What do you need to know to establish motion

Answer:

False

Explanation:

I learned it the hard way trust me T^T

Answer:

a) the plastic tube need to be 24.7 m long

b) the kilometer of copper wire required is 15.7

c) the resistance of this solenoid is 5538.8 2 ohms

Explanation:

Given the data in the question;

we determine the length of the plastic tube. assuming the solenoid is long.

the self inductance of a long solenoid is;

L = μ₀n²πr²l

μ₀ = 4π × 10⁻⁷ T-m/A

where

n = number of turns per unit length

r = radius of the solenoid = 8cm (as the diameter of the plastic hollow tube is 16 cm)

l = length of the solenoid or the length of the plastic tube

we find n = number of turns per unit length

given that, the copper wire to be wound around the solenoid is 0.79 mm in diameter

number of turns per meter = n = 1 / ( 0.79 × 10⁻³ m ) = 1265.8 turns/meter

So from our previous formula, we find l

L = μ₀n²πr²l

we substitute

1.0 H = (4π × 10⁻⁷ T-m/A)( 1265.8 )²(3.14)(0.08)² ( l)

1 = 0.04048 × l

l = 1 / 0.04048

l = 24.7 m

Therefore, the plastic tube need to be 24.7 m long

b)  

n = number of turns per unit length = 1265.8 turns/metre

so, the length of the plastic tube over which the copper wire is to be wound,

number of turns of copper wire required = n × l

= 1265.8 turns/meter × 24.7 m

= 31,265.26 turns

Now each turn of the copper wire is to be wound across the 18cm diameter of the plastic tube.

so for each turn length of copper wire required = 2π × r

= 2π × 0.08 m

= 0.5026548 m

So copper wire required for 31,265.26 turns will be;

⇒ 31,265.26 × 0.5026548 = 15715.63m = 15.7 km

Therefore, the kilometer of copper wire required is 15.7

c)

p = resistivity of copper = 1.68 × 10⁻⁸ ohm-m

Resistance = pl/a

where l is length of copper wire, a is cross sectional area;

diameter of copper wire is 0.79-mm

radius of copper wire is 0.79/2 = 0.395 mm = 0.000395 m

area of cross section of copper wire a = πr² = π( 0.00395)² = 4.9 × 10⁻⁷ m²

Resistance = pl/a

we substitute

Resistance = [(1.68 × 10⁻⁸ ohm-m)( 15715.63m )] / [ 4.9 × 10⁻⁷ m² ]

Resistance =  5538.8 2 ohms

Therefore,  the resistance of this solenoid is 5538.8 2 ohms

Answer:

Explanation:

(A)

Force is said to be the rate of change of momentum.

At the instant, the parallel component of the change of momentum is zero. SInce centripetal force always acts towards the center, hence the direction upwards ( along the positive y-axis).

[tex]\dfrac{dp}{dt} = F(+\hat y)[/tex]

[tex]= \dfrac{mv^2}{r}(+\hat y) \ \\ \\ = \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

(B)

The net force acting on the child is:

[tex]F_{net} = \dfrac{mv^2}{r}( + \hat y) \\ \\[/tex]

[tex]= \dfrac{28 \ kg \times (7 \ m/s)^2}{3.3 \ m }( + \hat y) \\ \\ = \mathbf{416 \ N ( + \hat y)}[/tex]

Answer:

The respiratory system provides oxygen for cells, while the circulatory system transports oxygen to cells.

Explanation:

so the answer is D

Answer:

Explanation:

Neutrinos are otherwise called leptons. They are principal particles. A lepton is a rudimentary half-spin molecule that doesn't go through solid reactions. Neutrinos are not usually charged and exceptionally light weighted so they once in a while interface with other matter. Neutrinos are light weighted. Their mass is around 10⁻⁷ kg. A neutrino possesses a small radius, too little to ever be estimated. A little span and very less mass make them imperceptible. Since neutrinos have next to no mass. they travel at almost the speed of light and thus they arrive at the outside of the Sun in only 2 seconds, dissimilar to photons which take convoluted ways to arrive at the Sun's surface in a huge number of years.  

The photon and neutrino, both were made in the Sun's center yet on various occasions. The neutrino is only a couple of minutes old though the photon is around 1,000,000 years of age. At the point when the photon was made in the Sun's center. it needed to venture out to the outside of the Sun. in any case, rather because of its hefty mass and cooperation with other matter, it headed out a crisscross way to the surface. Ordinarily, it was repulsed and it was sent back to the middle where it needed to begin once more. It required a large number of years for a photon to arrive at the outside of the Sun.

Nonetheless, when it arrived at the Sun's surface, it required just 8.8 minutes for the photon to arrive at Earth. The neutrino was anyway made only a couple of minutes prior in the Sun's center. Since it has an entirely irrelevant mass, little size, and no charge, it didn't interface with its environmental factors. So it just required 2 seconds for the neutrino to arrive at the Sun's surface. When it arrived at the Sun's surface, it arrived at the earth in about 8.8 minutes. with the photon. So both, photon and neutrino have various histories as the two of them were made at a hole of around 1,000,000 years.

Answer: Both False

Explanation:

Our Milky Way Galaxy is a spiral galaxy. Some spiral galaxies are what we call "barred spirals" because the central bulge looks elongated

Irregualuar glaxyices are all over  the place

Answer: 0.8 g/cm

Explanation:

p= m/V

= 4 kg/ 5 liter

= 0.8

Answer:

.8

Explanation:

give them brainliest

6.388 or 6.388468 cubic feet..

not sure

Answer:

151.12 million miles i actually just did this question on a test

Explanation:

Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

The term "revolution" describes how an item moves in its orbit around another object. For instance, the 24-hour day is created as the Earth rotates on its own axis.

The 365-day year is made possible by the Sun's rotation of the Earth. A planet spins around a satellite.

While orbiting the Sun, Mars travels at an average speed of 53,979 miles per hour, which equates to 86,871 km per hour.

Thus, Mars orbits—or completes one revolution—around the Sun every 686.98 Earth days, or once every 1.88 Earth years.

To learn more about One revolution, refer to the below link:

https://brainly.com/question/26289977

# SPJ2

Answer:

1. 2.5×10¯⁹ N

2. 3.33×10¯¹¹ m/s²

Explanation:

1. Determination of the force of attraction.

Mass of astronaut (M₁) = 75 Kg

Mass of spacecraft (M₂) = 125000 Kg

Distance apart (r) = 500 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Force of attraction (F) =?

The force of attraction between the astronaut and his spacecraft can be obtained as follow:

F = GM₁M₂ /r²

F = 6.67×10¯¹¹ × 75 × 125000 / 500²

F = 2.5×10¯⁹ N

Thus, the force of attraction between the astronaut and his spacecraft is 2.5×10¯⁹ N

2. Determination of the acceleration of the astronaut.

Mass of astronaut (m) = 75 Kg

Force (F) = 2.5×10¯⁹ N

Acceleration (a) of astronaut =?

The acceleration of the astronaut can be obtained as follow:

F = ma

2.5×10¯⁹ N = 75 × a

Divide both side by 75

a = 2.5×10¯⁹ / 75

a = 3.33×10¯¹¹ m/s²

Thus, the acceleration the astronaut is 3.33×10¯¹¹ m/s²

Answer:

In winter, the temperature of the water is very low which makes the soap dissolve in a very small amount. ... Since the water molecules move faster, they come in contact with soap more often causing it to dissolve faster

Hope it helps...

Have a great day : )

Answer:

OMG IM ON THE SAME QUESTION

Explanation:

Answer:

Heat loss per unit length = 642.358 W/m

The heat loss per unit length on a breezy day during 8 m/s speed is = 1760.205 W/m

Explanation:

From the information given:

Diameter D [tex]= 100 mm = 0.1 m[/tex]

Surface emissivity ε = 0.8

Temperature of steam [tex]T_s[/tex] = 150° C = 423K

Atmospheric air temperature [tex]T_{\infty} = 20^0 \ C = 293 \ K[/tex]

Velocity of wind V = 8 m/s

To calculate average film temperature:

[tex]T_f = \dfrac{T_s+T_{\infty}}{2}[/tex]

[tex]T_f = \dfrac{423+293}{2}[/tex]

[tex]T_f = \dfrac{716}{2}[/tex]

[tex]T_f = 358 \ K[/tex]

To calculate volume expansion coefficient

[tex]\beta= \dfrac{1}{T_f} \\ \\ \beta= \dfrac{1}{358} \\ \\ \beta= 2.79 \times 10^{-3} \ K^{-1}[/tex]

From the table of "Thermophysical properties of gases at atmospheric pressure" relating to 358 K of average film temperature; the following data are obtained;

Kinematic viscosity (v) = 21.7984 × 10⁻⁶ m²/s

Thermal conductivity k = 30.608 × 10⁻³ W/m.K

Thermal diffusivity ∝ = 31.244 × 10⁻⁶ m²/s

Prandtl no. Pr = 0.698

Rayleigh No. for the steam line is determined as follows:

[tex]Ra_{D} = \dfrac{g \times \beta (T_s-T_{\infty}) \times D_b^3}{\alpha\times v}[/tex]

[tex]Ra_{D} = \dfrac{9.8 \times (2.79 *10^{-3})(150-20) \times (0.1)^3}{(31.244\times 10^{-6}) \times (21.7984\times 10^{-6})}[/tex]

[tex]Ra_{D} = 5.224 \times 10^6[/tex]

The average Nusselt number is:

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(Ra_D)^{1/6}}{[ 1+ (0.559/Pr)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{0.387(5.224\times 10^6)^{1/6}}{[ 1+ (0.559/0.698)^{9/16}]^{8/27}} \Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + \dfrac{5.0977}{[ 1.8826]^{8/27}}\Big \}^2[/tex]

[tex]Nu_D = \Big \{ 0.60 + 4.226 \Big \}^2[/tex]

[tex]Nu_D = 23.29[/tex]

However, for the heat transfer coefficient; we have:

[tex]h_D = \dfrac{Nu_D\times k}{D_b} \\ \\ h_D = \dfrac{(23.29) \times (30.608 \times 10^{-3} )}{0.1}[/tex]

[tex]h_D = 7.129 \ Wm^2 .K[/tex]

Hence, Stefan-Boltzmann constant [tex]\sigma = 5.67 \times 10^{-8} \ W/m^2.K^4[/tex]

Now;

To determine the heat loss using the formula:

[tex]q'_b = q'_{ev} + q'_{rad} \\ \\ q'_b = h_D (\pi D_o) (T_t-T_{\infty})+\varepsilon(\pi D_b)\sigma (T_t^4-T_{\infty }^4)[/tex]

[tex]q'_b = (7.129)(\pi*0.1) (423-293) + (0.8) (\pi*0.1) (5.67 *10^{-8}) (423^4-293^4) \\ \\ q'_b = 291.153 + 351.205 \\ \\ \mathbf{q'_b = 642.258 \ W/m}[/tex]

Now; here we need to determine the Reynold no and the average Nusselt number:

[tex]Re_D = \dfrac{VD_b}{v } \\ \\ Re_D = \dfrac{8 *0.1}{21.7984 \times 10^{-6}} \\ \\ Re_D = 3.6699 \times 10^4[/tex]

However, to determine the avg. Nusselt no by using Churchill-Bernstein correlation, we have;

[tex]Nu_D = 0.3 + \dfrac{0.62 \times Re_D^{1/2}* Pr^{1/3}}{[1+(0.4/Pr)^{2/3}]^{1/4}} [1+ (\dfrac{Re_D}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = 0.3 + \dfrac{0.62 \times (3.6699*10^4)^{1/2}* (0.698)^{1/3}}{[1+(0.4/0.698)^{2/3}]^{1/4}} [1+ (\dfrac{3.669*10^4}{282000})^{5/8}]^{4/5}[/tex]

[tex]Nu_D = (0.3 +\dfrac{105.359}{1.140}\times 1.218) \\ \\ Nu_D = 112.86[/tex]

SO, the heat transfer coefficient for forced convection is determined as follows afterward:

[tex]h_D = \dfrac{Nu_{D}* k}{D_b} \\ \ h_D = \dfrac{112.86*30.608 *10^{-3}}{0.1} \\ \\ h_D = 34.5 \ W/m^2 .K[/tex]

Finally; The heat loss per unit length on a breezy day during 8 m/s speed is:

[tex]q'b = h_D (\pi D_b) (T_s-T_{\infty}) + \varepsilon (\pi D_b) \sigma (T_s^4-T_ {\infty}^4) \\ \\ q'b = (34.5) (\pi *0.1) (423-293) + (0.8) (\pi*0.1) (5.67*10^{-8}) (423^4 - 293^4) \\ \\ = 1409 +351.205 \\ \\ \mathbf{q'b = 1760.205 \ W/m}[/tex]

Answer:

Is where two or more inductors are “linked” so that voltage is induced in one coil proportional to the rate-of-change of current in another

Explanation:

When hot water is poured on the can in a bucket of cold water, the can crushes off means it gets unshaped

Answer:

The right solution is "126 Psi".

Explanation:

The given values are:

P₁ = 130 psig

i.e.,

   = [tex]130\times 6.894[/tex]

   = [tex]896.22 \ Kpa[/tex]

or,

   = [tex]896.22\times 10^3 \ Pa[/tex]

Z₂ = 10ft

    = 3.05 m

[tex]\delta[/tex] = 1000 kg/m³

According to the question,

Z₁ = 0

V₁ = V₂

As we know,

⇒  [tex]\frac{P_1}{\delta_g} +\frac{V_1^2}{2g} +Z_1=\frac{P_2}{\delta_g} +\frac{V_2^2}{2g} +Z_2[/tex]

On substituting the values, we get

⇒  [tex]\frac{P_1}{\delta_g} +0+0=\frac{P_2}{\delta_g} +0+Z_2[/tex]

⇒  [tex]\frac{896.22\times 10^3}{1000\times 9.8} =\frac{P_2}{1000\times 9.8} +3.05[/tex]

⇒  [tex]P_2=866330 \ P_a[/tex]

i.e.,

⇒       [tex]=866330\times 0.000145[/tex]

⇒       [tex]=126 \ Psi[/tex]

Answer:

[tex]1.199\ \mu\text{m}[/tex]

Explanation:

[tex]\lambda[/tex] = Wavelength = 600 nm

D = Distance of the light source from screen = 3 m

y = Distance of first order bright fringe from center = 4.84 mm

d = Distance between slits

m = Order = 1

We have the relation

[tex]y=\dfrac{D\lambda}{d}\\\Rightarrow d=\dfrac{D\lambda}{y}\\\Rightarrow d=\dfrac{3\times 600\times 10^{-9}}{4.84\times 10^{-3}}\\\Rightarrow d=0.0003719\ \text{m}[/tex]

From the question we have

[tex]y=\dfrac{\dfrac{1}{2}3\lambda}{d}\\\Rightarrow \lambda=\dfrac{2}{3}yd\\\Rightarrow \lambda=\dfrac{2}{3}\times 4.84\times 10^{-3}\times 0.0003719\\\Rightarrow \lambda=0.000001199\ \text{m}=1.199\ \mu\text{m}[/tex]

The required wavelength of light is [tex]1.199\ \mu\text{m}[/tex].

Answer:

Following are the solution to the given question:

Explanation:

The input linear polarisation was shown at an angle of [tex]2 \mu[/tex]. It's a very popular use of a half-wave plate. In particular, consider the case [tex]\mu = 45 \pm[/tex], at which the angle of rotation is [tex]90\pm[/tex]. HWP thereby provides a great way to turn, for instance, a linear polarised light that swings horizontally to polarise vertically. Illustration of action on event circularly polarized light of the half-wave platform. Customarily it is the slow axis of HWP that corresponds to either the rotation. Note that perhaps the vector of polarization is "double-headed," i.e., the electromagnetic current swinging back and forward in time. Therefore the turning angle could be referred to as the rapid axis to reach the same result. Please find the attached file.

Answer:

they are right it is a new moon

Explanation:

took the test

Answer:

If you have two batteries and they have precisely the same voltage then placing one backwards will effectively cancel out the voltages and no current will flow. However, batteries aren't like that. The slightest difference in voltages mean that current will flow.

Explanation:

Answer:

the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Explanation:

Given the data in the question;

At the first interface, a phase shift occurs as the incident light is in air that has less refractive index compare to the thin film of soap bubble.

At the second interface, no shift occurs,

condition for constructive interference;

t = ( m + 1/2) × λ/2n

where m = 0, 1, 2, 3 . . . . . .

now, the condition for the constructive interference;

t = mλ/2n

where t is the thickness of the soap bubble,  λ is the wavelength of light and n is the refractive index of soap bubble.

so the minimum thickness of the film which will enhance reflection of light will be;

t[tex]_{min[/tex] =  ( m + 1/2) × λ/2n

we substitute

t[tex]_{min[/tex] =  ( 0 + 1/2) × 711 /2(1.21)

t[tex]_{min[/tex] = 0.5 × 711/2.42

t[tex]_{min[/tex] = 0.5 × 293.80165

t[tex]_{min[/tex] = 146.9 nm

Therefore,  the minimum wall thickness that will enhance the reflection of light is 146.9 nm

Answer: 10.98 N

Explanation:

Given

mass of box is [tex]m=0.5\ slugs\approx 7.3\ kg[/tex]

The coefficient of kinetic friction is [tex]\mu= 0.4[/tex]

It is given that on the application of force box started moving with constant speed i.e. there is no net external force

[tex]\Rightarrow F+mg\sin 30^{\circ}=\mu mg\cos 30^{\circ}\\\Rightarrow F=mg(\mu \cos 30^{\circ}-\sin 30^{\circ})\\\Rightarrow F=7.3\times 9.8(0.866\times0 .4-0.5)=-10.98\ N\\[/tex]

The negative sign indicates direction of force is opposite i.e. it must be applied upwards

Answer:

The correct answer is A) It is upright and 1.5m behind the mirror

Explanation:

Your reflection must be upright, meaning vertical/erect, and the distance will be the exact same. Also, the reflected ray appears as if it had traveled from an object located behind the mirror.

Answer:

correct tysmm

Explanation:

Answer:

32.25 s

Explanation:

From the question,

P = W/t.............. Equation 1

Where P = Power, W = work done, t = time.

But

W = F×d................. Equation 2

Where F = force and d = distance

Substitute equation 2 into equation 1

P = F×d/t............... Equation 3

make t the subject of euqation 3

t = (F×d)/P............. Equation 4

Givn: F = 150 N, d = 4.3 m, P = 20 watts.

Substitute these values into equation 4

t = (150×4.3)/20

t = 32.25 s

Answer:

they were fast ⛷⛷

Answer:

the required minimum thickness is 179.19 nm

Explanation:

Given the data in the question;

Refraction index of air n₁ = 1

Refraction index of sheet n₂ = 1.49

Refraction index of film n₃ = 1.62

wavelength of green light λ = 534 nm

In the data given, n₂ > n₃

so, this case has no pie-phase shift,

the condition for constructive interference will be;

mλ = 2Ln₂

L = mλ / 2n₂

so we substitute

L = m(534) / 2( 1.49 )

L = m( 534/ 2.98 )

L = m( 179.19 nm )

so for minimum value of L,

let m = 1

such that, L[tex]_{min[/tex] will be;

L[tex]_{min[/tex] = 1 × ( 179.19 nm )

L[tex]_{min[/tex] = 179.19 nm

Therefore, the required minimum thickness is 179.19 nm