The mass of the second car is 1434.21 kg
<u>Explanation:</u>
Using law of conservation of momentum,
Given:
= 1090 kg
= 11 m/s
= 0
v = 4.75 m/s
We need to find 
When substituting the given values in the above equation, we get
Answer: The question has some details missing. here is the complete question ; Point charge 1.5 μC is located at x = 0, y = 0.30 m, point charge -1.5 μC is located at x = 0 y = -0.30m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 5.0 μC at x = 0.40 m, y = 0
Explanation:
- a) First of all find the distance between the two charges;
- x = 0, y = 0.30 and x = 0.40 m, y = 0
hence, the force F = 2Kq1q2cosθ /r²...............equation 1
but cosθ = y/r = 0.3/0.5
cosθ = 0.6
plugging back to equation 1;
F = 2 x 9 x 10^9 x 1.5 x 10^-6 x 5 x 10^-6 /0.5^2
F = 540 x 10^-3
Magnitude of Force = 0.54N
b) Direction is at angle 90
Absolutely false!! Creativity, curiosity encouraged in Scientific endeavors. Science is all about exploring knowledge, it just promotes intelligency!
In Short, Your Answer would be "False"
Hope this helps!
Answer:
áp dụng công thức v = s/t
s là dộ dài qduong
v là vận tốc
t là thời gian
xuyên suốt 2 câu hỏi đều dùng công thức này
Explanation:
Answer:
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
Explanation:
First, we must calculate the resultant force (
), in newtons, by vectorial sum:
![\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%20%5B%28-200%5C%2CN%29%5Ccdot%20%5Ccos%2060%5E%7B%5Ccirc%7D%2B%28400%5C%2CN%29%5Ccdot%20%5Ccos%2045%5E%7B%5Ccirc%7D%2B300%5C%2CN%5D%5C%2C%5Chat%7Bi%7D%20%2B%20%5B%28200%5C%2CN%29%5Ccdot%20%5Csin%2060%5E%7B%5Ccirc%7D%20%2B%20%28400%5C%2CN%29%5Ccdot%20%5Csin%2045%5E%7B%5Ccirc%7D-100%5C%2CN%5D%5C%2C%5Chat%7Bj%7D)
(1)
Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:
Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:
Where 
is the direction of the resultant force, in sexagesimal degrees.
The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.
