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Masja [62]
3 years ago
11

A 615.00 kg race car is uniformly traveling around a circular race track. It takes the race car 20.00 seconds to do one lap arou

nd the 80.00 m radius circular track. What is the frequency of the race car's circular motion?
Physics
1 answer:
koban [17]3 years ago
3 0

Answer:

The value is  f  =  0.05 \ Hz

Explanation:

From the question we are told that

     The mass of the car is  m  =  615 \  kg

      The period of the circular motion is  T  =  20 \  s

      The radius  is r =  80 \  m/s

Generally the frequency of the circular motion is  

       f  =   \frac{1}{T }

=>    f  =   \frac{1}{ 20  }

=>    f  =  0.05 \ Hz

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6 0
3 years ago
Which of the following best demonstrates the effect of static friction? A. A person pushing a couch and it slowly sliding across
Allisa [31]
<span> B. A person moving a ball through a stream of water</span>
5 0
2 years ago
A speedboat increases its speed from 14.5 m/s to 29.3 m/s in a distance of 172 m.
dalvyx [7]

Answer:

a. Acceleration, a = 1.88 m/s²

b. Time, t = 7.87 seconds.

Explanation:

Given the following data;

Initial velocity, U = 14.5m/s

Final velocity, V = 29.3m/s

Distance, S = 172m

a. To find the acceleration of the speedboat;

We would use the third equation of motion;

V² = U² + 2aS

Substituting into the formula

29.3² = 14.5² + 2a*172

858.49 = 210.25 + 344a

344a = 858.49 - 210.25

344a = 648.24

a = 648.24/344

Acceleration, a = 1.88 m/s²

b. To find the time;

We would use the first equation of motion;

V = U + at

29.3 = 14.5 + 1.88t

1.88t = 29.3 - 14.5

1.88t = 14.8

Time, t = 14.8/1.88

Time, t = 7.87 seconds.

6 0
3 years ago
A worker is thinking about two ways to get a box up 1.2 m onto a loading dock. He can use a force of 250 N to lift it straight u
harina [27]

Answer:

<em>The second option has a lower power output. P=30 W</em>

Explanation:

<u>Mechanical Power </u>

It is a physical magnitude that measures the rate a work W is done over time t.

\displaystyle P=\frac{W}{t}

Since W=F.d

\displaystyle P=\frac{F.d}{t}

The first option means the worker will lift the box by a distance of 1.2 meters in 3 seconds by applying 250 N of force. That produces a power of

\displaystyle P=\frac{(250). (1.2)}{3}=100\ Watt

The second option requires the worker applies 75 N of force and travel a distance of 4 meters for 10 seconds, thus the power is

\displaystyle P=\frac{(75). (4)}{10}=30\ Watt

The second option has a lower power output

7 0
3 years ago
Acceleration changes and objects...
brilliants [131]

Answer:

speed and direction

Explanation:

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2 years ago
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