Differentiate between a derived quantity and a derived unit.

A giant chorus of 1000 vocalists is singing the same note. Suddenly, 999 vocalists stop,leaving one soloist. By how many decibel

stiks02 [169]

Answer:

The decrease in decibels is 0.1 dB.

Explanation:

Let the intensity of one chorus is Io.

let the intensity of 1000 vocalist is dB.

The intensity of 1000 vocalist is 1000 Io.

$dB = 10log\frac{1000Io}{Io}=30$ ..... (1)

let the intensity of 999 vocalist is dB'.

$dB' = 10log\frac{999Io}{Io}=29.9$ ..... (2)

So, the change is

= dB - dB' = 30 - 29.9 = 0.1 dB

6 0

3 years ago

A solid conducting sphere with radius 0.75 m carries a net charge of 0.13 nC. What is the magnitude of the electric field at a p

Delicious77 [7]

Answer:

Explanation:

given that

Radius =0.75m

Cnet=0.13nC

a. Electric field inside the sphere located 0.5m from the center of the sphere.

The electric field located inside the sphere is zero.

b. The electric field located 0.25m beneath the sphere.

Since the radius is 0.75m

Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m

Then

E=kq/r2

K=9e9Nm2/C2

q=0.13e-9C

r=1m

Then,

E= 9e9×0.13e-9/1^2

E=1.17N/C. Q.E.D

3 0

3 years ago

Read 2 more answers

A car turns into a driveway that slopes upward at a 9 degree angle, car is moving at 6.5 m/s. If the driver lets the car coast,

expeople1 [14]

To put it in the simplest form, the automatic transmission has a torque converter that uses the transmission fluid to turn the other side of the torque converter. Ex. Putting two room fans facing each other and turning one on and wind flowing from one fan makes the second fan facing the rotating fan turn and the idling engine probably doesn't have enough power to hold it.

I guess more technically, in an automatic, the car selects a neutral gear (no gear) when it is at rest, and the brake pedal is pressed. Upon releasing the brake, the car will apply a small forward force that will hold the car steady on a very small gradient or propel it forwards slowly on a flat or downhill gradient. This force is only started when the brake is being released, as this is the indicator for the car to change into 1st gear.

In a situation where a car is rolling back down the hill it is facing up, there could be one or more of several situations at hand.

Most likely, it simply lacks power to hold on the gradient of the slope. In this case, you'll have to perform a hill start to maintain brake force until forward propulsion is enough to move the car forward. Hill starts are almost always necessary for manual transmission cars. Additionally, it could be the driver's pedal transfer from brake to accelerator is to slow. That would also account for some part of the slip.

Okay, what's missing here is the clutch. The clutch as I'm sure you're aware is a series of plates that connect to each other and transfer power due to friction. The amount of friction is adjustable depending on how much force is applied to hold the plates together (or apart). It is easier to see this in a manual car at low speeds, where the driver hovers the clutch actuation pedal around the "Friction Point". This point is the fine line between moving (increased friction between clutch plates holds them together more firmly, thus transferring more power) and staying stationary (clutch plates disengaged from each other). From the Friction Point, any further release of the clutch will cause the car to move forwards because the transmission is engaging with more of the engine's power. Depressing the clutch pedal back in will not have any effect, as it will just keep the clutch plates separate.

In an automatic car, this is all controlled by computer algorithms, determining how much the clutch should be engaged to reach a certain speed. Taking off from the lights on a hill for example will not necessarily register as any different to taking off on flat ground. The effect of this is that the car is assuming that is requires a certain number of revs and a predetermined clutch setting to accelerate smoothly. Due to the increased force the hill provides, the car will move backwards until the power again reaches a level that will overcome its slippage.

One other thing, is that clutches work both ways. Since only friction holds them together, the torque exerted by the wheels back through the drive-train to the clutch can cause the plates to slip when they are not completely engaged. This results in the wheels moving independently of the running engine, as the clutch is separating the forces they would exert on each other.

I think the key thing to note is that despite not having a pedal to operate it, Automatic cars still have a clutch - just one that relies on a computer to function.

If it's not the engine or the clutch which are both behaving as normal, and we're still assuming it's an auto, there could be a problem with the car's computer system, transmission, clutch or gearbox that is causing the slippage. I would assume this is less likely, but it might be worth checking if it happens to your car. your frickin' welcome

6 0

3 years ago

Read 2 more answers

Solve this physics for me <br>please with steps<br>

Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

a)

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

$[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\$

$[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] + [\frac{m^{2} }{s^{2} } ]$

$[\frac{m^2}{s^2} ]$

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

$v=\frac{dx}{dt} \\v = 0.75*2*t+5*t$

$(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]$

ii) replacing in the derivated equation.

$v=1.5*(4)+5\\v=11[m/s]$

iii) the average velocity is defined by the expresion v = x/t

$v = \frac{x-x_{0} }{t-t_{0} } \\$

$x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]$

2

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

$\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} } \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })$

b) One of the applications of the surface tension is the <u>capillarity</u> this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0

3 years ago

Stoichiometry is based on the law of conservation of

kotykmax [81]

The principles of stoichiometry are based upon the law of conservation of mass.

8 0

3 years ago