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3 years ago

11

Geologists know the past temperature of the planet based on rock and fossil evidence. Why do you think it is difficult to determ

ine what has caused climate change in the distant past?

1 answer:

Archy [21]3 years ago

5 0

Answer: Greenhouse gases affect how much of the Sun's energy the Earth loses back to space. Predicting global temperature change is hard, even though the principle sounds easy. In simple terms, energy reaches Earth from the Sun.

Explanation:

You might be interested in

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations:

$\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N$ (2)

$\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N$ (3)

The solution of this system is:

$F_{3,x} = 56.148\,N$ , $F_{3,y} = -116.884\,N$

The x-component of $F_{3}$ is 56.148 newtons.

5 0

3 years ago

As water rises through the troposphere and cools, it changes from a(n) ————- to a(n) ———— through the process of condensation; w

Hoochie [10]

Answer:

GAS, LIQUID, CLOUDS

Explanation:

Water can be in three states: solid, liquid and gas, passing from one to another depending on the pressure and temperature.

In this complementation exercise.

When the water cools down the GAS goes to the LIQUID state, these small drops unite and form the CLOUDS

4 0

3 years ago

What are the results of inserting a crimp connector into the crimp tool facing the wrong way ?

Margaret [11]

If you insert a crimp pin incorrectly, the ratcheted crimp tool will not sufficiently crimp the tabs. As a result, the wire may not fully conduct with the pin and the pin will be damaged.

<u>Explanation:</u>

The general theory for crimping all types of connectors is to strip a little bit of insulation off the wire. Then, put the connector into a suitably sized space in the jaws, insert the wire, and crimp it down. For non-ratcheting pliers, it's suggested the connector be re-crimped with the next smallest hole in the jaws.

A good crimp connection is gas tight and won't wick: it is sometimes referred to as a “cold weld”. Like the solder method, it can be used on solid or stranded conductors, and provides a good mechanical and electrical connection.

8 0

3 years ago

A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What f

Reil [10]

Answer:0.906 N/m

Explanation:

Given

time period $T=0.66 s$

mass $m=0.01 kg$

System can be considered as spring mass system

Time Period of spring mass system is given by

$T=2\pi \sqrt{\frac{m}{k}}$

squaring

$k=\frac{4\pi ^2m}{T^2}$

$k=\frac{4\times \pi ^2\times 0.01}{0.66^2}$

$k=0.906 N/m$

4 0

4 years ago

Someone plans to float a small, totally absorbing sphere 0.500 m above an isotropic point source of light,so that the upward rad

mote1985 [20]

Answer:

468449163762.0812 W

Explanation:

m = Mass = $\rhoV$

V = Volume = $\dfrac{4}{3}\pi r^3$

r = Distance of sphere from isotropic point source of light = 0.5 m

R = Radius of sphere = 2 mm

$\rho$ = Density = 19 g/cm³

c = Speed of light = $3\times 10^8\ m/s$

A = Area = $\pi R^2$

I = Intensity = $\dfrac{P}{4\pi r^2}$

g = Acceleration due to gravity = 9.81 m/s²

Force due to radiation is given by

$F=\dfrac{IA}{c}\\\Rightarrow F=\dfrac{\dfrac{P}{4\pi r^2}{\pi R^2}}{c}\\\Rightarrow F=\dfrac{PR^2}{4r^2c}$

According to the question

$F=mg\\\Rightarrow \dfrac{PR^2}{4r^2c}=\rho \dfrac{4}{3}\pi R^3g\\\Rightarrow P=\dfrac{16r^2\rho c\pi Rg}{3}\\\Rightarrow P=\dfrac{16\times 0.002\times 19000\times \pi\times 0.5^2\times 9.81\times 3\times 10^8}{3}\\\Rightarrow P=468449163762.0812\ W$

The power required of the light source is 468449163762.0812 W

4 0

3 years ago