. If the centripetal and thus frictional force between the tires and the roadbed of a car moving in a circularpath were reduced, what would happen?

Answers

Answer 1

Answer: In this particular case, where car is moving through curvature, so it is moving in circular motion, force acting on car is centripetal force which holds car not to fly out. Centripetal force is always pointed in the middle of circle. Here frictional force has role of centripetal force. If frictional force is to weak, car would fly out of curvutare.

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Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversion is Juan most likely doing? degrees Celsius to degrees Fahrenheit degrees Celsius to kelvins kelvins to degrees Celsius degrees Fahrenheit to degrees Celsius

Answers

Answer:

degrees Celsius to kelvins

Explanation:

A person walks first at a constant speed of 5.00 m/s along a straight line from point A to point B and then back along the line from B to A at a constant speed of 2.90 m/s. What is the average speed over the entire trip?

Answers

The Average Speed over the entire trip is 3.671 meters per second.

The person is moving in a straight line at a constant Speed. Let suppose that Distance between points A and B is $r$ , then we construct and use the Kinematic formulas for each stage of the travel to determine Times and Average Speed, based on the fact that Speed is inversely proportional to Time, we derive the resulting expression:

First Stage (from point A to point B):

$t_(1) = (r)/(v_(1))$ (1)

Second Stage (from point B to point A):

$t_(2) = (r)/(v_(2))$ (2)

Average Speed:

$t_(1)+t_(2) = (2\cdot r)/(\bar v)$ (3)

Where:

$t_(1)$ - Travelling time for the first stage, in seconds.

$t_(2)$ - Travelling time for the second stage, in seconds.

$r$ - Distance from A to B, in meters.

$v_(1)$ - Speed of the person in the first stage, in meters per second.

$v_(2)$ - Speed of the person in the second stage, in meters per second.

$\bar v$ - Average speed, in meters per second.

By applying (1) and (2) in (3), we derive an expression to determine the Average Speed:

$(r)/(v_(1)) + (r)/(v_(2)) = (2\cdot r)/(\bar v)$

$(v_(2)+v_(1))/(v_(1)\cdot v_(2)) = (2)/(\bar v)$

$\bar v = (2\cdot (v_(1)\cdot v_(2)))/(v_(2)+v_(1))$ (4)

If we know that $v_(1) = 5\,(m)/(s)$ and $v_(2) = 2.90\,(m)/(s)$ , the average speed over the entire trip is:

$\bar v = (2\cdot (v_(1)\cdot v_(2)))/(v_(2)+v_(1))$

$\bar v = (2\cdot \left(5\,(m)/(s) \right)\cdot \left(2.90\,(m)/(s) \right))/(2.90\,(m)/(s) + 5\,(m)/(s) )$

$\bar v = 3.671\,(m)/(s)$

The Average Speed over the entire trip is 3.671 meters per second.

Please see this question related to Average Speed for further details: brainly.com/question/19335778

This problem is a very interesting problem.
we know wacth:

Speed = distance / time.

Consider that:
distance from point A to point B=distance from point B to point A=d

We calculate the time to go from point A to point B.
time=distance / speed.
T₁=d / (5 m/s)

We calculate the time to go from point B to point A.
T₂=d / (2.9 m/s)

Therefore; the total time wil be: T₁+T₂
Total time=d/(5 m/s) + d / (2.9 m/s)
Least common multiple=(5 m/s)(2.9 m / s)=14.5 m²/s²
Total time=(2.9 m/s d + 5 m/s d)/ 14.5m²/s²
Total time=[(7.9 d) m/s] / (14.5 m²/s²)
Total time=7.9 d/ (14.5 m/s)

Now, the total distance will be =d+d=2d

Therefore:
Average speed=total distance / total time
Average speed=[(2d)] / [(7.9 d)/ (14.5 m/s)]
Average speed≈3.67 m/s.

Answer: the average speed will be ≈3.67 m/s.

A strip of land conneted to the shore, but then is eroded away by waves. it a column of land is left behind, it is called a

Answers

A stack or sea stack is a geological landform consisting of a steep and often vertical column or columns of rock in the sea near a coast, formed by wave erosion. Stacks are formed over time by wind and water, processes of coastal geomorphology.

It is called a Sea Stack.

You and a friend ride bicycles to school. Both of you start at the same instant from your house, you riding at 13 m/s and your friend riding at 18 m/s . During the trip your friend has a flat tire that takes him 13 min to fix. He then continues the trip at the same speed of 18 m/s . If the distance to school is 13 km , which of you gets to school first? By comparing the total journey time for you and your friend, this question can be answered.

Answers

V1 = 13 m/s
V2 = 18 m/s
L = 13 km = 13000 m
t0 = 13 min = 780 s

t1 = L/V1 = 13000/13 = 1000 seconds = 16.67 min
t2 = t0 + L/V2 = 780 + 13000/18 = 1500 s = 25 min
You will get to school first

How much force is needed to accelerate a 1000-kg car at a rate of 3 m/s2

Answers

In the given question we already know the mass of the car. The mass of the car is 1000 kg. The acceleration at which the car needs to reach is 3 m/s^2. Based on these information's, the required answer can be easily deduced. We already know that
Force = Mass * Acceleration
= (1000 * 3) newton
= 3000 newton
So the amount of force that is required to move give the car an acceleration of 3m/s^2 is 3000 Newton.

Britain and France signed an entente and became thea.
Central Forces.
c.
Central Powers.
b.
Allied Forces.
d.
Axis Powers.

Answers

im pretty sure that it is allied forces

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