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erastova [34]
3 years ago
11

A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to t

he northernmost point.
-What distance does the car travel? (In km)
-What is the car's displacement from its original position? (In km)
Physics
1 answer:
elixir [45]3 years ago
8 0

Answer:

(A) Distance will be equal to 1.75 km

(B) Displacement will be equal to 1.114 km

Explanation:

We have given circumference of the circular track = 3.5 km

Circumference is given by 2\pi r=3.5

r = 0.557 km

(a) It is given that car travels from southernmost point to the northernmost point.

For this car have to travel the distance equal to semi perimeter of the circular track

So distance will be equal to =\frac{3.5}{2}=1.75km

(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track

So displacement will be equal to d = 2×0.557 = 1.114 m

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In Applied Life Data Analysis (Wiley, 1982), Wayne Nelson presents the breakdown time of an insulating fluid between electrodes
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Answer:

18.375

Explanation:

Mean = total of measured values / no of elements of sample

Total = .15 + .89 +1.11+1.46 +2.78 + 3.12 + 4.30 + 4.59 + 4.92 + 6.42 + 7.20+8.04+8.21+12.13+31.86+32.53+33.82+36.60+ 72.99

= 273.03

Mean = 273 .03 / 19

= 14.37

( .15 - 14.37 )² +( 0.89 - 14.37 )²+( 1.11- 14.37 )²+( 1.46 - 14.37 )²+( 2.78 - 14.37 )²+( 3.12 - 14.37 )²+( 4.30 - 14.37 )²+( 4.59 - 14.37 )²+( 4.92 - 14.37 )²+( 6.42 - 14.37 )²+( 7.20 - 14.37 )²+( 8.04 - 14.37 )²+( 8.21 - 14.37 )²+( 12.13 - 14.37 )²+( 31.86 - 14.37 )²+( 32.53 - 14.37 )²+( 33.82 - 14.37 )²+( 36.60 - 14.37 )²+( 72.99 - 14.37 )²

= 202.20 +181.71+ 175.82 + 166.66+ 134.32 + 126.56 + 101.40 +95.64 + 89.30 +63.20 +51.40 +40.06 +37.94+5.01 +305.90 +329.78 +378.30 +494.17 +3436.30

=6415.67

Standard deviation =

\sqrt{\frac{6415.67}{19} }

= 18.375

5 0
3 years ago
A 450-N rightward force is used to drag a large box across the floor with a constant velocity of 1.2 m/s. The coefficient of fri
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Answer:

the mass of the box is 51.98 kg.

Explanation:

Given;

applied horizontal force, F = 450 N

coefficient of friction, μ = 0.795

constant velocity, v = 1.2 m/s

At constant velocity, the acceleration of the object is zero and the net force will be zero.

F_{Net} = F - F_k\\\\0 = F - F_k\\\\F_k = F\\\\\mu \ N = F\\\\\mu (mg) = F\\\\m = \frac{F}{\mu  g} \\\\m = \frac{405}{0.795 \ \times \ 9.8} \\\\m = 51.98 \ kg

Therefore, the mass of the box is 51.98 kg.

8 0
3 years ago
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UkoKoshka [18]

Answer:

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5 0
3 years ago
During the spin cycle of your clothes washer, the tub rotates at a steady angular velocity of 33.3 rad/s. Find the angular displ
Monica [59]

Answer:

The angular displacement of the tub will be 2580.75 rad or 411 revolutions.

Explanation:

We know the relation between angular velocity (\omega), the angular displacement (\theta) and time (t) is given by

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Given \omega = rad~s^{-1} and the time (t) required to complete a spin is 77.5 s.

Therefore, the required angular displacment (\theta) is

&& \theta = \omega \times t\\&or,& \theta = 33.3~rad~s^{-1} \times 77.5~s = 2580.75~rad

Also we know that,

&& 2 \pi~rad = 1~revolution\\&or,& 580.75~rad  = \dfrac{2580.75}{2~\pi} \approx 411~revolutions

So, the angular displacement of the tub will be 2580.75 rad or 411 revolutions.

3 0
3 years ago
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