Answer:
c. 
Explanation:
In order to calculate the gravitational force on the mass of the center, you take into account the following formula:
(1)
Furthermore, you take into account the components of the resultant vector.
By the illustration, you have that the force is given by:
![F_T=F_1+F_2+F_3+F_4\\\\F_1=\frac{Gm_1m}{r^2}[-cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_2=\frac{Gm_2m}{r^2}[cos45\°\hat{i}+sin45\°\hat{j}]\\\\F_3=\frac{Gm_3m}{r^2}[cos45\°\hat{i}-sin45\°\hat{j}]\\\\F_4=\frac{Gm_4m}{r^2}[-cos45\°\hat{i}-sin45\°\hat{j}]](https://tex.z-dn.net/?f=F_T%3DF_1%2BF_2%2BF_3%2BF_4%5C%5C%5C%5CF_1%3D%5Cfrac%7BGm_1m%7D%7Br%5E2%7D%5B-cos45%5C%C2%B0%5Chat%7Bi%7D%2Bsin45%5C%C2%B0%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_2%3D%5Cfrac%7BGm_2m%7D%7Br%5E2%7D%5Bcos45%5C%C2%B0%5Chat%7Bi%7D%2Bsin45%5C%C2%B0%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_3%3D%5Cfrac%7BGm_3m%7D%7Br%5E2%7D%5Bcos45%5C%C2%B0%5Chat%7Bi%7D-sin45%5C%C2%B0%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_4%3D%5Cfrac%7BGm_4m%7D%7Br%5E2%7D%5B-cos45%5C%C2%B0%5Chat%7Bi%7D-sin45%5C%C2%B0%5Chat%7Bj%7D%5D)
where:
m1 = m
m2 = 2m
m3 = m
m4 = 4m
m: mass at the center of the system
The distance r is:
You replace the values for all masses and sum the contributions of all forces:
![F_1=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[-\hat{i}+\hat{j}]=\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}+\hat{j}]\\\\F_2=\frac{\sqrt{2}}{2}\frac{2Gm^2}{(\frac{d^2}{2})}[\hat{i}+\hat{j}]=2\sqrt{2}\frac{Gm^2}{d^2}[\hat{i}+\hat{j}]\\\\F_3=\frac{\sqrt{2}}{2}\frac{Gm^2}{(\frac{d^2}{2})}[\hat{i}-\hat{j}]=\sqrt{2}\frac{Gm^2}{s^2}[\hat{i}-\hat{j}]\\\\F_4=\frac{\sqrt{2}}{2}\frac{4Gm^2}{(\frac{d^2}{2})}[-\hat{i}-\hat{j}]=4\sqrt{2}\frac{Gm^2}{d^2}[-\hat{i}-\hat{j}]\\\\F_T=-2\sqrt{2}\frac{Gm^2}{d^2}}[\hat{i}+\hat{j}]](https://tex.z-dn.net/?f=F_1%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cfrac%7BGm%5E2%7D%7B%28%5Cfrac%7Bd%5E2%7D%7B2%7D%29%7D%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5D%3D%5Csqrt%7B2%7D%5Cfrac%7BGm%5E2%7D%7Bd%5E2%7D%5B-%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_2%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cfrac%7B2Gm%5E2%7D%7B%28%5Cfrac%7Bd%5E2%7D%7B2%7D%29%7D%5B%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5D%3D2%5Csqrt%7B2%7D%5Cfrac%7BGm%5E2%7D%7Bd%5E2%7D%5B%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_3%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cfrac%7BGm%5E2%7D%7B%28%5Cfrac%7Bd%5E2%7D%7B2%7D%29%7D%5B%5Chat%7Bi%7D-%5Chat%7Bj%7D%5D%3D%5Csqrt%7B2%7D%5Cfrac%7BGm%5E2%7D%7Bs%5E2%7D%5B%5Chat%7Bi%7D-%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_4%3D%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%5Cfrac%7B4Gm%5E2%7D%7B%28%5Cfrac%7Bd%5E2%7D%7B2%7D%29%7D%5B-%5Chat%7Bi%7D-%5Chat%7Bj%7D%5D%3D4%5Csqrt%7B2%7D%5Cfrac%7BGm%5E2%7D%7Bd%5E2%7D%5B-%5Chat%7Bi%7D-%5Chat%7Bj%7D%5D%5C%5C%5C%5CF_T%3D-2%5Csqrt%7B2%7D%5Cfrac%7BGm%5E2%7D%7Bd%5E2%7D%7D%5B%5Chat%7Bi%7D%2B%5Chat%7Bj%7D%5D)
and the magnitude is:
c.
Answer:
96.5 km/h
Explanation:
The average velocity of the train is given by:
where
d is the displacement
t is the time taken
For this train, we have:
d = 55 km south (displacement is a vector, so we must also consider the direction)
Substituting into the equation, we find the average velocity:
<span>Mass doesn't change when the temperature
of the ball changes.
(Unless, of course, it gets so hot that it melts,
and part of it falls off
and rolls under the table.)</span>
Answer:
F = 39.2 N
Explanation:
Since, the object is in uniform motion. Therefore, the frictional force on object will be:
Frictional Force = μk N = μk mg
where,
μk = coefficient of kinetic friction = 0.2
m = mass of crate = 10 kg
g = 9.8 m/s²
Therefore,
Frictional Force = (0.2)(10 kg)(9.8 m/s²)
Frictional Force = 19.6 N
The horizontal component of force must be equal to this frictional force to continue the uniform motion:
F Sin 30° = 19.6 N
F = 19.6 N/Sin 30°
<u>F = 39.2 N</u>
