<em><u>1.car</u></em><em><u> </u></em><em><u>towing</u></em>
<em><u>2.pulling</u></em><em><u> </u></em><em><u>bucket</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>water</u></em>
<em><u>3.gym</u></em><em><u> </u></em><em><u>equipment</u></em><em><u> </u></em>
<em><u>4.crane</u></em><em><u> </u></em><em><u>machine</u></em>
<em><u>5.tug</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>war</u></em>
Answer:
A 75.1 N and a direction of 152° to the vertical.
B 85.0 N at 0° to the vertical.
Explanation:
A) The interaction partner of this normal force has what magnitude and direction?
The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>
B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?
Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.
Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.
<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>
In a slide, you’re potential energy (sitting on top of a slide) is transformed into kinetic energy when you slide down and then goes back to potential energy when you reach the bottom at a stop. No energy was destroyed nor created, just transformed from one state to another
Answer:
A. Using
Pgauge= Pmanometer
And we know that
Pgauge= deta(hpg)
So deta h = Pgauge/density x g
So
= 7.5(6894.76/1psig)/ 13.6*10^3*9.8m/s²)
= 387.9mm
So to find height of pipe connected to the pipe we say
= h -deta h
= 900-387.97mm
=512.02mm
B. We use manometry principle
Pgas+density xg(25*10^3)-PX density{h-(H-0.25)=0
So
Finally Pgas= 6.54psig