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nikklg [1K]
3 years ago
13

A man is pushing a heavy crate up an inclined plane (ramp) into the back of semi trailer. What can the man do to make it easier

to get the crate up the ramp?
A) Use gloves.
B) Make the ramp longer.
C) Make the ramp shorter.
D) Pull it instead of push it.
Physics
2 answers:
strojnjashka [21]3 years ago
6 0
Make the ramp longer

sertanlavr [38]3 years ago
5 0
C. Make the ramp shorter
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Once broken into parts curved motion can be worked as ________________ problems along both axes.
vovikov84 [41]

Answer:

projectile motion

Explanation:

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Ray Of Light [21]

The rotational equilibrium condition allows finding the response to the minimum force of the wind and what happens when changing the water for sand, in the system

  a) The minimum force of the wind that turns the system is Fw = 17.64 N

  b) The system resists much greater forces because the base has more mass

 Newton's Second Law can be applied to rotational motion in this case when the angular acceleration is zero we have the special case of rotational equilibrium

               Σ τ = 0

Where τ is the torque  

The reference system is a coordinate system with respect to which the torques are measured, in this case we will fix the system at the turning point, the junction of the base and the pole, we will assume that the counterclockwise rotations are positive.

For the torque the distance used is the perpendicular distance from the direction of the force to the axis of rotation, let's find this distance for each force

Wind force

         cos 15 = \frac{y_w}{2.35}

         y_w = 2.35 cos 15

Post Weight

        sin 15 = \frac{x_p}{2.00}

         xp = 2.0 sin 15

Base weight

         cos (90-15) = \frac{x_b}{0.25}

         xB = 0.25 cos 75

Let's substitute in the rotational equilibrium equation

     

          F_w \ y_w  + W_p \ x_p - W_b \ x_b = 0

a) To calculate the minimum wind force we substitute the given values

They indicate the weight of the post is W_p = 26.0 N and the weight of the base with water is W_b = 810 N

     F_w = \frac{W_b \ x_b - W_p \ x_p }{y_w}

     F_w = \frac{W_b \ 0.25 cos75 \ - W_p \ 2 sin 15}{2.35 cos 15}

       

Let's  calculate

     F_w = \frac{810 \ 0.25 \ cos75 \ - 26.0 \ 2 \ sin 15}{2.35 cos15}\\F_w = \frac{52.41 - 10.30}{2.3699}

     F_w = 17.64 N

b) The water is exchanged for sand.

In this case, as the density of the sand is greater than that of the water, the base will have more weight, so it will resist stronger winds before turning over.

Using the rotational equilibrium condition we can find the response to the minimum force of the wind and what happens when changing the water for sand,

  a) the minimum force of the wind that turns the system is Fw = 17.64 N

  b) the system resists much greater forces because the base has more mass

Learn more  here: brainly.com/question/7031958

3 0
3 years ago
Q
GalinKa [24]

Answer:

B

Explanation:

It would be diffrent if on a downward slope but assuming your going straight it would be the smallest student.

3 0
3 years ago
What would be the path of an object thrown in the air if there was no gravity?
alex41 [277]

Answer:D) in a straight line.

Hey

Newton's first law says that if an object is at rest it will stay at rest. But if it is moving it will continue moving in a straight line if there is no external force. If there where no gravity the object that you throw will keep going in a straight line according to Newton's first law.

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3 years ago
Read 2 more answers
When a charged capacitor is connected to earth, it discharges and the voltage V across it will decrease according to the equatio
algol [13]
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

For convenience write RC=D and solve the ODE dV/dt=-V/D which gives 
<span>V=Ae^(-t/D) by separating the variables. </span>
<span>When t=0, V=10 and this gives A=10, </span>
<span>when t=2, V=1 giving 1=10e^(-2/D) and you can find D by taking lns of </span>
<span>each side. D=RC, so R=D/C </span>
<span>You should get 2.90X10^5 ohms</span>
4 0
4 years ago
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