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Ber [7]
3 years ago
6

The Andromeda Galaxy is faintly visible to the naked eye in the constellation Andromeda. Suppose instead it were located in the

same direction in space as the center of the Milky Way Galaxy (but still at its current distance). How would it appear to the eye in that case?
a) We could not see it at all.
b) It would look about the same, but would be in the constellation Sagittarius instead of Andromeda.
c) It would be much brighter, because it would be illuminated by the many stars in the center of our galaxy.
d) It would look about the same, but it would be harder to pick out because its cloud-like appearance would make it blend in with the cloud-like appearance of the Milky Way in our sky.
Physics
1 answer:
Kaylis [27]3 years ago
8 0

Answer:

a) We could not see it at all.

Explanation:

The most distant object that can be seen is the andromeda galaxy, which we may have a slight view of. The andromeda galaxy is a large galaxy that along with the previous two is also part of the local group. Spiral-type galaxy that is approximately 250,000 light years in diameter (more than twice the diameter of the Milky Way!) And is about 2.9 million light years away from our galaxy. Because of its distance, we have difficulty visualizing this galaxy, we would have this difficulty even if the andromeda galaxy was in the center of the Milky Way, but maintaining its current distance. That is, even if the andromeda galaxy were located in the same direction in space as the center of the Milky Way (but still at its current distance), we could not see it at all.

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What is the acceleration of an object that goes from 45m/s to 10 m/s in 5 seconds?
kipiarov [429]

Answer:

\boxed {\boxed {\sf a= -7 \ m/s^2}}

Explanation:

Acceleration is the change in velocity over time.

a= \frac {v_f-v_i}{t}

The object accelerates <em>from</em> 45 meters per second <em>to </em>10 meters per second in 5 seconds. Therefore,

v_f=10 \ m/s \\v_i= 45 \ m/s \\t= 5 \ s

Substitute the values into the formula.

a= \frac{ 10 \ m/s - 45 \ m/s}{5 \ s}

Solve the numerator.

a= \frac { -35 \ m/s}{5 \ s}

Divide.

a= -7 \ m/s/s

a= -7 \ m/s^2

The acceleration of the object is -7 meters per square second. The acceleration is negative because the object's velocity decreases and the object slows down.

5 0
3 years ago
Starting from rest, a 2.3x10-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exerts
Harman [31]

Answer:

3.13 m/s

Explanation:

From the question,

Since the flea spring started from rest,

Ek = W................... Equation 1

Where Ek = Kinetic Energy of the flea spring, W = work done on the flea spring.

But,

Ek = 1/2mv²............ Equation 2

Where m = mass of the flea spring, v = flea's speed when it leaves the ground.

substitute equation 2 into equation 1

1/2mv² = W.................... Equation 3

make v the subject of the equation

v = √(2W/m)................. Equation 4

Given: W = 3.6×10⁻⁴ J, m = 2.3×10⁻⁴ kg

Substitute into equation 4

v = √[2×3.6×10⁻⁴ )/2.3×10⁻⁴]

v = 7.2/2.3

v = 3.13 m/s

Hence the flea's speed when it leaves the ground  = 3.13 m/s

4 0
3 years ago
What type of organization is used in a paragraph that lists similarities between two objects?
alekssr [168]
<span>A. Comparison 

</span>What type of organization is used in a paragraph that lists similarities between two objects? Comparison


NOT:
<span>B. Contrast 
C. Chronological order 
D. Cause and effect</span><span>
</span>
3 0
3 years ago
A block of gelatin is 120 mm by 120 mm by 40 mm when unstressed.
ycow [4]

Answer:

σ = 3.402 KPa ,  γ = 0.25 , G = 13.608 KPa

Explanation:

Given:-

- The dimension of gelatin block = ( 120 x 120 x 40 ) mm

- The applied force, F = 49 N

- The displacement of upper surface, x = 10 mm

Find:-

Find the shearing stress, shearing strain and  shear modulus.​

Solution:-

- The shear stress is the internal pressure created in an object opposing the applied action ( Force, moment, bending, or torque ).

- A force of F = 49 N was applied parallel to the top surface of the gelatin block.

- The shear effect results in a stress in the gelatin block.

- The formulation of stress ( σ ) is given below:

                        σ = F / A

Where,

           A : The surface area of the object that experiences the shear force.

- The top surface have the following dimensions:

          A = ( 0.120 )*( 0.120 ) = 0.0144 m^2

Therefore,

                     σ = 49 / 0.0144

                     σ = 3.402 KPa

- The shear strain ( γ ) is the measurement of change in dimension per unit depth of the block.

- The top surface undergoes a displacement of ( x ). The height of the top surface of the gelatin block is L = 40 mm.

Hence,

                    γ = x / L

                    γ = 10 / 40

                    γ = 0.25

- The shear modulus or the modulus of rigidity ( G ) is a material intrinsic property that signifies the amount of resistive stress to any cause of deformation.

- It is mathematically expressed as a ratio of shear stress  ( σ ) and shear strain ( γ ):

                   G =  σ / γ

                   G = 3.402 / 0.25

                   G = 13.608 KPa

7 0
3 years ago
Envision holding the end of a ruler with one hand and deforming it with the other. When you let go, you can see the oscillations
lapo4ka [179]

Answer: To increase the rigidity of the system you could hold the ruler at its midpoint so that the part of the ruler that oscillates is half as long as in the original experiment.

Explanation:

When a rule is displaced from its vertical position, it oscillates back and forth because of the restoring force opposing the displacement. That is, when the rule is on the left there is a force to the right.

By holding a ruler with one hand and deforming it with the other a force is generated in the opposite direction which is known as the restoring force. The restoring force causes the ruler to move back toward its stable equilibrium position, where the net force on it is zero. The momentum gained causes the ruler to move to the right leading to opposite deformation. This moves the ruler again to the left. The whole process is repeated until dissipative forces reduce the motion causing the ruler to come to rest.

The relationship between restoring force and displacement was described by Hooke's law. This states that displacement or deformation is directly proportional to the deforming force applied.

F= -kx, where,

F= restoring force

x= displacement or deformation

k= constant related to the rigidity of the system.

Therefore, the larger the force constant, the greater the restoring force, and the stiffer the system.

5 0
3 years ago
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