3. <span>The second piston will experience the same force as compared with the first. This is because since the </span>pressure is the same everywhere inside the fluid system,<span> the force is proportional to the surface area. We are told that both the first and the second piston have the same surface area, therefore, they will both experience the same force/pressure.
4. </span>The situation is much the same as number 3 above, with the exception that the second piston is twenty times larger than the first. Again, since the pressure is the same everywhere inside the fluid system, the force is proportional to the surface area. We are told that the second piston is 20 times larger than the first, therefore, the larger piston will experience 20 times larger the force of the small one.
6. The answer is TRUE. The <span>hydraulic </span>braking system<span> of most cars makes use of a vacuum servo (or booster), which is located between the </span>brake pedal<span> and the master cylinder piston. </span><span>This vacuum servo amplifies the force applied </span><span>from the </span>brake pedal<span>.</span>
<span>The surface charge density = q/A
So q = surface charge density x Area
The surface area of a sphere of radius R is 4*Pi*R^2. R = d/2 where d is diameter. This leaves us with 1.3/2 = 0.65. Area = 4 * pie * (0.65)^2 = 5.30998.
So the net charge q = 8.1 * 10^(-6) * 5.30998 = 42.47998 * 10^(-6)
The Total electric flux = Q/e_0 where , 8.854 Ă— 10â’12, e_0 is permitivity of free space.
So Flux = 42.47998 * 10^(-6) / 8.854 * 10(â’12) = 4.833 * 10^(-6 - (-12)) = 4.833 * 10^(6)</span>
Answer:
Negative
Explanation:
First law of thermodynamic also known as the law of conservation of energy states that the total energy of an isolated system is constant; energy can be transformed from one form to another, but can be neither created nor destroyed.
The first law relates relates changes in internal energy to heat added to a system and the work done by a system by the conservation of energy.
The first law is mathematically given as ΔU =
-
= Q - W
Where Q = Quantity of heat
W = Work done
From the first law The internal energy has the symbol U. Q is positive if heat is added to the system, and negative if heat is removed; W is positive if work is done by the system, and negative if work is done on the system.
Analyzing the pistol when it raises in isothermal and when it falls in isobaric state.The following can be said:
In the Isothermal compression of a gas there is work done on the system to decrease the volume and increase the pressure. For work to be done on the system it is a negative work done then.
In the Isobaric State An isobaric process occurs at constant pressure. Since the pressure is constant, the force exerted is constant and the work done is given as PΔV.If a gas is to expand at a constant pressure, heat should be transferred into the system at a certain rate.Isobaric is a fuction of heat which is Isothermal Provided the pressure is kept constant.
In Isobaric definition above it can be seen that " Heat should be transferred into the system ata certain rate. For heat to be transferred into the system work is deinitely been done on the system thereby favouring the negative work done.
Answer:
The initial velocity of the softball is 14.711 meters per second.
Explanation:
This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.
From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:
(Eq. 1)
Where:
- Initial height of the softball, measured in meters.
- Final height of the softball, measured in meters.
- Initial velocity of the softball, measured in meters per second.
- Time, measured in seconds.
- Gravitational acceleration, measured in meters per square second.
If we know that
,
and
, the initial velocity of the softball is:
![v_{o}\cdot (3\,s)+\frac{1}{2}\cdot (-9.807\,\frac{m}{s^{2}} )\cdot (3\,s)^{2} = 0](https://tex.z-dn.net/?f=v_%7Bo%7D%5Ccdot%20%283%5C%2Cs%29%2B%5Cfrac%7B1%7D%7B2%7D%5Ccdot%20%28-9.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%29%5Ccdot%20%283%5C%2Cs%29%5E%7B2%7D%20%3D%200)
![3\cdot v_{o} -44.132\,m= 0](https://tex.z-dn.net/?f=3%5Ccdot%20v_%7Bo%7D%20-44.132%5C%2Cm%3D%200)
![v_{o} = 14.711\,\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bo%7D%20%3D%2014.711%5C%2C%5Cfrac%7Bm%7D%7Bs%7D)
The initial velocity of the softball is 14.711 meters per second.