The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
At t =0, the velocity of A is greater than the velocity of B.
We are told in the question that the spacecrafts fly parallel to each other and that for the both spacecrafts, the velocities are described as follows;
A: vA (t) = ť^2 – 5t + 20
B: vB (t) = t^2+ 3t + 10
Given that t = 0 in both cases;
vA (0) = 0^2 – 5(0) + 20
vA = 20 m/s
For vB
vB (0) = 0^2+ 3(0) + 10
vB = 10 m/s
We can see that at t =0, the velocity of A is greater than the velocity of B.
Learn more: brainly.com/question/24857760
Read each question carefully. Show all your work for each part of the question. The parts within the question may not have equal weight. Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t= 0. For the interval 0 <t< 6 s, spacecraft A's velocity v A and spacecraft B's velocity vB as functions of t are given by the equations va (t) = ť^2 – 5t + 20 and VB (t) = t^2+ 3t + 10, respectively, where both velocities are in units of meters per second. At t = 6 s, the spacecrafts both turn off their engines and travel at a constant speed. (a) At t = 0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft B?
Answer:
Frequency = 3.19 * 10^14 Hz or 1/s
Explanation:
Relationship b/w frequency and wavelength can be expressed as:
C = wavelength * frequency, where c is speed of light in vacuum which is 3.0*10^8 m/s.
Now simply input value (but before that convert wavelength into meters to match the units, you do this by multiply it by 10^-9 so it will be 940*10^-9)
3.0 * 10^8 = Frequency * 940 x 10^-9
Frequency = 3.19 * 10^14 Hz or 1/s
Answer:
a. one-half as great
Explanation:
The power developed by the first lifter is one-half as great as that of the second person.
Power is defined as the rate at which work is done;
Power = 
Since the two lifters do the same work at different time, let us estimate their power;
P₁ = 
P₂ = 
We see that for P₁, power is half of the work done whereas in P₂ power is the same as the work done.
Therefore,
The power of the first weight lifter is one-half the second lifter.
Answer:
v = 29.4m/s
Explanation:
Since the ball is dropped at rest,
u = 0m/s
a = 9.81m/s²
Using
v = u + at
After 3 seconds,
v = 0 + (9.81)(3)
v = 29.4m/s
