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Mamont248 [21]
2 years ago
12

In the image, not to scale, which phase of the moon would you observe from earth?​

Physics
1 answer:
timofeeve [1]2 years ago
8 0

Answer:

The "new" moon phase is being shown because the moon shown is directly between the sun and earth and is not visible from the earth because of the brightess of the sun.

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A computer monitor uses 200 W of power. How much energy does it use in 10 seconds?
gavmur [86]

Answer:

<u>The correct answer is 0.556 Watts</u>

Explanation:

The computer monitor uses 200 Watts of power in an hour, that is the standard measure.

If we want to know, how much energy the computer monitor uses in one second, we will have to divide both sides of the equation into 3,600.

1 hour = 60 minutes = 3,600 seconds (60 x 60)

Energy per second = 200/3600

Energy per second = 0.0556 Watts

Therefore to calculate how much energy is used in 10 seconds, we do this:

Energy per second x 10

<u>0.0556 x 10 = 0.556 Watts</u>

<u>The computer monitor uses 0.556 Watts in 10 seconds</u>

3 0
3 years ago
Read 2 more answers
The process of digestion breaks down proteins into smaller molecules that are used to rebuild other proteins. During digestion,
Len [333]

The process of digestion breaks down proteins into smaller molecules that are used to rebuild other proteins. During digestion, proteins undergo chemical changes.

<u>Option: D</u>

<u>Explanation:</u>

Proteins are broken down into amino acids considered the foundational element. When issued, these tiny molecules can then be consumed into the bloodstream via the gut wall. An enzyme is a protein which can regulate biochemical response rate. An enzyme integrates a water molecule around the bond in enzymatic hydrolysis processes which allows it to split.

The bonds which hold together the amino acids are recognized as peptide bonds. A hydrolysis process comparable to that included in splitting up carbohydrates is required to break the peptide bonds within a protein. Enzymes identified as proteases are required for the protein to disintegrate.

7 0
3 years ago
14.A 90 kg quarterback gets tackled by being hit by a 120 kg lineman backwards
quester [9]

The acceleration of the quarterback and the lineman is 5.55m/s² and 4.16m/s² respectively in the same direction.

As, we know, the 120 Kg lineman is moving with a force of 500N.

His net acceleration will be in the same direction as his motion.

It is already known that, If M is the mass of the body and a is the acceleration of the body, then the force F on the body can be calculated by using the formula,

F = Ma.

The weight of the quarterback is 90 Kg. He is being hit by a force of 500N.

So, the acceleration can be calculated using the formula,

500N = 90kg x a

a = 5.55 m/s².

Now, the weight if the lineman is 120kg, the force applied by him is 500N.

So, from the formula, his acceleration A will be,

500N = 120Kg x A

A = 4.16 m/s².

both of them will have acceleration in the same direction,

To know more about Force, visit,

brainly.com/question/25239010

#SPJ9

8 0
10 months ago
What is the power of 10 when 0.00503 is written in scientific notation?
svetoff [14.1K]

Answer:

Negative 3

Explanation:

Bc scientific notation is the zeros either ahead or behind the actual numbers

4 0
3 years ago
Read 2 more answers
A prismatic bar AB of length L and solid circular cross section (diameter d) is loaded by a distributed torque of constant inten
Lyrx [107]

Answer:

a) the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

b) the angle of twist between the ends of the bar is 16tL² / πGd⁴  

Explanation:

Given the data in the question, as illustrated in the image below;

d is the diameter of the prismatic bar of length AB

t is the intensity of distributed torque

(a) Determine the maximum shear stress tmax in the bar

Maximum Applied torque  T_max = tL

we know that;

shear stress τ = 16T/πd³

where d is the diameter

so

τ_{max} = 16T_{max} /πd³

Therefore, the maximum shear stress τ_{max} the bar is 16T_{max} /πd³

(b) Determine the angle of twist between the ends of the bar.

let theta (\theta) be the angle of twist

polar moment of inertia I_p} = πd⁴/32

now from the second image;

lets length dx which is at distance of "x" from "B"

Torque distance x

T(x) = tx

Elemental angle twist = d\theta = T(x)dx / GI_{p}

so

d\theta = tx.dx / G(πd⁴/32)

d\theta = 32tx.dx / πGd⁴

so total angle of twist \theta will be;

\theta =  \int\limits^L_0  \, d\theta

\theta =  \int\limits^L_0  \, 32tx.dx / πGd⁴

\theta = 32t / πGd⁴  \int\limits^L_0  \, xdx

\theta = 32t / πGd⁴ [ L²/2]

\theta = 16tL² / πGd⁴  

Therefore,  the angle of twist between the ends of the bar is 16tL² / πGd⁴  

7 0
3 years ago
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