Best budget freestyle drone?

While discussing what affects the amount of pressure exerted by the brakes: Technician A says that the shorter the line, the mor

harina [27]

Answer:

Only Technician B is right.

Explanation:

The cylindrical braking system for a car works through the mode of pressure transmission, that is, the pressure applied to the brake pedals, is transmitted to the brake pad through the cylindrical piston.

Pressure applied on the pedal, P(pedal) = P(pad)

And the Pressure is the applied force/area for either pad or pedal. That is, P(pad) = Force(pad)/A(pad) & P(pedal) = F(pedal)/A(pedal)

If the area of piston increases, A(pad) increases and the P(pad) drops, Meaning, the pressure transmitted to the pad reduces. And for most cars, there's a pressure limit for the braking system to work.

If the A(pad) increases, P(pad) decreases and the braking force applied has to increase, to counter balance the dropping pressure and raise it.

This whole setup does not depend on the length of the braking lines; it only depends on the applied force and cross sectional Area (size) of the piston.

5 0

2 years ago

Phil wants to make a dark themed superhero movie. What could be his target demographic?

grin007 [14]

Answer:

D.

male, aged 18-25

5 0

2 years ago

Read 2 more answers

The current entering the positive terminal of a device is i(t)= 6e^-2t mA and the voltage across the device is v(t)= 10di/dtV.

liberstina [14]

Answer:

a) 2,945 mC

b) P(t) = -720*e^(-4t) uW

c) -180 uJ

Explanation:

Given:

i (t) = 6*e^(-2*t)

v (t) = 10*di / dt

Find:

( a) Find the charge delivered to the device between t=0 and t=2 s.

( b) Calculate the power absorbed.

( c) Determine the energy absorbed in 3 s.

Solution:

- The amount of charge Q delivered can be determined by:

dQ = i(t) . dt

$Q = \int\limits^2_0 {i(t)} \, dt = \int\limits^2_0 {6*e^(-2t)} \, dt = 6*\int\limits^2_0 {e^(-2t)} \, dt$

- Integrate and evaluate the on the interval:

$= 6 * (-0.5)*e^-2t = - 3*( 1 / e^4 - 1) = 2.945 C$

- The power can be calculated by using v(t) and i(t) as follows:

v(t) = 10* di / dt = 10*d(6*e^(-2*t)) /dt

v(t) = 10*(-12*e^(-2*t)) = -120*e^-2*t mV

P(t) = v(t)*i(t) = (-120*e^-2*t) * 6*e^(-2*t)

P(t) = -720*e^(-4t) uW

- The amount of energy W absorbed can be evaluated using P(t) as follows:

$W = \int\limits^3_0 {P(t)} \, dt = \int\limits^2_0 {-720*e^(-4t)} \, dt = -720*\int\limits^2_0 {e^(-4t)} \, dt$

- Integrate and evaluate the on the interval:

$W = -180*e^-4t = - 180*( 1 / e^12 - 1) = -180uJ$

6 0

3 years ago

2 Blocks 1 and 2 rest on rough surfaces with coefficient of frictions ¢1 and ¢2 respectively. The blocks

Amiraneli [1.4K]

Answer:

100N
25N

Explanation:

a) On the verge of tipping over, reaction acts at the corner A

When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at A must pass through B

tan b/2h, h b/ 2 θ µ = = ∴= k k ( µ )

b) When slippage occurs,

Block moves w/ const. velocity  equilibrium

Three-force member: reaction at C must pass through G

k tanθ µ =

tan x/ H/2 , x H/2

4 0

3 years ago

Convert 850 nm wavelength into frequency, eV, wavenumber, joules and ergs.

Sholpan [36]

Answer:

Frequency = $3.5294\times 10^{14}s^{-1}$

Wavenumber = $1.1765\times 10^6m^{-1}$

Energy = $2.3365\times 10^{-19}J$

Energy = 1.4579 eV

Energy = $2.3365\times 10^{-12}erg$

Explanation:

As we are given the wavelength = 850 nm

conversion used : $(1nm=10^{-9}m)$

So, wavelength is $850\times 10^{-9}m$

The relation between frequency and wavelength is shown below as:

$Frequency=\frac{c}{Wavelength}$

Where, c is the speed of light having value = $3\times 10^8m/s$

So, Frequency is:

$Frequency=\frac{3\times 10^8m/s}{850\times 10^{-9}m}$

$Frequency=3.5294\times 10^{14}s^{-1}$

Wavenumber is the reciprocal of wavelength.

So,

$Wavenumber=\frac{1}{Wavelength}=\frac{1}{850\times 10^{-9}m}$

$Wavenumber=1.1765\times 10^6m^{-1}$

Also,

$Energy=h\times frequency$

where, h is Plank's constant having value as $6.62\times 10^{-34}J.s$

So,

$Energy=(6.62\times 10^{-34}J.s)\times (3.5294\times 10^{14}s^{-1})$

$Energy=2.3365\times 10^{-19}J$

Also,

$1J=6.24\times 10^{18}eV$

So,

$Energy=(2.3365\times 10^{-19})\times (6.24\times 10^{18}eV)$

$Energy=1.4579eV$

Also,

$1J=10^7erg$

So,

$Energy=(2.3365\times 10^{-19})\times 10^7erg$

$Energy=2.3365\times 10^{-12}erg$

5 0

3 years ago