Answer:
The movement, or a tendency to move, toward a center of gravity, as in the falling of bodies to the earth is called gravitation
What distance does a police car travel if it is going 3.0 m/s for 20 seconds
Answer:
60 meters
Explanation:
If you are going 3 meters in a second, and you are traveling for 20 seconds, you have to multiply
3meters/second*20seconds
cross out the seconds and you have
3 meters*20
60 meters
Which layer is the igneous rock type?
options:
A
B
C
D
A Rocket took 5 minutes to travel the 185 km. How fast did the Rocket travel in kilometers per minute (km/min)?
Answer:
The rocket travelled 37km per minute
Explanation:
185÷5=37
An object on Earth weighs 600 N and on Mars 100 N. If its mass on Earth is approximately 60 kg what is the mass of the object on Mars?
Answer:
I think its
10 kg
plzz plzz plzz. mark as brainliest
If the mass of an object is approximately 60 kg on earth then its mass remains the same on Mars which is 60 kg because mass does not change from place to place.
What is mass?Mass in physics is a numerical representation of inertia, a fundamental property of all matter. What a body of matter is fundamentally is the resistance it offers to a change in its speed or location brought on by the application of a force. The more mass a body has, the less of a change a force exerts results. The kilogram is the unit of mass in the International System of Units, and it is defined by Planck's constant as being equal to 6.62607015 × 10⁻³⁴ joules second (SI).
According to the question, the weight of an object depends on gravity which is different from place to place so weight also varies from place to place but the mass is constant everywhere it does not change. So, if the mass of an object is 60 kg on earth then it remains the same on mars also.
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Please help me thank you
Fr+Cl=FeCl3
PLEASE HELP!!!!
1. Suppose a planet orbits a star at the same distance
as Earth (1 AU) but with a period of roughly 200
Earth days. The mass of the central star is:
a. less than the mass of our sun.
b. the same as the mass of our sun.
c.greater than the mass of our sun.
d. impossible to determine without
knowing the mass of the planet.
Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun. The answer is option C
According to Kepler's third law of planetary motion for planets and satellites, he described circular orbit about the same center body stating that:
The squared of the period is directly proportional to the cube of the radius. That is,
[tex]T^{2} \alpha r^{3}[/tex]
The orbital period formula is expressed as :
[tex]T^{2}[/tex] = 4[tex]\pi ^{2}[/tex] [tex]r^{3}[/tex] ÷ GM
Suppose a planet orbits a star at the same distance as Earth (1 AU) but with a period of roughly 200 Earth days. The mass of the central star will be greater than the mass of our sun based on the formula given above.
In a geostationary orbit, whatever the mass of the satellite may be, they will travel at the same speed in a particular orbit. mass M in the above formula becomes the mass of the sun or the center mass.
Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun.
Therefore, the correct answer is option C
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A 9.0V battery is connected across a 2.2kilo ohms and 6.8kilo omhs resistors connected in series. what is the potential differences across 2.2kilo ohms resistor.
Answer:
[tex]V_{2.2} =2.2V[/tex]
Fast and loose: that's a classic voltage divider. the drop from the i-th resistor is [tex]R_i \over {\sum R}[/tex] of the drop across the whole series. In our situation, it's [tex]2.2\cdot 10^3 \over {2.2\cdot10^3 + 6.8\cdot 10^3[/tex] of 9 V. By plugging numbers in a calculator, it's 22/90 of 9V, or 2.2V
With Ohm's Law
The series is equivalent of a single resistor of Resistance [tex]2.2\cdot 10^3+6.8\cdot 10^3 \Omega = 9.0 k\Omega[/tex]. By Ohm's first Law ([tex]V=Ri[/tex]) the current flowing through the resistor is [tex]9V = 9*10^3\Omega i \rightarrow i=1mA[/tex]. At this point, the drop across the first resistor is, again by Ohm's law[tex]V = 2.2 \cdot 10^3 \Omega \cdot 1\cdot 10^{-3} A = 2.2V[/tex]
