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anygoal [31]
3 years ago
14

If earth began to shrink, but its mass remained the same, what would happen to the value of g on earth's surface?

Physics
1 answer:
OverLord2011 [107]3 years ago
3 0
<span>The estimation of g on the surface of the earth relies upon mass and range, truth be told, g = GM/R^2 where G is the newton's gravitational constants, M is the mass of the earth and R the range . Should there be any occassion that would cause R to be diminished, the estimation of g would increment as the opposite square of R on the fact that R is divided, g would be 4 times the original value</span>
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1. What are the units of : pressure, force , area
White raven [17]

Pressure- Pascal(N/m^2)

Force- Newton

Area-m^2(Square metre)

<u>Explanation:</u>

  • Pressure is force continuous force applied to an object. The applied force is always perpendicular to the surface of the object.
  • Force is a push or pulls on an object by an external body where it changes its state from rest to motion.
  • Force is said to have both magnitude and direction.
  • Area is the portion that shows the scale of a 2D form or configuration in the level. The covering area is its analog on the 2D surface of a 3D object.
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collection of such objects we observe that they emit electromagnetic radiation of three different energies: 0.7 eV (infrared), 2
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4 0
3 years ago
Some of the highest tides in the world occur in the Bay of Fundy on the Atlantic Coast of Canada. At Hopewell Cape the water dep
Nady [450]

Answer:

(a) 1.939 m/h

(b) 0.926 m/h

(c) -0.315 m/h

(d) -1.21 m/h

Explanation:

Here, we have the water depth given by the function of time;

D(t) = 7 + 5·cos[0.503(t-6.75)]

Therefore, to find the velocity of the depth displacement with time, we differentiate the given expression with respect to time as follows;

D'(t) = \frac{d(7 + 5\cdot cos[0.503(t-6.75)])}{dt}

= 5×(-sin(0.503(t-6.75))×0.503

= -2.515×(-sin(0.503(t-6.75))

= -2.515×(-sin(0.503×t-3.395))

Therefore we have;

(a) at 5:00 AM = 5 -  0:00 = 5

D'(5) =  -2.515×(-sin(0.503×5-3.395)) = 1.939 m/h

(b) at 6:00 AM = 6 -  0:00 = 6

D'(5) =  -2.515×(-sin(0.503×6-3.395)) = 0.926 m/h

(c) at 7:00 AM = 7 -  0:00 = 7

D'(5) =  -2.515×(-sin(0.503×7-3.395)) = -0.315 m/h

(d) at Noon 12:00 PM = 12 -  0:00 = 12

D'(5) =  -2.515×(-sin(0.503×12-3.395)) = -1.21 m/h.

4 0
2 years ago
A tuning fork has a frequency of 280 Hz and the wavelength of the sound produced is 1.5 meters. Calculate the wave's frequency a
s2008m [1.1K]

Answer:

The answer would be 420 m/s

Explanation:

Look in attachment ⬇

I Hope this Helps!!!

3 0
3 years ago
A 4.0 m length of gold wire is connected to a 1.5 V battery, and a current of 4.0 mA flows through it. What is the diameter of t
sladkih [1.3K]

Explanation:

Given that,

Length of gold wire, l = 4 m

Voltage of battery, V = 1.5 V

Current, I = 4 mA

The resistivity of gold, \rho=2.44\times 10^{-8}\ \Omega-m

Resistance in terms of resistivity is given by :

R=\dfrac{\rho l}{A}

Also, V = IR

So,

\dfrac{V}{I}=\dfrac{\rho l}{A}

A is area of wire,

\dfrac{V}{I}=\dfrac{\rho l}{\pi r^2}, r is radius, r = d/2 (diameter=d)

\dfrac{V}{I}=\dfrac{\rho l}{\pi (d/2)^2}\\\\\dfrac{V}{I}=\dfrac{4\rho l}{\pi d^2}\\\\d=\sqrt{\dfrac{4\rho l I}{V\pi}} \\\\d=\sqrt{\dfrac{4\times 2.44\times 10^{-8}\times 4\times 4\times 10^{-3}}{1.5\times \pi}} \\\\d=18.2\ \mu m

Out of four option, near option is (C) 17 μm.

6 0
3 years ago
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