The blue line with triangles are a cold front. The red line with half circles are a warm front. The black H is high pressure. The black L is low pressure. The black dot is overcast skies. Finally the white circle with a line down and 4 lines off of that line is 40-knot wind
idk and dont understand
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Answer:
Explanation:
Given an LC circuit
Frequency of oscillation
f = 299 kHz = 299,000 Hz
AT t = 0 , the plate A has maximum positive charge
A. At t > 0, the plate again positive charge, the required time is
t =
t = 1 / f
t = 1 / 299,000
t = 0.00000334448 seconds
t = 3.34 × 10^-6 seconds
t = 3.34 μs
it will be maximum after integral cycle t' = 3.34•n μs
Where n = 1,2,3,4....
B. After every odd multiples of n, other plate will be maximum positive charge, at time equals
t" = ½(2n—1)•t
t'' = ½(2n—1) 3.34 μs
t" = (2n —1) 1.67 μs
where n = 1,2,3...
C. After every half of t,inductor have maximum magnetic field at time
t'' = ½ × t'
t''' = ½(2n—1) 1.67μs
t"' = (2n —1) 0.836 μs
where n = 1,2,3...
Answer:
The correct option is A
Explanation:
Firstly, it should be noted that the freezing point of a substance can be assumed to be melting point of that substance because a substance will normally change from liquid to solid (freezes) at the same point it changes from solid to liquid (melts). For example, water freezes at 0°C and also starts melting at 0°C.
Thus, the substance with the lowest melting point among the substances mentioned in the question is alcohol (ethanol) with the melting point of -114°C. Hence, <u>ethanol also has the lowest freezing point thereby freezing at the lowest temperature.</u>
