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bearhunter [10]
2 years ago
11

If a young protostar with a disk is rotating and shrinking. how much faster is it rotating after its size has decreased by a fac

tor of 4? (enter an integer, as in it will be rotating ________ times faster). (3 pts)
Physics
1 answer:
maks197457 [2]2 years ago
5 0
In this system we have the conservation of angular momentum: L₁ = L₂
We can write L = m·r²·ω

Therefore, we will have:
m₁ · r₁² · ω₁ = m₂ · r₂² · ω₂

The mass stays constant, therefore it cancels out, and we can solve for ω<span>₂:
</span>ω₂ =  (r₁/ r₂)² · ω<span>₁
     
Since we know that r</span>₁ = 4r<span>₂, we get:
</span>ω₂ =  (4)² · ω<span>₁
     = 16 </span>· ω<span>₁

Hence, the protostar will be rotating 16 </span><span>times faster.</span>
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What conclusion can be derived by comparing the central tendencies of the two data sets?
zhannawk [14.2K]
The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest
3 0
2 years ago
550 J of work must be done to compress a gas to half its initial volume at constant temperature. How much work must be done to c
Over [174]

Answer:

The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J

Explanation:

The given variables are

Work done = 550 J

Volume change = V₂ - V₁ = -0.5V₁

Thus the product of pressure and volume change = work done by gas, thus

P × -0.5V₁ = 500 J

Hence -PV₁ = 1000 J

also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore  P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂

Also to compress the gas by a factor of 11 we have

P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11  = 909.091 J of work

7 0
2 years ago
The image shows positions of the earth and the moon in which region would an astronaut feel the lightest
trapecia [35]

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

5 0
2 years ago
Read 2 more answers
A parallel-plate capacitor is constructed of two square plates, size L×L, separated by distance d. The plates are given charge ±
vampirchik [111]

Answer: A) 2 B) 4 C) 1

Explanation:

The Electric field from a parallel-plate capacitor  is given by:

A) E=Q/(L^2 * ε0) so if we put a charge double the final electric field is double that the original.

B) from the above expression for the electric field,  If the size of the plate is double, then the E final is four times weaker that the original.

C) If the distante between plates is doubled the final electric field is the same that initial.

3 0
2 years ago
If the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal
iVinArrow [24]

Answer:

xcritical = d− m1 /m2 ( L /2−d)

Explanation: the precursor to this question will had been this

the precursor to the question can be found online.

ff the mass of the block is too large and the block is too close to the left end of the bar (near string B) then the horizontal bar may become unstable (i.e., the bar may no longer remain horizontal). What is the smallest possible value of x such that the bar remains stable (call it xcritical)

. from the principle of moments which states that sum of clockwise moments must be equal to the sum of anticlockwise moments. aslo sum of upward forces is equal to sum of downward forces

smallest possible value of x such that the bar remains stable (call it xcritical)

∑τA = 0 = m2g(d− xcritical)− m1g( −d)

xcritical = d− m1 /m2 ( L /2−d)

6 0
2 years ago
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