PHYSICS COLLEGE

Find the net downward force on the tank's flat bottom, of area 1.60 m2 , exerted by the water and air inside the tank and the air outside the tank. Assume that the density of water is 1.00 g/cm3

Answers

Answer 1

Answer:

Answer:

The net downward force on the tank is $1.85*10^(5)\ N$

Explanation:

Given that,

Area = 1.60 m²

Suppose the design of a cylindrical, pressurized water tank for a future colony on Mars, where the acceleration due to gravity is 3.71 meters per second per second. The pressure at the surface of the water will be 150 K Pa , and the depth of the water will be 14.4 m . The pressure of the air in the building outside the tank will be 88.0 K Pa.

We need to calculate the net downward force on the tank

Using formula of formula

$F=(P+\rho* g* h-P_(out))A$

Where, P = pressure

g = gravity at mars

h = height

A = area

Put the value into the formula

$F=(150*10^3+1.00*10^3*3.71*14.4-88.0*10^(3))*1.60$

$F=1.85*10^(5)\ N$

Hence, The net downward force on the tank is $1.85*10^(5)\ N$

Answer 2

Answer:

Final answer:

The net downward force on the tank's flat bottom can be found by calculating the pressure at the bottom of the container.

Explanation:

Since the density is constant, the weight can be calculated using the density:

w = mg = pVg = pAhg.

The pressure at the bottom of the container is therefore equal to atmospheric pressure added to the weight of the fluid divided by the area.

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Beverage can is thrown upward and then falls back down to the floor. As usual, a y axis extends upward (positive direction). Which of the following best describes the acceleration of the can during its free flight?a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
b) +9.8 m/s^2 throughout
c) -9.8 m/s^2 throughout
d) zero throughout
e) +9.8 m/s^2, then momentarily zero, then -9.8 m/s^2

Answers

a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s

The acrylic plastic rod is 20 mm long and 15 mm in diameter. If an axial load of 300 N is applied to it, determine the change in its length and the change in its diameter. Eₚ = 2.70 GPa, vₚ = 0.4.

Answers

Given Information:

diameter = d = 15 mm

Length = L = 20 mm

Axial load = P = 300 N

Eₚ = 2.70x10⁹ Pa

vₚ = 0.4

Required Information:

Change in length = ?

Change in diameter = ?

Answer:

Change in length = 0.01257 mm

Change in diameter = -0.003772 mm

Explanation:

Stress is given by

σ = P/A

Where P is axial load and A is the area of the cross-section

A = 0.25πd²

A = 0.25π(0.015)²

A = 0.000176 m²

σ = 300/0.000176

σ = 1697792.8 Pa

The longitudinal stress is given by

εlong = σ/Eₚ

εlong = 1697792.8/2.70x10⁹

εlong = 0.0006288 mm/mm

The change in length can be found by using

δ = εlong*L

δ = 0.0006288*20

δ = 0.01257 mm

The lateral stress is given by

εlat = -vₚ*εlong

εlat = -0.4*0.0006288

εlat = -0.0002515 mm/mm

The change in diameter can be found by using

Δd = εlat*d

Δd = -0.0002515*15

Δd = -0.003772 mm

Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm

How does the mass of an object affect its motion through the air?

Answers

The motion of an object through the air does not affect by its mass. The rate of fall of objects does not depend upon the mass.

What are free fall and air resistance?

Free fall is a motion of a body in which gravity is the only force acting upon it. An object moving upwards might not be considered to be falling. But if the object is under the effect of the force of gravity, it is said to be in free fall.

Free fall is a type of motion in which the force acting upon an object is only gravity. Objects are not encountering a significant force of airresistance as they are only falling under the sole influence of gravity. All objects under such conditions will fall with the same rate of acceleration, regardless of their masses.

As an object falls through the air, have gone through some degree of air resistance. Air resistance is the collisions of the object's leading surface with molecules present in the air. The two most common factors that have a direct effect on the amount of air resistance are the cross-sectional area of the object and the speed of the object.

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Mass does not affect the speed of falling objects assuming there is only gravity acting on it. For example Both Bullets will strike the ground at the same time. The horizontal force applied does not affect the downward motion of the bullets — only gravity and friction (air resistance), which is the same for both bullets
(Sorry if u don’t get what I mean)

The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?

Answers

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery.Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.

I hope it helps you!

You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 90.3 minutes, what is the half-life of this substance?

Answers

Answer : The half-life of this substance will be, 45 minutes.

Explanation :

First we have to calculate the value of rate constant.

Expression for rate law for first order kinetics is given by:

$k=(2.303)/(t)\log(a)/(a-x)$

where,

k = rate constant = ?

t = time passed by the sample = 90.3 min

a = initial amount of the reactant = 400

a - x = amount left after decay process = 100

Now put all the given values in above equation, we get

$k=(2.303)/(90.3min)\log(400)/(100)$

$k=1.54* 10^(-2)\text{ min}^(-1)$

Now we have to calculate the half-life of substance, we use the formula :

$k=(0.693)/(t_(1/2))$

$1.54* 10^(-2)\text{ min}^(-1)=(0.693)/(t_(1/2))$

$t_(1/2)=45min$

Therefore, the half-life of this substance will be, 45 minutes.

Three children want to play on a see saw that is 6 meters long and has a fulcrum in the middle. Two of the children are twins and weigh 40 kg each and sit on the same side at a distance 2m and 3m away from the fulcrum. The other child weighs 80 kg. How far away should he sit from the fulcrum so that the see saw is balanced

Answers

Given Information:

mass of child 1 = m₁ = 40 kg

distance from fulcrum of child 1 = d₁ = 2 m

mass of child 2 = m₂ = 40 kg

distance from fulcrum of child 2 = d₂ = 3 m

mass of child 3 = m₃ = 80 kg

Required Information:

distance from fulcrum of child 3 = d₃ = ?

Answer:

distance from fulcrum of child 3 = 2.5 m

Explanation:

In order to balance the see-saw, the moment of force should be same on both sides of the fulcrum.

Since 2 children are sitting on one side and only 1 on the other side

F₁d₁ + F₂d₂ = F₃d₃

Where Force is given by

F = mg

m₁gd₁ + m₂gd₂ = m₃gd₃

m₁d₁ + m₂d₂ = m₃d₃

Re-arrange the equation for d₃

m₃d₃ = m₁d₁ + m₂d₂

d₃ = (m₁d₁ + m₂d₂)/m₃

d₃ = (40*2 + 40*3)/80

d₃ = 2.5 m

Therefore, the child on the other side should sit 2.5 m from the fulcrum so that the see-saw remains balanced.

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