Answer:
for abt 36 seconds
Explanation: cuz thats how long it was
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?
You can use fixture wires: For installation in luminaires where they are enclosed and protected and not subject to bending and twisting and also can be used to connect luminaires to their branch circuit conductors.
<h3>What are some uses of fixture wires?</h3>
Fixture wires are flexible conductors that are used for wiring fixtures and control circuits. There are some special uses and requirements for fixture wires and no fixture can be smaller than 18 AWG
In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.
To know more about fixture wires, refer
brainly.com/question/26098282
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Answer:
conduction.
Explanation:
Hoped I helped! Im Eve btw have a great day and consider marking this brainliest if you do thank you in advanced!
Let say the point is inside the cylinder
then as per Gauss' law we have
here q = charge inside the gaussian surface.
Now if our point is inside the cylinder then we can say that gaussian surface has charge less than total charge.
we will calculate the charge first which is given as
now using the equation of Gauss law we will have
now we will have
Now if we have a situation that the point lies outside the cylinder
we will calculate the charge first which is given as it is now the total charge of the cylinder
now using the equation of Gauss law we will have
now we will have
