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wolverine [178]
3 years ago
6

1. Astronomical observatories have been available since ancient times, and many cultures set aside special sites for astronomica

l observations. The thing modern observatories have that was missing from these older observatories until about 1610 was: a. a dark site, where lights did not get in the way b. graduate students or other assistants for sharing the astronomer’s work c. a method of keeping good records d. a way of measuring from what general direction in the sky a beam of light was coming e. telescopes
Physics
1 answer:
nydimaria [60]3 years ago
4 0

Answer:

e telescopes

Explanation:

may i be marked brainliest?

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What is an ampere?<br><br> NO GUESSING! ANSWERS ONLY!
Igoryamba

Answer:

the SI base unit of electrical current

7 0
3 years ago
Read 2 more answers
The wavelength of an electromagnetic waves is __________ also known as the period. the number of waves that pass a given point i
Andreas93 [3]

The sentence can be completed as follows:

The wavelength of an electromagnetic waves is the spatial distance between two successive troughs.


Note that the wavelength of a wave can also measured as the spatial distance between two successive crests of the wave. Also note that the second part of the sentence ("also known as the period") is not true, because period is another thing (in fact, the period is the time interval between two successive troughs).



6 0
3 years ago
A: How far did she travel? <br><br><br><br> B: How long did she take?
allochka39001 [22]
A. 60 miles
B. 5 hours

Unless you are looking for slope, in which case the answer is different
7 0
3 years ago
If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock com
Lelu [443]

Answer:

<em> B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.</em>

<em></em>

Explanation:

This is the complete question

A person will feel a shock when a current of greater than approximately 100μ A flows between his index finger and thumb. If the resistance of dry skin is 200 times larger than the resistance of wet skin, how do the maximum voltages without shock compare in each scenario?

A) The voltage on dry skin needs to be 200 times smaller than the voltage on wet skin.

B) The voltage on dry skin needs to be 200 times larger than the voltage on wet skin.

C) The voltage on dry skin is the same as the voltage on wet skin.

D) The voltage on dry skin needs to be 40,000 times larger than the voltage on wet skin.

Ohm's law states that electric current is proportional to voltage and inversely proportional to resistance.

the equation is written as

V = IR

Where V is the voltage

I is the current

R is the resistance

for this case, the current I is 100μ A = 100 x 10^16 A

resistance of wet skin = R

resistance of dry skin = 200R

for the wet skin, voltage will be

V = IR = 100*10^{-6} R

for dry skin, voltage will be

V = IR = 100*10^{-6}*200R = 0.02R

Comparing both voltages

0.02R ÷  100*10^{-6} R  = 200

<em>this means that the voltage on the wet skin should be 200 times lesser than the voltage on the dry skin or the voltage on the dry skin should be 200 times more than the voltage on the wet skin.</em>

4 0
4 years ago
The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
madam [21]

Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

  • E = hf  
  • E = h.v/λ
  • λ = h/(mv)
  • E = (1/2)mv²

Where:

  • E = kinetic energy of the particle
  • λ = de-Broglie wavelength
  • m = mass of the particle
  • v = speed of the particle
  • h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

       \dfrac{m_\alpha}{m_p}=4

For the kinetic energies you find:

          \dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}

            \dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4

Thus:

           \dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}

          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

5 0
3 years ago
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