Which of the following energy sources is in no way derived from the Sun?

A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .

sp2606 [1]

How much work in J does the string do on the boy if the boy stands still?

answer: None. The equation for work is W = force x distance. Since the boy isn't moving, the distance is zero. Anything times zero is zero 
--------------------------------------... 
How much work does the string do on the boy if the boy walks a horizontal distance of 11m away from the kite? 

answer: might be a trick question since his direction away from the kite and his velocity weren't noted. Perhaps he just set the string down and walked away 11m from the kite. If he did this, it is the same as the first one...no work was done by the sting on the boy. 

If he did walk backwards with no velocity indicated, and held the string and it stayed at 30 deg the answer would be: 
4.5N + (boys negative acceleration * mass) = total force1 
work = total force1 x 11 meters 
--------------------------------------... 

How much work does the string do on the boy if the boy walks a horizontal distance of 11m toward the kite? 

answer: same as above only reversed: 
4.5N - (boys negative acceleration * mass) = total force2 
work = total force2 x 11 meters

6 0

3 years ago

A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

Suppose that an object is moving along a vertical line. Its vertical position is given by the equation L(t) = 2t3 + t2-5t + 1, w

Tatiana [17]

Answer:

The average velocity is

$266\frac{m}{s},274\frac{m}{s}$ and $117\frac{m}{s}$ respectively.

Explanation:

Let's start writing the vertical position equation :

$L(t)=2t^{3}+t^{2}-5t+1$

Where distance is measured in meters and time in seconds.

The average velocity is equal to the position variation divided by the time variation.

$V_{avg}=\frac{Displacement}{Time}$ = Δx / Δt = $\frac{x2-x1}{t2-t1}$

For the first time interval :

t1 = 5 s → t2 = 8 s

The time variation is :

$t2-t1=8s-5s=3s$

For the position variation we use the vertical position equation :

$x2=L(8s)=2.(8)^{3}+8^{2}-5.8+1=1049m$

$x1=L(5s)=2.(5)^{3}+5^{2}-5.5+1=251m$

Δx = x2 - x1 = 1049 m - 251 m = 798 m

The average velocity for this interval is

$\frac{798m}{3s}=266\frac{m}{s}$

For the second time interval :

t1 = 4 s → t2 = 9 s

$x2=L(9s)=2.(9)^{3}+9^{2}-5.9+1=1495m$

$x1=L(4s)=2.(4)^{3}+4^{2}-5.4+1=125m$

Δx = x2 - x1 = 1495 m - 125 m = 1370 m

And the time variation is t2 - t1 = 9 s - 4 s = 5 s

The average velocity for this interval is :

$\frac{1370m}{5s}=274\frac{m}{s}$

Finally for the third time interval :

t1 = 1 s → t2 = 7 s

The time variation is t2 - t1 = 7 s - 1 s = 6 s

Then

$x2=L(7s)=2.(7)^{3}+7^{2}-5.7+1=701m$

$x1=L(1s)=2.(1)^{3}+1^{2}-5.1+1=-1m$

The position variation is x2 - x1 = 701 m - (-1 m) = 702 m

The average velocity is

$\frac{702m}{6s}=117\frac{m}{s}$

5 0

3 years ago

A closed cylinder with a 0.15-m radius ends is in a uniform electric field of 300 n/c, perpendicular to the ends. the total flux

bixtya [17]

The total flux through the cylinder is zero.

In fact, the electric flux through a surface (for a uniform electric field) is given by:

$\Phi = E A \cos \theta$

where

E is the intensity of the electric field

A is the surface

$\theta$ is the angle between the direction of E and the perpendicular to the surface, whose direction is always outwards of the surface.

We can ignore the lateral surface of the cylinder, since the electric field is parallel to it, therefore the flux through the lateral surface of the cylinder is zero (because $\theta=90^{\circ}$ and $\cos \theta=0$ ).

On the other two surfaces, the flux is equal and with opposite sign. In fact, on the first surface the flux will be

$\Phi_1 = E \pi r^2$

where r is the radius, and where we have taken $\theta=0^{\circ}$ since the perpendicular to the surface is parallel to the direction of the electric field, so $\cos \theta=1$ . On the second surface, however, the perpendicular to the surface is opposite to the electric field, so $\theta=180^{\circ}$ and $\cos \theta=-1$ , therefore the flux is

$\Phi_2 = -E \pi r^2$

And the net flux through the cylinder is

$\Phi = \Phi_1 + \Phi_2 = E \pi r^2 - E \pi r^2=0$

4 0

3 years ago

How to find resultant force?

lys-0071 [83]

Assuming you're working in a 3D cartesian coordinate system, i.e. each point in space has an x, y, and z coordinate, you add up the forces' x/y/z components to find the resultant force.

8 0

3 years ago