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Korvikt [17]
2 years ago
12

Kon'nichiwa~please help me with this question!!​

Physics
1 answer:
andrezito [222]2 years ago
4 0

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

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I don't understand how to find the frequency and period of wavelengths. Can anyone help me?
nordsb [41]

Frequency of a wave =

                 (speed of the wave) divided by (wavelength)
or
                 (1) divided by (period of the wave) .


Period of the wave =

                   (wavelength) divided by (speed of the wave)
or
                   (1) divided by (frequency of the wave) .
3 0
3 years ago
The fact that some asteroids cluster in what are called asteroid families is probably the result of:
alisha [4.7K]

Answer:

The families are thought to form as a result of collisions between asteroids. In many or most cases the parent body was shattered, but there are also several families which resulted from a large cratering event which did not disrupt the parent body (e.g. the Vesta, Pallas, Hygiea, and Massalia families).

8 0
2 years ago
A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.
elena-14-01-66 [18.8K]

Answer:

Torque = –207.4 Nm

Explanation:

Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)

Torque = I × α

α = angular acceleration

I = moment of inertia

I = MR² for a circular hoop

Torque = 3.2×5.4×(– 12)

Torque = –207.4 Nm

6 0
3 years ago
Read 2 more answers
A filamentary conductor is formed into an equilateral triangle with sides of length carrying current i . find the magnetic field
arsen [322]

magnetic field due to a finite straight conductor is given by

B = \frac{\mu_0 i}{4\pi r}(sin\theta_1 + sin\theta_2)

here since it forms an equilateral triangle so we will have

\theta_1 = \theta_2 = 60 degre

also the perpendicular distance of the point from the wire is

r = \frac{a}{2\sqrt3}

now from the above equation magnetic field due to one wire is given by

B = \frac{\mu_0 i}{4\pi \frac{a}{2\sqrt3}(sin60 + sin60)

B = \frac{\mu_0 i*2\sqrt3}{4\pi a}(\sqrt3)

B = \frac{3\mu_0 i}{2\pi a}

now since in equilateral triangle there are three such wires so net magnetic field will be

B = \frac{9\mu_0 i}{2\pi a}

5 0
3 years ago
A current relationship is one that
Alex
C is the best answer
hope it helped
8 0
3 years ago
Read 2 more answers
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