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slega [8]
4 years ago
14

The sun is directly overhead at the equator on what day of the year

Physics
2 answers:
Aleks [24]4 years ago
8 0

Equinox...

The Sun is directly overhead at "high-noon" on the equator twice per year, at the two equinoxes. Spring (or Vernal) Equinox is usually March 20, and Fall (or Autumnal) equinox is usually September 22.

Phoenix [80]4 years ago
7 0
Both equinoxes ... September 21 and March 21.
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what is the energy of a single photon of ultraviolet light with a wavelength of 30.0 nm? 1 nm=10^-9 m
Arada [10]
From the planks equation
E=hv
V= c/ wave length
V= 3×10^8/30×10^-9
=1×10^16
E= hv
6.63×10^-34×1×10^16
= 6.63×10^-18
4 0
3 years ago
How is the pressure of a gas related to its concentration of particles?
solniwko [45]
As the volume of gas decreases, the pressure concentration of particles will increase.
7 0
3 years ago
A .5kg football is thrown with a velocity of 15m/s to the right. A stationary receiver catch the bail and bring it to rest in 0.
storchak [24]
<h3>Answer:</h3>

375 N

<h3>Explanation:</h3>

Topic tested: Newton's Law of motion

The question is testing on the application of Newton's second Law of motion.

We are given;

  • Mass of the football = 5 Kg
  • Initial velocity of the football, Vf = 15 m/s
  • Time taken to bring the ball to rest = 0.2 s
  • Final velocity, Vo = 0 m/s ( since the ball went to rest)

We are required to determine the force exerted to bring it to rest.

  • According to the Newton's second law of motion the resultant force and rate of change in momentum are directly proportion.
  • Therefore;

F=\frac{(MVf-MVo)}{t}

Thus;

F=\frac{(5kg.5m/s)-(5kg.0m/s)}{0.2s}

F=375Newtons

Force = 375 N

Hence, the force exerted on the ball by the receiver was 375 N

5 0
4 years ago
When mass m is tied to the bottom of a long, thin wire suspended from the ceiling, the wire's second-harmonic frequency is 180 h
Katena32 [7]
The frequency of the nth-harmonic of a string is given by
f_n =  \frac{n}{2L}  \sqrt{ \frac{T}{\mu} }
where n is the number of the harmonic, L is the length of the string, T the tension and \mu the linear density. 

In our problem, since the mass m is tied to the string, the tension is equal to the weight of the object tied:
T=mg
Substituting into the first formula, we have
f_n =  \frac{n}{2L}  \sqrt{ \frac{mg}{\mu} }

In our problem we have n=2 (second harmonic). In the previous equation, the only factor which is not constant between the first and the second part of the problem is m, the mass. So, we can rewrite everything as
f_2 = K  \sqrt{m}
where we called 
K= \frac{2}{2L}  \sqrt{ \frac{g}{\mu} }

In the first part of the problem, the mass of the object is m and f_2 = 180 Hz. So we can write 
180 Hz = K  \sqrt{m}

When the mass is increased with an additional 1.2 kg, the relationship becomes
270 Hz = K \sqrt{(m+1.2 Kg)}

By writing K in terms of m in the first equation, and subsituting into the second one, we get
180 Hz  \sqrt{ \frac{m+1.2 Kg}{m} }=270 Hz
and solving this, we find
m=0.96 kg


5 0
3 years ago
10. Complete each of the following radioactive decay equations.
alukav5142 [94]

1. Alpha

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(because 231 + 4 = 235)

The other number below the 4 is 2 - helium

(because 90 + 2 = 92)

2. Beta

Missing numbers are zero and one. Same as the one below it.

3. Beta

11 is the number above the C

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5 is the number next to the B

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Basically, the missing numbers must balance on both side of the equation; creating the elements you've started of with, meaning if you where to add the right side of the equation up - you should end up with the same protons and mass numbers you began with.

Hope this helps!

6 0
1 year ago
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