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3 years ago

Jupiter is made of gas(like Saturn, Uranus and Neptune). What would happen to the strength of gravity if you

Physics

1 answer:

garik1379 [7]3 years ago

7 0

Answer:

a) The strength of gravity decreases if one moved away from Jupiter

b) The strength of gravity increases if one fell into Jupiter

Explanation:

The gravitational attraction is given by Newton law of gravitation as follows;

$Force \ (strength) \ of \ gravity = \dfrac{G \times M \times m}{R^2}$

Where;

G = The universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)

M = The mass of Jupiter

m = The mass of the nearby body

R = The distance between the centers of Jupiter and the body

From the equation, we have that the gravitational strength varies inversely with the square of the separation distance between two bodies

Therefore, as one moves away, R increases, and the strength of gravity reduces

Similarly as the body falls into Jupiter, R, reduces the gravitational strength increases.

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3 years ago

Water cycle

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3 years ago

Read 2 more answers

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago

If the intensity of a loud car horn is 0.005 W/m^2 when you are 2 meters away from the source. Calculate the sound intensity lev

nekit [7.7K]

Answer:

Explanation:

We have given the intensity of the loud car horn $I=0.005w/m^2$

We know that $I_O=10^{-12}w/m^2$

Now the sound intensity level is given by $\beta =10log\frac{I}{I_0}=10log\frac{0.005}{10^{-12}}=96.98dB$ , which is nearly equal to 97

So the sound intensity level will be 97 dB

So option (c) will be the correct option

4 0

3 years ago