<span>The gravity pulls itself into a sphere also known as </span><span><span>hydrostatic equilibrium. This gives the planet a sphere-like shape.
Hope that helps. -UF aka Nadia</span> </span>
Answer:
The current in second wire is 5.0 A.
(B) is correct option.
Explanation:
Given that,
Current in first wire = 3.7 A
Distance = 8.0 cm
We need to calculate the magnetic field due to the current carrying wire
Using formula of magnetic field
Where, I = current
r = distance
Put the value into the formula
For first wire
...(I)
For second wire,
The distance is 8-3.7 = 4.3 cm
...(II)
The magnetic field in both the wires,
From equation (I) and (II)
Hence, The current in second wire is 5.0 A.
<u>Complete Question:</u>
A hockey player swings her hockey stick and strikes a puck. According to Newton’s third law of motion, which of the following is a reaction to the stick pushing on the puck?
A. the puck pushing on the stick
.
B. the stick pushing on the player
.
C. the player pushing on the stick
.
D. the puck pushing on the player.
<u>Correct Option:</u>
According to Newton’s third law of motion the puck pushing on the stick is a reaction to the stick pushing on the puck.
<u>Option: A</u>
<u>Explanation:</u>
As when the hockey exert force on the puck (which is a flat ball basically used in ice hockey) then this action by hockey will receive equal and opposite reaction by puck. Thus when the stick is pushing on the this flat ball, then puck also push the stick. This is understood by newton's third law pf motion, where action and reaction forces are subject of discussion, displaying their is pair of forces applied among the interacting objects. This form is observed more practically in life and very frequent.
Here's a formula that's simple and useful, and if you're really in
high school physics, I'd be surprised if you haven't see it before.
This one is so simple and useful that I'd suggest memorizing it,
so it's always in your toolbox.
This formula tells how far an object travels in how much time,
when it's accelerating:
Distance = (1/2 acceleration) x (Time²).
D = 1/2 A T²
For your student who dropped an object out of the window,
Distance = 19.6 m
Acceleration = gravity = 9.8 m/s²
D = 1/2 G T²
19.6 = 4.9 T²
Divide each side by 4.9 : 4 = T²
Square root each side: 2 = T
When an object is dropped in Earth gravity,
it takes 2 seconds to fall the first 19.6 meters.
it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]
