Answer:
Explanation:
STEP 1
<u>Given</u>
Radius of cylinder = r = 25cm, 2.5m
mass = 27kg
cylinder is mounted so as to rotate freely about a horizontal axis that is parallel to and 60cm to the central logitudinal axis of the cylinder
height = 0.6m
<u>part 1</u>
The cylinder is mounted so as to rotate freely about a horizontal axis tha is paralle to 60cm from the central longitudinal axis of then cylinder. The rotational inertia of the cylinder about the axis of rotation is given by
<em>I = Icm + mh²</em>
<em>∴ I = 1/2mr² + mh² = 1/2x27x (0.5)² + 20 x (0.6)²</em>
<em>I=13.09kg.m²</em>
where
<em>I</em>cm is the rotational inertia of the cylinder about its central axis
m is the mass of the cylinder
h is the distance between the axis of the rotation and the central axis of the cylinder
r is the radius of the cylinder
<em> </em><em> I=13.09kg.m²</em>
<em>part2</em>
<em>from the conservation of the total mechanical energy of the meter stick, the change in gravitational potential energyof the meter stick plus the change in kinetic energy must be zero</em>
<em>Δk + Δu = 0</em>
<em>1/2 </em>I(w²-w²) = Ui-Uf
1/2 x 13.09w² = mgh
∴w=√20 x 9.8 x 0.6/(1/2 x 13.09) =117.6/6.5
w=18.09rad/s
Answer:
Approximately 
, if this gas is an ideal gas, and that the quantity of this gas stayed constant during these changes.
Explanation:
Let 
and 
denote the pressure of this gas before and after the changes.
Let 
and 
denote the volume of this gas before and after the changes.
Let 
and 
denote the temperature (in degrees Kelvins) of this gas before and after the changes.
Let 
and 
denote the quantity (number of moles of gas particles) in this gas before and after the changes.
Assume that this gas is an ideal gas. By the ideal gas law, the ratios 
and 
should both be equal to the ideal gas constant, 
.
In other words:
.
.
Combine the two equations (equate the right-hand side) to obtain:
.
Rearrange this equation for an expression for , the temperature of this gas after the changes:
.
Assume that the container of this gas was sealed, such that the quantity of this gas stayed the same during these changes. Hence: 
, 
.
![\begin{aligned} T_2 &= \frac{P_2}{P_1} \cdot \frac{V_2}{V_1} \cdot \frac{n_1}{n_2}\cdot T_1 \\[0.5em] &= \frac{1.67\; \rm atm}{1.12\; \rm atm} \times \frac{11.2\; \rm m^{3}}{15.7\; \rm m^{3}} \times 1 \times 245\; \rm K \\[0.5em] &\approx 261\; \rm K\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%20T_2%20%26%3D%20%5Cfrac%7BP_2%7D%7BP_1%7D%20%5Ccdot%20%5Cfrac%7BV_2%7D%7BV_1%7D%20%5Ccdot%20%5Cfrac%7Bn_1%7D%7Bn_2%7D%5Ccdot%20T_1%20%5C%5C%5B0.5em%5D%20%26%3D%20%5Cfrac%7B1.67%5C%3B%20%5Crm%20atm%7D%7B1.12%5C%3B%20%5Crm%20atm%7D%20%5Ctimes%20%5Cfrac%7B11.2%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%7B15.7%5C%3B%20%5Crm%20m%5E%7B3%7D%7D%20%5Ctimes%201%20%5Ctimes%20245%5C%3B%20%5Crm%20K%20%5C%5C%5B0.5em%5D%20%26%5Capprox%20261%5C%3B%20%5Crm%20K%5Cend%7Baligned%7D)
.
Answer:
Explanation:
Possible causes for overcurrent include short circuits, excessive load, incorrect design, an arc fault, or a ground fault. Fuses, circuit breakers, and current limiters are commonly used overcurrent protection (OCP) mechanisms to control the risks
please
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The correct answer is a blackhole.
A blackhole is the only object in which nothing, not even light can escape.
Hope this helps!
When two stars are bound together gravitationally and orbit a common mass, theyre known as B. BINARY STARS.
