Conserve natural resources, energy and landfill space.
Answer:
The bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
Explanation:
Given;
density of the liquid, ρ = 1500 kg/m³
frequency of the wave, F = 410 Hz
wavelength of the sound, λ = 7.80 m
The speed of the wave is calculated as;
v = Fλ
v = 410 x 7.8
v = 3,198 m/s
The bulk modulus of the liquid is calculated as;
![V = \sqrt{\frac{B}{\rho} } \\\\V^2 = \frac{B}{\rho}\\\\B = V^2 \rho\\\\B = (3,198 \ m/s)^2 \times 1500 \ kg/m^3\\\\B = 1.534 \ \times 10^{10} \ N/m^2](https://tex.z-dn.net/?f=V%20%3D%20%5Csqrt%7B%5Cfrac%7BB%7D%7B%5Crho%7D%20%7D%20%5C%5C%5C%5CV%5E2%20%3D%20%5Cfrac%7BB%7D%7B%5Crho%7D%5C%5C%5C%5CB%20%3D%20V%5E2%20%5Crho%5C%5C%5C%5CB%20%3D%20%283%2C198%20%5C%20m%2Fs%29%5E2%20%5Ctimes%201500%20%5C%20kg%2Fm%5E3%5C%5C%5C%5CB%20%3D%201.534%20%5C%20%5Ctimes%2010%5E%7B10%7D%20%5C%20N%2Fm%5E2)
Therefore, the bulk modulus of the liquid is 1.534 x 10¹⁰ N/m²
Acceleration
Explanation:
Acceleration is a physical quantity that expresses the change in the velocity of a body per unit of time.
Acceleration = ![\frac{V - U }{T}](https://tex.z-dn.net/?f=%5Cfrac%7BV%20-%20U%20%7D%7BT%7D)
V is the initial velocity
U is the final velocity
T is the time
It is has a unit of m/s²
Learn more:
Acceleration brainly.com/question/3820012
#learnwithBrainly
Answer:
Explanation:
Single-phase transformers can operate to either increasing or decreasing the voltage applied to the primary winding. When a transformer is used to “increase” the voltage on the secondary winding with respect to the primary, it is called a Step-up transformer
Answer:
Part a)
![T = 2\sqrt{\frac{R}{3g}}](https://tex.z-dn.net/?f=T%20%3D%202%5Csqrt%7B%5Cfrac%7BR%7D%7B3g%7D%7D)
Part b)
![v_x = \frac{\sqrt{3Rg}}{2}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7B%5Csqrt%7B3Rg%7D%7D%7B2%7D)
Part c)
![v_y = \sqrt{Rg/3}](https://tex.z-dn.net/?f=v_y%20%3D%20%5Csqrt%7BRg%2F3%7D)
Part d)
![v = \frac{1}{2}\sqrt{13Rg}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Csqrt%7B13Rg%7D)
Part e)
![\theta_i = 33.7 degree](https://tex.z-dn.net/?f=%5Ctheta_i%20%3D%2033.7%20degree)
Part f)
![H = \frac{13R}{8}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7B13R%7D%7B8%7D)
Part g)
![X = \frac{13R}{4}](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7B13R%7D%7B4%7D)
Explanation:
Initial speed of the launch is given as
initial speed = ![v_i](https://tex.z-dn.net/?f=v_i%20)
angle =
degree
Now the two components of the velocity
![v_x = v_i cos\theta_i](https://tex.z-dn.net/?f=v_x%20%3D%20v_i%20cos%5Ctheta_i)
similarly we have
![v_y = v_i sin\theta_i](https://tex.z-dn.net/?f=v_y%20%3D%20v_i%20sin%5Ctheta_i)
Part a)
Now we know that horizontal range is given as
![R = \frac{v_i^2 (2sin\theta_icos\theta_i)}{g}](https://tex.z-dn.net/?f=R%20%3D%20%5Cfrac%7Bv_i%5E2%20%282sin%5Ctheta_icos%5Ctheta_i%29%7D%7Bg%7D)
maximum height is given as
![H = \frac{R}{6} = \frac{v_i^2 sin^2\theta_i}{2g}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7BR%7D%7B6%7D%20%3D%20%5Cfrac%7Bv_i%5E2%20sin%5E2%5Ctheta_i%7D%7B2g%7D)
so we have
![v_i sin\theta = \sqrt{Rg/3}](https://tex.z-dn.net/?f=v_i%20sin%5Ctheta%20%3D%20%5Csqrt%7BRg%2F3%7D)
time of flight is given as
![T = \frac{2v_isin\theta_i}{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2v_isin%5Ctheta_i%7D%7Bg%7D)
![T = \frac{2\sqrt{Rg/3}}{g}](https://tex.z-dn.net/?f=T%20%3D%20%5Cfrac%7B2%5Csqrt%7BRg%2F3%7D%7D%7Bg%7D)
![T = 2\sqrt{\frac{R}{3g}}](https://tex.z-dn.net/?f=T%20%3D%202%5Csqrt%7B%5Cfrac%7BR%7D%7B3g%7D%7D)
Part b)
Now the speed of the ball in x direction is always constant
so at the peak of its path the speed of the ball is given as
![R = v_x T](https://tex.z-dn.net/?f=R%20%3D%20v_x%20T)
![R = v_x 2\sqrt{\frac{R}{3g}}](https://tex.z-dn.net/?f=R%20%3D%20v_x%202%5Csqrt%7B%5Cfrac%7BR%7D%7B3g%7D%7D)
![v_x = \frac{\sqrt{3Rg}}{2}](https://tex.z-dn.net/?f=v_x%20%3D%20%5Cfrac%7B%5Csqrt%7B3Rg%7D%7D%7B2%7D)
Part c)
Initial vertical velocity is given as
![v_y = v_i sin\theta_i](https://tex.z-dn.net/?f=v_y%20%3D%20v_i%20sin%5Ctheta_i)
![v_i sin\theta = \sqrt{Rg/3}](https://tex.z-dn.net/?f=v_i%20sin%5Ctheta%20%3D%20%5Csqrt%7BRg%2F3%7D)
Part d)
Initial speed is given as
![v = \sqrt{v_x^2 + v_y^2}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7Bv_x%5E2%20%2B%20v_y%5E2%7D)
so we will have
![v = \sqrt{Rg/3 + 3Rg/4}](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7BRg%2F3%20%2B%203Rg%2F4%7D)
![v = \frac{1}{2}\sqrt{13Rg}](https://tex.z-dn.net/?f=v%20%3D%20%5Cfrac%7B1%7D%7B2%7D%5Csqrt%7B13Rg%7D)
Part e)
Angle of projection is given as
![tan\theta_i = \frac{v_y}{v_x}](https://tex.z-dn.net/?f=tan%5Ctheta_i%20%3D%20%5Cfrac%7Bv_y%7D%7Bv_x%7D)
![tan\theta_i = \frac{\sqrt{Rg/3}}{\sqrt{3Rg}/2}](https://tex.z-dn.net/?f=tan%5Ctheta_i%20%3D%20%5Cfrac%7B%5Csqrt%7BRg%2F3%7D%7D%7B%5Csqrt%7B3Rg%7D%2F2%7D)
![\theta_i = 33.7 degree](https://tex.z-dn.net/?f=%5Ctheta_i%20%3D%2033.7%20degree)
Part f)
If we throw at same speed so that it reach maximum height
then the height will be given as
![H = \frac{v^2}{2g}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7Bv%5E2%7D%7B2g%7D)
![H = \frac{13R}{8}](https://tex.z-dn.net/?f=H%20%3D%20%5Cfrac%7B13R%7D%7B8%7D)
Part g)
For maximum range the angle should be 45 degree
so maximum range is
![X = \frac{v^2}{g}](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7Bv%5E2%7D%7Bg%7D)
![X = \frac{13R}{4}](https://tex.z-dn.net/?f=X%20%3D%20%5Cfrac%7B13R%7D%7B4%7D)