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Jet001 [13]
3 years ago
7

The higher the amplitude of the wave, the greater its intensity and the greater its loudness.

Physics
1 answer:
lyudmila [28]3 years ago
4 0
True,when you turn the volume up on your  television , you're actually turning up the amplitude<span>!

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Mary walked north from her home to Sheila's home, which is 4.0 kilometers away. Then she turned right and walked another 3.0 kil
bixtya [17]
The answer is speed: 4.7 km/h, velocity: 3.3 km/h.

Distances and time are given:
d1 =  4 km
d2 = 3 km
d3 = 5 km
t = 1.5 h

The speed can be expressed as a distance (d) divided by time (t). The average speed (s) is total distance travelled divided by time:
s = (d1 + d2)/t = (4+3)/1.5 = 7/1.5 = 4.7 km/h

The average velocity (v) is total displacement (d₁) from the starting point divided by time. Since Mary's starting point was home, and she walked to the supermarket, which is 5.0 kilometers from her own home, her displacement is 5 km:
v = d₁/t = 5/1.5 = 3.3 km/h

8 0
4 years ago
Which is not a characteristic of a heat pump?
kramer
The answer is c. it requires no works
5 0
3 years ago
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The significant feature of a Cepheid variable is that there is a relationship between two intrinsic parameters, one of which can
Ghella [55]

Answer:

Period of brightness variation and luminosity.

Explanation:

The Cepheid variables are used as distance indicators. This requires estimation of periods and (usually) intensity-mean magnitudes in order to establish a period—apparent luminosity relation. It is particularly important for the techniques employed to be as accurate and efficient as possible.

5 0
3 years ago
A physical property of halite (table salt) is its _____.
nexus9112 [7]
Taste: salty
color: varies. ex: white, clear, purple, yellow, etc.
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4 0
3 years ago
You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh
spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

\theta = \frac{1.22 \lambda }{D}

Here,

D is diameter of the eye

D = \frac{1.22 (539nm)}{5.11 mm}

D= 1.287*10^{-4}m

The angle that relates the distance between the lights and the distance to the lamp is given by,

Sin\theta = \frac{d}{L}

For small angle, sin\theta = \theta

sin \theta = \frac{d}{L}

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle sin \theta = \theta

Therefore,

L = \frac{d}{sin\theta}

L = \frac{0.691m}{1.287*10^{-4}}

L = 5367m

Therefore the distance is 5.367km.

4 0
4 years ago
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