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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?
In thermodynamics, work of a system at constant pressure conditions is equal to the product of the pressure and the change in volume. It is expressed as follows:
W = P(V2 - V1)
W = 1.3x10^5 (2x6 - 6 )
<span>W = 780000 J
</span>
Hope this answers the question. Have a nice day.
As soon as you let go of it it is at its max speed because gravity is constantly pulling it down
Answer:
L = 1.15 m
Explanation:
The diffraction phenomenon is described by the equation
a sin θ = m λ
Where a is the width of the slit, λ the wavelength and m is an integer, the order of diffraction is left.
The diffraction measurements are made on a screen that is far from the slit, and the angles in the experiment are very small, let's use trigonometry
tan θ = y / L
tan θ = sint θ / cos θ≈ sin θ
We substitute in the first equation
a (y / L) = m λ
The first maximum occurs for m = 1
The distance is measured from the center point of maximum, which coincides with the center of the slit, in this case the distance is the total width of the central maximum, so the distance (y) measured from the center is
y = 1.15 / 2 = 0.575 cm
y = 0.575 10⁻² m
Let's clear the distance to the screen (L)
L = a y / λ
Let's calculate
L = 115 10⁻⁶ 0.575 10⁻² / 575 10⁻⁹
L = 1.15 m
