Explanation:
When m=<em>mass</em>
G=<em>a</em><em>c</em><em>c</em><em>e</em><em>l</em><em>e</em><em>r</em><em>a</em><em>t</em><em>i</em><em>o</em><em>n</em><em> </em><em>d</em><em>u</em><em>e</em><em> </em><em>t</em><em>o</em><em> </em><em>gravity</em>
<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>H</em><em>=</em><em>h</em><em>e</em><em>i</em><em>g</em><em>h</em><em>t</em>
<em>U</em><em>s</em><em>i</em><em>n</em><em>g</em><em> </em><em>f</em><em>o</em><em>r</em><em>m</em><em>u</em><em>l</em><em>a</em>
<em>M</em><em>g</em><em>h</em>
<em>(</em><em>M</em><em>=</em><em>6</em><em>, </em><em>g</em><em>=</em><em>10</em><em>,</em><em>h</em><em>=</em><em>?</em><em>) </em>
6×10×h
=60joules
Answer:
Explanation:
A proton of charge
q=+1.609×10^-19C
Orbit a radius of 12cm
r=0.12m
Magnetic Field of 0.31T
Angle between velocity and field is 90°
a. Because the magnetic force F supplies the centripetal force Fc.
The magnitude of the magnetic force F on a charge q moving at a speed v in a magnetic field of strength B is given by
F = qvB sin θ
And the centripetal force is given as
Fc=mv²/r
Where m is mass of proton
m=1.673×10^-27kg
Then, F=Fc
qvB sin θ=mv²/r
qBSin90=mv/r
rqB=mv
Then, v=rqB/m
v=0.12×1.609×10^-19×0.31/1.673×10^-23
v=3577692.78m/s
v=3.58×10^6m/s
b. Since,
F=qVBSin90
F=1.609×10^-19×3.58×10^6×0.31
F=1.785×10^-13 N.
Answer:
V=2.8 ml
Explanation:
volume of the cube is it would be 20.3 - 17.5 ml so 2.8 ml.
Answer:
Part a: When the road is level, the minimum stopping sight distance is 563.36 ft.
Part b: When the road has a maximum grade of 4%, the minimum stopping sight distance is 528.19 ft.
Explanation:
Part a
When Road is Level
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is this case is 0 as the road is level
Substituting values

So the minimum stopping sight distance is 563.36 ft.
Part b
When Road has a maximum grade of 4%
The stopping sight distance is given as

Here
- SSD is the stopping sight distance which is to be calculated.
- u is the speed which is given as 60 mi/hr
- t is the perception-reaction time given as 2.5 sec.
- a/g is the ratio of deceleration of the body w.r.t gravitational acceleration, it is estimated as 0.35.
- G is the grade of the road, which is given as 4% now this can be either downgrade or upgrade
For upgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% upgrade is 528.19 ft.</em>
For downgrade of 4%, Substituting values

<em>So the minimum stopping sight distance for a road with 4% downgrade is 607.59 ft.</em>
As the minimum distance is required for the 4% grade road, so the solution is 528.19 ft.
Answer:
Agriculture has been the cause of significant modification of landscapes throughout the world. Tillage of land changes the infiltration and runoff characteristics of the land surface, which affects recharge to ground water, delivery of water and sediment to surface-water bodies, and evapotranspiration.
Explanation: