Answer:
Explanation:
Important here is to know that due north is a 90 degree angle, due east is a 0 degree angle, and due south is a 270 degree angle. Then we find the x and y components of each part of this journey using the sin and cos of the angles multiplied by each magnitude:
Add them all together to get the x component of the resultant vector, V:
Do the same to find the y components of the part of this journey:
Add them together to get the y component of the resultant vector, V:
One thing of import to note is that both of these components are positive, so the resultant angle lies in QI.
We find the final magnitude:
and, rounding to 2 sig dig's as needed:
1.0 × 10² m; now for the direction:
58°
Answer:
The water level rises more when the cube is located above the raft before submerging.
Explanation:
These kinds of problems are based on the principle of Archimedes, who says that by immersing a body in a volume of water, the initial water level will be increased, raising the water level. That is, the height in the container with water will rise in level. The difference between the new volume and the initial volume of the water will be the volume of the submerged body.
Now we have two moments when the steel cube is held by the raft and when it is at the bottom of the pool.
When the cube is at the bottom of the water we know that the volume will increase, and we can calculate this volume using the volume of the cube.
Vc = 0.45*0.45*0.45 = 0.0911 [m^3]
Now when a body floats it is because a balance is established in the densities, the density of the body and the density of the water.
![Ro_{H2O}=R_{c+r}\\where:\\Ro_{H2O}= water density = 1000 [kg/m^3]\\Ro_{c+r}= combined density cube + raft [kg/m^3]](https://tex.z-dn.net/?f=Ro_%7BH2O%7D%3DR_%7Bc%2Br%7D%5C%5Cwhere%3A%5C%5CRo_%7BH2O%7D%3D%20water%20density%20%3D%201000%20%5Bkg%2Fm%5E3%5D%5C%5CRo_%7Bc%2Br%7D%3D%20combined%20density%20cube%20%2B%20raft%20%5Bkg%2Fm%5E3%5D)
Density is given by:
Ro = m/V
where:
m= mass [kg]
V = volume [m^3]
The buoyancy force can be calculated using the following equation:
![F_{B}=W=Ro_{H20}*g*Vs\\W = (200+730)*9.81\\W=9123.3[N]\\\\9123=1000*9.81*Vs\\Vs = 0.93 [m^3]](https://tex.z-dn.net/?f=F_%7BB%7D%3DW%3DRo_%7BH20%7D%2Ag%2AVs%5C%5CW%20%3D%20%28200%2B730%29%2A9.81%5C%5CW%3D9123.3%5BN%5D%5C%5C%5C%5C9123%3D1000%2A9.81%2AVs%5C%5CVs%20%3D%200.93%20%5Bm%5E3%5D)
Vs > Vc, What it means is that the combined volume of the raft and the cube is greater than that of the cube at the bottom of the pool. Therefore the water level rises more when the cube is located above the raft before submerging.
I'll just give you the link for it but count it as my answer. http://www.differencebetween.com/difference-between-leptons-and-vs-hadrons/
Answer:
a) 0.0288 grams
b) 
Explanation:
Given that:
A typical human body contains about 3.0 grams of Potassium per kilogram of body mass
The abundance for the three isotopes are:
Potassium-39, Potassium-40, and Potassium-41 with abundances are 93.26%, 0.012% and 6.728% respectively.
a)
Thus; a person with a mass of 80 kg will posses = 80 × 3 = 240 grams of potassium.
However, the amount of potassium that is present in such person is :
0.012% × 240 grams
= 0.012/100 × 240 grams
= 0.0288 grams
b)
the effective dose (in Sieverts) per year due to Potassium-40 in an 80- kg body is calculate as follows:
First the Dose in (Gy) = 
= 
= 
Effective dose (Sv) = RBE × Dose in Gy
Effective dose (Sv) = 
Effective dose (Sv) =
Answer:
t = 5.19 s
Explanation:
We have,
Height of the cliff is 132 m
It is required to find the time taken by the ball to fall to the ground. Let t is the time taken. So, using equation of kinematics as :
So, it will take 5.19 seconds to fall to the ground.
