Answer:
The first part of the question is asking about BUOYANT FORCE or UPTHRUST.
Upthrust =TRUE WEIGHT-APPARENT WEIGHT
TRUE WEIGHT=mg
TRUE weight=50kg×10m/s²
=500N
upthrust=500N-380N
FB=120N
volume of the rock=mass/density.
since the granite is completely submerged, the volume of the displaced liquid will be equal to the volume of the body.
upthrust=Vdg
120N=V×1000kg/m³×10m/s²
120N=V×10000kg/m²s²
120/10000=V
v=0.012m³
please mark brainliest, hope it helped
Answer:
3.28 m
3.28 s
Explanation:
We can adopt a system of reference with an axis along the incline, the origin being at the position of the girl and the positive X axis going up slope.
Then we know that the ball is subject to a constant acceleration of 0.25*g (2.45 m/s^2) pointing down slope. Since the acceleration is constant we can use the equation for constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a * t^2
X0 = 0
V0 = 4 m/s
a = -2.45 m/s^2 (because the acceleration is down slope)
Then:
X(t) = 4*t - 1.22*t^2
And the equation for speed is:
V(t) = V0 + a * t
V(t) = 4 - 2.45 * t
If we equate this to zero we can find the moment where it stops and begins rolling down, that will be the highest point:
0 = 4 - 2.45 * t
4 = 2.45 * t
t = 1.63 s
Replacing that time on the position equation:
X(1.63) = 4 * 1.63 - 1.22 * 1.63^2 = 3.28 m
To find the time it will take to return we equate the position equation to zero:
0 = 4 * t - 1.22 * t^2
Since this is a quadratic equation it will have to answers, one will be the moment the ball was released (t = 0), the other will eb the moment when it returns:
0 = t * (4 - 1.22*t)
t1 = 0
0 = 4 - 1.22*t2
1.22 * t2 = 4
t2 = 3.28 s
1. In 2010-2011, the temperature decreased. In 2011-present, the temperature started increasing. From 390 to 405+
2. As the more carbon dioxide increases the temperature also increases with it.
Huh huh what? ¿Can’t you translate?
Answer:
The tension is 
Explanation:
The free body diagram of the question is shown on the first uploaded image From the question we are told that
The distance between the two poles is 
The mass tied between the two cloth line is 
The distance it sags is 
The objective of this solution is to obtain the magnitude of the tension on the ends of the clothesline
Now the sum of the forces on the y-axis is zero assuming that the whole system is at equilibrium
And this can be mathematically represented as
To obtain 
we apply SOHCAHTOH Rule
So 
![\theta = tan^{-1} [\frac{opp}{adj} ]](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7Bopp%7D%7Badj%7D%20%5D)
![= tan^{-1} [\frac{1}{7}]](https://tex.z-dn.net/?f=%3D%20tan%5E%7B-1%7D%20%5B%5Cfrac%7B1%7D%7B7%7D%5D)
