ANSWER
T₂ = 10.19N
EXPLANATION
Given:
• The mass of the ball, m = 1.8kg
First, we draw the forces acting on the ball, adding the vertical and horizontal components of each one,
In this position, the ball is at rest, so, by Newton's second law of motion, for each direction we have,
![\begin{gathered} T_{1y}-F_g=0_{}_{}_{} \\ T_2-T_{1x}=0 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T_%7B1y%7D-F_g%3D0_%7B%7D_%7B%7D_%7B%7D%20%5C%5C%20T_2-T_%7B1x%7D%3D0%20%5Cend%7Bgathered%7D)
The components of the tension of the first string can be found considering that they form a right triangle, where the vector of the tension is the hypotenuse,
![\begin{gathered} T_{1y}=T_1\cdot\cos 30\degree \\ T_{1x}=T_1\cdot\sin 30\degree \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20T_%7B1y%7D%3DT_1%5Ccdot%5Ccos%2030%5Cdegree%20%5C%5C%20T_%7B1x%7D%3DT_1%5Ccdot%5Csin%2030%5Cdegree%20%5Cend%7Bgathered%7D)
We have to find the tension in the horizontal string, T₂, but first, we have to find the tension 1 using the first equation,
![T_1\cos 30\degree-m\cdot g=0](https://tex.z-dn.net/?f=T_1%5Ccos%2030%5Cdegree-m%5Ccdot%20g%3D0)
Solve for T₁,
![T_1=\frac{m\cdot g}{\cos30\degree}=\frac{1.8kg\cdot9.8m/s^2}{\cos 30\degree}\approx20.37N](https://tex.z-dn.net/?f=T_1%3D%5Cfrac%7Bm%5Ccdot%20g%7D%7B%5Ccos30%5Cdegree%7D%3D%5Cfrac%7B1.8kg%5Ccdot9.8m%2Fs%5E2%7D%7B%5Ccos%2030%5Cdegree%7D%5Capprox20.37N)
Now, we use the second equation to find the tension in the horizontal string,
![T_2-T_1\sin 30\degree=0](https://tex.z-dn.net/?f=T_2-T_1%5Csin%2030%5Cdegree%3D0)
Solve for T₂,
![T_2=T_1\sin 30\degree=20.37N\cdot\sin 30\degree\approx10.19N](https://tex.z-dn.net/?f=T_2%3DT_1%5Csin%2030%5Cdegree%3D20.37N%5Ccdot%5Csin%2030%5Cdegree%5Capprox10.19N)
Hence, the tension in the horizontal string is 10.19N, rounded to the nearest hundredth.
We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.
I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.
Answer: The answer is A for sure
Explanation:
That is, there will be no acceleration. If you are sitting at rest in a chair and the upward push of the chair is equal to the downward pull of gravity, you will stay at rest in the chair. ... You now have an unbalanced force acting on you and therefore, according to Newton's First Law, your motion is going to change.
Plz give brainlest :)
The circuit is no longer closed.
Answer:
The time it will take for the car to reach a velocity of 28 m/s is 7 seconds
Explanation:
The parameters of the car are;
The acceleration of the car, a = 4 m/s²
The final velocity of the car, v = 28 m/s
The initial velocity of the car, u = 0 m/s (The car starts from rest)
The kinematic equation that can be used for finding (the time) how long it will take for the car to reach a velocity of 28 m/s is given as follows;
v = u + a·t
Where;
v = The final velocity of the car, v = 28 m/s
u = The initial velocity of the car = 0 m/s
a = The acceleration of the car = 4 m/s²
t = =The time it will take for the car to reach a velocity of 28 m/s
Therefore, we get;
t = (v - u)/a
t = (28 m/s - 0 m/s)/(4 m/s²) = 7 s
The time it will take for the car to reach a velocity of 28 m/s, t = 7 seconds.