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2 years ago

A diode for which the forward voltage drop is 0.7 V at 1.0 mA is operated at 0.5 V. What is the value of the current

Physics

1 answer:

Vesnalui [34]2 years ago

4 0

Answer:

the current value is $0.335 \mu A$

Explanation:

The computation of the value of the current is given below:

$z_i = I_s e^{\frac{0.7}{ut} }= 10^{-3}\\\\Z_z = I_s e^{\frac{0.5}{ut} }\\\\\frac{Z_z}{Z_i}= \frac{Z_z}{10^{-3}} = e^{\frac{0.5\times 0.7}{0.025} }\\\\= 0.335 \mu A$

Hence, the current value is $0.335 \mu A$

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A soccer player applies a force of 48.4 N to a soccer ball while kicking it. If the ball has

AlekseyPX

Answer:

C. 110 m/s2

Explanation:

Force = Mass x Acceleration

Since we have the force and the mass, we can rearrange this equation to solve for acceleration by dividing both sides by mass:

Force/Mass = (Mass x Acceleration)/Mass

Acceleration = Force/Mass

Now we just have to plug in our values and calculate!

Acceleration = 48.4/0.44

Acceleration = 110m/s/s

It is option C. 110 m/s2

Hope this helped!

6 0

2 years ago

Read 2 more answers

The speed at which a light aircraft can take off is 120 km/h. (A) What is the minimum constant acceleration required for the pla

kondor19780726 [428]

Answer:

A) a = 2.31[m/s^2]; B) t = 14.4 [s]

Explanation:

We can solve this problem using the kinematic equations, but firts we must identify the data:

Vf= final velocity = take off velocity = 120[km/h]

Vi= initial velocity = 0, because the plane starts to move from the rest.

dx= distance to run = 240 [m]

$v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\$

To find the time we must use another kinematic equation.

$v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]$

7 0

2 years ago

What is the approximate velocity of the object at 5 seconds ? .

Gnoma [55]

Answer:

do you have an image?

Explanation:

3 0

3 years ago

A car with mass 1500 kg moves with constant velocity of 36 m/s. The driver sees a group of cows in front and he immediately step

Crazy boy [7]

From laws of motion:

$S = ( \frac{v + u}{2} ) \times t$
Where S is the distance/displacement (as you would call it) which is unknown
v = final velocity which is 0m/s (this is because the car stops)
u = initial velocity which is 36m/s (from the data given)
t = time taken for the distance to be covered and it is 6s

Substitute the values, hence:
$S = ( \frac{0 + 36}{2} ) \times 6$
$S = (18) \times 6 \\ \\ S = 108m$

But this is merely the distance he travelled in the 6 seconds he was trying to stop the car.

Therefore, the distance between the car and the cows = 160-108
Distance = 52m

6 0

3 years ago

By what factor is the intensity of sound at a rock concert louder than that of a whisper when the two intensity levels are 120 d

Kipish [7]

Answer:

The intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

Explanation:

Given;

rock concert sound intensity level, β₁ = 120 dB

whisper sound intensity level, β₂ = 20 dB

The sound intensity level is given as;

$\beta = 10Log(\frac{I}{I_o} )\\\\$

where;

I₀ is the threshold sound intensity of hearing = 10⁻¹² W/m²

I is the sound intensity

Intensity of sound at rock concert ;

$120 = 10Log(\frac{I}{10^{-12}} )\\\\12 = Log(\frac{I}{10^{-12}} )\\\\10^{12} = \frac{I}{10^{-12}}\\\\I = 10^{12} * 10^{-12}\\\\I = 10^0\\\\I = 1 \ W/m^2$

The intensity of sound of a whisper;

$20 = 10Log(\frac{I}{10^{-12}} )\\\\2 = Log(\frac{I}{10^{-12}} )\\\\10^{2} = \frac{I}{10^{-12}}\\\\I = 10^{2} * 10^{-12}\\\\I = 10^{-10} \ W/m^2\\\\$

Determine the factor by which the intensity of sound at a rock concert louder than that of a whisper

$\frac{I_{Concert}}{I_{whisper}} = \frac{1}{10^{-10}} \\\\\frac{I_{Concert}}{I_{whisper}} = 1 * 10^{10}\\\\I_{Concert} = 1 * 10^{10}*I_{whisper}$

Therefore, the intensity of sound at a rock concert is louder than that of a whisper by a factor of 1 x 10¹⁰

3 0

2 years ago