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°.
Exothermich cools and endo heats so it heats
Answer:
a) Tc = 750 [N] ;b) See the explanation below.
Explanation:
To solve this problem, we first need a graphical explanation of this, as well as knowing the corresponding questions. Therefore, a search was carried out in google, in the attached image we will find a graphical description of the problem.
b)
The solution of this type of problem corresponds to the use of Newton's third law, applying static which tells us that the sum of the forces in a system in equilibrium without movement must be equal to zero.
a)
In this way we can find by means of a sum of forces on the y axis equal to zero:
- 850 - 450 + 550 + Tc = 0
Tc = 750 [N]
Answer:
0.8214 m/s^2
Explanation:
Fnet= Fpushed - Ffriction
Fpushed = 12.7N Ffriction = 8.33N
Fnet = 12.7N - 8.33N = 4.37N
Fnet= mass(acceleration)
Fnet = 4.37N mass = 5.32 kg
4.37N = 5.32 kg(acceleration)
acceleration= 0.8214 m/s^2
Taking specific heat of lead as 0.128 J/gK = c
We have energy of ball at 7.00 meter height = mgh = 
When leads gets heated by a temperature ΔT energy needed = mcΔT
= 
ΔT
Comparing both the equations
= ΔT
ΔT = 0.536 K
Change in temperature same in degree and kelvin scale
So ΔT = 0.536 
