Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
Answer:
Addition reactions with benzenes lead to the loss of aromaticity.
Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism.
Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions.
Explanation:
Answer:
kinetic energy
Explanation:
a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.
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?
Constant acceleration of plane = 3m/s²
a) Speed of the plane after 4s
Acceleration = speed/time
3m/s² = speed/4s
S = 12m/s
The speed of the plane after 4s is 12m/s.
b) Flight point will be termed as the point the plane got initial speed, u, 20m/s
Find speed after 8s, v
a = 3m/s²
from,
a = <u>v</u><u> </u><u>-</u><u> </u><u>u</u>
t
3 = <u>v</u><u> </u><u>-</u><u> </u><u>2</u><u>0</u>
8
24 = v - 20
v = 44m/s
After 8s the plane would've 44m/s speed.
