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bearhunter [10]
3 years ago
6

The greenhouse effect keeps earth warmer than it would normally be. True False

Physics
1 answer:
MariettaO [177]3 years ago
6 0

Answer:

The World Is Pure Enough as it is that we don't melt or (lose air) \_/

Explanation:

All people don't have to be concerned about is a disruption of the plethora of gases we already have in our atmosphere and trees need C02 in order to breath and that's what cows produce that's what we breath out and what coal plants and fossil fuels produce the more C02 the more trees produce oxygen because trees breathe in C02 and pop out Oxygen and plus power plants don't produce C02 they produce water vapor so (ALL IN ALL C02 Will not cause the Ice Caps to melt because trees already have that issue covered that's why the scientists said they fixed the global warming issue to keep people from whining about a non-existent problem. so basically its true The greenhouse effect does keep the earth warm halfly because of the hot gases that get trapped in the atmosphere like C02.True

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The closeness of measured values to an accepted value is the BLANK of the data ?
wlad13 [49]
When you refer to how close a measured value is to a standard, accepted or known value, you are talking about the ACCURACY of the data. This is the definition of accuracy when it comes to engineering and other fields of science. 

Accuracy is usually associated or with the term precision, as their definitions are often interchanged.
3 0
3 years ago
If an object moving at 12 m/s has a kinetic energy of 72 j what is the mass
EleoNora [17]
Simply, apply the formula E = \frac{1}{2}mv^{2} and insert the values of m = mass, v = velocity and E = Energy.
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5 0
3 years ago
For copper, ρ = 8.93 g/cm3 and M = 63.5 g/mol. Assuming one free electron per copper atom, what is the drift velocity of electro
viktelen [127]

Answer:

V_d = 1.75 × 10⁻⁴ m/s

Explanation:

Given:

Density of copper, ρ = 8.93 g/cm³

mass, M = 63.5 g/mol

Radius of wire = 0.625 mm

Current, I = 3A

Area of the wire, A = \frac{\pi d^2}{4} = A = \frac{\pi 0.625^2}{4}

Now,

The current density, J is given as

J=\frac{I}{A}=\frac{3}{ \frac{\pi 0.625^2}{4}}= 2444619.925 A/mm²

now, the electron density, n = \frac{\rho}{M}N_A

where,

N_A=Avogadro's Number

n = \frac{8.93}{63.5}(6.2\times 10^{23})=8.719\times 10^{28}\ electrons/m^3

Now,

the drift velocity, V_d

V_d=\frac{J}{ne}

where,

e = charge on electron = 1.6 × 10⁻¹⁹ C

thus,

V_d=\frac{2444619.925}{8.719\times 10^{28}\times (1.6\times 10^{-19})e} = 1.75 × 10⁻⁴ m/s

4 0
3 years ago
Read 2 more answers
If the car’s speed decreases at a constant rate from 64 mi/h to 30 mi/h in 3.0 s, what is the magnitude of its acceleration, ass
mixas84 [53]

Answer:3.874 m/s^2

Explanation:

Given

Car speed decreases at a constant rate from 64 mi/h to 30 mi/h

in 3 sec

60mi/h \approx 26.8224m/s

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we know acceleration is given by =\frac{velocity}{Time}

a=\frac{15.1994-26.8224}{3}

a=-3.874 m/s^2

negative indicates that it is stopping the car

Distance traveled

v^2-u^2=2as

\left ( 15.1994\right )^2-\left ( 26.8224\right )^2=2\left ( -3.874\right )s

s=\frac{488.419}{2\times 3.874}

s=63.038 m

7 0
3 years ago
Which material would you take, steel or bricks to design a bridge? Why?
Vesna [10]

Answer:

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Explanation:

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4 0
2 years ago
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