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love history [14]
3 years ago
6

A 2200 kilogram car is accelerating at 3.4 m/s/s. what is the NET force?

Physics
1 answer:
sdas [7]3 years ago
7 0
We Know, F = m*a
F = 2200 * 3.4
F = 7480 Kg m/s²

So, your final answer is 7480
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Use the magnitude acceleration formula .

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3 years ago
If he leaves the ramp with a speed of 31.0 m/s and has a speed of 29.5 m/s at the top of his trajectory, determine his maximum h
raketka [301]

Answer:

The maximum height reached is 4.63 m.

Explanation:

Given:

Initial speed of the man (u) = 31.0 m/s

Speed at the top of trajectory (u_x) = 29.5 m/s

Acceleration due to gravity (g) = 9.8 m/s²

When the man reaches the top of the trajectory, the vertical component of velocity becomes zero and hence only horizontal component of velocity acts on him.

Also, since there is no net force acting in the horizontal direction, the acceleration is zero in the horizontal direction from Newton's second law. Thus, the horizontal component of velocity always remains the same.

So, speed at the top of trajectory is nothing but the horizontal component of initial velocity.

Now, initial velocity can be rewritten in terms of its components as:

u^2=u_x^2+u_y^2

Where, u_x\ and\ u_y are the initial horizontal and vertical velocities of the man.

Now, plug in the given values and simplify. This gives,

(31.0)^2=(29.5)^2+u_y^2\\\\961=870.25+u_y^2\\\\u_y^2=961-870.25\\\\u_y^2=90.75\ m^2/s^2--------1

Now, we know that, for a projectile motion, the maximum height is given as:

H=\frac{u_y^2}{2g}

Plug in the value from equation (1) and 9.8 for 'g' to solve for 'H'. This gives,

H=\frac{90.75}{2\times 9.8}\\\\H=4.63\ m

Therefore, the maximum height reached is 4.63 m.

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3 years ago
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Good morning dear...

Have a beautiful and joyful day ahead.

6 0
2 years ago
Bus starts from rest if the acceleration of the bus is 0.5 MS squared what will be the velocity at the end of two minutes and wh
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Explanation:

Given that,

Initial speed of the bus, u = 0

Acceleration of the bus, a = 0.5 m/s²

Let v is the velocity at the end of 2 minutes. The change in velocity divided by time equals acceleration.

So,

a=\dfrac{v-u}{t}\\\\v=u+at\\\\v=0+0.5\times 120\\\\v=60\ m/s

Let d is the distance cover during that time. So,

v^2-u^2=2ad\\\\d=\dfrac{v^2-u^2}{2a}\\\\d=\dfrac{(60)^2}{2\times 0.5}\\\\d=3600\ m

So, the final speed is 60 m/s and the distance covered during that time is 3600 m.

4 0
3 years ago
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