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3 years ago

Does the atmosphere contain 21% nitrogen in 78% oxygen

Physics

1 answer:

Alex777 [14]3 years ago

7 0

Answer:

An atmosphere is the layers of gases surrounding a planet or other celestial body. Earth's atmosphere is composed of about 78% nitrogen, 21% oxygen, and one percent other gases

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An atom has a charge of 1+. If the atoms nucleus caintains 23 protons, the atom must have how many electrons

Dmitry [639]

If your atom has a charge of +1, it means you have 1 more proton than electrons. This is because protons have a positive charge and electrons have a negative charge. Because you have 23 protons, and a charge of +1, you have 22 electrons.

7 0

3 years ago

An unknown charged particle passes without deflection through crossed electric and magnetic fields of strengths 187,500 V/m and

Kobotan [32]

Answer:

Explanation:

Given that

The electric fields of strengths E = 187,500 V/m and

and The magnetic fields of strengths B = 0.1250 T

The diameter d is 25.05 cm which is converted to 0.2505m

The radius is (d/2)

= 0.2505m / 2 = 0.12525m

The given formula to find the magnetic force is $F_{ma}=BqV---(i)$

The given formula to find the electric force is $F_{el}=qE---(ii)$

The velocity of electric field and magnetic field is said to be perpendicular

Electric field is equal to magnectic field

Equate equation (i) and equation (ii)

$Bqv=qE\\\\v=\frac{E}{B}$

$v=\frac{187500}{0.125} \\\\v=15\times10^5m/s$

It is said that the particles moves in semi circle, so we are going to consider using centripetal force

$F_{ce}=\frac{mv^2}{r}---(iii)$

magnectic field is equal to centripetal force

Lets equate equation (i) and (iii)

$Bqr=\frac{mv^2}{r} \\\\\frac{q}{m}=\frac{v}{Br} \\\\\frac{q}{m} =\frac{15\times 10^5}{0.125\times0.12525} \\\\=\frac{15\times10^5}{0.015656} \\\\=95808383.23\\\\=958.1\times10^5C/kg$

Therefore, the particle's charge-to-mass ratio is $958.1\times10^5C/kg$

To identify the particle

Then 1/ 958.1 × 10⁵ C/kg

The charge to mass ratio is very close to that of a proton, which is about 1*10^8 C/kg

Therefore the particle is proton.

8 0

4 years ago

An arrow is shot upward from the roof of a building 75 m high with an initial speed of 40 m/s. If g = 10 m/s2 , what total dista

Elenna [48]

Answer:

60m

Explanation:

According to one of the equation of motions, v² = u²+2as where;

S is the distance

u is the initial velocity

v is the final velocity

a is the acceleration

Since the arrow is shot upwards, the body will experience a negative acceleration due to gravity i.e a = -g

Therefore our equation will become;

v² = u² - 2gS

Given u = 40m/s, g = 10m/s², S = 75m

Substituting to get the final velocity of the arrow we will have;

v² = 40²-2(10)(75)

v² = 1600 - 1500

v² = 100

v = √100

v = 10m/s

Total distance traveled is speed of the object × time taken

Total distance traveled = 10 × 6

= 60m

The arrow has therefore traveled 60m after 6seconds

3 0

3 years ago

A spacecraft at rest has moment of inertia of 100 kg-m^2 about an axis of interest. If a 1 newton thruster with a 1 meter moment

snow_lady [41]

Answer:

18 radians

Explanation:

The computation is shown below:

As we know that

Torque = Force × Moment arm

= 1N × 1M

= 1N-M

Torque = $I\alpha$

$\alpha = \frac{torque}{I}\\\\= \frac{1}{100}\\\\= 0.01 rad/s^2$

Now

$\theta = w_ot + \frac{1}{2} \alpha t^2\\\\w_o = 0\\\\\theta = 0 \times 60 + \frac{1}{2} \times 0.01 \times 60^2\\\\= 18\ radians$

Here t = 1 minutes = 60 seconds

3 0

3 years ago

4 This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive

zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

B1 at 20km/h

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0

3 years ago