Answer:
Both will turn red
Both will turn blue
X will turn red and Y will turn blue
X will turn blue and Y will turn red
Answer:
The x-coordinate of the particle is 24 m.
Explanation:
In order to obtain the x-coordinate of the particle, you have to apply the equations for Two Dimension Motion
Xf=Xo+Voxt+0.5axt²(I)
Yf=Yo+Voyt+0.5ayt² (II)
Where Xo, Yo are the initial positions, Xf and Yf are the final positions, Vox and Voy are the initial velocities, ax and ay are the accerelations in x and y directions, t is the time.
The particle starts from rest from the origin, therefore:
Vox=Voy=0
Xo=Yo=0
Replacing Yf=12, Yo=0 and Voy=0 in (I) and solving for t:
12=0+(0)t+ 0.5(1.0)t²
12=0.5t²
Dividing by 0.5 and extracting thr squareroot both sides:
t=√12/0.5
t=√24 = 2√6
Replacing t=2√6, ax=2.0,Xo=0 and Vox=0 in (I) to obain the x-coordinate:
Xf=0+0t+0.5(2.0)(2√6)²
Xf= 24 m
Answer:
A = m³/s³ = [L]³/[T]³ = [L³T⁻³]
B = m³s = [L³T]
Explanation:
We have the equation:
V = At³ + B/t
where, the dimensions of each variable are as follows:
V = m³ = [L]³
t = s = [T]
substituting these in equation, we get:
m³ = A(s)³ + B/s
for the homogeneity of the equation:
A(s)³ = m³
<u>A = m³/s³ = [L]³/[T]³ = [L³T⁻³]</u>
Also,
B/s = m³
<u>B = m³s = [L³T]</u>
Answer:
Explanation:
Given that,
Hot reservoir temperature is
TH = 58°C
TH = 58 + 273 = 331 K
Cold reservoir temperature
TC = -17°C
TC = -17°C + 273 = 256K
In 24minutes 590 J of heat was removed from hot reservoir
t = 24mins
t = 24 × 60 = 1440 seconds
Then,
Power is given as
P = E / t
P = 590 / 1440
P = 0.41 W
Since this removed from the hot reservoir, then, QH = 0.41W
We want to find heat expelled to the cold reservoir QX
Efficiency is given as
η = 1 - TH/TC = 1 - QH/QC
1 - TH/TC = 1 - QH/QC
TH/TC = QH/QC
Make QC subject of formula
QC = QH × TC / TH
QC = 0.41 × 256 / 331
QC = 0.317 W
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