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11Alexandr11 [23.1K]
2 years ago
10

Why is the moon tilted​

Physics
1 answer:
butalik [34]2 years ago
8 0
It’s cause by the fact that the earth is tilted 23.5 degrees
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If a molecule has four hybrid sp3 orbitals, it can be concluded that the molecule has a
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If a molecule is found to have four hybrid sp3 then we say that it has a <span>tetrahedral molecular geometry. thsi is the correct term to define this type of molecules. So your option would be C. Hope this is useful</span>
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3 years ago
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A stationary siren creates an 894 Hz
valentina_108 [34]

Answer:

12.3 m/s

Explanation:

The Doppler equation describes how sound frequency depends on relative velocities:

fr = fs (c + vr)/(c + vs),

where fr is the frequency heard by the receiver,

fs is the frequency emitted at the source,

c is the speed of sound,

vr is the velocity of the receiver,

and vs is the velocity of the source.

Note: vr is positive if the receiver is moving towards the source, negative if away.

Conversely, vs is positive if the receiver is moving away from the source, and negative if towards.

Given:

fs = 894 Hz

fr = 926 Hz

c = 343 m/s

vs = 0 m/s

Find: vr

926 = 894 (343 + vr) / (343 + 0)

vr = 12.3

The speed of the car is 12.3 m/s.

5 0
2 years ago
A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60
Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

=1.2cm

5 0
3 years ago
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What minerals attracts iron nails?
Helga [31]
A lodestone is a naturally magnetized piece of the mineral magnetite. They are naturally occurring magnets, which can attract iron. The property of magnetism was first discovered in antiquity through lodestones.
3 0
3 years ago
A spherical, conducting shell of inner radius r1= 10 cm and outer radius r2 = 15 cm carries a total charge Q = 15 μC . What is t
lutik1710 [3]

a) E = 0

b) 3.38\cdot 10^6 N/C

Explanation:

a)

We can solve this problem using Gauss theorem: the electric flux through a Gaussian surface of radius r must be equal to the charge contained by the sphere divided by the vacuum permittivity:

\int EdS=\frac{q}{\epsilon_0}

where

E is the electric field

q is the charge contained by the Gaussian surface

\epsilon_0 is the vacuum permittivity

Here we want to find the electric field at a distance of

r = 12 cm = 0.12 m

Here we are between the inner radius and the outer radius of the shell:

r_1 = 10 cm\\r_2 = 15 cm

However, we notice that the shell is conducting: this means that the charge inside the conductor will distribute over its outer surface.

This means that a Gaussian surface of radius r = 12 cm, which is smaller than the outer radius of the shell, will contain zero net charge:

q = 0

Therefore, the magnitude of the electric field is also zero:

E = 0

b)

Here we want to find the magnitude of the electric field at a distance of

r = 20 cm = 0.20 m

from the centre of the shell.

Outside the outer surface of the shell, the electric field is equivalent to that produced by a single-point charge of same magnitude Q concentrated at the centre of the shell.

Therefore, it is given by:

E=\frac{Q}{4\pi \epsilon_0 r^2}

where in this problem:

Q=15 \mu C = 15\cdot 10^{-6} C is the charge on the shell

r=20 cm = 0.20 m is the distance from the centre of the shell

Substituting, we find:

E=\frac{15\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.20)^2}=3.38\cdot 10^6 N/C

4 0
3 years ago
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