Answer:
Epithelial tissue and Muscle tissue
Explanation:
Answer:
d₂ = 1.466 m
Explanation:
In this case we must use the rotational equilibrium equations
Στ = 0
τ = F r
we must set a reference system, we use with origin at the easel B and an axis parallel to the plank
, we will use that the counterclockwise ratio is positive
+ W d₁ - w_cat d₂ = 0
d₂ = W / w d₁
d₂ = M /m d₁
d₂ = 5.00 /2.9 0.850
d₂ = 1.466 m
Part (a): Velocity of the snowball
By conservation of momentu;
m1v1 + m2v2 = m3v3,
Where, m1 = mass of snowball, v1, velocity of snowball, m2 = mass of the hat, v2 = velocity of the hat, m3 = mass of snowball and the hat, v3 = velocity of snowball and the hut.
v2 = 0, and therefore,
85*v1 + 0 = 220*8 => v1 = 220*8/85 = 20.71 m/s
Part (b): Horizontal range
x = v3*t
But,
y = vy -1/2gt^2, but y = -1.5 m (moving down), vy =0 (no vertical velocity), g = 9.81 m/s^2
Substituting;
-1.5 = 0 - 1/2*9.81*t^2
1.5 = 4.905*t^2
t = Sqrt (1.5/4.905) = 0.553 seconds
Then,
x = 8*0.553 = 4.424 m
Speed is the rate at which something covers a distance; velocity is the same but it takes into account whether it goes forwards or backwards; and acceleration is the rate of an increase in speed.
Answer:
The force on one side of the plate is 3093529.3 N.
Explanation:
Given that,
Side of square plate = 9 m
Angle = 60°
Water weight density = 9800 N/m³
Length of small strip is
![y=\dfrac{\Delta y}{\sin60}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B%5CDelta%20y%7D%7B%5Csin60%7D)
![y=\dfrac{2\Delta y}{\sqrt{3}}](https://tex.z-dn.net/?f=y%3D%5Cdfrac%7B2%5CDelta%20y%7D%7B%5Csqrt%7B3%7D%7D)
The area of strip is
![dA=\dfrac{9\times2\Delta y}{\sqrt{3}}](https://tex.z-dn.net/?f=dA%3D%5Cdfrac%7B9%5Ctimes2%5CDelta%20y%7D%7B%5Csqrt%7B3%7D%7D)
We need to calculate the force on one side of the plate
Using formula of pressure
![P=\dfrac{dF}{dA}](https://tex.z-dn.net/?f=P%3D%5Cdfrac%7BdF%7D%7BdA%7D)
![dF=P\times dA](https://tex.z-dn.net/?f=dF%3DP%5Ctimes%20dA)
On integrating
![\int{dF}=\int_{0}^{9\sin60}{\rho g\times y\times 6\sqrt{3}dy}](https://tex.z-dn.net/?f=%5Cint%7BdF%7D%3D%5Cint_%7B0%7D%5E%7B9%5Csin60%7D%7B%5Crho%20g%5Ctimes%20y%5Ctimes%206%5Csqrt%7B3%7Ddy%7D)
![F=9800\times6\sqrt{3}(\dfrac{y^2}{2})_{0}^{9\sin60}](https://tex.z-dn.net/?f=F%3D9800%5Ctimes6%5Csqrt%7B3%7D%28%5Cdfrac%7By%5E2%7D%7B2%7D%29_%7B0%7D%5E%7B9%5Csin60%7D)
![F=9800\times6\sqrt{3}\times(\dfrac{(9\sin60)^2}{2})](https://tex.z-dn.net/?f=F%3D9800%5Ctimes6%5Csqrt%7B3%7D%5Ctimes%28%5Cdfrac%7B%289%5Csin60%29%5E2%7D%7B2%7D%29)
![F=3093529.3\ N](https://tex.z-dn.net/?f=F%3D3093529.3%5C%20N)
Hence, The force on one side of the plate is 3093529.3 N.