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den301095 [7]
3 years ago
10

A 782.10 kg car is brought from 7.60 m/s to 3.61 m/s over a time period of 4.23 seconds. What is the average force the car exper

iences during that time?
Physics
1 answer:
alexandr402 [8]3 years ago
4 0

Answer:

–735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

Explanation:

The following data were obtained from the question:

Mass (m) of car = 782.10 kg

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Force (F) =?

Next, we shall determine the acceleration of the car. This can be obtained as follow:

Initial velocity (u) = 7.60 m/s

Final velocity (v) = 3.61 m/s

Time (t) = 4.23 s

Acceleration (a) =?

a = (v – u) / t

a = (3.61 – 7.60) / 4.23

a = –3.99 / 4.23

a = –0.94 m/s²

Finally, we shall determine the force experienced by the car as shown below:

Mass (m) of car = 782.10 kg

Acceleration (a) = –0.94 m/s²

Force (F) =?

F = ma

F = 782.10 × –0.94

F = –735.17 N

The negative sign indicate that the force is acting in opposition direction to the car.

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Juli2301 [7.4K]

Answer:

showm

Explanation:

Consider a dipole having magnetic moment 'm' is placed in magnetic field \vec{B} then the torque exerted by the field on the dipole is

\tau = m\times B

\tau=mBsin\alpha

Now to rotate the dipole in the field to its final position the work required to be done is

U=\int \tau d\alpha

U=\int mBsin\alpha d\alpha

U= -mBcos\alpha

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Minimum energy mB is for the case when m is anti parallel to B.

Minimum energy -mB is for the case when m is parallel to B.

5 0
3 years ago
Marissaâs car accelerates uniformly at a rate of +2.60 m/s². How long does it take for Marissaâs car to accelerate from a speed
DIA [1.3K]

Answer:

The time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds.

Explanation:

Given that,

Acceleration of the car, a=+2.6\ m/s^2

Initial speed of the car, u = 24.6 m/s

Final speed of the car, v = 26.8 m/s

We need to find the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s. The acceleration of an object is given by :

t=\dfrac{v-u}{a}

t=\dfrac{(26.8-24.6)\ m/s}{2.6\ s}

t = 0.84 seconds

So, the time taken by the car to accelerate from a speed of 24.6 m/s to a speed of 26.8 m/s is 0.84 seconds. Hence, this is the required solution.                                    

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The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}

Then, the final velocity will be

v^{2} = 2 * 3 * 3600 + 0 = 21600\\ \\v=\sqrt{21600}=146.969 m/s

Thus,  the final velocity of the truck is found as 146.969 m/s.

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