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Kaylis [27]
3 years ago
11

It appears as though the moon disappears and the sun comes out in the daytime. Based on what you know, explain why this is happe

ning in the daytime.
Earth is rotating toward the sun.


Earth is rotating away from the sun.

The moon is rotating away from the sun.

The moon is rotating towards the sun.
Physics
2 answers:
atroni [7]3 years ago
8 0
The Earth is rotating toward the sun and away from the moon. Because of this, part of the Earth’s hemisphere sees daylight while the other part is not able to.
zimovet [89]3 years ago
6 0

Answer:

The Earth is toward the sun

Explanation:

DONT LISTEN TO ME I AM A CHILD AND I JUST GUESSED

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The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally. If the sheets original widt
devlian [24]
The original width was 94.71 cm 
<span>The area decreased 33.1% </span>

<span>The equation for the final size is </span>
<span>2X^2 = 1.2 m^2 </span>
<span>X^2 - 0.6 m^2 </span>
<span>X^2 = 10000 * .6 cm </span>
<span>X = 77.46 cm (this is the width) </span>

<span>The length is 2 * 77.46 = 154.92 cm </span>

<span>The original length was 154.92 + 34.5 = 189.42 cm </span>
<span>The original width was 189.42 / 2 = 94.71 cm </span>

<span>The original area was 94.71 * 189.92 = 17939.9 cm^2 </span>
<span>The new area is 79.46 * 154.92 = 12000.1 cm^2 </span>

<span>The difference between the original and current area is 17939.9 - 12000.1 = 5939.86 cm^2 </span>

<span>The percentage the area decreased is 5939.86 ' 17939.9 = 33.1%</span>
6 0
3 years ago
At a certain instant, coil A is in a 10-T external magnetic field and coil B is in a 1-T external magnetic field. Both coils hav
Neporo4naja [7]

Answer:

A) coil A

Explanation:

According to Faraday, Induced emf is given as;

E.M.F = ΔФ/t

ΔФ = BACosθ

where;

ΔФ  is change in magnetic flux

θ is the angle between the magnetic field, B, and the normal to the loop of area A

A is the area of the loop

B is the magnetic field

From the equation above, induced emf depends on the strength of the magnetic field.

Both coils have the same area and are oriented at right angles to the field.

Coil A has a magnetic field strength of 10-T which is greater than 1 T of coil B, thus, coil A will have a greater emf induced in it.

7 0
4 years ago
Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
uysha [10]

Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

3 0
3 years ago
A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part
mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

4 0
2 years ago
A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on
Juliette [100K]

Answer:

k = 5\times 10^{4}\ N/m

b = 0.707\times 10^{3}

t = 7.1\times 10^{- 5}\ s

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

v^{2} = u^{2} + 2gh

where

u = initial velocity = 0

g = 10m/s^{2}

Thus

v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s

Force on the mass is given by:

F = mg = 1000\times 10 = 10000 N = 10\ kN

Also, we know that the spring force is given by:

F = - kx

Thus

k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m

Now, to find the damping constant b, we know that:

F = - bv

b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s

4 0
3 years ago
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