F = M x G Find the force of gravity acting upon a 1500Kg Hippopatamus.

Place the yellow, 25g ball at 60 cm on the ramp. How far did the box move? Measure from the left side of the box.

Anna007 [38]

Answer: approximately 13-15 centimeters

3 0

3 years ago

If the distance from a light source triples, how does light intensity change? The intensity will be 3x greater. The intensity wi

Tcecarenko [31]

Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

$I\alpha \frac{1}{r^{2} }$

Now according to the question the distance is increased by three times than,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }$

Therefore,

$\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }$

Therefore the intensity will become 1/9 times to the initial intensity.

3 0

3 years ago

The reflective surface of a CD consists of spirals of equally spaced grooves. If you shine a laser pointer on a CD, each groove

Ipatiy [6.2K]

Answer:

d = 1.55 * 10⁻⁶ m

Explanation:

To calculate the distance between the adjacent grooves of the CD, use the formula, $d = \frac{m \lambda}{sin(A_{m}) }$ ..........(1)

The fringe number, m = 1 since it is a first order maximum

The wavelength of the green laser pointer, $\lambda$ = 532 nm = 532 * 10⁻⁹ m

Distance between the central maximum and the first order maximum = 1.1 m

Distance between the screen and the CD = 3 m

$A_{m}$ = Angle between the incident light and the diffracted light

From the setup shown in the attachment, it is a right angled triangle in which

$sin(A_{m}) = \frac{opposite}{Hypotenuse} \\sin(A_{m}) =\frac{1.1}{\sqrt{1.1^{2}+3^{2}}}$

$sin(A_{m} ) = 0.344\\A_{m} = sin^{-1} 0.344\\A_{m} = 20.14^{0}$

Putting all appropriate values into equation (1)

$d = \frac{1* 532*10^{-9} }{0.344 }\\d = 0.00000155 m\\d = 1.55 * 10^{-6} m$

3 0

3 years ago

What's the difference between electromagnetic waves and electromagnetic spectrum?

White raven [17]

The difference between the two is, well for one

Spectrum: The entire range that the "waves" could be such, as visible light, x-ray's and so on.

Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.

It may confuse you but it makes sense to me (Sorry)

4 0

3 years ago

Help with the two questions above? Correct answers?

LenKa [72]

(6) first choice: the frequency appears higher and wavelength is shorter.

The car approaches a stationary observer and so the sound will appear to have shorter wavelength. This creates an effect of its siren to sound with higher frequency than it would do if both were stationary.

(7) The Doppler formula for frequency in the case of a stationary observer and source approaching it is as follows:

$f_O = \frac{v}{v-v_s}\cdot f= \frac{343\frac{m}{s}}{(343-25)\frac{m}{s}}\cdot 400Hz \approx 431Hz$

The wavelength is then

$\lambda = \frac{343\frac{m}{s}}{431Hz}\approx 0.80 m$

The third choice "0.80m; 431Hz" is correct

7 0

3 years ago