<span>Scientists have documented that the current level of carbon dioxide in the atmosphere is increasing.</span>
I'm pretty sure the answer would be D
Given :
An object 50 cm high is placed 1 m in front of a converging lens whose focal length is 1.5 m.
To Find :
the image height (in cm).
Solution :
By lens formula :
Here, u = - 100 cm
f = 150 cm
Now, magnification is given by :
Therefore, the image height is 3 m or 300 cm.
<span>6.20 m/s^2
The rocket is being accelerated towards the earth by gravity which has a value of 9.8 m/s^2. Given the total mass of the rocket, the gravitational drag will be
9.8 m/s^2 * 5.00 x 10^5 kg = 4.9 x 10^6 kg m/s^2 = 4.9 x 10^6 N
Add in the atmospheric drag and you get
4.90 x 10^6 N + 4.50 x 10^6 N = 9.4 x 10^6 N
Now subtract that total drag from the thrust available.
1.250 x 10^7 - 9.4 x 10^6 = 12.50 x 10^6 - 9.4 x 10^6 = 3.10 x 10^6 N
So we have an effective thrust of 3.10 x 10^6 N working against a mass of 5.00 x 10^5 kg. We also have N which is (kg m)/s^2 and kg. The unit we wish to end up with is m/s^2 so that indicates we need to divide the thrust by the mass. So
3.10 x 10^6 (kg m)/s^2 / 5.00 x 10^5 kg = 0.62 x 10^1 m/s^2 = 6.2 m/s^2
Since we have only 3 significant figures in our data, the answer is 6.20 m/s^2</span>
