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AveGali [126]
3 years ago
6

An astronaut on a spacewalk accidentally drops a tool, and it floats away.Which of the following objects is exerting a gravitati

onal force on the floating tool? Choose all that apply
the astronaut
the Moon
the Earth
the Sun
Physics
2 answers:
castortr0y [4]3 years ago
4 0

Answer:

All of them are correct

Explanation:

They are all correct select all

a_sh-v [17]3 years ago
3 0

Gravitational force is given as

F = \frac{Gm_1m_2}{r^2}

so here gravitational force will be given by above formula where the force between two objects depends on the mass of two objects and distance between them.

Now here since the tool is at finite distance from Astronaut and Moon, Sun, Earth

So all the above will exert gravitational force on the tool but the magnitude of force will be different as all of the above are of different masses and situated at different distance.

So all options are correct here

The Astronaut

The Moon

The Sun

The Earth

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You recently purchased a home four blocks from the busy tollway. On a typical evening, the decibel level resulting from tollway
inn [45]

Answer:

74dB

Explanation:

Sound intensity is inversely proportional to the square of the distance. Thus, a house four times closer hears a sound 16x more intense.

The dB scale is logarithmic, so 16x louder than 62 dB is equal to

62 + 10*log(16) ~= 62 + 12 = 74dB

5 0
3 years ago
Many meteorites appear to have formed very early in the solar system's history. How do these meteorites support our theory about
tigry1 [53]

Answer:

The appearance and composition of meteorites is what we would expect if metal and rock condensed and increased as our theory suggests.

Explanation:

The planets and most of their satellites were formed by accretion of matter that accumulated around the largest pieces of the proto-nebula. After a chaotic succession of collisions, mergers and reconstruction processes, they acquired a size similar to the current one and moved until they were in the positions we know.

The area closest to the Sun was too warm to retain light materials. That is why the inner planets are small and rocky, while the outer ones are large and gaseous. The evolution of the Solar System has not stopped, but, after the initial chaos, most of the materials are now part of bodies located in more or less stable orbits.

Any theory that attempts to explain the formation of the Solar System should take into account that the Sun rotates slowly and only has 1 percent of the angular momentum, but it has 99.9% of its mass, while the planets have 99% of the angular momentum and only 0.1% of the mass. One of the explanations argues that, at first, the Sun was much colder; The density of its materials was slowing its rotation, while warming, until a certain balance was achieved.

6 0
3 years ago
Consider a long, closely wound solenoid with 10,000 turns per meter. What current, in amperes, is needed in the solenoid to prod
Makovka662 [10]

Answer:

0.4344A

Explanation:

From Ampere's law, it can be shown that the magnetic field B inside a long solenoid is

B= \mu_0NI

Where

B= Magnetic field strenght at distance d

I= current

\mu_0 =Permeability of free space (4\pi*10^{-7} Tm/A)

N= Number of loops

Our values are defined as follow,

N=10000

B=5.25*10^{-5}T

B'=5.25*10^{-5} * 104 = 5.46*10^{-3}T

As a current required to become 104 times the Earth's magnetic field is required, we use B '

B'= \mu_0NI

5.46*10^{-3}=4\pi*10^{-7}*10000*I

I=\frac{5.46*10^{-3}}{4\pi*10^{-7}*10000}

I=0.4344A

<em>Therefore is needed 0.4344A in the solenoid to produce a magnetic field inside the solenoid, near its center, that is 104 times the Earth's magnetic field.</em>

4 0
2 years ago
A very long line of charge with charge per unit length +8.00 μC/m is on the x-axis and its midpoint is at x = 0. A second very l
Ainat [17]

Answer:

a. E=6.5*10^{5}N/C

b. E=-4.7*10^{4}N/C

Explanation:

We can use the result of applying the Gauss theorem to a infinite line of charge

E=\frac{\lambda}{2\pi \epsilon_0 r}

with r the perpendicular distance to the line of charge. The total field will be the sum of the contribution to the field by each line of charge.

E=E1+E2

we have

+8.00 μC/m

-6.00 μC/m

y=1m

(a) y2=0.2m

E=\frac{8*10^{-6}\frac{C}{m}}{2\pi (0.2m)(8.85*10^{-12})\frac{C^2}{Nm^2}}+\frac{-6*10^{-6}\frac{C}{m}}{2\pi (0.8m)(8.85*10^{-12})\frac{C^2}{Nm^2}}\\\\E=7.9*10^{5}N/C-1.3*10^{5}N/C=6.5*10^{5}N/C

(b) y3=0.616m

E=\frac{8*10^{-6}\frac{C}{m}}{2\pi (0.616m)(8.85*10^{-12})\frac{C^2}{Nm^2}}+\frac{-6*10^{-6}\frac{C}{m}}{2\pi (0.384m)(8.85*10^{-12})\frac{C^2}{Nm^2}}\\\\E=2.33*10^{5}N/C-2.8*10^{5}N/C=-4.7*10^{4}N/C

In this case the field is directed toward the line of charge placed in y=1m

8 0
3 years ago
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