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3 years ago

How long have advances in astronomy been occurring? Only for the past few years

Physics

1 answer:

solniwko [45]3 years ago

3 0

I'd have to say that the list of choices doesn't go far enough.

Advances in Astronomy have been occurring for at least the past two millennia (2000 years). Maybe longer.

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The terminal velocity of a person falling in air depends upon the weight and the area of the person facing the fluid. Find the t

AVprozaik [17]

Answer:

v=115 m/s

v=414 km/h

Explanation:

Given data

$A_{area}=0.140m^{2}\\ p_{air}=1.21 kg/m^{3}\\ m_{mass}=80kg$

To find

Terminal velocity (in meters per second and kilometers per hour)

Solution

At terminal speed the weight equal the drag force

$mg=1/2*C*p_{air}*v^{2}*A_{area}\\ v=\sqrt{\frac{2*m*g}{C**p_{air}*A_{area}} }\\ Where C=0.7\\v=\sqrt{\frac{2*9.8*80}{1.21*0.14*0.7} }\\ v=115m/s$

For speed in km/h(kilometers per hour)

To convert m/s to km/h you need to multiply the speed value by 3.6

$v=(115*3.6)km/h\\v=414km/h$

5 0

3 years ago

The radioactive 60co isotope is used in nuclear medicine to treat certain types of cancer. Calculate the wavelength and frequenc

Ivanshal [37]

1. Frequency: $3.23\cdot 10^{20} Hz$

The energy given is the energy per mole of particles:

$E=1.29\cdot 10^{11} J/mol$

1 mole contains a number of Avogadro of particles, $N_A$ , equal to

$N_A=6.022\cdot 10^{23}$ particles

So, by setting the following proportion, we can calculate the energy of a single photon:

$1.29 \cdot 10^{11} J/mol : 6.022 \cdot 10^{23} ph/mol = E_1 : 1 ph\\E_1 = \frac{(1.29\cdot 10^{11} J/mol)(1 ph)}{6.022\cdot 10^{23} ph/mol}=2.14\cdot 10^{-13} J$

This is the energy of a single photon; now we can calculate its frequency by using the formula:

$E_1 = hf$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck's constant

f is the photon frequency

Solving for f, we find

$f=\frac{E_1}{h}=\frac{2.14\cdot 10^{-13} J}{6.63\cdot 10^{-34} Js}=3.23\cdot 10^{20} Hz$

2. Wavelength: $9.29\cdot 10^{-13} m$

The wavelength of the photon is given by the equation:

$\lambda=\frac{c}{f}$

where

$c=3\cdot 10^8 m/s$

is the speed of the photon (the speed of light). Substituting,

$\lambda=\frac{3 \cdot 10^8 m/s}{3.23\cdot 10^{20} Hz}=9.29\cdot 10^{-13} m$

6 0

3 years ago

You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 294.0 kPa when the ther

Travka [436]

Answer: 361° C

Explanation:

Given

Initial pressure of the gas, P1 = 294 kPa

Final pressure of the gas, P2 = 500 kPa

Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K

Final temperature of the gas, T2 = ?

Let us assume that the gas is an ideal gas, then we use the equation below to solve

T2/T1 = P2/P1

T2 = T1 * (P2/P1)

T2 = (100 + 273) * (500 / 294)

T2 = 373 * (500 / 294)

T2 = 373 * 1.7

T2 = 634 K

T2 = 634 K - 273 K = 361° C

5 0

3 years ago

Consider the model above. It represents the electrical force. As r increases, the attractive force decreases. How would this mod

aivan3 [116]

Answer:

As we keep on increasing the radius the value of the gravitation force of attraction decreases and as we decrease the radius the gravitation force increases.

Explanation:

Like the coulombs law of electrostatics, the law of gravitation also depends inversely on the square of the value of r. Therefore, as we keep on increasing the value of r the value of the gravitation force decreases and as we decrease the value of the r the value of gravitation force increases.

Gravitation Force= $\frac{Gm_{1}m_{2} }{r^{2}}$

Coulombs's Law= $\frac{Kq_{1}q_{2} }{r^{2}}$

6 0

3 years ago

Read 2 more answers

The molar enthalpy of fusion for water is 6.008 kj/mol. what quantity of energy is released when 253g of liquid water freezes? (

e-lub [12.9K]

During freezing, energy is released by the mass of water without change in temperature. Such energy will also be required if the same mass of water has to be melted.

Then,

Number of moles = mass/molar mass = 253/18.02 =14.04 moles

Energy released = moles*molar enthalpy of fusion = 14.04*6.008 = 84.35 kJ

6 0

3 years ago