A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to
2 answers:
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Answer:
i) The motion of both particles are shown on the same speed-time curve included
ii) Approximately 19.5 seconds
Explanation:
We are given that;
Initial velocity of particle, P = 140 m/s
Start time of particle P = 6 s before start time of particle Q
Position of particle Q when velocity is 25 m/s = 125 m
Therefore, from the equation of motion, we have for particle Q;
v² = u² + 2·a·s
Where:
v = Final velocity = 25 m/s
u = Initial velocity = 0 m/s
a = Acceleration
s = Distance covered = 125 m
Therefore;
25² = 0² + 2×a×125
Which gives a = 25²/(2×125) = 2.5 m/s²
The time taken for particle Q to reach 125 m is found from the relation;
s = u·t + 1/2·a·t²
Where:
t = Time of journey
Therefore;
125 = 0×t + 1/2×2.5×t²
Which gives 125 = 1.25 × t²
Hence, t² = 125/1.25 = 100
t = √(100) = 10 s
The equation for particle Q is v = 0 + 2.5×t
Hence, since particle P starts deceleration 6 seconds before the commencement of motion of particle Q, the amount of seconds after the commencement of deceleration of the first particle P that it takes for particle P to come to rest is found as follows;
Hence, at t = 6 + 10 = 16 seconds particle P speed = 25 m/s
From the equation of motion, for particle P (decelerating) we have
v = u - a·t
Where:
v = 25 m/s
u = 140 m/s
t = 16 s
Hence, 25 = 140 - a×16
∴ 16·a = 140 - 25 = 115
a = 115/16 = 7.1875 m/s²
Therefore, the time it takes before particle P comes to rest is found from the same equation of motion, where v = 0 as follows;
v = u - a·t
0 = 140 - 7.1875 × t
∴7.1875·t = 140
t = 140/7.1875 = 19.48 s ≈ 19.5 seconds.
Answer:
a)6.34 x W/m²
b)1.37 x W/m²
c) see explanation.
Explanation:
a)The relation of intensity'I' of the radiation and area 'A' is given by:
I= P/A
where P= power of sunlight i.e 3.9 x J
and the area of the sun is given by,
A= 4π => 4π
A=6.15 x m²
= 3.9 x
/ 6.15 x
=><u> 6.34 x </u>
<u>W/m²</u>
b) First determine the are of the sphere in order to determine intensity at the surface of the virtual space
A= 4π
Now R= 1.5 x m
A= 4π x 1.5 x =>2.83 x
m²
The power that each square meter of Earths surface receives
= 3.9 x
/2.83 x
=><u>1.37 x </u>
<u> W/m²</u>
<u />
c) in part (b), by assuming the shape of the wavefront of the light emitted by the Sun is a spherical shape so each point has the same distance from the source i.e sun on the wavefront.