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3 years ago

13

In the real world, orbiting objects such as satellites and stations do not need to be constantly accelerating. What is different

about the real world satellites and the theoretical satellites we are discussing here?

1 answer:

JulijaS [17]3 years ago

8 0

What do you mean by theoretical satellites? When you mean real world satellites are you talking about the ones in space because I'm not sure what else you talking about.. I wanna help but could you explain plz??

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If the N pole of a magnet is brought near to an unmagnetized iron, nail, then the nail becomes

Andreyy89

Answer:

The nail also magnatized .

5 0

2 years ago

A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless, vertical axle. A

Flura [38]

Answer:

The mass of the wheel is 2159.045 kg

Explanation:

Given:

Radius $r = 0.330$

m

Force $F = 290$ N

Angular acceleration $\alpha = 0.814 \frac{rad}{s^{2} }$

From the formula of torque,

Γ $= I\alpha$ (1)

Γ $= rF$ (2)

$rF = I \alpha$

Find momentum of inertia $I$ from above equation,

$I = \frac{rF}{\alpha }$

$I = \frac{0.330 \times 290}{0.814}$

$I = 117.56$ $Kg. m^{2}$

Find the momentum inertia of disk,

$I = \frac{1}{2} Mr^{2}$

$M = \frac{2I}{r^{2} }$

$M = \frac{2 \times 117.56}{(0.330)^{2} }$

$M = 2159.045$ Kg

Therefore, the mass of the wheel is 2159.045 kg

8 0

4 years ago

86 Km/h convertir a (m/min)

gogolik [260]

Answer: 1433.3 m/min

Explanation:

For 86 Km/h converted to a (m/min), convert kilometers to meters, and hour to minutes

So, 86 Km/h means 86 kilometers per 1 hour

- If 1 kilometer = 1000 metres

86 kilometers = 86 x 1000 = 86,000m

- If 1 hour = 60 minute

1 hour = 60 minutes

In m/min: (86,000m / 60 minute)

= 1433.3 m/min

Thus, 86 Km/h convert to 1433.3 m/min

5 0

3 years ago

Alfred wegener was the one who made contiental drift theory ?

navik [9.2K]

Yep. he discovered that coastline from south america and africa fit together like a puzzle, which later became a part of the continential drift theory

4 0

3 years ago

If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

sdas [7]

Data:
$f_{2} = 42 Hz$
n (Wave node)
V (Wave belly)
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

Wire 1 → 1º Harmonic or Fundamental rope → n = 1

$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

If, We have:
V = 42L
Soon:
$f_{1} = \frac{V}{2L}$
$f_{1} = \frac{42L}{2L}$
$\boxed{f_{1} = 21 Hz}$

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0

3 years ago

Read 2 more answers