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3 years ago

How far away is the furthest artificial satellite orbiting earth?

1 answer:

8 0

Real-time distance and velocity data is provided by NASA and JPL. At a distance of 153.88 AU (23.020 billion km; 14.304 billion mi) from Earth as of September 5, 2021, it is the most distant artificial object from Earth. The probe made successful flybys of Jupiter, Saturn, and Saturn's largest moon, Titan.

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When a potential difference of 154 V is applied to the plates of a parallel-plate capacitor, the plates carry a surface charge d

tester [92]

Answer:

$4.7\mu m$

Explanation:

We are given that

Potential difference=V=154 V

Surface charge density= $\sigma=29nC/m^2=29\times 10^{-5} C/m^2$

Using $1 nC/cm^2=10^{-5} C/m^2$

We know that

$C=\frac{\epsilon_0A}{d}=\frac{Q}{V}=\frac{\sigma A}{V}$

$d=\frac{\epsilon_0V}{\sigma}$

Where

$\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Using the formula

$d=\frac{8.85\times 10^{-12}\times 154}{29\times 10^{-5}}$

$d=4.7\times 10^{-6} m=4.7\mu m$

Using $1\mu m=10^{-6} m$

Hence, the spacing between the plates= $4.7\mu m$

3 0

3 years ago

Will give brainliest to right answer!

viva [34]

The correct answer is D.

8 0

4 years ago

A 2.00-m long uniform beam has a mass of 4.00 kg. The beam rests on a fulcrum that is 1.20 m from its left end. In order for the

Shalnov [3]

Answer:

x ’= 1,735 m, measured from the far left

Explanation:

For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.

Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive

They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,

the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar

x_{cm} = 1.2 -1

x_ {cm} = 0.2 m

Σ τ = 0

w₁ 1.2 + mg 0.2 - W₂ x = 0

x = $\frac{m_1 g\ 1.2 \ + m g \ 0.2}{M_2 g}$

x = $\frac{m_1 \ 1.2 \ + m \ 0.2 }{M_2}$

let's calculate

x = $\frac{2.9 \ 1.2 \ + 4 \ 0.2 }{8.00}$ 2.9 1.2 + 4 0.2 / 8

x = 0.535 m

measured from the pivot point

measured from the far left is

x’= 1,2 + x

x'= 1.2 + 0.535

x ’= 1,735 m

8 0

3 years ago

In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl

irga5000 [103]

Answer:

$1.5048\times 10^{-23}\ Am^2$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

r = Radius of circle = $5.7\times 10^{-11}\ m$

v = Velocity of proton = $3.3\times 10^6\ m/s$

Magnetic moment is given by

$M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2$

The magnetic moment associated with this motion is $1.5048\times 10^{-23}\ Am^2$

5 0

3 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$