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vichka [17]
4 years ago
10

Electrons are in constant motion around the nucleus because of

Physics
1 answer:
TEA [102]4 years ago
8 0

because of the continnuity of quaternion energy.

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Which projectiles will be visibly affected by air resistance when they fall
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Balloon
this could be it
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A flat disk, a solid sphere, and a hollow sphere each have the same mass m and radius r. The three objects are arranged so that
Aloiza [94]

Answer:

b)

Explanation:

Rotational inertia of the sphere: \frac{2}{5}mr^2

Rotational inertia of hollow sphere: \frac{2}{3}mr^2

Rotational inertia of flat disk: \frac{1}{2}mr^2

The largest value is 2/3 mr², therefore, hollow sphere's inertia is largest.

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A 500-n crate needs to be lifted 1 meter vertically in order to get it into the back of a pickup truck. what gives the crate a g
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The weight an weight of the truck 
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4 years ago
Read 2 more answers
A 1056-hertz tuning fork is struck at the same time as a note on the piano and you hear 2 beats/second. You tighten the piano st
elena-14-01-66 [18.8K]

Answer:

The frequency of the piano string is <em>1059 Hz</em>.

Explanation:

The frequency beat (fb), 2 beats/second, is the absolute difference between the frequency of the tuning fork (1056 Hz) and the frequency of the piano string.

As the piano string gets tightened, the frequency beat becomes 3 beats/second.

Therefore,

fb = fb = fpiano - ftuning fork\\ 3 Hz=fpiano-1056Hz\\ fpiano=1056Hz+3Hz\\ fpiano=1059Hz

6 0
3 years ago
A carnot heat engine receives 600 kj of heat from a source of unknown temperature and rejects 175 kj of it to a sink at 20°c. de
oee [108]

(b) 71%

The thermal efficiency of a Carnot heat engine is given by:

\eta = \frac{W}{Q_{in}}

where

W is the useful work done by the engine

Q_{in} is the heat in input to the machine

In this problem, we have:

Q_{in}=600 kJ is the heat absorbed

W=600 kJ-175 kJ=425 kJ is the work done (175 kJ is the heat released to the sink, therefore the work done is equal to the difference between the heat in input and the heat released)

So, the efficiency is

\eta = \frac{425 kJ}{600 kJ}=0.71 = 71\%

(a) 737^{\circ}C

The efficiency of an engine can also be rewritten as

\eta = 1-\frac{T_C}{T_H}

where

T_C is the absolute temperature of the cold sink

T_H is the temperature of the source

In this problem, the temperature of the sink is

T_C = 20^{\circ}C + 273=293 K

So we can re-arrange the equation to find the temperature of the source:

T_H = \frac{T_C}{1-\eta}=\frac{293 K}{1-0.71}=1010 K\\T_H = 1010 K - 273=737^{\circ}C

7 0
3 years ago
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