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lana66690 [7]
3 years ago
11

What is critical to being an active listener?

Physics
1 answer:
Lelechka [254]3 years ago
5 0
Showing appreciation for what is being said is the answer. for example making eye contact with the speaker, not interrupting, and nodding for assurance is being an active listener
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Is it possible to determine from the data in the graphs whether the sensor attached to cart a is actually plugged into ch a and
ladessa [460]
<span>It is possible to determine which cart to which ch is connected if the graph would show electrical charge, ie, amps or voltage. If the graph showed a series circuit diagram this would also allow determination. Bottom line is that a correct graph data will show the requested information.</span>
4 0
3 years ago
The metric unit for temperature is _______________.<br> a. fahrenheitb. celsiusc. secondd. liter
Nikitich [7]
The metric unit for temperature is celsius.
8 0
3 years ago
Explain the law of conservation of energy. Give a specific example using kinetic and potential energy that shows how energy is c
Ede4ka [16]

Answer:

As you said you already know, energy cannot be created or destroyed.

Explanation:

You cannot gain energy or lose energy, it can only be converted. So if you start on a 3m high hill and go down it, your potential energy is equal to mgh, and if you get to the bottom of the hill, your KE would be equal to your PE at the top, and when you start going up another hill again, the maximum height you can reach is 3m, because energy cannot be created or destroyed, and your mass and gravitational acceleration are the same, so therefore you can only reach the same height you started from due to the conservation of energy.

7 0
3 years ago
A sound wave has a frequency of 295 Hz and travels the length of a football field, 91.4 m in 0.506 s. What is the period of the
dem82 [27]

Answer:1/295 seconds

Explanation:

Period=1/frequency

Period=1/295 seconds

7 0
3 years ago
A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

5 0
3 years ago
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