Answer:
M = 9.9 x 10⁶ Solar masses
Explanation:
Here the centripetal force is given by the gravitational force between star and the object:
[tex]Gravitational\ Force = Centripetal\ Force \\\frac{mv^2}{r} = \frac{GmM}{r^2}\\\\M = \frac{v^2r}{G}[/tex]
where,
M = Mass of Object = ?
v = orbital speed of star = 1400 km/s = 1400000 m/s
G = Universal Gravittaional Constant = 6.67 x 10⁻¹¹ N.m²/kg²
r = distance between star and object = (26 light-days)(2.59 x 10¹³ m/1 light-day) = 6.735 x 10¹⁴ m
Therefore,
[tex]M = \frac{(1400000\ m/s)^2(6.735\ x\ 10^{14}\ m)}{6.67\ x \ 10^{-11}\ N.m^2/kg^2}[/tex]
M = (1.97 x 10³⁷ kg)(1 solar mass/ 1.989 x 10³⁰ kg)
M = 9.9 x 10⁶ Solar masses
PLEASE HELP MEEEEEEE:((((
Here's the solution,
A.)
we know,
[tex]power = \dfrac{work \: done}{time} [/tex]
so,
=》
[tex]600 = \dfrac{work \: done}{10} [/tex]
=》
[tex]work \: done = 600 \times 10[/tex]
=》
[tex]work \: done = 6000 \: \: joules[/tex]
B.)
[tex]work \: done = force \times displacement[/tex]
so,
=》
[tex]6000 = force \times10[/tex]
=》
[tex]force = \dfrac{6000}{10} [/tex]
=》
[tex]force = 600 \: \: Newtons[/tex]
What is another way to describe the vector 100 m/s down
Answer:
Describe how one-dimensional vector quantities are added or subtracted.
At the gym, a man pulls a bar on a machine that works the muscles of the upper back. It takes him 0.5 seconds to raise 30 kilograms of weights a vertical distance of 0.5 meters. Which of these exerts the same power output? (Estimate g as 10 m/s2.)
A) lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds
B) lifting 45 kilograms a distance of 2.4 meters in 3.0 seconds
C) leg pressing 45 kilograms a distance of 0.5 meters in 0.5 seconds
D) bench pressing 30 kilograms a distance of 0.5 meter in 1.5 seconds
Answer: Lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds
Explanation:
Got it wrong and that’s the answer
Lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds exerts the same power output. Hence, option (A) is correct.
What is power?The quantity of energy moved or converted per unit of time is known as power in physics. The watt, or one joule per second, is the unit of power in the International System of Units.
Power is also referred to as activity in ancient writings. A scalar quantity is power.
Power required for lifting 30 kilograms of weights a vertical distance of 0.5 meters in 0.5 second = (30×9.8×0.5)/0.5 watt = 294 watt.
Power required for lifting 25 kilograms a distance of 2.4 meters in 2.0 seconds= (25×9.8×2.4)/2.0watt = 294 watt.
Power required for lifting 45 kilograms a distance of 2.4 meters in 3.0 seconds = (45×9.8×2.4)/3.0watt = 352.8 watt.
Power required for lifting 45 kilograms a distance of 0.5 meters in 0.5seconds = (45×9.8×0.5)/0.5watt = 441 watt.
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both of yall are right and wrong its b and d
Answer:
Yup
Explanation:
what change do you need to make when comparing objects going upward against gravity versus downward with gravity?
A. Different amount of time in motion
B. Different length of displacement
C. Different sign + or - on the acceleration
D. Different changes to velocity
If an object is working against gravity, that means gravity must have the opposite sign of the object (this is why we see objects slow down at their peak height and return to us if we throw them in the air!). If the object is working with gravity, though, they must both be moving in the same direction, therefore gravitational acceleration would be the same sign as the velocity (positive or negative depending on how you define your axes).
Therefore, your answer should be C. Different sign + or - on the acceleration.
I hope this helps!
Q.2. Assertion: When distance between two bodies is doubled and also mass of each body is doubled. then the gravitational force between them remains the same. Reason: According to Newton's law of gravitation, force is directly proportional to the product of mass of the two bodies and inversely proportional to square of the distance between them. ent to
Answer:
True:
F1 = G M1 M2 / R^2
F2 = G (2 M1) * (2 M2) / (2 R)^2
F1 = F2
Pls help :(( I need help!! Its physics! motion and forces!
Answer: Pedaling your bike : acceleration :: applying the brakes : inertia.
The reason I think this to be the answer to the analogy is because there is energy and work used in both processes (and the unit focuses on forces); gravity is constant and does not change whether one pedals or applies brakes. And I do not think it's deceleration, as deceleration tends to equate to acceleration within the physics perspective.
Edit: I should also add that since you clarified that your unit is motion and forces, Newtons 1st law is the law of inertia. The way to change an objects motion for it to slow down is by applying an additional force. That resistance the bike experiences to slow is the process of inertia. Inertia happens in order to accelerate an object (either by slowing it down, or speeding it up): i.e., the resistance to change.
During launches, rockets often discard unneeded parts. A certain rocket starts from rest on the launch pad and accelerates upward at a steady 3.45m/s2 . When it is 230m above the launch pad, it discards a used fuel canister by simply disconnecting it. Once it is disconnected, the only force acting on the canister is gravity (air resistance can be ignored).
1- How high is the rocket when the canister hits the launch pad, assuming that the rocket does not change its acceleration?
2- What total distance did the canister travel between its release and its crash onto the launch pad?
Answer:
Explanation:
Time elapsed to reach the height of 230 m be t
s = ut + 1/2 at²
230 = .5 x 3.45 t²
t = 11.55 s
velocity at height of 230 m
v = u + at
= 0 + 3.45 x 11.55 = 39.84 m/s
This velocity will be attained by canister .
time to reach zero velocity at the top position t
v = u - gt
0 = 39.84 - 9.8 t
t = 4.06 s
height travelled by canister during this 4.06 s
v² = u² - 2gH
0 = 39.84² - 2 X 9.8 H
H = 80.98 M
Total height attained by canister = 80.98 + 230 = 310.98 m
Time of fall by canister t
s = 1/2 gt²
310.98 = .5 x 9.8 t²
t = 7.97 s
Total time taken by canister to reach the ground after its release from rocket
= 4.06 + 7.97 = 12.03 s
Distance travelled by rocket in 12.03 s
s = ut + 1/2 a t²
= 39.84 x 12.03 + .5 x 3.45 x 12.03²
= 479.27 + 249.64
= 728.91 m
height of rocket required = 230 + 728.91
= 958.91 m
2 )
Distance travelled by canister between its release and fall on the ground
= 80.98 + 80.98 + 230
= 391.96 m.
HELPPP PLEASEEEEE, BRIANLEST WILL BE GIVEN ON CORRECT!
Answer:
a. 25000J or 25KJ
b.2500w
Explanation:
a. since we say work is force applied in the same line or straight line
W=F(newtons )*D(metres)
b. power is work over time
Answer:
25000 J work with 2500 W power
Explanation:
work done = force . displacement = 500 N . 50 m= 25000 J
Power= work done / time taken = 25000 J / 10 s = 2500 W
PLEASE HELPPP ASAP, I'LL GIVE BRAINLEST
Answer:
a) Joules
b) W (symbol for work) (J is symbol for Joules)
c) seconds
d) watts
Two protons are held a distance d apart. The electrostatic force and the gravitational force that one proton exerts on the other are Fe and Fg, respectively. Which of the following correctly compares the magnitude and direction of these forces?
Fe > Fg, opposite direction
a. Fe > Fg, same direction
b. Fe < Fg opposite direction
c. Fe < Fg same direction
Answer:
Explanation:
charge on each proton = 1.6 x 10⁻¹⁹ C
mass of proton = 1.67 x 10⁻²⁷ kg
Electrostatic force of repulsion Fe = 9 x 10⁹ x ( 1.6 x 10⁻¹⁹ )² / d²
= 23.04 x 10⁻²⁹ / d²
Gravitational force of attraction = G Mm / d²
M = m = 1.67 x 10⁻²⁷ kg
Gravitational force of attraction Fg = 6.67 x 10⁻¹¹ x ( 1.67 x 10⁻²⁷ )² / d²
= 18.60 x 10⁻⁻⁶⁵ / d²
So Fg is far less than Fe and former is attractive , later is repulsive .
Fe > Fg, opposite direction , is the answer .
I cant solve this problem, and our teacher said that this would be in the test we'll have tomorrow, can someone help me?
A body of m = 6.8kg is launched with a speed of 7.5 m / s towards the top of an inclined plane of 15 ° with respect to the horizontal. in the absence of friction, what displacement does it make before reversing the direction of motion?
Answer:
d = 11.1 m
Explanation:
Since the inclined plane is frictionless, this is just a simple application of the conservation law of energy:
[tex] \frac{1}{2} m {v}^{2} = mgh[/tex]
Let d be the displacement along the inclined plane. Note that the height h in terms of d and the angle is as follows:
[tex] \sin(15) = \frac{h}{d} \\ or \: h = d \sin(15) [/tex]
Plugging this into the energy conservation equation and cancelling m, we get
[tex] {v}^{2} = 2gd \sin(15)[/tex]
Solving for d,
[tex]d = \frac{ {v}^{2} }{2g \sin(15) } = \frac{ {(7.5 \: \frac{m}{s}) }^{2} }{2(9.8 \: \frac{m}{ {s}^{2} })(0.259)} \\ = 11.1 \: m[/tex]
At what separation will two charges, each of magnitude 6.0 μC, exert a force of 0.70 N on
each other?
The two charges, each of magnitude 6.0 μC, exert a force of 0.70 N at separation of 1.47 meters.
What is electric force?Electric force is the attracting or repulsive interaction between any two charged things. Similar to any force, Newton's laws of motion describe how it affects the target body and how it does so. One of the many forces that affect objects is the electric force.
We know that electric force can be defined as:
Force: F = kQq/r²
0.70 = 9.0 × 10⁹ × (6.0×10⁻⁶)²/r²
r = 1.47 meter.
Hence, the separation between them is 1.47 meters.
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Can anyone help with this?
Answer:
Explanation:
Add a horizontal line. An object that is not moving has no change in distance with respect to time.
Two parallel plates are charged
with 4.98*10^-7 C of charge.
What must the area of the plates
be to create an electric field of
8720 N/C?
(Unit = m^2)
Answer:
6.45
Explanation:
Got it right on Acellus
PLEASE HURRY I GIVE BRAINLIST and extra points !!!!!!
1 What happens to the energy from the light when the material changes?
2 What happens to the energy from the light when the material does not change?
Answer:
2 stays the same 1 it will go bad or go good depends on which material you use
A brick is thrown upward from the top of a building at an angle of 10° to the horizontal and with an initial speed of 15 m/s. If the brick is in flight for 2.9 s, how tall is the building? any help would be greatly appreciated
The total height of the building from which the brick is thrown is 33.60 meters.
What is Speed?
Speed is the rate of change of position of an object in any direction with respect to time. Speed is measured as the ratio of distance to the time in which the distance was covered. The SI unit of speed is m/s (meter per second).
The time of flight of the object from the building is:
t = 2μ sinθ/ g
t = 2 × 15 × sin10/ 9.8
t = 30 × 0.175/ 9.8
t = 0.535 seconds
Total time to cover the distance = 2.9 seconds
Total time = 2.9 - 0.535
Total time = 2.365
μg = μ sinθ
h = 15 × sin10 × 2.365 + 1/2 × 9.8 × (2.365)²
h = 2.625 × 2.365 + 1/2 × 9.8 × (2.365)²
h = 6.20 + 4.9 × 5.59
h = 6.20 + 27.40
h = 33.60 meters
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It takes one year for ______.
a.earth to circle the sun one time
b.earth to rotate on its axis one time
c. The sun to circle earth one time
d.the sun to rotate on its axis one time
Answer: A. Earth to circle the sun one time.
A physics instructor wants to project a spectrum of visible light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating with 500 lines per mm, projecting a pattern on a screen 2.4 m behind the grating. How wide is the spectrum corresponding to m=1?
Answer:
औडंठः की दुनिया कार्टून दर्शन होली है और यह बियर का अच्छा विकल्प है। विकल्प है। यो कुरा मेरो आफ्ना लागि विचार र त्यहाँको बडेमाको बिल्डिङ भित्र म जिन्दगी एक यात्रा हो कहाँबाट शुरु भइ कहाँ पुगेर अन्त हुन्छ
The "Little Boy" nuclear bomb that was dropped on
Hiroshima had a yield of 6.9 x 10^13 J. In order to
release this much energy, kg of the uranium-
235 used in the bomb was converted into energy.
Answer:My gues is about 1/3 kg
Explanation:
I don't know physics
How does heat transfer?
Part E
Once you complete your outline, write a 500- to 750-word paper using word processing software. Add a works cited page at the end to give credit to your sources. Submit your completed paper along with this activity to your teacher for evaluation.
A Voyage to Proxima Centauri
Proxima Centauri is the second closest star to our solar system, which makes scientists curious to know more about it. What do we know about Proxima Centauri? How did we discover this information? Is it possible for humans to travel to this star? In this task, you will research and write a 500- to 750 word paper that answers these questions. Follow these steps to complete your research and writing. This guide about the research process can help.
Estimated time to complete: 3 hours
Part A
The goal of your paper is to describe the history of the discovery and research of Proxima Centauri. Another goal is to find the possibilities of traveling to this star. Some questions that your paper should answer are:
When was Proxima Centauri discovered?
How was Proxima Centauri discovered?
How have scientists researched Proxima Centauri?
What technologies have they used? What types of data do these technologies collect?
Have spacecraft ever reached this star?
What are the limitations of sending spacecraft to Proxima Centauri?
What accommodations would humans need to travel to Proxima Centauri?
Proxima Centauri otherwise known as Alpha Centauri C, is the closest star to our solar system, located about 4.24 light-years away. It was discovered in 1915 by Robert Innes, the director of the Union Observatory in South Africa. Innes identified Proxima Centauri as a possible member of the Alpha Centauri system after noticing that the star had a similar proper motion to Alpha Centauri A and B.
Proxima Centauri was first observed through telescopes, which allowed scientists to study the star's spectra and estimate its distance from Earth. In the following decades, scientists continued to study Proxima Centauri through telescopes, including the Hubble Space Telescope, which provided high-resolution images of the star.
In 2016, the European Southern Observatory announced the discovery of an exoplanet orbiting Proxima Centauri, called Proxima Centauri b. This discovery was made using the radial velocity method, which measures the star's small wobbles caused by the gravitational pull of an orbiting planet.
What are the major challenges facing the discovery of Proxima Centauri?Despite the advances in our understanding of Proxima Centauri, no spacecraft have ever reached this star. The distance between Proxima Centauri and Earth is so vast that it would take tens of thousands of years to reach the star using current propulsion technology.
One potential solution to this problem is the use of interstellar travel, which would allow humans to travel to Proxima Centauri within a human lifetime. There are several proposed methods for interstellar travel, including using a starship propelled by fusion engines or a massive light sail pushed by a beam of lasers. However, these technologies are still in the theoretical stage and have not yet been developed.
There are also many other challenges that must be overcome in order for humans to travel to Proxima Centauri. For example, humans would need to find a way to protect themselves from the high levels of radiation that they would be exposed to during the journey. They would also need to find a way to provide enough food, water, and other resources to sustain themselves for the duration of the trip.
Overall, while it is theoretically possible for humans to travel to Proxima Centauri, it would be a massive undertaking that would require significant technological advancements and the development of new solutions to many challenges.
Sources of Works Cited include:
"Proxima Centauri." Wikipedia, Wikimedia Foundation, 5 Jan. 2021,
"Proxima Centauri b." Wikipedia, Wikimedia Foundation, 20 Nov. 2020,
"Interstellar Travel." Wikipedia, Wikimedia Foundation, 4 Jan. 2021,
Therefore, the correct answer is as given above
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when a resistor is connected to a battery, current flows through the resistor. If the voltage of the battery is doubled, the current will be?
Resistors limits the current/amps but keep the voltage the same.
What happens if you connect the resistor to the battery?You don't want to attach them together with a wire since there would be too much current loose through, such that the battery heats up till it blows or something.
Current is directly corresponding to voltage; a doubling of the voltage will double the current. But the current is also inversely comparable to the resistance; a doubling of the resistance will halve the current.
So we can conclude that Resistance residue is the same, but power becomes four times i.e. multiplication. If the current is I and the voltage is V,
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Critical angle of glass is 42 .what does it mean?
Answer:
i think..its fraction that its have multiple fractions on it..if you minus the 397 000-355 it should be 381+ so i say if you get the 5 multiply it by 9!! so you will get it!
Explanation:
HOPE IT HELPS!!
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN. The elongation of the rod is delta = 1.40 mm and its diameter becomes df = 19.9837 mm. Determine the modulus of elasticity (E) and the modulus of elasticity in shear (G). Assume the material does not yield.
Answer:
a) the modulus of elasticity (E) is 68.22 GPa
b) the modulus of elasticity in shear is 25.28 GPa
Explanation:
Given the data in the question;
First we determine the cross-sectional area A of the road;
A = π/4 × d²
given that; diameter d = 20 mm
we substitute
A = π/4 × ( 20 mm )²
A = π/4 × 400 mm
A = 314.159 mm²
Next, we find, the normal stress of the rod σ
σ = P / A
given that load p = 50 kN = 50,000 Newton
we substitute
σ = 50,000 N / 314.159 mm²
σ = 159.155 N/mm²
σ = 159.155 Mpa
Next, is the normal strain ε
ε = δ / L
given that; change in length δ = 1.40 mm and Length of rod L = 600 mm
we substitute
ε = 1.40 mm / 600 mm
ε = 0.002333
Now, we find the modulus of elasticity;
we know that;
σ = Eε
modulus of elasticity E = σ / ε
we substitute
E = 159.155 Mpa / 0.002333
E = 68219.031 MPa
E = ( 68219.031 / 1000 ) GPa
E = 68.22 GPa
Therefore, the modulus of elasticity (E) is 68.22 GPa
b)
we know that the Poisson's ratio v is;
v = -(ε[tex]_{lat[/tex] / ε )
v = - (((d[tex]_f[/tex] - d)/d) / ε )
where d[tex]_f[/tex] is the final diameter of the rod ( 19.9837 mm ) and ε[tex]_{lat[/tex] is the lateral strain
so we substitute
v = - ((( 19.9837 - 20 ) / 20 ) / 0.002333 )
v = - (( -0.0163 / 20 ) / 0.002333 )
v = - ( - 0.000815 / 0.002333 )
v = - ( - 0.3493 )
v = 0.3493
So, our shear modulus will be;
G = E / 2( 1 + v )
we substitute
G = 68.22 GPa / 2( 1 + 0.3493 )
G = 68.22 GPa / 2( 1.3493 )
G = 68.22 GPa / 2.6986
G = 25.28 GPa
Therefore, the modulus of elasticity in shear is 25.28 GPa
The magnitude of the charge on an electron compared to the magnitude of the charge on a proton (3 points)
a) is greater
b) is smaller
c) is the same
Option C is correct.
The magnitude of the charge on an electron compared to the magnitude of the charge on a proton is same.
We have a electron and a proton.
We have to estimate the fact whether the magnitude of the charge on an electron compared to the magnitude of the charge on a proton is -
a) is greater
or
b) is smaller
or
c) is the same
What is the Electric charge ?The sufficiency or deficiency of electrons is called electric charge.
According to the question -
The charge on a electron is - q(e) = - 1.6 x [tex]10^{-19}[/tex] Coulombs
The charge on a proton is - q(p) = + 1.6 x [tex]10^{-19}[/tex] Coulombs
But, according to the property of magnitude :
| + x| = + x
| - x| = - (- x) = + x
In the similar manner -
|q(e)| = |q(p)| = 1.6 x [tex]10^{-19}[/tex] coulombs
Hence, the magnitude of the charge on an electron compared to the magnitude of the charge on a proton is same.
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______is the amount of matter in an object. It does not change regardless
of location
__________which one
Friction
Mass
Weight
Answer:
Mass
Explanation:
The weight of an object is dependent upon the mass and the value of acceleration due to gravity:
W = mg
Since the value of acceleration due to gravity (g) changes from planet to planet. Hence, the weight of an object also changes with location.
Friction depends upon the material of surfaces and the weight.
F = (coefficient of friction)W
Since weight changes with location. Therefore, the friction also changes with location.
The mass of an object is the quantity of matter contained in the body. It remains constant anywhere in the universe.
Hence, the correct option is:
Mass
30. How do you make a conclusion?
A solenoid is connected to a battery as shown in the figure, and a bar magnet is placed nearby. What is the direction of the magnetic force that this solenoid exerts on the bar magnet? (Hint: Think of the solenoid as a bar magnet, and identify what would be its north and south poles.)
According to the right-hand rule, the magnetic field of the solenoid will be directed towards the left direction.
What is Right-hand rule?Right Hand Thumb Rule is the rule which is used If a current carrying conductor is imagined to be held in the right hand direction such that the thumb always points along the direction of the current flow, then the direction of the wrapped fingers in the hand will give the direction of the magnetic field lines of the bar magnet.
By the application of the right-hand rule, the magnetic field of the solenoid will be directed towards the left hand direction.
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How much force is needed to accelerate a 100 kilogram car 5 meters per second?
The force needed to accelerate the mass of 100 kg is 500 N.
What is force?Force is the product of mass and acceleration. The S.I unit of force is Newton (N).
To calculate the amount of force needed to accelerate the mass of 100 kg, we use the formula below.
Formula:
F = ma.................. Equation 1Where:
F = Forcem = Mass a = AccelerationFrom the question,
Given:
m = 100 kga = 5 m/s²Substitute these values into equation 1
F = 100×5F = 500 NHence, the force is 500 N.
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