Answer:
a)
, b) ![\omega = 0\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%200%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
Explanation:
The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:
![\tau = F \cdot r](https://tex.z-dn.net/?f=%5Ctau%20%3D%20F%20%5Ccdot%20r)
![\tau = m\cdot a \cdot r](https://tex.z-dn.net/?f=%5Ctau%20%3D%20m%5Ccdot%20a%20%5Ccdot%20r)
![\tau = m \cdot \alpha \cdot r^{2}](https://tex.z-dn.net/?f=%5Ctau%20%3D%20m%20%5Ccdot%20%5Calpha%20%5Ccdot%20r%5E%7B2%7D)
Where
is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:
![\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20%5Cfrac%7B170%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20-%20200%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%7D%7B10%5C%2Cs%7D)
![\alpha = -3\,\frac{rad}{s^{2}}](https://tex.z-dn.net/?f=%5Calpha%20%3D%20-3%5C%2C%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%7D)
Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:
a) t = 50 s.
![\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)](https://tex.z-dn.net/?f=%5Comega%20%3D%20200%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20-%20%5Cleft%283%5C%2C%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%20%5Ccdot%20%2850%5C%2Cs%29)
![\omega = 50\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%2050%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
b) t = 100 s.
Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:
![t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B0%5C%2C%5Cfrac%7Brad%7D%7Bs%7D-200%5C%2C%5Cfrac%7Brad%7D%7Bs%7D%20%7D%7B%5Cleft%28-3%5C%2C%5Cfrac%7Brad%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%7D)
![t = 66.667\,s](https://tex.z-dn.net/?f=t%20%3D%2066.667%5C%2Cs)
Since
, then the angular velocity is equal to zero. Therefore:
![\omega = 0\,\frac{rad}{s}](https://tex.z-dn.net/?f=%5Comega%20%3D%200%5C%2C%5Cfrac%7Brad%7D%7Bs%7D)
Answer:
V = 26.95 cm³
Explanation:
Density is given by the formula :
ρ = m÷V
Density = mass ÷ Volume
Given both density and mass we rearrange, substitute and solve for Volume :
Rearranging the equation to make Volume the subject :
ρ = m÷V
ρV = m
V = m÷ ρ
Now substitute :
V = 45 ÷ 1.67
V = 26.9461077844
Take 2 decimal places as the density is 2 decimal places :
V = 26.95
Units will be cm³ as it is volume
Hope this helped and have a good day
Answer:
The ballon will brust at
<em>Pmax = 518 Torr ≈ 0.687 Atm </em>
<em />
<em />
Explanation:
Hello!
To solve this problem we are going to use the ideal gass law
PV = nRT
Where n (number of moles) and R are constants (in the present case)
Therefore, we can relate to thermodynamic states with their respective pressure, volume and temperature.
--- (*)
Our initial state is:
P1 = 754 torr
V1 = 3.1 L
T1 = 294 K
If we consider the final state at which the ballon will explode, then:
P2 = Pmax
V2 = Vmax
T2 = 273 K
We also know that the maximum surface area is: 1257 cm^2
If we consider a spherical ballon, we can obtain the maximum radius:
![R_{max} = \sqrt{\frac{A_{max}}{4 \pi}}](https://tex.z-dn.net/?f=R_%7Bmax%7D%20%3D%20%5Csqrt%7B%5Cfrac%7BA_%7Bmax%7D%7D%7B4%20%5Cpi%7D%7D)
Rmax = 10.001 cm
Therefore, the max volume will be:
![V_{max} = \frac{4}{3} \pi R_{max}^3](https://tex.z-dn.net/?f=V_%7Bmax%7D%20%3D%20%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20R_%7Bmax%7D%5E3)
Vmax = 4 190.05 cm^3 = 4.19 L
Now, from (*)
![P_{max} = P_1 \frac{V_1T_2}{V_2T_1}](https://tex.z-dn.net/?f=P_%7Bmax%7D%20%3D%20P_1%20%5Cfrac%7BV_1T_2%7D%7BV_2T_1%7D)
Therefore:
Pmax= P1 * (0.687)
That is:
Pmax = 518 Torr
Answer:
Explanation:
All substances have characteristic physical and chemical properties. Physical properties are those that can be observed with the senses without changing the identity of the substance. Chemical properties describe how a substance can be changed into a new substance. Physical and chemical properties, such as color, density, boiling point, solubility, conductivity, and flammability, A. are always different between substances. B. depend on the amount of the substance. C. do not depend on the amount of the substance. D. have the same values for all substances.
Ask for details Follow Report by S27754738 2 hours ago
Answers
Answer:
(a) r = 1.062·R
= ![\frac{531}{500} R_E](https://tex.z-dn.net/?f=%5Cfrac%7B531%7D%7B500%7D%20R_E)
(b) r = ![\frac{33}{25} R_E](https://tex.z-dn.net/?f=%5Cfrac%7B33%7D%7B25%7D%20R_E)
(c) Zero
Explanation:
Here we have escape velocity v
given by
and the maximum height given by
![\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20v%5E2-%5Cfrac%7BGM%7D%7BR_E%7D%20%3D%20-%5Cfrac%7BGM%7D%7Br%7D)
Therefore, when the initial speed is 0.241v
we have
v =
so that;
v² =
v² = ![{\frac{0.116162\times GM}{R_E} }](https://tex.z-dn.net/?f=%7B%5Cfrac%7B0.116162%5Ctimes%20GM%7D%7BR_E%7D%20%7D)
is then
![\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%7B%5Cfrac%7B0.116162%5Ctimes%20GM%7D%7BR_E%7D%20%7D-%5Cfrac%7BGM%7D%7BR_E%7D%20%3D%20-%5Cfrac%7BGM%7D%7Br%7D)
Which gives
or
r = 1.062·R
(b) Here we have
![K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}](https://tex.z-dn.net/?f=K_i%20%3D%200.241%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20m%20%5Ctimes%20v_e%5E2%20%3D%200.241%5Ctimes%20%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%20m%20%20%5Ctimes%20%5Cfrac%7B2GM%7D%7BR_E%7D%20%3D%20%5Cfrac%7B0.241mGM%7D%7BR_E%7D)
Therefore we put
in the maximum height equation to get
![\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}](https://tex.z-dn.net/?f=%5Cfrac%7B0.241%7D%7BR_E%7D%20-%5Cfrac%7B1%7D%7BR_E%7D%20%3D-%5Cfrac%7B1%7D%7Br%7D)
From which we get
r = 1.32·R
(c) The we have the least initial mechanical energy, ME given by
ME = KE - PE
Where the KE = PE required to leave the earth we have
ME = KE - KE = 0
The least initial mechanical energy to leave the earth is zero.