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>
The uses of fossil fuel causes:
- <em>A</em><em>l</em><em>l</em><em> </em><em>o</em><em>f</em><em> </em><em>t</em><em>h</em><em>e</em><em> </em><em>a</em><em>b</em><em>o</em><em>v</em><em>e</em>
<u>F</u><u>o</u><u>s</u><u>s</u><u>i</u><u>l</u><u> </u><u>f</u><u>u</u><u>e</u><u>l</u><u>s</u><u> </u><u>b</u><u>u</u><u>r</u><u>n</u><u> </u><u>a</u><u>n</u><u>d</u><u> </u><u>i</u><u>n</u><u>c</u><u>r</u><u>e</u><u>a</u><u>s</u><u>e</u><u>s</u><u> </u><u>i</u><u>n</u><u> </u><u>g</u><u>l</u><u>o</u><u>b</u><u>a</u><u>l</u><u> </u><u>t</u><u>e</u><u>m</u><u>p</u><u>e</u><u>r</u><u>a</u><u>t</u><u>u</u><u>r</u><u>e</u><u> </u><u>b</u><u>e</u><u>c</u><u>a</u><u>u</u><u>s</u><u>e</u><u> </u><u>i</u><u>t</u><u> </u><u>e</u><u>m</u><u>i</u><u>t</u><u>s</u><u> </u><u>g</u><u>r</u><u>e</u><u>e</u><u>n</u><u>h</u><u>o</u><u>u</u><u>s</u><u>e</u><u> </u><u>g</u><u>a</u><u>s</u><u>e</u><u>s</u><u>.</u><u> </u><u>A</u><u>n</u><u>d</u><u> </u><u>t</u><u>h</u><u>e</u><u> </u><u>e</u><u>x</u><u>t</u><u>r</u><u>e</u><u>m</u><u>e</u><u> </u><u>c</u><u>l</u><u>i</u><u>m</u><u>a</u><u>t</u><u>e</u><u> </u><u>c</u><u>h</u><u>a</u><u>n</u><u>g</u><u>e</u><u> </u><u>d</u><u>e</u><u>c</u><u>r</u><u>e</u><u>a</u><u>s</u><u>e</u><u>s</u><u> </u><u>b</u><u>i</u><u>o</u><u>d</u><u>i</u><u>v</u><u>e</u><u>r</u><u>s</u><u>i</u><u>t</u><u>y</u><u> </u><u>and</u><u> </u><u>last</u><u> </u><u>but</u><u> </u><u>not</u><u> </u><u>the</u><u> </u><u>least</u><u> </u><u>it</u><u> </u><u>increases</u><u> </u><u>pollution</u><u> </u><u>because</u><u> </u><u>burning</u><u> </u><u>of</u><u> </u><u>fuels</u><u> </u><u>emit</u><u> </u><u>gases</u><u> </u><u>like</u><u> </u><u>Carbon-dioxide</u><u> </u><u>etc</u><u>.</u><u>.</u><u>.</u><u>~</u>
The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.
To solve this problem it is necessary to apply the concepts related to energy conservation.
In this case the kinetic energy is given as

Where,
m = mass
v= Velocity
In the case of heat lost energy (for all 4 wheels) we have to

m = mass
Specific Heat
= Change at temperature
For conservation we have to





Therefore the temperature rises in each of the four brake drums around to 47°C