All categories

2 years ago

In __________ waves, the motion of the particles in a medium is along the direction of the wave (parallel). *

Physics

1 answer:

True [87]2 years ago

7 0

Longitude is the answer

You might be interested in

The diameter of a baseball is 7.4 cm and its mass is 0.15 kg. a) If a pitcher throws the baseball at a velocity of 44.3 m/s (100

Tema [17]

Answer:

drag force $F_D = 1.5 \ N$

Velocity (V) = 40.169 m/s

Explanation:

The drag force $F_D$ is given by the formula:

$F_D = C_D * \frac{1}{2}* \rho * V^2*A$

where:

$C_D$ = drag coefficient depending on the Reynolds number

Reynolds number Re = $\frac{\rho *V*D}{ \mu}$

Let's Assume that the air is in room temperature at 25 °C ; Then

density of the air $\rho$ = 1.1845 kg/m³

viscosity of fluid or air $\mu$ = 1.844 × 10⁻⁵ kg/ms

diameter of the baseball D = 7.4 cm

Velocity V = 44.3 m/s

Replacing them into the equation of Reynolds number ; we have :

$Re = \frac{1.1845 \ kg/m^3*44.3 m/s*0.074 m}{1.844*10^{-5}kg/ms}\\\\Re = 2.1*10^5$

A = Projected Area

From the diagram attached below which is gotten from NASA for baseball;

the drag coefficient which depends on Reynolds number is read as:

$C_D$ = 0.3

Projected Area A = $\frac{\pi D^2}{4}$

A = $\frac{\pi 0.074^2}{4}$

A = 0.0043 m²

Finally, drag force is then calculated as ;

$F_D = C_D * \frac{1}{2}* \rho*V^2*A\\\\F_D = 0.3* \frac{1}{2}*1.1845 \ kg/m^3*(44.3 \ m/s) ^2*0.0043 m^2\\\\F_D = 1.5 \ N$

$- F_D = ma$

since acceleration a = $\frac{dV}{dt}$

Then;

$-F_D = m \frac{dV}{dt}$

Also;

velocity (V) = $\frac{dx}{dt}$

Then;

$- F_D = \frac{md_2x}{dt^2}$

$\frac{d_2x}{dt^2} = \frac {- F_D}{m}$

$F_D = 1.5 \ N\\m = 0.15 \ kg$

Then;

$\frac{d_2x}{dt^2} = \frac {- 1.5 }{0.15}$

$\frac{d_2x}{dt^2} =- 10$

Integrating the above equation ; we have :

$\frac{dx}{dt}= - 10 t + C\\$

when time (t) = 0 ; then $\frac{dx}{dt}= V = 44.3$

44.3 = - 10 × 0 + C

C = 44.3

$\frac{dx}{dt}= V = -10 t + 44.3$

Time (t) =

$\frac{distance }{velocity} \\\\= \frac{18.3 m}{44.3 m/s}\\\\= 0.413 s$

∴ Velocity ; $\frac{dx}{dt}= V = - 10t +44.3$

$\frac{dx}{dt}= V = - 10(0.413 s) +44.3$

Velocity (V) = 40.169 m/s

6 0

3 years ago

You have a graduated cylinder that you use to measure volume. The cylinder

ad-work [718]

Answer:

D. 21 ml

Explanation:

Since, the cylinder is marked and graduated in the intervals if 1 ml. Therefore, the values between two consecutive ml, such as between 30 ml and 31 ml can not be determined. Because, we do not have any scale in between the ml. So, the least count of this instrument is 1 ml. This graduated cylinder can give the answers to zero decimal places, accurately. And it can not determine any decimal value due to its graduating or the marking limitation. So, all the options given, contain a decimal value, except for the option D. In option D there is no decimal value, hence it is a correct answer.

D. <u>21 ml</u>

5 0

3 years ago

Read 2 more answers

Calculate the x-component of the electric field produced by the charge distribution Q at points on the positive x-axis where x&g

Alexxx [7]

Answer:

Explanation:

From the image attached below:

It is required to determine the electric field because of the line charge distribution of positive charge distributed uniformly from x = 0 to x = a at point x from O;

where x > a

Assume we chose an element dy at a distance from the point.

Then, the change on $dy = \dfrac{Q}{a} \times dy$

The electric field due to this dy length is $\dfrac{kdq}{y^2 }= \dfrac{kQ dy }{ay^2}$

Thus, the total electric field $= \dfrac{kQ}{a}\int \limits ^{x}_{x-a} \dfrac{dy}{y^2}$

$= \dfrac{kQ}{a} \Big [ \dfrac{1}{y}\Big ]^{x-a}_{x}$

$=\dfrac{kQ}{a}\Big [\dfrac{1}{x-a}-\dfrac{1}{x}\Big ]$

$E=\dfrac{kQ}{a}\Big (\dfrac{x-(x-a)}{(x-a)x}\Big )$

$E=\dfrac{kQa}{a(x-a)x}$

Hence, the electric field $E=\dfrac{kQ}{(x-a)x} \ \ \ where; x>a$

8 0

3 years ago

Salmon often jump waterfalls to reach their

erik [133]

Answer:

6.35 m/s

Explanation:

The motion of the salmon is equivalent to that of a projectile, which consists of two independent motions:

- A horizontal motion with constant speed

- A vertical motion with constant acceleration ( $g=-9.8 m/s^2$ , acceleration of gravity)

The horizontal velocity of the salmon is given by:

$v_x = u cos \theta$

where

u = ? is the initial speed

$\theta=32^{\circ}$ is the angle of projection

Then the horizontal distance covered by the salmon after a time t is given by

$d=v_x t =(u cos \theta) t$

Or equivalently, the time taken to cover a distance d is

$t=\frac{d}{u cos \theta}$ (1)

Along the vertical direction, the equation of motion is

$h = (u sin \theta) t + \frac{1}{2}gt^2$ (2)

where

$u sin \theta$ is the initial vertical velocity

If we substitute (1) into (2), we get:

$h = (u sin \theta) \frac{d}{cos \theta} + \frac{1}{2}g(\frac{d}{ ucos \theta})^2=d tan \theta + \frac{gd^2}{2u^2 cos^2 \theta}$

We now that in order to reach the breeding grounds, the salmon must travel a distance of

d = 2.02 m

reaching a height of

h = 0.574 m

Substituting these data into the equation and solving for u, we find the initial speed that the salmon must have:

$u =\sqrt{ \frac{gd^2}{2(h-d tan \theta) cos^2 \theta}}=\sqrt{\frac{(-9.8)(2.02)^2}{2(0.574-(2.02)(tan 32))(cos^2(32))}}=6.35 m/s$

8 0

3 years ago

A vector of magnitude 10 has an angle with the positive x axis (east) of 120 degrees. what are its components

Andrews [41]

<span>The angle with the positive x-axis is 120 degrees. We can assume that this angle is measured counterclockwise from the positive x-axis. We can find the x-component of the vector. x-component = 10 cos(120) = -5 We can find the y-component of the vector. y-component = 10 sin(120) = 8.66 The x-component of the vector is -5 and the y-component of the vector is 8.66</span>

3 0

3 years ago