = 92 %

(b) Let v be the volume of ice below the mercury.

By the principle of flotation

Buoyant force = weight of ice block

Volume immersed x density of mercury x g = Total volume of ice block x  

                                                                      density of ice x g

v x 13600 x g = V x 920 x g

v / V = 0.0676

% of volume immersed in water = v/V x 100 = 0.0676 x 100 = 6.76 %

4 0
3 years ago
Superconductors have no measurable resistance. true or false?
trapecia [35]
True, they have zero electrical resistance.
7 0
3 years ago
A solid cylinder of mass 7 kg and radius 0.9 m starts from rest at the top of a 20º incline. It is released and rolls without sl
ohaa [14]

Answer:

157.8 J

Explanation:

m = mass of the cylinder = 7 kg

h = height difference in top and bottom of the incline = 2.3 m

g = acceleration due to gravity = 9.8 m/s²

TE = Total Energy at the bottom

PE = Gravitational potential energy at the top

Using conservation of energy

Total Energy at the bottom = Gravitational potential energy at the top  

TE = PE

TE = m g h

TE = (7) (9.8) (2.3)

TE = 157.8 J

7 0
3 years ago
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