Hi there!

We know that:

U = Potential Energy (J)
K = Kinetic Energy (J)
E = Total Energy (J)
At 10m, the total amount of energy is equivalent to:
U + K = 50 + 50 = 100 J
To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:
100J = mgh
We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².
50J = 10(10)m
m = 1/2 kg
Now, solve for height given that E = 100 J:
100J = 1/2(10)h
100J = 5h
<u>h = 20 meters</u>
Answer:
A- Astronomical body
C- Galaxy
D- Comet
B- Moon
Hope this helps you! Have a great day!
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
Answer:
a) Eₓ = - A y + 2B x
, b) Ey = -Ax –C
, c) Ez = 0
, d) The correct answer is 3
Explanation:
The electric field and the electric power are related
E = - dV / ds
a) Let's find the electric field on the x axis
Eₓ = - dV / dx
dV / dx = A y - B 2x
Eₓ = - A y + 2B x
b) calculate the electric field on the y-axis
Ey = - dV / dy
dV / dy = A x + C
Ey = -Ax –C
c) the electric field on the z axis
dv / dz = 0
Ez = 0
.d) at which point the electric field is zero
Since the electric field is a vector quantity all components must be zero
X axis
0 = = - A y + 2B x
y = 2B / A x
Axis y
0 = -Ax –C
.x = -C / A
We substitute this value in the previous equation
.y = 2B / A (-C / A)
.y = 2 B C / A2
The correct answer is 3
Answer:
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