Which layer of the sun's atmosphere are you looking at when you look at me in image of the Sun
1 answer:
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(a) +2.60 m/s
The motion of the fish dropped by the seagul is a projectile motion, which consists of two independent motions:
- a horizontal uniform motion, at constant speed
- a vertical motion, at constant acceleration (acceleration of gravity, , downward)
In this part we are only interested in the horizontal motion. As we said the horizontal component of the fish's velocity does not change, therefore its value when the fish reaches the ocean is equal to its initial value, which is the speed at which the seagull was flying (because it was flying horizontally):
(b) -17.2 m/s
The vertical component of the fish's velocity instead follows the equation:
where
is the initial vertical velocity, which is zero
is the acceleration of gravity
t is the time
Since the fish reaches the ocean at t = 1.75 s, we can substitute this time into the formula to find the final vertical velocity:
where the negative sign indicates the direction (downward).
(c)
The horizontal component of the fish's velocity would increase
The vertical component of the fish's velocity would stay the same.
As we said from part (a) and (b):
- The horizontal component of the fish's velocity is constant during the motion and it is equal to the initial velocity of the seagull -> so if the seagull's initial speed increases, the horizontal velocity of the fish will increase too
- The vertical component of the fish's velocity does not depend on the original speed of the seagull, therefore it is not affected.
Answer:
Explanation:
First of all we shall calculate electric field near charge 2Q .
electric field due to charge Q = K x Q / (5 x 10⁻² )²
E₁ = KQ / 25 x 10⁻⁴ = KQ x 10⁴ / 25 . It is acting along positive x axis
E₁ = KQ x 10⁴ i / 25
Similarly electric field due to charge 3Q near 2Q
= 3KQ x 10⁴ i / 25 . It is acting along y-axis
E₂ = 3KQ x 10⁴ j / 25
Similarly electric field due to charge 4Q near 2Q
= 4KQ x 10⁴ j / (25 x 2 )
= 2 KQ x 10⁴ / 25 . It is acting acting along north east direction
unit vector in north east direction = ( i + j )/ √2
So E₃ can be represented by
E₃ = 2 KQ x 10⁴ ( i + j ) / 25 x √2
Total field = KQ x 10⁴ i / 25 + 3KQ x 10⁴ j / 25 + 2 KQ x 10⁴ ( i + j ) / ( 25 x √2 )
= KQ x 10⁴ [ i + 3 j + √2 i + √2 j ) / 25
= 400 KQ ( 2.414 i + 4.414 j ) N / C
Force on 2Q = Field x charge = 400 KQ ( 2.414 i + 4.414 j ) x 2Q N
= 800 KQ² ( 2.414 i + 4.414 j ) N
= 800 x 9 x 10⁹ x ( 2.5 x 10⁻⁶ )² x 2.414 x ( i + 2 j ) N
= 108.63 ( i + 2 j ) N .
Magnitude of this force
= 108.63 x √5
= 243 N approx .
