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3 years ago

A car accelerates on a horizontal road due to the force exerted by :

1 answer:

8 0

I am absolutely sure that if a car accelerates on a horizontal road due to the force exerted by road. I consider the last option from the list represented above to be the correct one because the <span>road oses as a nice example of a frictional force. Hope you will agree with me and find it helpful! Regards!</span>

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Calculate the time needed for a 0.600 kg hammer to reach the surface of the Earth

USPshnik [31]

The time needed for the hammer to reach the surface of the Earth is 3.54 s.

<h3>Time of motion of the hammer</h3>

The time of motion is calculated as follows;

t = √(2h/g)

where;

h is height of fall
g is acceleration due to gravity

t = √(2 x 10 / 1.6)

t = 3.54 s

Thus, the time needed for the hammer to reach the surface of the Earth is 3.54 s.

Learn more about time of motion here: brainly.com/question/2364404

#SPJ1

8 0

2 years ago

Keeping the mass at 1.0 kg and the velocity at 10.0 m/s, record the magnitude of centripetal acceleration for each given radius

Paha777 [63]

Answer:

The centripetal acceleration for the first radius; 2.0 m = 50 m/s²

The centripetal acceleration for the second radius; 4.0 m = 25 m/s²

The centripetal acceleration for the third radius; 6.0 m = 16.67 m/s²

The centripetal acceleration for the fourth radius; 8.0 m = 12.5 m/s²

The centripetal acceleration for the fifth radius; 10.0 m = 10 m/s²

Explanation:

Given;

mass of the object, m = 1 kg

velocity of the object, v = 10 m/s

different values of the radius, 2.0 m 4.0 m 6.0 m 8.0 m 10.0 m

The centripetal acceleration for the first radius; 2.0 m

$a_c = \frac{v^2}{r} \\\\a_c_1= \frac{(10)^2}{2} \\\\a_c_1= 50 \ m/s^2$

The centripetal acceleration for the second radius; 4.0 m

$a_c_2= \frac{(10)^2}{4} \\\\a_c_2= 25 \ m/s^2$

The centripetal acceleration for the third radius; 6.0 m

$a_c_3= \frac{(10)^2}{6} \\\\a_c_3= 16.67 \ m/s^2$

The centripetal acceleration for the fourth radius; 8.0 m

$a_c_4= \frac{(10)^2}{8} \\\\a_c_4= 12.5 \ m/s^2$

The centripetal acceleration for the fifth radius; 10.0 m

$a_c_5= \frac{(10)^2}{10} \\\\a_c_5= 10 \ m/s^2$

6 0

3 years ago

A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 40N for 3.0s. What impulse acts on the boat

jarptica [38.1K]

Answer:

120 Ns

Explanation:

The impulse exerted on an object is given by:

$I=Ft$

where

F is the force applied

t is the time taken

In this problem, we have:

F = 40 N

t = 3.0 s

So, the impulse acting on the boat is

$I=Ft=(40 N)(3.0 s)=120 Ns$

8 0

3 years ago

Here is a graph of speed vs time. If the object is moving to the east, which BEST describes the speed and velocity of the graph?

san4es73 [151]

Answer: A JUST DID IT ON USA TESTPREP

Explanation:

5 0

3 years ago

Help pls it’s urgent giving brainiest!!

Lorico [155]

Answer:

2nd no. positive.... 3rd no. negitive......

3 0

2 years ago