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3 years ago

A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Physics

1 answer:

Pie3 years ago

6 0

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

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A 60kg student traveling in a 1000kg car with a constant velocity has a kinetic energy of 1.2 x 10^4 J. What is the speedometer

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Answer:

17.64 km/h

Explanation:

mass of car, m = 1000 kg

Kinetic energy of car, K = 1.2 x 10^4 J

Let the speed of car is v.

Use the formula for kinetic energy.

$K = \frac{1}{2}mv^{2}$

By substituting the values

$1.2\times 10^{4} = \frac{1}{2}\times 1000\times v^{2}$

v = 4.9 m/s

Now convert metre per second into km / h

We know that

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So, $v = \left (\frac{4.9}{1000} \right )\times \left ( \frac{3600}{1} \right )$

v = 17.64 km/h

Thus, the reading of speedometer is 17.64 km/h.

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3 years ago

What happens to the resistance of a wire as it gets wider

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it snaps

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3 years ago

A chef sanitized a thermometer probe and then checked the temperature of minestrone soup being held in a hot-holding unit. the t

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Answer: Option (A) is the correct answer.

Explanation:

A corrective action is defined as the action with the help of which a person can avoid a difficulty or problem that he/she was facing earlier.

For example, when the chef checked the temperature of soup using thermometer then it was 120 but his operation's critical limit was 135.

So, to avoid this problem he heated the soup to 165 at 15 seconds following which he got the result as desired.

Therefore, reheating the soup was his corrective action.

Thus, we can conclude that reheating the soup was the corrective action.

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3 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

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3 years ago