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BARSIC [14]
3 years ago
7

If the moon rotated on it's axis same speed as earth would we see different sides of the moon

Physics
1 answer:
SSSSS [86.1K]3 years ago
6 0
Yes, because the moon rotates at a different rate than what it usually do.
You might be interested in
g When the movable mirror of the Michelson interferometer is moved a small distance X while making a measurement, 246 fringes ar
zhenek [66]

Answer:

X = 69.1 x 10⁻⁶ m = 69.1 μm

Explanation:

The relationship between the motion of the moveable mirror and the fringe count of the Michelson's Interferometer is given by the following formula:

d = mλ/2

where,

d = distance moved by the mirror = X = ?

m = No. of Fringes counted = 246

λ = wavelength of light entering interferometer = 562 nm = 5.62 x 10⁻⁷ m

Therefore,

X = (246)(5.62 x 10⁻⁷ m)/2

Therefore,

<u>X = 69.1 x 10⁻⁶ m = 69.1 μm</u>

5 0
3 years ago
Find the magnitude of the torque that acts on the molecule when it is immersed in a uniform electric field of 6.19×105 N/C with
Ivan

Answer:

\tau=5.81\times 10^5p\ N-m

Explanation:

We have,

Electric field, E=6.19\times 10^5\ N/C

The electric dipole vector at an angle of 69.9 degrees from the direction of the field.

The torque acting on a molecule is given by :

\tau=p\times E\\\\\tau=pE\sin\theta

p is electric dipole moment

\tau=p\times 6.19\times 10^{5}\times \sin (69.9)\\\\\tau=5.81\times 10^5p\ N-m

So, the magnitude of the torque acting on the molecule is 5.81\times 10^5p\ N-m.

7 0
4 years ago
A 20-kg object is subjected to three forces which produce an acceleration a = -8 m.s^-2 i + 6.0 m.s^-2 j on the object. Two of t
lana66690 [7]

Answer:

F₃ = -151 N i + 96 N j

Explanation:

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Forces acting on the object

F₁= 3.0 N i + 16.0 N j

F₂ = -12.0 N i+ 8.0 N j

F₃ = F₃x N i +F₃ y N j

x component of the net force on the object

Fx=F₁x+F₂x+F₃ x

Fx = 3.0 N-12.0 N +F₃x

Fx = F₃x - 9 N

y component of the net force on the object

Fy=F₁y+F₂y+F₃ y

Fy =16.0 N+ 8.0 N +F₃y

Fy = F₃y + 24 N

Newton's second law to the object:

a = -8 m/s² i + 6.0 m/s² j

∑Fx = m*ax    m=20 kg , ax = -8 m/s²

F₃x - 9 = 20 *(-8)

F₃x = -160+9

F₃x = -151 N

∑Fy = m*ay    m=20 kg , ay = 6 m/s²

F₃y + 24 =20*( 6 )

F₃y =120 - 24

F₃y = 96 N

F₃ = -151 N i + 96 N j

7 0
3 years ago
When an object is placed just outside the focal length of a concave mirror, the image is
swat32

Answer:

O larger than the object and real

Explanation:

As we know by the formula of mirror

\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}

here we know that

d_o = (f_o + x)

so we have

\frac{1}{d_i} = \frac{1}{f_o} - \frac{1}{f_o + x}

so we have

d_i = \frac{(f_o)(f_o + x)}{x}

so magnification is given as

M = -\frac{d_i}{d_o}

M = - \frac{f_o}{x}

so here we have

|M| > 1

so image will be larger than object and real

8 0
3 years ago
Arightward force of 302 N is applied to a 28.6-kg crate to accelerate it across the floor. What will its acceleration be?
Musya8 [376]

Answer:

a \approx 10.6 m/s²

Explanation:

Since F = ma (Force = mass * acceleration), acceleration would be...

a = F/m

a = 302 N/28.6

a \approx 10.6 m/s²

8 0
3 years ago
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