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3 years ago

During a fission each time an atom divides ___.

2 answers:

7 0

It releases a lot of energy so I would say, the answer IS NOT half. More energy than what is released though? I would say the Third option

garri49 [273]3 years ago

3 0

“More energy is released”

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An airplane is heading due south at a speed of 690 km/h . A) If a wind begins blowing from the southwest at a speed of 90 km/h (

Afina-wow [57]

Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position

Explanation:

Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where

see fig 1

Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then

90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,

this because we have an isosceles triangle, then the cathetus length is

hypotenuse/ $\sqrt{2}$

using Pythagoras, here the hypotenuse is 90, then the cathetus are of length

90/ $\sqrt{2}$ km/h= 63,6396 km/h.

Now the total speed of the plane is

690km/h south + 63,6396 km/h north +63,6396 km/h east,

this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.

then using again Pythagoras we get the magnitude of the total speed it is

$\sqrt{626,3604 ^2+63,6396^2} km/h=629,5851km/h$ ,

the direction is calculated with respect to the south using trigonometry, we know the

sin x= cathetus opposed / hypotenuse,

then

x= $sin^{-1}'frac{63,6396}{629,5851}$ =5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form

626,6396 km/h south + 63,6396km/h east.

Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get

Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,

since 11 min=11/60 hours=0.1833 hours.

Then the total detour from the expected position, the one it should have without the influence of the wind, we get

Xdetour=[/tex]\sqrt{2* 11,6672x^{2} }[/tex] = 16,5km at 45 degrees from east pointing north

The situation is sketched as follows see fig 2

4 0

4 years ago

A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 2.3 cm from the axis of rotation. (a) Calcul

VLD [36.1K]

Answer:

a) $a_{r} = 0.275\,\frac{m}{s^{2}}$ , b) $\mu_{s} = 0.028$ , c) $\mu_{s} = 0.036$

Explanation:

a) The linear acceleration of the watermelon seed is:

$a_{r} = \omega^{2}\cdot r$

$a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)$

$a_{r} = 0.275\,\frac{m}{s^{2}}$

b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:

$\Sigma F = \mu_{s}\cdot m\cdot g = m\cdot a$

$a = \mu_{s}\cdot g$

$\mu_{s} = \frac{a}{g}$

$\mu_{s} = \frac{0.275\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$\mu_{s} = 0.028$

c) Angular acceleration experimented by the turntable is:

$\alpha = \frac{\omega-\omega_{o}}{\Delta t}$

$\alpha = \frac{3.456\,\frac{rad}{s}-0\,\frac{rad}{s} }{0.36\,s}$

$\alpha = 9.6\,\frac{rad}{s^{2}}$

The tangential acceleration experimented by the watermelon seed is:

$a_{t} = \left(9.6\,\frac{rad}{s^{2}} \right)\cdot (0.023\,m)$

$a_{t} = 0.221\,\frac{m}{s^{2}}$

The linear acceleration experimented by the watermelon seed is:

$a = \sqrt{a_{t}^{2}+a_{r}^{2}}$

$a = \sqrt{\left(0.221\,\frac{m}{s^{2}} \right)^{2}+\left(0.275\,\frac{m}{s^{2}} \right)^{2}}$

$a = 0.353\,\frac{m}{s^{2}}$

The minimum coefficient of static friction is:

$\mu_{s} = \frac{0.353\,\frac{m}{s^{2}} }{9.807\,\frac{m}{s^{2}} }$

$\mu_{s} = 0.036$

4 0

3 years ago

The 400-turn primary coil of a step-down transformer is connected to an ac line that is 120 V rms. The secondary coil voltage is

KengaRu [80]

Explanation:

The given data is as follows.

$V_{1}$ = 120 V, $V_{2}$ = 6.50 V

$N_{1}$ = 400, $N_{2}$ = ?

Formula to calculate the number of turns in secondary coil are as follows.

$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$

Putting the given values into the above formula as follows.

$\frac{N_{1}}{N_{2}} = \frac{V_{1}}{V_{2}}$

$\frac{400}{N_{2}} = \frac{120}{6.50}$

$N_{2}$ = 21.66

= 22 (approx)

Thus, we can conclude that number of turns in the secondary coil are 22.

3 0

3 years ago

What Global Positioning system signals can you tell

natulia [17]

<span>Global Positioning system(GPS), It is a satellite which broadcast "Microwave" signal to determine Location or Velocity.

Hope this helps!</span>

8 0

4 years ago

If a textbook weighs 37.2 N on Venus, and the book has a mass of 4.4 kg, what is the strength of gravity on Venus?

Romashka [77]

$F=ma$

Where F is force, m is mass and a is acceleration.

$37.2N=4.4kg\cdot a\Longrightarrow a=\dfrac{37.2N}{4.4kg}=\boxed{8.45\dfrac{m}{s^2}}$

Hope this helps.

r3t40

4 0

3 years ago