how does x-rays tube work
A)Compton effect
b)Photoelectric phenomenon
c) cathod and anode ​

The distance between the aliens traveling in group and their launchpad in kilometers is 40.23km.

Given that data in the question;

Note that;

1 miles = 1.609 kilometers

Now, the distance in kilometers will be;

distance = 25 miles

distance = ( 25 × 1.609 )kilometers

distance = 40.23km

Therefore, the distance between the aliens traveling in group and their launchpad in kilometers is 40.23km.

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Answer:

real depth / apparent depth = R.I.

Real depth = 1.5 x 1 = 1.5 m

Answer:

gravity

Explanation:

as the earth rotates on an axis, it causes an effect known as centripetal acceleration with is an acceleration that pulls objects towards the center of the object. in planets, we call this Gravity

Answer:

See below

Explanation:

The initial downward velocity, the acceleration of gravity and air friction all play a part

v² = u² + 2as

v² = 4² + 2 × 1.2 × 20

v² = 16 + 2 × 12/10 × 20

v² = 16 + 48

v² = 64

v = 8 ms - 1

ii)

V = u + at

8 = 4 + 1.2 × t

8 - 4 = 1.2t

4 = 12/10 t

4 × 10 /12 = t

3.33 = t

Average velocity = Displacement/Time

20/3.33

6.006 ms - 1

If the current is held constant, an increase in voltage will result in an increase in resistance.

hope it helps...!!!

F=m.a

a = F/m

a = 90/65

a = 1.385 m/s²

Answer:

solution

here,

Explanation:

weight = 90N

mass=65kg

acceleration due to gravity (g)=??

Now,

w=m×g

90=65×g

90/65=g

:g=1.38m/s^2

Answer:

Q = c M ΔT      where c is the heat capacity and M the mass present

Q2 / Q1 = M2 / M1    since the other factors are the same

M = ρ V     where ρ is the density

M = ρ Π (d / 2)^2           where d is the diameter of the sphere

M2 / M1 = (2 D/2)^2 / (D/2)^2 = 4

It will take 4Q heat to heat the second sphere

Answer:

Here is the answer

Here is the answerExplanation:

Here is the answerExplanation:Bay of Pigs Invasion is the answer

Sonar is used to determine the speed of an object. The object is moving at 8.4146 m/s.

The frequency of any wave is defined as the number of cycles per second.

The frequency of the sound received is

f' = f x [(V+Vo /V-Vo)]

where f = 40kHz, f' = 42kHz, V =345 m/s, then the velocity of the observer will be

(V+Vo /V-Vo) = 42/40 =1.05

On rearranging the doppler shift formula, we have

Vo = 0.05 x 345 /2.05

Vo = 8.4146 m/s.

Thus, the object is moving at 8.4146 m/s.

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Answer:

option C...

resistance training is the correct answer

The coefficient of linear expansion of the metal at the given length is determined as  3.75 x 10⁻⁴/⁰C.

The coeffcient of linear expansion of the metal is calculated as follows;

ΔL = αLΔT

α = ΔL/LΔT

Where;

α = (1.5)/(20 x 200)

α = 3.75 x 10⁻⁴/⁰C

The complete question is below:

A metallic axle with a diameter of 20mm has to be fitted into a wheel which has a circular hole at its center with diameter of 18.5mm. To fit the axle into the hole, the axle is dipped into liquid nitrogen at a temperature of -200°C. Calculate the coefficient of linear expansion of the metal.

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Answer:

Gravitational force is an attraction between masses. The greater the size of the masses, the greater the size of the gravitational force (also called the gravity force). The gravitational force weakens rapidly with increasing distance between masses

The distance traveled by the object in the first 1 second is determined as 0.98 m.

The distance traveled by the object is calculated as follows;

F - μkmgcosθ = ma

0 - μkmgcosθ = ma

- μkgcosθ = a

-(0.4)(9.8)cos60 = a

- 1.96 m/s² = a

s = ut + ¹/₂at²

s = 0(1) + ¹/₂(-1.96)(1)²

s = -0.98 m

|s| = 0.98 m

Thus, the distance traveled by the object in the first 1 second is determined as 0.98 m.

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The kind of atoms or molecules are present and their relative abundance can be determined by observing wavelengths.

Wavelength is the distance between adjacent crest or trough of the wave.

The light travels in the form of wave and particle. It behaves like both. Light ray consist energy of the photons. The kind of atoms or molecules present behave like particle and they have particular wavelength.

Thus, by observing wavelength , we can determine the kind of atoms or  molecules are present and their relative abundance.

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Answer:

A

Explanation:

d = 1/2 a t^2 = 1/2 (4)(28^2) = 1568 m  ~ 1.6 x 10^3

Answer:

[tex]\huge\boxed{\sf f=0.55 \ Hz}[/tex]

Explanation:

Given Data:

Length = l = 820 mm = 0.82 m

Acceleration due to gravity = g = 9.8 ms⁻²

Required:

Frequency = f = ?

Formula:

[tex]\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} }[/tex]

Solution:

[tex]\displaystyle f =\frac{1}{2 \pi} \sqrt{\frac{g}{l} } \\\\Put\ the\ givens\\\\f=\frac{1}{2 \pi} \sqrt{\frac{9.8}{0.82} }\\\\ f = 0.159 \times \sqrt{11.95} \\\\f=0.159 \times 3.457\\\\f=0.55 \ Hz\\\\\rule[225]{225}{2}[/tex]

Answer:

What is the interaction between two different species in an ecosystem called?

Interactions between different species in a community are called interspecific interactions—inter- means "between." Different types of interspecific interactions have different effects on the two participants, which may be positive (+), negative (-), or neutral (0).

Explanation:

Answer: i'll tell them to be careful. However, the most common is using a drill bit. Take a step close to the shore and listen for cracking. If there is cracking, do not risk going out on the ice. If there is no cracking, take a cordless drill and check the thickness using an extended drill bit.

Explanation:

?????????!!!!!!!??????

Answer:

Animals follow the seasons very closely. Changes in the length of the day or temperature affect the animals' hormone production and cause them to behave according to the season. For that reason, animals instinctively know when it's time for them to mate, build a nest, fly south or stock up on food for the winter.

Answer:

low.

Explanation:

it makes the flow easier if its lower because, the resistance isn’t pushing as hard against the flow which means that the flow is easier.

When q₂ is place between q₁ and q₃, the net force on q₂ is 0.486N.

The electrostatic force of attraction or repulsion is proportional to the product of magnitude of charges and inversely proportional to the square of distance between them.

Force on q₂ due to q₁

F₁₂ = kq₁q₂ / r₁₂²

Substitute the values provided , we get

F₁₂ = 9 x 10⁹ x 5 x 10⁻⁶ x 2.5 x 10⁻⁶ / (0.5)²

F₁₂ = 0.414 N

Force on q₂ due to q₃ placed at distance 0.25m

F₂₃ =kq₂q₃ / r₂₃²

Substitute the values, we get

F₂₃ =  9 x 10⁹ x 2.5 x 10⁻⁶ x 2.5 x 10⁻⁶ / (0.25)²

F₂₃ = 0.9N

The net force F =F₂₃ -F₁₂

F =0.9 - 0.414 = 0.486 N

Thus, the net force of q₂ is 0.486 N.

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Answer:

0.45

Explanation:

trust me

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[tex] \large \sf \underline{Question:}[/tex]

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[tex] \large \sf \underline{Answer:}[/tex]

[tex] \qquad \qquad \huge \bold{Option \: D}[/tex]

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[tex] \large \sf \underline{Explanation:}[/tex]

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Answer:

P V = N R T     ideal gas equation

P2 / P1 = T2 / T1       where the other variables are constant

P2 = (T2 / T1) * P1 = (313 / 293) * 40 psi = 42.7 psi

Hi there!

We can begin by deriving the magnetic field strength for a circular loop at its center.

Using Biot-Savart's Law:
[tex]dB = \frac{\mu_0}{4\pi }\frac{idl \times \^r}{r^2}[/tex]

μ₀ = Permeability of free space (Tm/A)

dl = Differential length element

r = radius (m)

We can rather use 'ds' to represent a differential arc length since we are finding the magnetic field for a circular loop.

We must begin by dealing with the cross-product. Luckily, this derivation is simple since the radius vector is ALWAYS perpendicular to the path of integration along the loop's circumference. (ONLY if we are finding the field for the direct center.) Thus, since sin(90) = 1, we can get rid of the cross-product.

Therefore:

[tex]dB = \frac{\mu_0}{4\pi }\frac{ids}{r^2}[/tex]

Now, we can integrate with respect to ds.

[tex]= \frac{\mu_0}{4\pi }\int\limits^{2\pi r}_0 {\frac{i}{r^2}} \, ds\\\\ = \frac{\mu_0}{4\pi }{\frac{i * 2\pi r}{r^2}} \, ds[/tex]

Simplify:

[tex]B = \frac{\mu_0 i_{circle}}{2r}\\[/tex]

Now, for a straight conductor (assuming of infinite length), we know the magnetic field strength equation to be:

[tex]B = \frac{\mu_0 i_{rod}}{2\pi r}[/tex]

Set the two equal.

[tex]\frac{\mu_0 i_{circle}}{2r}= \frac{\mu_0 i_{rod}}{2\pi r}[/tex]

Cancel out '2', μ₀, and 'r':
[tex]i_{circle}}= \frac{ i_{rod}}{\pi }[/tex]

Plug in the given values and solve.

[tex]40\pi = i_{rod}\\\\i_{rod} = \boxed{125.663 A}[/tex]