Answer:
5.25 m
Explanation:
Given;
The height equation h;
h=-x^2+3x+3
Where;
h = the height above water
x = horizontal distance from the end of the board
The maximum height is at h' = 0, when change in h with respect to change in x is equal to zero.
differentiating the equation h.
dh/dx = h' = -2x + 3 = 0
Solving for x;
2x = 3
x = 3/2
Substituting into the function h;
h max = -x^2+3x+3
h max = -(3/2)^2 + 3(3/2) +3 = -9/4 +9/2 +3 = 9/4 + 3 =
h max = 21/4 = 5.25 m
It would be 12,000 because newton’s third 2nd law states F=ma (force=matter x acceleration) so 30x400 would be your force .
please mark brainliest and i hope this helps!
There are two force acting on an object that is being lifted. (1) the weight of the car, (2) the upward force. The difference of these force should be equal to the product of the mass and the acceleration. (This is the content of Newton's 2nd Law of Motion). If we let the lifting force be F,
F - (830)(9.8) = (830)(3.8)
The value of F from the equation is 11288 N.
Answer:
The acceleration is in 2 D as in between east and south.
Explanation:
mass, m = 50 kg
acceleration, a = 0.25 m/s^2 horizontal
acceleration of elevator, a' = 1 m/s^2 downwards
When a person on the ground the resultant acceleration of the person with respect to the ground is between east and south direction so the path os parabolic in nature. It graph is shown below:
Answer:
Explanation:
Let the charge particle have charge equal to +q .
force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along - y axis. ( negative of y axis )
force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.
F = q ( v i x B k ) , ( velocity is along x direction and magnetic field is along z axis. )
= (Bqv) - j
= - Bqv j
The force will be along - ve y - direction .
If we take charge as negative or - q
force due to electric field will be along - y axis .
magnetic force = F = -q ( v i x B k )
= + Bqv j
magnetic force will be along + y axis
So it is difficult to find out the nature of charge on the particle from this experiment.
