The force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.
<h3>What is force?</h3>
Force is defined as the product of mass and acceleration of an object.
Friction is defined as the force that resists the movement of an object over another.
Therefore, the force exerted on the tires of a car that directly accelerate it along a road is exerted by the road friction.
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The car has an initial velocity
of 23 m/s and a final velocity
of 0 m/s. Recall that for constant acceleration,

The car stops in 7 s, so the acceleration is


The crate’s acceleration is gravity= 9.81m/s^2
The ball’s acceleration is also gravity=9.81m/s^2
Answer:
Explanation:
Inital KE = (1/2) m v^2 = (1/2) * 1500 * 50^2 = 1,875,000 J
Final KE = (1/2) * 1500 * 100^2 = 7,500,000 J
But ,
4 * 1875000 = 7500000
so the KE has increased by 4 times.
The free-body diagram of the forces acting on the flag is in the picture in attachment.
We have: the weight, downward, with magnitude

the force of the wind F, acting horizontally, with intensity

and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):


By dividing the second equation by the first one, we get

From which we find

which is the angle of the rope with respect to the horizontal.
By replacing this value into the first equation, we can also find the tension of the rope: