Answer:
a) 20.29N
b) 19.42N
c) 15N
Explanation:
To find the magnitude of the resultant vector you can consider an axis in the middle of the vector, from which you can calculate the components of the vectors by using the angles given:
a) for 30°
![F_1=[9cos(15\°)\hat{i}+9sin(15\°)\hat{j}]N\\\\F_1=[8.69\hat{i}+2.32\hat{j}]N\\\\F_2=[12cos(15\°)\hat{i}-12sin(15\°)\hat{j}]N\\\\F_2=[11.59\hat{i}-3.10\hat{j}]N\\\\F=F_1+F_2=20.28N\hat{i}-0.78N\hat{j}\\\\|F|=\sqrt{(20.28N)^2+(0.78N)^2}=20.29N](https://tex.z-dn.net/?f=F_1%3D%5B9cos%2815%5C%C2%B0%29%5Chat%7Bi%7D%2B9sin%2815%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_1%3D%5B8.69%5Chat%7Bi%7D%2B2.32%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B12cos%2815%5C%C2%B0%29%5Chat%7Bi%7D-12sin%2815%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B11.59%5Chat%7Bi%7D-3.10%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF%3DF_1%2BF_2%3D20.28N%5Chat%7Bi%7D-0.78N%5Chat%7Bj%7D%5C%5C%5C%5C%7CF%7C%3D%5Csqrt%7B%2820.28N%29%5E2%2B%280.78N%29%5E2%7D%3D20.29N)
F = 20.29N
b) for 45°
![F_1=[9cos(22.5\°)\hat{i}+9sin(22.5\°)\hat{j}]N\\\\F_1=[8.31\hat{i}+3.44\hat{j}]N\\\\F_2=[12cos(22.5\°)\hat{i}-12sin(22.5\°)\hat{j}]N\\\\F_2=[11.08\hat{i}-4.59\hat{j}]N\\\\F=F_1+F_2=19.39N\hat{i}-1.15\hat{j}\\\\|F|=\sqrt{(19.39N)^2+(1.15N)^2}=19.42N](https://tex.z-dn.net/?f=F_1%3D%5B9cos%2822.5%5C%C2%B0%29%5Chat%7Bi%7D%2B9sin%2822.5%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_1%3D%5B8.31%5Chat%7Bi%7D%2B3.44%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B12cos%2822.5%5C%C2%B0%29%5Chat%7Bi%7D-12sin%2822.5%5C%C2%B0%29%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF_2%3D%5B11.08%5Chat%7Bi%7D-4.59%5Chat%7Bj%7D%5DN%5C%5C%5C%5CF%3DF_1%2BF_2%3D19.39N%5Chat%7Bi%7D-1.15%5Chat%7Bj%7D%5C%5C%5C%5C%7CF%7C%3D%5Csqrt%7B%2819.39N%29%5E2%2B%281.15N%29%5E2%7D%3D19.42N)
F= 19.42N
c) for 90°
for this case you can consider that the direction of both vectors are the y and x axis of the Cartesian plane:

F=15N
Answer:
See explanation
Explanation:
Range of a projectile is given by; U^2sin2θ/g
When θ = 45°
The maximum range is obtained as;
Rmax = U^2/g
Maximum range on earth = RME = U^2/g
Maximum range in the moon= RMM =U^2/g/6
Maximum range in Mars = RMMA = U^2/0.38g
a)
RME/RMM = U^2/g/U^2/g/6
RME/RMM = 6
But RME = 3 m
RMM = 3 * 6 = 18 m
b)
RME/RMMA =U^2/g/U^2/0.38g
RME/RMMA = 1/0.38
But RME = 3 m
RMMA = 3/0.38
RMMA = 7.9 m
Answer:
a)
b)
Explanation:
First, we need to obtain the linear momentum of the photons of wavelength 350nm.
We are going to use the following formula:

So the linear momentum is given by:

Having the linear momentum of the photon, we can calculate the speed of the hydrogen molecule to have the same momentum, we can use the classic formula for that:


The mass of the hydrogen molecule is given by:


What we've done here is to use the molecular weight of the hydrogen, and covert it kilograms, we had to multiply by two because the hydrogen molecule is found in pairs.
so:
