Moving an object up an inclined plane<span> requires </span>less<span>force </span>than<span> lifting it straight up, at a cost of an increase in the distance moved. The </span>mechanical advantage<span>of an </span>inclined plane<span>, the factor by which the force is reduced, is equal to the ratio of the length of the sloped surface to the height it spans.</span>
To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.
By definition Newton's second law is described as
F= ma
Where,
m= mass
a = Acceleration
Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,
In the case of mass A,
Making summation of Torques in the Pulley we have to
Replacing the values previously found,
Replacing with our values
PART B) Ignoring the moment of inertia the acceleration would be given by
Therefore the error would be,
Answer:
Th-234, Pa-234
Explanation:
In alpha decay, 2 protons and 2 neutrons are lost. So U-238 would become Th-234.
In beta decay, one neutron is turned into a proton. So Th-234 would become Pa-234.
<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>
