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3 years ago

Substances released into the air are known as

Physics

2 answers:

snow_lady [41]3 years ago

5 0

Emissions is the correct awnser :)

irakobra [83]3 years ago

3 0

Your answer is Emissions

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What is scientific evidence and when do we have enough?

Mandarinka [93]

Scientific evidence is evidence that is backed up by science and you have enough when you can understandingly prove your point or hypothesis

4 0

3 years ago

What do you call the procedure that helps you determine the volume of an irregularly shaped object, while using a graduated cyli

Lena [83]

Water displacement. You fill a graduated cylinder with an amount of water, place the object inside the graduated cylinder, and then measure the new water level. The change in volume of the water is the volume of the object, assuming the object was completely submerged.

8 0

3 years ago

A satellite has a mass of 5832 kg and is in a circular orbit 4.13 × 105 m above the surface of a planet. The period of the orbit

elena55 [62]

Answer:

W = 28226.88 N

Explanation:

Given,

Mass of the satellite, m = 5832 Kg

Height of the orbiting satellite from the surface, h = 4.13 x 10⁵ m

The time period of the orbit, T = 1.9 h

= 6840 s

The radius of the planet, R = 4.38 x 10⁶ m

The time period of the satellite is given by the formula

$T = 2\pi \sqrt{\frac{(R+h)^{3} }{R^{2} g} }$ second

Squaring the terms and solving it for 'g'

g = 4 π² $\frac{(R+h)^{3} }{R^{2}T^{2} }$ m/s²

Substituting the values in the above equation

g = 4 π² $\frac{(4.38X10^6+4.13X10^5)^{3} }{(4.38X10^6)^{2}X6840^{2}}$

g = 4.84 m/s²

Therefore, the weight

w = m x g newton

= 5832 Kg x 4.84 m/s²

= 28226.88 N

Hence, the weight of the satellite at the surface, W = 28226.88 N

8 0

3 years ago

15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers

timofeeve [1]

The percent difference between two numbers $x$ and $y$ is given by

$\dfrac{|y-x|}x \times 100\%$

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from $x$ to $y$ or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

$\dfrac{|6-3|}3 \times 100\% = 100\%$

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

$\dfrac{3-6}3\times100\%=-50\%$

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take $x$ from the left column and $y$ from the right column.

$\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%$

$\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%$

$\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%$

3 0

2 years ago

Discuss the factors that are used to determine power.

True [87]

<span>The factors that are used to determine power are:
Voltage,current and the power factor.

</span><span>Power = Voltage x Current x K
Watts = Volts x Amps x Power Factor</span>

5 0

3 years ago

Read 2 more answers