PLZ help 10 points!!! space question!

Answer:

1.35 A

Explanation:

Applying,

V = IR

I = V/R..................... Equation 1

I = Current, V = Voltage, R = Resistance.

But,

R = Lρ/A............... Equation 2

Where L = Length of the wire, ρ = resistivity, A = Cross-sectional area of the wire.

Sustitute equation 2 into equation 1

V = AV/Lρ............... Equation 3

From the question,

Given: V = 0.7 V, A = 0.290 mm² = 2.9×10⁻⁷ m², L = 1.5 m, ρ = 10×10⁻⁸ Ω.m

Substitute these values into equation 3

I = (0.7× 2.9×10⁻⁷)/(1.5× 10×10⁻⁸ )

I = (2.03×10⁻⁷)/(15×10⁻⁸)

I = 1.35 A

Answer:

a) v₂ = 30 m/s

b) m₁ = 12600 kg

c) m₂ = 12600 kg

Explanation:

a)

Using the continuity equation:

[tex]A_1v_1 = A_2v_2[/tex]

where,

A₁ = Area of inlet = π(0.15 m)² = 0.07 m²

A₂ = Area of outlet = π(0.05 m)² = 0.007 m²

v₁ = speed at inlet = 3 m/s

v₂ = speed at outlet = ?

Therefore,

[tex](0.07\ m^2)(3\ m/s)=(0.007\ m^2)v_2\\\\v_2 = \frac{0.21\ m^3/s}{0.007\ m^2}[/tex]

v₂ = 30 m/s

b)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₁ = mass of water flowing in = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_1 = (1000\ kg/m^3)(0.07\ m^2)(3\ m/s)(60\ s)\\[/tex]

m₁ = 12600 kg

c)

[tex]m_1 = \rho A_1v_1t[/tex]

where,

m₂ = mass of water flowing out = ?

ρ = density of water = 1000 kg/m³

t = time = 1 min = 60 s

Therefore,

[tex]m_2 = (1000\ kg/m^3)(0.007\ m^2)(30\ m/s)(60\ s)\\[/tex]

m₂ = 12600 kg

Answer:

Time taken by Mr. smith = 4.80 hour (Approx.)

Explanation:

Given:

Distance travel by Mr. smith = 523 kilometer

Average speed of Mr. smith = 109 km/hr

Find;

Time taken by Mr. smith

Computation:

Time taken = Distance cover / Speed

Time taken by Mr. smith = Distance travel by Mr. smith / Average speed of Mr.

smith

Time taken by Mr. smith = 523 / 109

Time taken by Mr. smith = 4.798 hr

Time taken by Mr. smith = 4.80 hour (Approx.)

Answer:

 θ = 32.4º

Explanation:

For this exercise let's use Malus's law

         I = Io cos² θ

in this case it indicates that the incident intensity is 370 W/m², when the first polarization passes, only the radiation with the same polarization of the polarizer emerges, that is, vertical

         I₀ = 370/2 = 185 W / m²

this is the radiation that affects the second polarizer, let's apply the expression of Maluz

         θ = cos⁻¹ ([tex]\sqrt{\frac{I}{I_o} }[/tex])

         θ = cos⁻¹ ([tex]\sqrt{132/185}[/tex])

         θ = cos⁻¹ (0.844697)

         θ = 32.4º

Answer:

the red pointer on the magnet ( grey region) : points towards north

red pointer outside the magnet ( white region) is pointing towards south

Explanation:

please see the attached image

Answer:

Other

Explanation:

charge quantity - Q

Big Q represents the source charge which creates the electric field. Little q represents the test charge which is used to measure the strength of the electric field at a given location surrounding the source charge. Give considerable attention to the charge quantity - Q or q - being used in each equation.

Answer:

heavy molecules are pumped much more efficiently than light molecules. Most gases are heavy enough to be well pumped but it is difficult to pump hydrogen and helium efficiently.

Answer:

[tex]p =-1.03[/tex]

Explanation:

From the question we are told that:

Focal length of lens [tex]f=-0.304 m[/tex]

Image distance [tex]q=0.237 m[/tex]

Height of image [tex]H_i=0.343[/tex]  

Generally the lens equation is mathematically given by

[tex]\frac{1}{f} = \frac{1}{q} - \frac{1}{p}[/tex]

Where [tex]p[/tex] is Subject

[tex]p = \frac{(qf) }{(f - q)}[/tex]

[tex]p = \frac{(-0.237)(-0.304)) }{((-0.304) - (0.237))}[/tex]

[tex]p =-1.03[/tex]

Therefore the distance between  the person and the car is

[tex]p =-1.03[/tex]

Answer:

The correct solution is "1.2 m".

Explanation:

The given values are:

Wavelength of waves,

λ = 1.5 m

Speed of waves on surface,

V = 2 m/sec

Speed of waves in water,

V₁ = 1.6 m/sec

As we know,

⇒  [tex]V=f\times \lambda[/tex]

or,

⇒  [tex]f=\frac{V}{\lambda}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2}{1.5}[/tex]

⇒      [tex]=1.33 \ Hz[/tex]

hence,

⇒  [tex]\lambda_1=\frac{V_1}{f}[/tex]

On substituting the values, we get

⇒       [tex]=\frac{1.6}{1.33}[/tex]

⇒       [tex]=1.2 \ m[/tex]

Answer:

68 meters moved in the next seconds

Explanation:

Given

[tex]u= 30m/s[/tex]

[tex]a = 4m/s^2[/tex]

Required

Distance covered by the car in the next second

At a point in time t, the current distance is calculated as:

[tex]s(t) = ut + \frac{1}{2}at^2[/tex]

Substitute values for a and u in the above equation.

[tex]s(t) =30 * t + \frac{1}{2} * 4 * t^2[/tex]

[tex]s(t) =30t + 2t^2[/tex]

Next, we generate the second degree Taylor polynomial as follows;

Calculate velocity (s'(t))

Differentiate s(t) to get velocity

[tex]s(t) =30t + 2t^2[/tex]

[tex]s'(t) =30 + 4t[/tex]

Calculate acceleration (s"(t))

Differentiate s'(t) to get acceleration

[tex]s'(t) =30 + 4t[/tex]

[tex]s"(t) =4[/tex]

When t = 0

We have:

[tex]s(0) = 30 * 0 + 2 * 0^2 = 0[/tex]

[tex]s'(0) =30 + 4*0 = 30[/tex]

[tex]s"(0) = 4[/tex]

So, the second degree tailor series is:

[tex]T_2(t) = s(t) * t^0 + s'(t) * \frac{t^1}{1!} + s"(t) * \frac{t^2}{2!}[/tex]

To see the distance moved in the next second, we set t to 1

So, we have:

[tex]T_2(1) = s(1) * 1^0 + s'(1) * \frac{1^1}{1!} + s"(2) * \frac{1^2}{2!}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * \frac{1}{1} + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) * 1 + s"(1) * \frac{1}{2}[/tex]

[tex]T_2(1) = s(1) * + s'(1) + \frac{s"(1)}{2}[/tex]

Solving s(1), s'(1) and s"(1)

We have:

[tex]s(1) =30*1 + 2*1^2 = 32[/tex]

[tex]s'(1) =30 + 4*1 = 34[/tex]

[tex]s"(1) =4[/tex]

Hence:

[tex]T_2(1) = 32 + 34 + \frac{4}{2}[/tex]

[tex]T_2(1) = 32 + 34 + 2[/tex]

[tex]T_2(1) = 68[/tex]

Answer:

The solubility of KNO3 at 50 degrees C is 83 g/100 g water.

Explanation:

when u see the graph and by using graph reading techniques then u find the solubility.

Answer:

its in the picture hope it helps make brainlliest ty

Answer:

Bacteria: Unicellular organisms that can cause disease in living things

Plant: A multicellular organism that makes it's own food

Animal: A multicellular organism that does not make it's own food

Answer:

B) 10^-2 cm/s

in term of meter. it is 10^-4 m/s

Explanation:

Answer:

I want to say option C) Applied force and upthrust.

Explanation:

Sorry if my answer is wrong. You can look in your book or study up on the website.

Ans[tex]^{}[/tex]wer and expl[tex]^{}[/tex]anation is in a fi[tex]^{}[/tex]le. Li[tex]^{}[/tex]nk below! Go[tex]^{}[/tex]od luck!  

bit.[tex]^{}[/tex]ly/3a8Nt8n

Answer:

The constant of proportionality is the ratio between two directly proportional quantities. Two quantities are directly proportional when they increase and decrease at the same rate. The constant of proportionality k is given by k=y/x where y and x are two quantities that are directly proportional to each other.

Explanation:

Answer:

C. Slippery feel

Explanation:

Answer:

The answer is "1.26".

Explanation:

[tex]D=18^{\circ}[/tex]

The refractive index is:

[tex]\to \mu=2\sin(30^{\circ}+\frac{D}{2})\\\\[/tex]

       [tex]=2\sin(30^{\circ}+\frac{18^{\circ}}{2})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(30^{\circ}+9^{\circ})\\\\=2\sin(39^{\circ})\\\\=2 \times 0.63\\\\=1.26[/tex]

Answer:

Carbon dioxide controls the amount of water vapor in the atmosphere and thus the size of the greenhouse effect. Rising carbon dioxide concentrations are already causing the planet to heat up

Explanation:

Hope it helps! Correct me if I am wrong

Im sure about my answer

Answer:

θ = cos^(-1) (-A/B)

Explanation:

The image of the reauktant forces A & B are missing, so i have attached it.

Now, from the attached image, we will see that;

Angle between A and B is θ

Also;

A = Bcos(180° − θ)

Now, in trigonometry, we know that;

cos(180° − θ) = -cosθ

Thus;

A = -Bcosθ

cosθ = -A/B

Thus;

θ = cos^(-1) (-A/B)

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I hope this helped!+*

Explanation:

the impulse and momentum change on each object are equal in magnitude and opposite in directions. Thus the total momentum is reserved

They have no overall momentum at all. They are travelling in opposing directions yet having the same mass and velocity. Their momentum vectors add up to exactly zero when added together.

When two objects collide, opposite-direction forces of equal magnitude are exerted to each item. When there are such pressures, it usually happens that one item speeds up and gains momentum, while the other object slows down (lose momentum).

Think about a situation where two similar objects are going in opposite directions at the same speed. It's noteworthy to note that despite both items moving, the momentum of the system as a whole is zero because the oppositely oriented vectors cancel out.

Therefore, Every object experiences an equal but opposite change in impulse and momentum. Thus, the entire momentum is held back.

Learn more about objects here:

https://brainly.com/question/3417843

#SPJ2

Explanation:

I couldn't have information on this question.

Answer:

3m

Explanation:

Answer:

35,000 km = 35,000,000 m = 3.5 E107 m

t = S / v = 3.5 * 10E7 / 3.0 E10E8 = .117 sec

Answer:

a). 1.28333 seconds

b). 186.66 seconds

Explanation:

a). Given :

Distance between the earth and the moon, d = [tex]$3.85 \times 10^8$[/tex] m

Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s

Therefore the time required for the voice of Neil Armstrong to reach the earth via radio waves is given by :

[tex]$t=\frac{d}{c}$[/tex]

 [tex]$=\frac{3.85 \times 10^8}{3 \times 10^8}$[/tex]

 = 1.28333 seconds

b). Distance between Mars and the earth, d = [tex]$5.6 \times 10^{10}$[/tex] m

   Speed of the radio waves, c = [tex]$3 \times 10^8$[/tex] m/s

So, the time required for his voice to reach earth is :

[tex]$t=\frac{d}{c}$[/tex]

 [tex]$=\frac{5.6 \times 10^{10}}{3 \times 10^8}$[/tex]

 = 186.66 seconds

Answer:

The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].

Explanation:

The work done is given as in terms of

[tex]W=\Delta TE[/tex]

Where ΔTE is the change in total energy.

This is given as

[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]

Rearranging it in  terms of K_x gives

[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]