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Novay_Z [31]
3 years ago
14

What would MOST LIKELY happen if the amount of incoming solar energy decreased without a change in the amount of reflection or o

utgoing radiation?
Physics
2 answers:
777dan777 [17]3 years ago
8 0
<h2><u><em>Well, you see, that depends. </em></u></h2><h2><u><em>The firsy thing we have to tak intp account is the angle at witch the sun's rays hit the earth, and that fact can make all the difference, seeing as it does discriminate against seasons. It's more likely that i the winter, a more drastic effect would talk.</em></u></h2><h2 /><h2 /><h2 /><h2>oωo</h2>
Licemer1 [7]3 years ago
5 0

The Earth would heat up.

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If the mass of a portion is 1.67 x 10-27kg and the mass of an electron is 9.11 x 10-31kg, calculate the force of gravitation bet
emmainna [20.7K]

Answer:

1.38 into 10 power minus 58 Newton is the answer

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3 years ago
• What value is used to identify each element in the periodic table?​
bearhunter [10]

Answer:

protons

Explanation:

5 0
3 years ago
The vertical component of the projectile motion of an object depends on which of these? initial velocity or angel of trajectory
SVEN [57.7K]
It depends on both of them.

In fact, the projectile begins its motion with an initial velocity of v_0 and an angle of \alpha. On the y-axis (vertical direction), it is an accelerated motion with acceleration equal to -g (gravitational acceleration). The vertical velocity of the projectile at any time t is given by
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3 0
3 years ago
A cube has a drag coefficient of 0.8. What would be the terminal velocity of a sugar cube 1 cm on a side in air ( = 1.2 kg/mº)?
anzhelika [568]

0.495 m/s

Explanation

the formula for the terminal velocity is given by:

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ \text{where} \\  \end{gathered}

m is the mass

g is 9.81 m/s²

ρ is density

A is area

C is the drag coefficient

then

Step 1

Let's find the mass

\begin{gathered} \sigma=\frac{m}{v} \\ m=\sigma\cdot v \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(0.01m)^3 \\ \text{mass}=(2\cdot10^3\frac{\operatorname{kg}}{m^3})\cdot(1\cdot10^{-6}) \\ \text{mass}=2\cdot10^{-3}\operatorname{kg} \\ \text{mass}=0.002\text{ kg } \\ \text{Area}=(0.01\text{ m}\cdot0.01m)=0.0001m^2 \end{gathered}

now, replace

\begin{gathered} v=\sqrt[]{\frac{2mg}{\sigma AC}} \\ v=\sqrt[]{\frac{2(0.002kg)(9.81\text{ }\frac{m}{s^2})}{(2\cdot10^3\frac{\operatorname{kg}}{m^3})(0.0001m^2)0.8}} \\ v=\sqrt[]{\frac{0.03924\frac{\operatorname{kg}m}{s^2}}{0.16\frac{\operatorname{kg}}{m^{}}}} \\ v=\sqrt[]{0.2452\frac{m^2}{s^2}} \\ v=0.495\text{ m/s} \end{gathered}

hence, the answer is 0.495 m/s

3 0
1 year ago
A train travels 90 kilometers in 1 hour, and then 82 kilometers in 5 hours. What is it’s average speed?
PIT_PIT [208]
Distance is 90km,time 1 hr
distance 2 is 82km,time 5 hrs
average speed=total distance travelled/total time taken.
90+82=172/5+1=6
average speed=28.7km over hour
4 0
3 years ago
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