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sveticcg [70]
4 years ago
12

An object's potential energy is set it cannot change is this true

Physics
2 answers:
zepelin [54]4 years ago
8 0
The energy achieved I think
beks73 [17]4 years ago
4 0
The energy is archived, I think
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Help wit these questions someone.
romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, \frac{1}{Req}  = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}

R₂ + 2 = 12

R₂ = 10Ω

<h3>Q8.</h3>

\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}

\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}

Req = 1.7Ω

7 0
3 years ago
How much force is required to accelerate a 5 kg mass at 20 m/s^2
Studentka2010 [4]

Hello!

\large\boxed{F = 100N}

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:

F = 5 · 20

F = 100N

6 0
3 years ago
Which is NOT one of the four principals of Dalton’s atomic theory?
Gre4nikov [31]

Answer:

Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>

Explanation:

Hope it helps..

Just correct me if I'm wrong, okay?

But ur welcome!!

(;ŏ﹏ŏ)(◕ᴗ◕✿)

8 0
3 years ago
Read 2 more answers
What led astronomers to believe in the existence of dark matter?
Tju [1.3M]
It’s either C or D. let me know but I would do C!
3 0
3 years ago
An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte
telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2

a. The angular acceleration is Torque divided by moments of inertia:

\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2

b. 5 revolution would be equals to 10\pi rad, or 31.4 rad. Since the engine just got started

\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5

\omega = \sqrt{1893.5} = 43.5 rad/s

c. Work done during the first 5 revolution would be torque times angular displacement:

W = T*\theta = 1930 * 31.4 = 60633 J

d. The time it takes to spin the first 5 revolutions is

t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s

The average power output is work per unit time

P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

P_i = T*\omega = 1930*43.5=83983 W or 84 kW

7 0
3 years ago
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