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3 years ago

8

Using the same cost and time estimates, consider the time-effectiveness of each engineer.

1 answer:

sammy [17]3 years ago

4 0

Answer:

Camilla

Explanation:

I got it right on edge. :)

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A printed circuit board (PCB) is supported by a chassis that is attached to a vibrating motor. The board is 1.6 mm thick, 200 mm

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Explain what is the capacity of a battery in your own words

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3 years ago

Read 2 more answers

A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimet

Gnom [1K]

Answer:

d. 37 °C

Explanation:

$m_{m}$ = mass of lump of metal = 250 g

$c_{m}$ = specific heat of lump of metal = 0.25 cal/g°C

$T_{mi}$ = Initial temperature of lump of metal = 70 °C

$m_{w}$ = mass of water = 75 g

$c_{w}$ = specific heat of water = 1 cal/g°C

$T_{wi}$ = Initial temperature of water = 20 °C

$m_{c}$ = mass of calorimeter = 500 g

$c_{c}$ = specific heat of calorimeter = 0.10 cal/g°C

$T_{ci}$ = Initial temperature of calorimeter = 20 °C

$T_{f}$ = Final equilibrium temperature

Using conservation of heat

Heat lost by lump of metal = heat gained by water + heat gained by calorimeter

$m_{m} c_{m} (T_{mi} - T_{f}) = m_{w} c_{w} (T_{f} - T_{wi}) + m_{c} c_{c} (T_{f} - T_{ci}) \\(250) (0.25) (70 - T_{f} ) = (75) (1) (T_{f} - 20) + (500) (0.10) (T_{f} - 20)\\T_{f} = 37 C$

6 0

3 years ago

When the current in a toroidal solenoid is changing at a rate of 0.0240 A/s , the magnitude of the induced emf is 12.4 mV . When

Gemiola [76]

Answer:

The number of turns in the solenoid is 230.

Explanation:

Given that,

Rate of change of current, $\dfrac{dI}{dt}=0.0240\ A/s$

Induced emf, $\epsilon=12.4\ mV=12.4\times 10^{-3}\ V$

Current, I = 1.5 A

Magnetic flux, $\phi=0.00338\ Wb$

The induced emf through the solenoid is given by :

$\epsilon=L\dfrac{dI}{dt}$

or

$L=\dfrac{\epsilon}{(di/dt)}$ ........(1)

The self inductance of the solenoid is given by :

$L=\dfrac{N\phi}{I}$ .........(2)

From equation (1) and (2) we get :

$\dfrac{\epsilon}{(di/dt)}=\dfrac{N\phi}{I}$

N is the number of turns in the solenoid

$N=\dfrac{\epsilon I}{\phi (dI/dt)}$

$N=\dfrac{12.4\times 10^{-3}\times 1.5}{0.00338 \times 0.024}$

N = 229.28 turns

or

N = 230 turns

So, the number of turns in the solenoid is 230. Hence, this is the required solution.

3 0

3 years ago