According to the properties you provided, that should be metals.
Answer:
There isn't enough information to solve. Is this related to a graph? The initial and final velocities are needed. The expression for solving is noted under Explanation.
Explanation:
Given final velocity, initial velocity and displacement, one can solve for the acceleration using:
a=v2−u22s,
where v is final velocity (m/sec), u is initial velocity (m/sec) and s it the distance travelled (in m).
Answer:
0.546
Explanation:
From the given information:
The force on a given current-carrying conductor is:
where the length usually in negative (x) direction can be computed as
Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:
where;
current I = 7.0 A
F = 546 × 10⁻³ T/mT
F = 0.546
Answer:
Explanation:
Given that,
Diameter of pipe
d = 20cm
Then, radius =d/2 =20/2
r = 10cm =0.1m
The speed at the bottom is
Vi = 3m/s
Speed at the top Vf?
At the bottom the cube is at a height of 0m
Then, y1 = 0m
At the top the cube is at a height which is the same as the diameter of the pipe
y2 = 0.2m
Now, let us consider, the energy conservation equation , which is the sum of kinetic energy and gravitational potential energy, given by,
K2 + U2 = K1 + U1
½m•Vf² + m•g•y2 = ½m•Vi² + m•g•y1
Divide all through by m
½•Vf² + g•y2 = ½•Vi² + g•y1
Since y1 = 0
So we have,
½•Vf² + g•y2 = ½•Vi²
½•Vf² = ½•Vi² — g•y2
Multiply through by 2
Vf² = Vi² —2g•y2
Vf = √(Vi²—2g•y2)
g is a constant =9.81m/s2
Vf = √(3²—2×9.81×0.2)
Vf = √(9—0.981)
Vf = √8.019
Vf = 2.83m/s
The speed of the ice cubes at the top of the pipe is 2.83m/s