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Mariulka [41]
1 year ago
8

What is the acceleration of a 2000 kg truck that has an engine exert 400 N of force?

Physics
1 answer:
amid [387]1 year ago
6 0

The acceleration of a 2000 kg truck that has an engine exerts 400 N of force <u>at 0.2 m/s².</u>

<u />

Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.

Calculation:-

mass = 2000 kg

Force = 400 N

acceleration = F/mass

                      =  400/2000

                      = <u>0.2 m/s²</u>

<u />

Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.

Learn more about acceleration here:- brainly.com/question/29110429

#SPJ9

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A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
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Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

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  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

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MaRussiya [10]

Answer:

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