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dedylja
4 days ago
8

The path of a meteor passing Earth is affected by its gravitational force and falls to Earth's surface. Another meteor of the sa

me mass falls to Jupiter's surface due to its gravitational force. What statement accurately compares these two events? A) The meteors fall at the same rate since they have the same mass. B) The meteor falls to Earth faster due to its greater gravitational force. C) The meteor falls to Jupiter faster due to its greater gravitational force. D) The meteor falls to Jupiter at a faster rate due to its thinner atmosphere.
Physics
2 answers:
serg [3.2K]4 days ago
8 0
The right choice is option (C). Explanation: The scenario describes how a meteor on a trajectory toward Earth is influenced by Earth's gravitational pull, causing it to descend to the planet's surface. Another meteor of equivalent mass encounters Jupiter's gravity, thus falling onto its surface. According to Newton's universal law of gravitation, all particles in the universe exert gravitational forces on one another, which are directly proportional to their masses and inversely proportional to the square of the distance between them. Jupiter, being the most massive planet in our solar system, has a gravitational force that is 2.4 times stronger than that of Earth. Thus, a person weighing 100 pounds on Earth would weigh 240 pounds on Jupiter. Consequently, the correct answer is (C): the meteor falls to Jupiter quicker due to its stronger gravitational force.
Yuliya22 [2.9K]4 days ago
6 0
For USATest prep, the answer is C.
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The rod measures 450mm in length, while the disk has a radius of 75mm. An upward-supporting pin holds the assembly in place when Θ=0, and there exists a torsional spring with a constant of k=20N m/rad at the pin. One end of the rod connects to the pin, while the other connects to the disk.


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1 month ago
To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj
kicyunya [2911]

Response:

1.138\times 10^{-3}V

Details:

Utilizing Faraday's Newmann Lenz law enables the assessment of the induced emf within the loop:

\epsilon=\frac{d\phi}{dt}

where:

d\Phi- represents the change in magnetic flux

dt- symbolizes the change over time.

#The magnetic flux linked to the coil can be represented as:

\Phi=NBA \ Cos \theta

Where:

N represents the number of loops.

A denotes the area for each loop (A=\pi r^2=\pi \times 5^2=78.5398).

B indicates the strength of the magnetic field.

\theta=15\textdegree represents the angle between the magnetic field direction and the normal to the loop's area.

\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}

\frac{dB}{dt}-=0.0250T/s is indicated as the rate of magnetic field increase.

#Plugging in values into the emf equation:

=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V

Thus, the induced emf is 1.138\times 10^{-3}V

4 0
1 month ago
Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg
Maru [2979]

1) 9.18 s

During the initial phase, the rocket accelerates at

a_1=13.5 m/s^2

over a time interval of

t_1=3.50 s

The final velocity after this period is found using the SUVAT formula:

v_1=u+a_1t_1

with initial velocity u = 0. Substituting a1 and t1 gives:

v_1=(13.5)(3.50)=47.3 m/s

Afterward, the rocket decelerates uniformly at

a_2 = -5.15 m/s^2

until it stops, meaning the final velocity is

v_2 = 0

Again using the SUVAT formula,

v_2 = v_1 + a_2 t_2

and knowing v_1 = 47.3 m/s, solve for t2, the time until the rocket halts:

t_2 = -\frac{v_1}{a_2}=-\frac{47.3}{5.15}=9.18 s

2) 299.9 m

Calculate distances covered in both phases.

Distance in the first phase:

d_1 = ut_1 + \frac{1}{2}a_1 t_1^2

Substituting values from part 1,

d_1 = 0 + \frac{1}{2}(13.5) (3.50)^2=82.7 m

Distance in the deceleration phase:

d_2= v_1 t_2 + \frac{1}{2}a_2 t_2^2

Substituting known values,

d_2 = (47.3)(9.18) + \frac{1}{2}(-5.15) (9.18)^2=217.2 m

Total distance traveled is

d = 82.7 m + 217.2 m = 299.9 m

6 0
1 month ago
Two infinite parallel surfaces carry uniform charge densities of 0.20 nC/m2 and -0.60 nC/m2. What is the magnitude of the electr
Yuliya22 [2968]

Answer:

The electric field strength, E = 45.19 N/C

Explanation:

It is indicated that,

Surface charge density on the first surface, \sigma_1=0.2\ nC/m^2=0.2\times 10^{-9}\ C/m^2

Surface charge density on the second surface, \sigma=-0.6\ nC/m^2=-0.6\times 10^{-9}\ C/m^2

The electric field at a location between the two surfaces can be calculated as:

E=\dfrac{\sigma}{2\epsilon_o}

E=\dfrac{\sigma1-\sigma_2}{2\epsilon_o}

E=\dfrac{0.2\times 10^{-9}-(-0.6\times 10^{-9})}{2\times 8.85\times 10^{-12}}

Consequently, E = 45.19 N/C

Therefore, the electric field's magnitude at a point between both surfaces is 45.19 N/C.

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The result will be 21.6, but rounding yields 22J.
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28 days ago
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