Answer:
488.6KN
Explanation:
Hello!
the first step to solve this problem we must find the pressure exerted at the bottom of the tank (P) which is the sum of the external air pressure (P1 = 92kPa), the pressure inside the tank (P2 = 100kPa) and the pressure due to the weight of the water (P3), taking into account the above we have the following equation
P=P1+P2+P3
to find the pressure at the bottom of the tank due to the weight of the water we use the following equation
where
α=density=1 g/cm^3=1000kg/M^3
H=height=14.1m
g=gravity=3.71m/s^2
solving
P3=(1000)(14.1)(3.71)=52311Pa=52.3kPa
P=P1+P2+P3
P=100kPa+92kPa+52.3kPa=244.3kPa
finally to solve the problem we remember that the pressure is the force exerted on the area
From the description of the substances, the best answer would be option A. Substance D would generally function as a solvent because it has the largest composition in a solution. A solvent in a solution has the highest percent composition among all components involved. Having a composition of more than 50% would mean that a substance is the solvent.
Answer:
1.3m/s
Explanation:
Data given,
Mass,m=1.0kg,
Amplitude,A=0.10m,
Frequency,f=2.0Hz.
From the equation of a simple harmonic motion, the displacement of the object at a given time is define as
we can express the velocity by the derivative of the displacement,
Hence
at equilibrium, the velocity becomes
Hence if we substitute values we arrive at
Answer:
true
Explanation:
Newton is the measure of the force with turns to be gravity multiplying the mass. Thus, the forces acts on the particles in the direction of the movement of the particles
Answer:
The value is 
Explanation:
From the question we are told that
The period of the asteroid is 
Generally the average distance of the asteroid from the sun is mathematically represented as
![R = \sqrt[3]{ \frac{G M * T^2 }{4 \pi} }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7BG%20M%20%2A%20T%5E2%20%7D%7B4%20%5Cpi%7D%20%7D)
Here M is the mass of the sun with a value
G is the gravitational constant with value 
![R = \sqrt[3]{ \frac{6.67 *10^{-11} * 1.99*10^{30} * [5.55 *10^{9}]^2 }{4 * 3.142 } }](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B6.67%20%2A10%5E%7B-11%7D%20%20%2A%201.99%2A10%5E%7B30%7D%20%2A%20%5B5.55%20%2A10%5E%7B9%7D%5D%5E2%20%7D%7B4%20%2A%203.142%20%7D%20%7D)
=> 
Generally
So
=> 
=>
