Which of the following best describes the upper respiratory tract?

A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant

atroni [7]

Answer:

The time constant is $\tau = 1.265 s$

Explanation:

From the question we are told that

the time take to charge is $t = 2.4 \ s$

The mathematically representation for voltage potential of a capacitor at different time is

$V = V_o - e^{-\frac{t}{\tau} }$

Where $\tau$ is the time constant

$V_o$ is the potential of the capacitor when it is full

So the capacitor potential will be 100% when it is full thus $V_o =$ 100% = 1

and from the question we are told that the at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

V = 0.85

Hence

$0.85 = 1 - e^{-\frac{2.4}{\tau } }$

$- {\frac{2.4}{\tau } } = ln0.15$

$\tau = 1.265 s$

7 0

2 years ago

Calculate the speed of a gamma ray with a frequency of 3.0 x 10^19 Hz and a wavelength of 1.0 x 10^-11 m.

Ulleksa [173]

Answer:

Speed of gamma rays = 3 x 10⁸ m/s

Explanation:

Given:

Frequency of gamma ray = 3 x 10¹⁹ Hz

Wavelength of gamma rays = 1 x 10⁻¹¹ meter

Find:

Speed of gamma rays

Computation:

Velocity = Frequency x wavelength

Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays

Speed of gamma rays = [3 x 10¹⁹][1 x 10⁻¹¹]

Speed of gamma rays = 3 x [10¹⁹⁻¹¹]

Speed of gamma rays = 3 x [10⁸]

Speed of gamma rays = 3 x 10⁸ m/s

6 0

3 years ago

Find V. in the circuit of the following figure:

Angelina_Jolie [31]

Answer:

where is the figure

Explanation:

8 0

2 years ago

An object is thrown directly downward from the top of a very tall building. The speed of the object just as it is released is 17

Softa [21]

Answer:

distance cover is = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

$t_1 = 3.32 sec$

$t_2 = 5.08 sec$

from equation of motion we know that

$d_1 = vt_1 + \frac{1}{2} gt_1^2$

where d_1 is distance covered in time t1

so $d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2$ =

$d_1 = 110.78 m$

$d_2 = vt_2 + \frac{1}{2} gt_2^2$

where d_2 is distance covered in time t2

$d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2$

$d_2 = 213.31 m$

distance cover is = 213.31 - 110.78 = 102.53 m

3 0

3 years ago

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i

evablogger [386]

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 $\frac{m}{s}$

To find:

the total time it is in the air = ?

Formula used:

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = $\frac{v-u}{a}$

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = $\frac{0-8}{-9.8}$

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

4 0

2 years ago