Answer:
1) The greatest height attained by the ball equals 20.387 meters.
2) The time it takes for the ball to reach 15 meters approximately equals 1 second.
Explanation:
The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.
thus using third equation of kinematics we obtain the height attained as

where
'v' is the final speed of the ball
'u' is the initial speed of the ball
'a' is the acceleration that the ball is under which in this case equals 9.81 
's' is the distance it covers
Thus for maximum height applying the values in the equation we get

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

Answer:
<em>-z axis</em>
Explanation:
According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.
Answer:
B) 18.5 m/
s west
Explanation:
Scalar quantity has MAGNITUDE only
Vector quantity has MAGNITUDE and DIRECTION
Hope this is correct and helpful
HAVE A GOOD DAY!
Answer:
T
Explanation:
= magnitude of current in each wire = 2.0 A
= length of the side of the square = 4 cm = 0.04 m
= length of the diagonal of the square =
a =
(0.04) = 0.057 m
= magnitude of magnetic field by wires at A and C


T
= magnitude of magnetic field by wire at B


T
Net magnitude of the magnetic field at D is given as



T
Answer:
Q=185.84C
Explanation:
We have to take into account the integral

In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4

but, for symmetry:
![Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C](https://tex.z-dn.net/?f=Q%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%284x%2B4y%2B4x%5E2%2B4y%5E2%29%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E%7B2%7D%5B4x%5Csqrt%7B4-x%5E2%7D%2B2%284-x%5E2%29%2B4x%5E2%5Csqrt%7B4-x%5E2%7D%2B%5Cfrac%7B4%7D%7B3%7D%284-x%5E2%29%5E%7B3%2F2%7D%5Ddx%5C%5C%5C%5CQ%3D4%5B46.46%5D%3D185.84C)
HOPE THIS HELPS!!