PHYSICS COLLEGE

What causes rain? a.air becomes filled with water vapor
b.water vapor condenses on dust particles
c. dust particles can no longer support water droplets

Answers

Answer 1

Answer:

Answer:

the awnser is A

Explanation:

Brainliest Please

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A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?

Answers

The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".

Angle and Force

According to the question,

Angle, a = 3.7 degrees or,

Sin a = 0.064

Force, F = 46 tons

We know the relation,

Parallel (tangential), $F_t$ = F Sin a

By substituting the values,

= 46 × 0.064

= 2.9 tons

Thus the response above is appropriate answer.

Find out more information about Force here:

brainly.com/question/25239010

Answer:

2.9tons

Explanation:

Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:

parallel ("tangential"): F_t = F sin a

perpendicular ("normal"): F_n = F cos a

At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:

F sin a ~~ (46 tons)*0.064 ~~ 2.9tons

Also see attached file

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the kinetic energy received in the above situation?

Answers

Answer:

The compression is $√(2) \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_(potential) = (1)/(2) k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_(kinetic) = (1)/(2) m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$(1)/(2) m v_1^2 = (1)/(2) k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * ((1)/(2) m v_1^2) = (1)/(2) k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * ((1)/(2) k d^2) = (1)/(2) k D^2$

$D^2 = (2 \ ((1)/(2) k d^2))/((1)/(2) k)$

$D^2 = 2 \ d^2$

$D = √(2 \ d^2)$

$D = √(2) \ d$

And this is the compression we are looking for

Answer:

$d'=√(2) d$

Explanation:

By hooke's law we have that the potential energy can be defined as:

$U=(kd^(2) )/(2)$

Where k is the spring constant and d is the compression distance, the kinetic energy can be written as

$K=(mv^(2) )/(2)$

By conservation of energy we have:

$(mv^(2) )/(2)=(kd^(2) )/(2)$ (1)

If we double the kinetic energy

$2((mv^(2) )/(2))=(kd'^(2) )/(2)$ (2)

where d' is the new compression, now if we input (1) in (2) we have

$2((kd^(2) )/(2))=(kd'^(2) )/(2)$

$2((d^(2) )/(2))=(d'^(2) )/(2)$

$d'=√(2) d$

At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature is 3.72 x 10-3 atm/K. Calculate the molar latent enthalpy of boiling of iron at 3330 K:

Answers

The molar latent enthalpy of boiling of iron at 3330 K is ΔH = 342 $*$ 10^3 J.

Explanation:

Molar enthalpy of fusion is the amount of energy needed to change one mole of a substance from the solid phase to the liquid phase at constant temperature and pressure.

d ln p = (ΔH / RT^2) dt

(1/p) dp = (ΔH / RT^2) dt

dp / dt = p (ΔH / RT^2) = 3.72 $*$ 10^-3

(p) (ΔH) / (8.31) (3330)^2 = 3.72 $*$ 10^-3

ΔH = 342 $*$ 10^3 J.

A tuning fork vibrates at 15,660 oscillations every minute. What is the period (in seconds) of one back and forth vibration of the tuning fork?

Answers

We are given:

The tuning fork vibrates at 15660 oscillations per minute

Period of one back-and forth movement:

the given data can be rewritten as:

1 minute / 15660 oscillations

60 seconds / 15660 oscillations (1 minute = 60 seconds)

dividing the values

0.0038 seconds / Oscillation

Therefore, one back and forth vibration takes 0.0038 seconds

A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the spring is compressed by 0.097 m. After the block is released from rest, it travels upward and then leaves the spring. To what maximum height above the point of release does it rise?

Answers

Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

Given that,

Mass of block = 0.221 kg

Spring constant k = 5365 N/m

Distance x = 0.097 m

We need to calculate the height

Using stored energy in spring

$U=(1)/(2)kx^2$ ...(I)

Using gravitational potential energy

$U' =mgh$ ....(II)

Using energy of conservation

$E_(i)=E_(f)$

$U_(i)+U'_(i)=U_(f)+U'_(f)$

$(1)/(2)kx^2+0=0+mgh$

$h=(kx^2)/(2mg)$

Where, k = spring constant

m = mass of the block

x = distance

g = acceleration due to gravity

Put the value in the equation

$h=(5365*(0.097)^2)/(2*0.221*9.8)$

$h=11.653\ m$

Hence, The maximum height above the point of release is 11.653 m.

The surface pressure of the atmosphere is about 14.7 psi (pounds per square inch). How many pounds per square yard does that amount to

Answers

Answer:

14.7 psi is equal to 19051.2 pounds per square yard.

Explanation:

Dimensionally speaking, a square yard equals 1296 square inches. Therefore, we need to multiply the atmospheric pressure by 1296 to obtain its equivalent in pounds per square yard. That is:

$p = 14.7\,(lbf)/(in^(2))* 1296\,(in^(2))/(yd^(2))$