Ω = 2.81
A = 0.232
k = 29.8

x = A cos(ωt + Ф)

at t = 0:
x = A = A cos(ωt + Ф) = A cos(Ф)
Ф = 0

at t = 1.42, with Ф = 0:
x = A cos(ωt)

U = 1/2 k x² = 1/2 k [A cos(ωt)]²

4 0

3 years ago

A hair dryer is basically a duct of constant diameter in which a few layers of electric resistors are placed. A small fan pulls

nekit [7.7K]

Given:

density of air at inlet, $\rho_{a} = 1.20 kg/m_{3}$

density of air at inlet, $\rho_{b} = 1.05 kg/m_{3}$

Solution:

Now,

$\dot{m} = \dot{m_{a}} = \dot{m_{b}}$

$\rho_{a} A v_{a} = \rho _{b} Av_{b}$ (1)

where

A = Area of cross section

$v_{a}$ = velocity of air at inlet

$v_{b}$ = velocity of air at outlet

Now, using eqn (1), we get:

$\frac{v_{b}}{v_{a}} = \frac{\rho_{a}}{\rho_{b}}$

$\frac{v_{b}}{v_{a}} = \frac{1.20}{1.05}$ = 1.14

% increase in velocity = $1.14\times 100$ =114%

which is 14% more

Therefore % increase in velocity is 14%

5 0

3 years ago

Read 2 more answers

An autotransformer is used to reduce the voltage of a 100-kilovolt amp, 480-volt secondary of an isolated type transformer, to s

tigry1 [53]

Answer:

42KVA

Explanation:

Given data

High Voltage (HV)= 480V

Low Voltage (LV)= 277V

Fo find

Size of transformer=?

Solution

To find the size of transformer here we use the co-ratio.The Co-ratio is given as:

Co-Ratio= (HV - LV)/HV

where

HV is High Voltage

LV is Low Voltage

Now put the values we get

Co- Ratio=(480-277)/480=.42

So the size of transformer is 42KVA

5 0

3 years ago