Answer:
w=255
Explanation:
The change in internal energy is given by the first law:
ΔE = Q - w
where ΔE is the change in internal energy of the system
q is the heat added to the system
w is the work done *by* the system on the surroundings
So, for the first phase of this process:
ΔE = Q - w
Q=160J
w=309J
ΔE = 160J - 309J = -149J
To bring the system back to its initial state after this, the internal energy must change by +149J (the system myst gain back the 149 J of energy it lost). We are told that the system loses 106 J of heat in returning to its initial state, so the work involved is given by:
ΔE = Q - w
+149J = -106J - w
255J = -w
w = -255J
Answer:
I believe its forward or south :D
Explanation:
What is meant by the number of complete oscillation made by an oscillating body in 10 seconds is 500 complete oscillations is that the frequency of the oscillating body is 50Hz
<h3>
What is frequency?</h3>
Frequency of an oscillating body can be defined as the number of complete oscillations per unit time
Frequency is measured in hertz (Hz).
Also, the frequency of 1Hz is one oscillation per second.
frequency = number of oscillations/ time taken
frequency = 500/10 = 50 Hz
Thus, what is meant by the number of complete oscillation made by an oscillating body in 10 seconds is 500 complete oscillations is that the frequency of the oscillating body is 50Hz
Learn more about frequency here:
brainly.com/question/1199084
#SPJ1
-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
Given mass= 1kg
Weight on earth = mg(gravity of earth) = 9.8N
weight on moon = mg(gravity of moon)= 1.62N
weight on outer space mg(gravity outer space = 0) = 0N
