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xxMikexx [17]
3 years ago
14

Which statement is true of equinoxes? They occur in June and December. Days and nights are equal in length everywhere. The lengt

h of daylight in the Arctic and Antarctic circles is 24 hours. The sun’s vertical rays are striking either 23.5°S or 23.5°N.
Physics
2 answers:
Zanzabum3 years ago
8 0
<span>Days and nights are equal in length everywhere.(gradpoint)</span>
Nana76 [90]3 years ago
5 0

Answer: Days and nights are equal in length everywhere.

Explanation:

Equinox means equal night. There are two equinox - vernal equinox and autumnal equinox. When the sun crosses celestial equator, the sun rays fall directly over equator at noon. Thus, there is equal length of day and night in the two hemisphere. It occurs around March 22 and September 22.

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A rocket passes you at a speed of 0.85c, and you measure its length to be 35.0 m. What is its measured length when at rest?
viktelen [127]

Answer:

66.4 m

Explanation:

To solve the problem, we can use the length contraction formula, which states that the length observed in the reference frame moving with the object (the rocket) is given by

L=L_0 \sqrt{1-(\frac{v}{c})^2}

where

L_0 is the proper length (the length measured from an observer at rest)

v is the speed of the object (the rocket)

c is the speed of light

Here we know

v = 0.85c

L = 35.0 m

So we can re-arrange the equation to find the length of the rocket at rest:

L_0 = \frac{L}{\sqrt{1-(\frac{v}{c})^2}}=\frac{35.0}{\sqrt{1-(\frac{0.85c}{c})^2}}=66.4 m

8 0
3 years ago
El coeficiente de variación de la resistencia con la temperatura del carbón es -0.0005/°c.Si la resistencia de una resistencia d
Doss [256]

Answer:

 R (120) = 940Ω

Explanation:

The variation in resistance with temperature is linear in metals

           ΔR (T) = R₀ α ΔT

where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC

let's calculate

            ΔR = 1000 (-0,0005) (120-0)

            ΔR = -60 Ω

            ΔR = R (120) + R (0) = -60

            R (120) = -60 + R (0)

            R (120) = -60 + 1000

            R (120) = 940Ω

3 0
3 years ago
A 3kg horizontal disk of radius 0.2m rotates about its center with an angular velocity of 50rad/s. The edge of the horizontal di
Lyrx [107]

Answer:

D

Explanation:

From the information given:

The angular speed for the block \omega = 50 \ rad/s

Disk radius (r) = 0.2 m

The block Initial velocity is:

v = r \omega \\ \\  v = (0.2  \times 50) \\ \\  v= 10 \ m/s

Change in the block's angular speed is:

\Delta _{\omega} = \omega - 0 \\ \\ = 50 \ rad/s

However, on the disk, moment of inertIa is:

I= mr^2 \\ \\ I = (3 \times 0.2^2) \\ \\ I = 0.12 \ kgm^2

The time t = 10s

∴

Frictional torques by the wall on the disk is:

T = I \times (\dfrac{\Delta_{\omega}}{t}) \\ \\ = 0.12 \times (\dfrac{50}{10})  \\ \\ =0.6 \ N.m

Finally, the frictional force is calculated as:

F = \dfrac{T}r{}

F= \dfrac{0.6}{0.2} \\ \\ F = 3N

8 0
3 years ago
A ball is falling from the sky down to the left. What two forces are being used?
suter [353]

Answer:

push and pull

Explanation:

i beleve good luck❤️❤️

4 0
2 years ago
Read 2 more answers
The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?
vitfil [10]

The expression of the electric flux is

\Phi = \frac{Q}{\epsilon_0}

Here,

Q = Total charge enclosed in the closed surface

\epsilon_0 = Permittivity due to free space

Rearranging to find the charge,

Q = \epsilon_0 \Phi

Replacing with our values we have finally

Q = (8.85*10^{-12}F\cdot m^{-1})(1.84*10^3 N\cdot m^2/C)

Q = 1.6284*10^{-8} C (\frac{10^9nC}{1C})

Q = 0.1684nC

The charge enclosed by the box is 0.1684nC

The sign of the charge can be decided by using the direction of the flux. The charge enclosed by the cube can be calculated by using the electric flux and the permitivity of free space.

7 0
3 years ago
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