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ser-zykov [4K]
3 years ago
8

A car with an initial velocity of O m/s and a mass of 1500 kg reaches a velocity of 15 m/s in 5

Physics
2 answers:
umka21 [38]3 years ago
6 0

The acceleration of the car is 3 m / s^{2}. And the force required to reach the acceleration is 4500 N.

<u>Explanation:</u>

Equations of Motion describe different relations between motion parameters. One of them is,

                    v_{f}=v_{i}+a t

And, F=m \times a , from Newton’s laws

So, given,

Initial velocity of the car v_{i} = 0  

Mass of the car, m = 1500 kg

Velocity acquired v_{f} = 15 m/s

Time taken to acquire velocity t = 5 seconds.

Now, since  v_{f}=v_{i}+a t, To find the acceleration, rearrange the above equation, we get as follows,

               a=\frac{\left(v_{f}-v_{i}\right)}{t}

Then, by applying the given data in above equation, we get

                a=\frac{(15-0)}{5}=\frac{15}{5}=3 m / s^{2}

Now, find the force to reach 3 m / s^{2} acceleration as follows,

                F=m \times a=1500 \times 3=4500 N

Flura [38]3 years ago
3 0

Answer:the answer can be find out by the following steps;

Explanation:

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Answer:

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Explanation:

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x1=d1*cos(60), y1=d1*sin(60)\\\\ x2=x1 , y2=y1-d2\\\\ x3=x2-d3*cos(30) , y3=y2+d3*sin(30)

Using the Pitagoras theorem, the distance from (x3,y3) to the start point can be calculated as:

d=\sqrt{x3^{2} +y3^{2} }

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A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho
creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano (m) = 190 kg

Inclined angle (\theta) = 18 degree

Considering gravity, g = 9.8 ms^-^2

And

Using, sin(18) =0.30 and cos(18)=0.95

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

a.

When the man pushes it parallel to the incline.

Balancing the forces as  \sum F=0 .

⇒ F+mgsin(\theta) =0

⇒ F=-mgsin(\theta)

⇒ Here it is negative as the force is acting downward.

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b.

When the force is parallel to the floor.

⇒ Fcos(\theta)=mgsin(\theta)

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⇒ Plugging the values.

⇒ F=\frac{190\times 9.8\times sin(18)}{cos(18)}

⇒ F=605 N

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The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

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