Gravitational potential energy=mass*gravitational acceleration*height
Kinetic energy = 0.5*mass*velocity²
So with the given data
K.E 0.5*1*x²=12.5 v²=12.5÷(0.5*1)
v=√12.5÷(0.5*1) v=5
GPE mass*gravitational acceleration*height
1*9.81*h=98
h=98÷(9.81*1)
h= 9.98 m
Answer:
Explanation:
Given
radius 
Charge on surface 
Point Charge inside sphere 
Electric Field at 
Treating Surface charge as Point charge and applying Gauss law

where A=surface area up to distance r



Answer:
R2 = 10.31Ω
Explanation:
For two resistors in parallel you have that the equivalent resistance is:
(1)
R1 = 13 Ω
R2 = ?
The equivalent resistance of the circuit can also be calculated by using the Ohm's law:
(2)
V: emf source voltage = 23 V
I: current = 4 A
You calculate the Req by using the equation (2):

Now, you can calculate the unknown resistor R2 by using the equation (1):

hence, the resistance of the unknown resistor is 10.31Ω
Answer:
q₁ = + 1.25 nC
Explanation:
Theory of electrical forces
Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
Known data
q₃=5 nC
q₂=- 3 nC
d₁₃= 2 cm
d₂₃ = 4 cm
Graphic attached
The directions of the individual forces exerted by q1 and q₂ on q₃ are shown in the attached figure.
For the net force on q3 to be zero F₁₃ and F₂₃ must have the same magnitude and opposite direction, So, the charge q₁ must be positive(q₁+).
The force (F₁₃) of q₁ on q₃ is repulsive because the charges have equal signs ,then. F₁₃ is directed to the left (-x).
The force (F₂₃) of q₂ on q₃ is attractive because the charges have opposite signs. F₂₃ is directed to the right (+x)
Calculation of q1
F₁₃ = F₂₃

We divide by (k * q3) on both sides of the equation



q₁ = + 1.25 nC
Answer:
Bounce 1 , pass 3, emb2
Explanation:
(By the way I am also doing that question on College board physics page) For the Bounce arrow, since it bumps into the object and goes back, it means now it has a negative momentum, which means a larger momentum is given to the object. P=mv, so the velocity is larger for the object, and larger velocity means a larger kinetic energy which would result in a larger change in the potential energy. Since K=0.5mv^2=U=mgh, a larger potential energy would have a larger change in height which means it has a larger angle θ with the vertical line. Comparing with the "pass arrow" and the "Embedded arrow", the embedded arrow gives the object a larger momentum, Pi=Pf (mv=(M+m)V), it gives all its original momentum to the two objects right now. (Arrow and the pumpkin), it would have a larger velocity. However for the pass arrow, it only gives partial of its original momentum and keeps some of them for the arrow to move, which means the pumpkin has less momentum, means less velocity, and less kinetic energy transferred into the potential energy, and means less change in height, less θangle. So it is Bounce1, pass3, emb2.