Answer:
Less than.
Explanation:
We have the positive charged metal sphere and we have to determine the electric field at a point near to it. In order to find that if we bring the positive test charge at that point then as we know that "like charges repel" so their electric field lines will repel each other resulting in a weaker electric field.
However if we bring the negative test charge at that point then of course there will be attraction and also the the electric field lines will direct from the positive to negative resulting in a stronger electric field between them. So there will be larger electric field then before.
"In this case, It can be concluded that electric field will be less than it was at this point before the test charge was present."
Answer:
H = 1/2 g t^2 where t is time to fall a height H
H = 1/8 g T^2 where T is total time in air (2 t = T)
R = V T cos θ horizontal range
3/4 g T^2 = V T cos θ 6 H = R given in problem
cos θ = 3 g T / (4 V) (I)
Now t = V sin θ / g time for projectile to fall from max height
T = 2 V sin θ / g
T / V = 2 sin θ / g
cos θ = 3 g / 4 (T / V) from (I)
cos θ = 3 g / 4 * 2 sin V / g = 6 / 4 sin θ
tan θ = 2/3
θ = 33.7 deg
As a check- let V = 100 m/s
Vx = 100 cos 33.7 = 83,2
Vy = 100 sin 33,7 = 55.5
T = 2 * 55.5 / 9.8 = 11.3 sec
H = 1/2 * 9.8 * (11.3 / 2)^2 = 156
R = 83.2 * 11.3 = 932
R / H = 932 / 156 = 5.97 6 within rounding
Answer:
t=62 s
Explanation:
Applying the conservation of linear momentum formula:
the initial velocity is zero, we can calculate the man's mass using the gravitational force formula:
now:
That is 0.13m/s due south.
because there is no friction, the man will maintain a constant velocity, so:
Answer:
R = 6i - 4j + 10k
Explanation:
R = (2 + 1 + 3)i + (-3 + 1 - 2)j + (4 + 2 + 4)k
It would mostly depend on its weight
