Answer:
ANULAR ECLIPSE
Explanation:
ANULAR ECLIPSE. Because the moon is very far, only a portion of the sun would be obscured, and then only the moon's outer ring will be viewable; this is called the anular eclipse.The ring of fire marks the maximum stage of an annular solar eclipse.
That is the Sigma Symbol. It’s the addition of a sequence of numbers; the result is the sum of the total. if numbers are added sequentially from left to right, any intermediate result is a partial sum, prefix sum, or running total of the summation
Answer:
t_{out} =
t_{in}, t_{out} = 
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is

The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D / 
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point

= D / 
D = v_{sg 2} t_{in}
with the distance is the same we can equalize

t_{out} = t_{in}
t_{out} =
t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D / 
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} = 
Answer:
(a) 3.807 s
(b) 145.581 m
Explanation:
Let Δt = t2 - t1 be the time it takes from the moment when the motorcycle starts to accelerate until it catches up with the car. We know that before the acceleration, both vehicles are travelling at a constant speed. So they would maintain a distance of 58 m prior to the acceleration.
The distance traveled by car after Δt (seconds) at
speed is

The distance traveled by the motorcycle after Δt (seconds) at
speed and acceleration of a = 8 m/s2 is


We know that the motorcycle catches up to the car after Δt, so it must have covered the distance that the car travels, plus their initial distance:





(b)


Answer:
Explanation:
Given that
The mass of the body is 0.04kg
M=0.04kg
The radius of the paths is 0.6m
r=0.6m
The normal force exerted at A is 3.9N
Fa=3.9N
The normal force exerted at B is 0.69N
Fb=0.69N
Then work done by friction from point A to B will be the change in K.E
W=∆K.E+P.E
So we need to know the velocity at both point A and B
Then since the centripetal force is given as
Ft=mv²/r
Then,
For point A
Fa=mv²/r
3.9=0.04v²/0.6
3.9=0.0667v²
v²=3.9/0.0667
v²=58.5
v=√58.5
v=7.65m/s
Va=7.65m/s
Now at point B
Fb=mv²/r
0.69=0.04v²/0.6
0.69=0.0667v²
v²=0.69/0.0667
v²=10.35
v=√10.35
v=3.22m/s
Vb=3.22m/s
Then, the work done is
W=∆K.E+P.E
P.E is given as mgh
The height will be 2R =1.2m
P.E=mgh
P.E=0.04×9.81×1.2
P.E=0.471J
Final kinetic energy at B minus initial kinetic energy at A
W=K.Eb-K.Ea
K.E is given as 1/2mv²
W=1/2m(Vb²-Va²) +P.E
W=0.5×0.04(3.22²-7.65²) +0.471
W=0.5×0.04×(-48.1541) +0.471
W=-0.96+0.471
W=-0.49J
work was done on the block by friction during the motion of the block from point A to point B is 0.49J.
Friction opposes motions and that is why the work done is negative