Pressure depends upon

Vector A is 3 m long and vector B is 4 m long. The length

LUCKY_DIMON [66]

Answer:d

Explanation:

Given

Magnitude of Vector A is 3 m

and Magnitude of vector B is 4 m

So the maximum value of resultant can be 7 m when both are at an angle of $0^{\circ}$

and its minimum value can be 1 m when both are at angle of $180 ^{\circ}$

So the resultant must lie between 1 m to 7 m

6 0

3 years ago

PLEASE HELP!! I PROMISE I'LL GIVE BRAINLIEST!!

andrew-mc [135]

like a black cloud

Explanation:

Define: Signals

Before going too much further, we should talk a bit about what a signal actually is, electronic signals specifically (as opposed to traffic signals, albums by the ultimate power-trio, or a general means for communication). When one speaks of analog one often means an electrical context, however mechanical, pneumatic, hydraulic, and other systems may also convey analog signals.

An analog signal uses some property of the medium to convey the signal's information. Any information may be conveyed by an analog signal, often such a signal is a measured change in physical phenomena, such as sound, light, temperature, position, or pressure.

For example, in sound recording, changes in air pressure (that is to say, sound) strike the diaphragm of a microphone which causes related changes in a voltage or the current in an electric circuit. The voltage or the current is said to be an "analog" of the sound.

4 0

3 years ago

Steam flows steadily through an adiabatic turbine. The inlet conditions of the steam are 4 MPa, 500◦C, and 80 m/s, and the exit

Cerrena [4.2K]

Answer:

a) ΔEC=-23.4kW

b)W=12106.2kW

c) $A=0.01297m^2$

Explanation:

A)

The kinetic energy is defined as:

$\frac{m*vel^2}{2}$ (vel is the velocity, to differentiate with v, specific volume).

The kinetic energy change will be: Δ ( $\frac{mvel^2}{2}$ )= $\frac{m*vel_2^2}{2}-\frac{m*vel_1^2}{2}$

Δ ( $\frac{mvel^2}{2}$ )= $\frac{m}{2}*(vel_2^2-vel_1^2)$

Where 1 and 2 subscripts mean initial and final state respectively.

Δ( $\frac{mvel^2}{2}$ )= $\frac{12\frac{kg}{s}}{2}*(50^2-80^2)\frac{m^2}{s^2}=-23400W=-23.4kW$

This amount is negative because the steam is losing that energy.

B)

Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).

$H_1+\frac{m*vel_1^2}{2}=H_2+\frac{m*vel_2^2}{2}+W\\W=m*(h_1-h_2)+\frac{m}{2} *(vel_1^2-vel_2^2)$