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weeeeeb [17]
3 years ago
10

A car is traveling at a speed of 38.0 m/s on an interstate highway where the speed limit is 75.0 mi/h. Is the driver exceeding t

he speed limit?
Physics
1 answer:
lidiya [134]3 years ago
6 0

Answer:

Yes, the car driver is exceeding the given limit.

Explanation:

<u>Given:</u>

  • Speed of the car, v = 38.0 m/s.
  • Speed limit of the highway, \rm v_o=75.0\ mi/h.

<h2><u>Converting the speed limit from mi/h to m/s:</u></h2>

We know,

1 mi = 1.60934 km.

1 km = 1000 m.

Therefore, 1 mi = 1.60934 × 1000 m = 1609.34 m.

1 hour = 60 minutes.

1 minute = 60 seconds.

Therefore, 1 hour = 60 × 60 seconds = 3600 seconds.

Using these values,

\rm 1\ \dfrac{mi}{h}=\dfrac{1609.34\ m}{3600\ s}=0.447\ m/s.

Therefore,

\rm v_o = 75.0\ mi/h=75.0\times 0.447=33.52\ m/s.

Clearly,

\rm v_o

which means, the car driver is exceeding the given speed limit.

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. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio
Reil [10]

Answer:

Please see below as the answer is self-explanatory.

Explanation:

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6 0
3 years ago
A cannonball is fired at a 45.0° angle and an initial velocity of 670 m/s. Assume no air resistance. What is the vertical compon
elixir [45]

Answer:

<h3>473.8 m/s; 473.8 m/s</h3>

Explanation:

Given the initial velocity U = 670m/s

Horizontal velocity Ux = Ucos theta

Vertical component of the cannon velocity Uy = Usin theta

Given

U = 670m/s

theta = 45°

horizontal component of the cannonball’s velocity = 670 cos 45

horizontal component of the cannonball’s velocity = 670(0.7071)

horizontal component of the cannonball’s velocity = 473.757m/s

Vertical component of the cannonball’s velocity = 670 sin 45

Vertical component of the cannonball’s velocity  = 670 (0.7071)

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2 years ago
A cross-country skier can complete a 7.5 km race in 45 min. What is the skier’s average speed?
kap26 [50]

Answer:

Speed = 10 km/hr

Explanation:

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Therefore,

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<u></u>

<u>Hope it helps!</u>

3 0
3 years ago
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