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jeyben [28]
3 years ago
7

Draw a net force arrow on the picture below. What is the net force? State the direction.

Physics
2 answers:
zalisa [80]3 years ago
6 0

Answer:

Net force=0 N

Explanation:

The drawing will look like the attached image, with no net force on the object. We stablished a reference system located in the center of the body, with + axis take right and upward for x and y. All forces are on the x axis, there is one positive force of 7 N (pointing right) and to negative forces of -4 and -3 N (pointing left) if we add the vectors in the x axis we have that

+7-4-3=0 N

The net force is then 0 newton and the drawing has no net force vector then

exis [7]3 years ago
5 0

Answer: No arrow; no net force.

Explanation: -7 + 7= 0  same thing here

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Answer:

The following factors could be responsible for size of an Impact Crater:

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5 0
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Two bodies made of the same material have same external dimension and appearance, but one is compact solid and other is hollow.
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Answer:

Different

Explanation:

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3 years ago
What makes the planets gravity?
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2 years ago
A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.
wlad13 [49]

Answer:

12m/s

Explanation:

v_f=v_o+at

Let's call the velocity that the car maintains for 10 seconds v_f_1, and the final velocity v_f_2.

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Hope this helps!

5 0
3 years ago
A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels
Crank

Answer:

v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }

Explanation:

The average velocity is total displacement divided by time:

v_{avg} =\dfrac{D_{tot}}{t}

And in the case of vertical v_{avg}

v_{avg}=\dfrac{y_{tot}}{t}

where y_{tot} is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height H with initial velocity v_0 is given by:

y=H+v_0t-\dfrac{1}{2} gt^2

The time it takes for the rock to reach maximum height is when y'(t)=0, and it is

t=\frac{v_0}{g}

The vertical distance it would have traveled in that time is

y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2

y_{max}=\dfrac{2gH+v_0^2}{2g}

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

y_{down}=\dfrac{2gH+v_0^2}{2g}+H

Therefore, the total displacement throughout the rock's journey is

y_{tot}=y_{max}+y_{down}

y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H

\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

t=\dfrac{v_0}{g},

and it should take the rock the same amount of time to return to the roof, and it takes another t_0 to go from the roof of the building to the ground; therefore,

t_{tot}=2\dfrac{v_0}{g}+t_0

where t_0 is the time it takes the rock to go from the roof of the building to the ground, and it is given by

H=v_0t_0+\dfrac{1}{2}gt_0^2

we solve for t_0 using the quadratic formula and take the positive value to get:

t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

Therefore the total time is

t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH}  }{g}

\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH}  }{g}}

Now the average velocity is

v_{avg}=\dfrac{y_{tot}}{t}

v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }

\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }

5 0
3 years ago
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