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3 years ago

What is the value of the composite constant (Gme,/r2e) to be multiplied by the mass of the object mo, in equation below:

1 answer:

8 0

To solve this problem we will apply the definitions given in Newtonian theory about the Force of gravity, and the Force caused by weight. Both will be defined below, and in equal equilibrium condition to clear the variable concerning acceleration due to gravity. Finally, with the values provided in the statement, it will be replaced.

The equation for the gravitational force between the Earth and the object on the surface of the Earth is

$F_g = \frac{Gm_em_o}{r^2_e}$

Where,

G = Universal gravitational constant

$m_e$ = Mass of Earth

$r_e$ = Distance between object and center of earth

$m_o$ = Mass of Object

The equation for the gravitational pulling force on the object due to gravitational acceleration is

$F_g = m_o g$

Equation the two expression we have

$m_o g = \frac{Gm_em_o}{r_e^2}$

$g = \frac{Gm_e}{r_e^2}$

This the acceleration due to gravity which is composite constant.

Replacing with our values we have then

$g = \frac{(6.67*10^{-11}N\cdot m^2/kg^2)(5.98*10^{24}kg)}{6378km(\frac{10^3m}{1km})^2}$

$g = 9.8m/s^2$

The value of composite constant is $9.8m/s^2$ . Here, the composite constant is nothing but the acceleration due to gravity which is constant always.

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mario62 [17]

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6 0

3 years ago

A stretched string has a mass per unit length of 5.00 g/cm and a tension of 10.0 N. A sinusoidal wave on this string has an ampl

8_murik_8 [283]

Answer:

0.12 mm ; 140.50 rad/m ; 628.32 rad/sec ; +

Explanation:

Given the wave equation of the form :

y(x, t) = ym sin(kx ± ωt)

Mas per unit length (u) = 5 g/cm = (5÷1000)kg / 0.01m) = 0.005kg/0.01m = 0.5kg/m

Tension, T = 10 N

Amplitude, A = 0.12 mm

Frequency, F = 100 Hz

Comparing with the general wave equation :

y = Asin(kx ± ωt)

A = amplitude = ym = 0.12 mm

2.) k = 2π / λ

Recall :

v = fλ

v = sqrt(T/u) = sqrt(10/0.5) = sqrt(20) = 4.472

λ = v/ f = 4.472 / 100 = 0.04472

Hence,

k = (2 * π) / 0.04472

k = 140.50 rad/m

3.) Angular frequency, ω

ω = 2πf = 2 * 3.14 * 100 = 628.32 rad/sec

4.) sign is +ve

Direction of wave propagation as given is in the negative x axis

4 0

2 years ago

Calculate the height from from which a body is released from rest if its velocity just before hitting the ground is30m\s

Kamila [148]

Answer:

height = 45 m

Explanation:

Initial velocity = 0 m/s

Final velocity = 30 m/s

Acceleration due to gravity = +10m/s^2 ( because ball is going down )

Using the eqn of motion

${v}^{2} - {u}^{2} = 2gh$

where v= final velocity ; u = initial velocity ; g = acceleration due to gravity ; h = height

So,

${30}^{2} - {0}^{2} = 2 \times 10 \times h$

$= > 20h = 900$

$= > h = \frac{900}{20} = 45$

5 0

3 years ago

Which configuration would produce an electric current? A) Rotate a coil of copper wire between two magnets. B) Connect a wire be

FrozenT [24]

the answer is (A) the movement of the magnet relative to the coil

4 0

3 years ago

Read 2 more answers

As viewed from above in this picture, what direction will the current be in the coil of wire that will cause the loop to rotate

Gala2k [10]

Answer:

When viewed from above, the current in the coil should point towards the top-right corner of the picture.

Explanation:

The current in this coil have only two possible directions: clockwise or counter-clockwise. However, since the diagram shows the coil from above, not from a cross-section, just saying clockwise or counter-clockwise might be ambiguous. The statement that the current is directed towards the top-right corner of the picture is equivalent to saying that when viewed from the lower-right corner of this diagram, the current in the coil is moving clockwise.

Note that at the center of this picture, the current is parallel to the magnetic field- there will be no force on the coil at that position. On the other hand, (also when viewed from above,) at the top-right corner and the lower-left corner of the coil, the current in the coil will be perpendicular to the magnetic field. That's where the force on the coil will be the strongest.

With that in mind, apply the right-hand rule to find the direction of the force on the coil in each of the two possibilities.

Assume that when viewed from above, the current is flowing towards the top-right corner of the picture. Consider the wire near the top-right corner of this coil (as viewed above on this picture.) The current will be going into the picture into the magnetic field. By the right-hand rule, the current on the wire near that point should be pointing towards the bottom of this picture. (Point fingers on the right hand in the direction of the current $I$ . Rotate the right hand such that when curling the fingers, they point in the direction of the magnetic field $B$ . The direction of the right thumb should now point in the direction of the force on the wire $F$ .)

Based on the same assumption, the current in the wires near the bottom left corner of this coil will be pointing out of the picture. By the right hand rule, the magnetic force on the coil in that region should be pointing towards the top of this picture. Combing these two forces, the coil would indeed be rotating around the center of this picture in the direction shown in the diagram.

It can also be shown that if the current points towards the bottom left corner of the picture when viewed from above, the coil will be rotating about the center of this picture in the opposite direction.

7 0

3 years ago