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lawyer [7]
3 years ago
7

A car rounds a flat curve and experiences a centripetal force directed toward the center of the curve and perpendicular to the d

irection of its motion. If the driver of the car doesn't hit the brake OR the gas, how will the curve affect the motion of the car?
Physics
2 answers:
lara31 [8.8K]3 years ago
8 0

Answer:

The answer is The car's speed will remain constant and its direction will change.

Explanation:

The answer is The car's speed will remain constant and its direction will change. The applied force is perpendicular to the direction of motion, so the speed of the car won't change. Going into a curve doesn't necessarily change the speed of the object.

Usimov [2.4K]3 years ago
3 0
<h2>Answer:</h2>

If a car is rounding a flat curve, it experiences a centripetal force that pulls it towards the center of the circle it is rotating in.

Now,

The centripetal force can be balanced by the centrifugal force caused due to the acceleration of the body at the high speed which counters the centripetal force and in turn <u>prevents the car from slipping down the curve.</u>

So,

If the car doesn't hit the gas then the <em><u>car will fall down from the curve</u></em> as the Centripetal force will exceed the Centrifugal force of the car.

However, if the car doesn't hit the brake then the <em><u>car will maintain it's position on the flat curve</u></em> track as the centrifugal force will counter the effect of centripetal force directed towards the center.

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A traditional unit of length in Japan is the ken (1 ken 1.97 m). What are the ratios of (a) square kens to square meters and (b)
Roman55 [17]
If 1 ken is 1.97 meter, then 1 square ken is 3.8809 square meters, and one cubic ken is 7.645373. As for the cylindrical tank, the volume of it would be 10.835 times the radius of the cylinder time 1.97^2 times pi. As you didn't specify the radius, I can't give the exact answer but that would be how to get it.
7 0
3 years ago
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A 50.0 kg driver is riding at 35.0 m/s in her red sports car when she must suddenly slam on the brakes to avoid
egoroff_w [7]

Answer:

3500N

Explanation:

Given parameters:

Mass of driver  = 50kg

Speed  = 35m/s

Time  = 0.5s

Unknown:

Average force the seat belt exerts on her = ?

Solution:

The average force the seat belt exerts on her can be deduced from Newton's second law of motion.

   F = mass x acceleration

So;

     F  = mass x  \frac{change in velocity }{time}

  F  = 50 x \frac{35}{0.5}    = 3500N

8 0
3 years ago
Formed through longshore drift<br> a. sea stack<br> b. sandbar<br> c. spit<br> d. headland
ANTONII [103]
<h3>Answer;</h3>

<em><u>Sand Spit or Spit </u></em>

<h3><u>Explanation;</u></h3>
  • <em><u>Long shore drift is the process that occurs when a sheet of water moves on and off the beach, in other words the swash and back swash</u></em>, thus capturing and transporting sediment on the beach back out to the sea.
  • <em><u>Sandbar</u></em> is normally formed when the sandspit stretches across a bay and connects the two sides. <em><u>Headland</u></em> is a high piece of land that extends out onto the sea. <em><u>Sea stacks </u></em>on the other hand results from the collapsing of the roof of the arch.
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3 years ago
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Does air resistance affect the motion of a falling object differently when the initial velocity of the object is greater?
DIA [1.3K]
Yes. Even greater. Air resistance or drag becomes harder the faster an object goes. This is why when cars reach their max speed they don't accelerate as fast, because they are pushing harder against the wind. If I take a tennis ball and shoot it down a bottomless pit, a 400 kph, the drag will slow the ball down till it reaches terminal velocity. 
4 0
2 years ago
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A child of mass 27 kg swings at the end of an elastic cord. At the bottom of the swing, the child's velocity is horizontal, and
snow_tiger [21]

Answer:

The magnitude of the rate of change of the child's momentum is 794.11 N.

Explanation:

Given that,

Mass of child = 27 kg

Speed of child in horizontal = 10 m/s

Length = 3.40 m

There is a rate of change of the perpendicular component of momentum.

Centripetal force acts always towards the center.

We need to calculate the magnitude of the rate of change of the child's momentum

Using formula of momentum

\dfrac{dp}{dt}=F

\dfrac{dP}{dt}=\dfrac{mv^2}{r}

Put the value into the formula

\dfrac{dP}{dt}=\dfrac{27\times10^2}{3.40}

\dfrac{dP}{dt}=794.11\ N

Hence, The magnitude of the rate of change of the child's momentum is 794.11 N.

7 0
3 years ago
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