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Simora [160]
3 years ago
7

As the air on the surface of the earth warms what happens to the density of the air

Physics
1 answer:
Arada [10]3 years ago
4 0
The density of the air it will become foggy and and become smoky
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Which of the following is true about radiation from the sun?
Lelu [443]
><span>It can travel through vacuum. 

The rays must travel in the vacuum of space between Earth's atmosphere and the sun.</span>
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3 years ago
How much energy from the sun actually reaches the corn answer?
Serhud [2]
The energy from the sun that reaches the corn is about two billionths.
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Read 2 more answers
A third baseman makes a throw to first base 40.5 m away. The ball leaves his hand with a speed of 30.0 m/s at a height of 1.4 m
Ugo [173]

Answer:

When the ball goes to first base it will be 4.23 m high.

Explanation:

Horizontal velocity = 30 cos17.3 = 28.64 m/s

   Horizontal displacement = 40.5 m

   Time  

         t=\frac{40.5}{28.64}=1.41s          

   Time to reach the goal posts 40.5 m away = 1.41 seconds

Vertical velocity = 30 sin17.3 = 8.92 m/s

    Time to reach the goal posts 40.5 m away = 1.41 seconds

    Acceleration = -9.81m/s²

    Substituting in s = ut + 0.5at²

             s = 8.92 x 1.41 - 0.5 x 9.81 x 1.41²= 2.83 m

    Height of throw = 1.4 m

    Height traveled by ball = 2.83 m

    Total height = 2.83 + 1.4 = 4.23 m

    When the ball goes to first base it will be 4.23 m high.

8 0
3 years ago
This problem has been solved!
lana66690 [7]

Answer:

Charge on each metal sphere will be 8\times 10^{8}C

Explanation:

We have given number of electron added to metal sphere A n=10^{12}electron

As both the spheres are connected by rod so half -half electron will be distributed on both the spheres.

So electron on both the spheres =\frac{10^{12}}{2}=5\times 10^{11}electron

We know that charge on each electron e=1.6\times 10^{-19}C

So charge on both the spheres will be equal to q=1.6\times 10^{-19}\times 5\times 10^{11}=8\times 10^{8}C

So charge on each metal sphere will be equal to 8\times 10^{8}C

6 0
2 years ago
A cart for hauling ore out of a gold mine has a mass of 413 kg, including its load. the cart runs along a straight stretch of tr
Andrei [34K]
Given from the problem :mass m = 413 kg;coefficient of friction u = 0.0163;acceleration due to gravity g = 9.8 m/s2;inclined angle @1 = 14.3;inclined angle  @2 = 4.69;distance travelled d = 175 m;applied fore F = 410 N;  the component of the force from the donkey in the direction of motion isF2 = F1 [email protected]= 397.2964498768165 N
Fy = N - mg [email protected] = 0N = mg [email protected] = 4037.964151113007 NFx = F2 - mg [email protected] - f = mahere f = u N=65.8188156631420141

F2 - mg [email protected] - f = maa = F2 - mg [email protected] - f/ m=0.31923412183075155 m/s^2
work done by donkeyW = F2 d=69526.8787284428875 J
8 0
3 years ago
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