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4 years ago

What would be the correct answer for this?

Physics

2 answers:

Anna007 [38]4 years ago

5 0

Why ask us the question if you are right.

vazorg [7]4 years ago

3 0

You already got it right................................

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The evil team of Ms. Moonstruck and Mr. Luny intend to deflect the Moon from its orbit around Earth by pulling it with a force o

BabaBlast [244]

Answer:

92867kg

Explanation:

DATA:

mass of the moon ' $m_1$ '= 7.36 x $10^2^2$ kg

mass of the mini black hole ' $m_2$ '=?

universal gravitational constant= 6.67x $10^-^1^1$ Nm²/kg²

distance 'r'= 7.47× $10^6$ m

force 'F'= 8170N

Newton's Law of Universal Gravitation is given by:

$F=G\frac{m_1m_2}{r^2}$

$m_2=$ $\frac{(8170)(7.47*10^6)^2}{(6.67*10^-^1^1)(7.36*10^2^2)}$

$m_2=$ 92867kg

8 0

3 years ago

What percentage of the intensity gets through both polarizers?

sp2606 [1]

Answer:

if the two polarizers have the same direction the transmitted light is 50% of the incident and if the two polarizers are at 90º the transmitted light is zero

Explanation:

The incident light is generally random, that is, it does not have a polarization plane, when the first polarized stops by half, this already polarized light arrives at the second polarizer and the causticity passes

I = I₀ cos² θ

therefore if the two polarizers have the same direction the transmitted light is 50% of the incident and if the two polarizers are at 90º the transmitted light is zero

8 0

3 years ago

A policeman in a stationary car measures the speed of approaching cars by means of an ultrasonic device that emits a sound with

Ymorist [56]

Answer:

the frequency of the beats is 8.7687 kHz

Explanation:

Given the data in the question;

The frequency for stationary source and moving observer is;

f' = f( 1 +( v_observer/v_sound))

we know that, speed of sound in dry air = 343 m/s

so we substitute

f' = 41.2 kHz( 1 + (33.0 m/s / 343 m/s) ) = 41.2kHz( 1 + 0.0962) = 41.2kHz(1.0962)

f' = 45.1634 kHz

Now the frequency for stationary observer and moving source with frequency f' will be

f" = f'( 1 / (1 - ( v_observer/v_sound)))

45.1634 kHz( 343 / 343 - 33)

we substitute

f" = 45.1634 kHz( 1 / (1 - (33.0 m/s / 343 m/s)))

f" = 45.1634 kHz( 1 / (1 - 0.0962))

f" = 45.1634 kHz( 1 / 0.9038 )

f" = 45.1634 kHz( 1.1064 )

f" = 49.9687 kHz

Now the beat frequency will be;

f_beat = f' - f

we substitute

f_beat = 49.9687 kHz - 41.2 kHz

f_beat = 8.7687 kHz

Therefore, the frequency of the beats is 8.7687 kHz

7 0

3 years ago

A large aquarium of height 5 m is filled with fresh water to a depth of D = 1.80 m. One wall of the aquarium consists of thick p

Damm [24]

To solve the problem we will first start considering the Pressure given the hydrostatic definition of the product between the density, the gravity and the depth. We will define the area where the liquid acts and later we will use the definition of the force as a product between the pressure and the area to calculate the force given in the two depths. The gauge pressure at the depth x will be

$(P-P_a)= \rho g x$