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cluponka [151]
3 years ago
10

What is the function of a power source in a circuit?ill give brainliest btw

Physics
1 answer:
In-s [12.5K]3 years ago
7 0
Electrical energy hope this helps <3
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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

7 0
3 years ago
A rocket ship is accelerating at 200 m/s2, its mas is 135,000,000 kg. What is the force generated by this acceleration?
Rina8888 [55]

Acceleration does NOT "generate" force.  Acceleration NEEDS force to make it happen.  Without force ... provided by something else ... acceleration can't happen.

The force NEEDED to accelerate a mass with a certain acceleration is

Force needed = (mass) times (acceleration)

For the rocket ship in the question,

Force = (135,000,000 kg) times (200 m/s²)

Force = (135,000,000 x 200) kg-m/s²

<em>Force = 27 Giga-Newtons  </em>(27,000,000,000 Newtons)


The gas-generator cycle F-1 rocket engine, developed in the US by Rocketdyne in the late 1950s, was used in the Saturn V rocket, the main launch vehicle of NASA's Apollo moon lander program .  Five F-1 engines were used in the first stage of each Saturn V.  

==> The thrust of each F-1 engine at full throttle is 7,770 kilo-Newtons.  

It would take <em>3,475 </em>of these F-1 rocket engines, running full-throttle, to provide the force calculated in the answer to this question.  If you didn't have 3,475 F-1 rocket engines, then you couldn't accelerate 135,000,000 kg at 200 m/s².

(And by the way ... the mass of each F-1 engine is 8,400 kg.  So 3,475 engines alone account for 22% of the mass you're trying to accelerate.  And don't even get me started about the mass of the FUEL you'd need to carry.)

5 0
3 years ago
Place the following into scientific notation...(.000635)<br> *
AleksAgata [21]
6.35x10^-4 OR 6.3x10-4 (if only one decimal number is allowed)
6 0
3 years ago
A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s
belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, \omega_i=114\ rev/min = 11.93\ rad/s

Final angular speed of the wheel, \omega_f=0

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

I=mr^2

I=49\times (0.73)^2

I=26.11\ kg-m^2

We know that the work done is equal to change in kinetic energy.

W=\Delta E

W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)

W=-\dfrac{1}{2}\times 26.11\times (11.93^2)

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

P=\dfrac{W}{t}

P=\dfrac{1858.05\ J}{22\ s}

P =84.45 watts

Hence, this is the required solution.

4 0
3 years ago
A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
2 years ago
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