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2 years ago

Why the sea does not get any fuller?

2 answers:

8 0

The sea is getting fuller.Global warming is causing a rise in a sea level.This is partly due to melting ice in Greenland and Antarctica,and partly due to thermal expansion of the water.The rise so far is about nine inches and the rate has increased to about one inch every seven years.

MrRissso [65]2 years ago

6 0

Answer:

The sea is getting fuller. you just cant see it because it rises by small amounts a time. Global warming is causing a rise in sea level. This is partly due to melting of ice in Greenland and Antarctica, and partly due to thermal expansion of the water. The rise so far is about nine inches and the rate has increased to about one inch every seven years.

Explanation:

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the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?

Inessa [10]

Via the half-life equation:

$A_{final}=A_{initial}(\frac{1}{2})^{\frac{t}{h}}$

Where the time elapse is 11,460 year and the half-life is 5,730 years.

$A_{final}=A_{initial}(\frac{1}{2})^\frac{11460}{5730} \\\\A_{final}=A_{initial}\frac{1}{4} \\\\A_{final}=\frac{1}{4}A_{initial}$

Therefore after 11,460 years the amount of carbon-14 is one fourth (1/4) of the original amount.

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3 years ago

Describe asexual reproduction

Brrunno [24]

“Asexual reproduction is a type of reproduction by which offspring arise from a single organism, and inherit the genes of that parent only; it does not involve the fusion of gametes, and almost never changes the number of chromosomes.” -Wikipedia

4 0

3 years ago

Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o

VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

$5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }$

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so, we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

$K=Celsius+273.15$

So, converting we have:

$T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K$

Therefore, substituting the given information and calculating, we have:

$HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )$

$HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW$

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0

3 years ago

What is the magnitude of the momentum change of two gallons of water (inertia about 7.3 kg ) as it comes to a stop in a bathtub

aliya0001 [1]

We know that the change in momentum is equals to the product of force and time that is impulse ( $F \times t$ ). Therefore, we need to determine the value of that the water is in air by using the second equation of motion,

$s=ut+\frac{1}{2} gt^2$

Here, u is initial velocity which is zero.

$s= \frac{1}{2} gt^2 \\\\ t = \sqrt{\frac{2s}{g} }$ .

Thus, impulse

$= F \times \sqrt{\frac{2s}{g} }$

From Newton`s second law,

$F =mg$

Therefore, impulse

$= mg \times \sqrt{\frac{2s}{g} } = m \sqrt{2gs}$

Given, $m = 7.3 kg$ and $s = 2.0 m$

Substituting these values, we get

Change in momentum = impulse

$= 7.3 \ kg \sqrt{2 \times 9.8 \ m/s^2 \times 2.0 \ m } = 45 .7 \ Ns$ .

8 0

2 years ago

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

4 0

3 years ago