The number of years required for the asteroid to make one revolution about the sun will be 4.55 years
What is the meaning of Kepler's third law?
According to Kepler's Third Law, the planets' squared orbital periods and cubes of their semi-major axes are directly inversely related. According to Kepler's Third Law, a planet's period of orbiting the Sun grows exponentially with its orbital radius.
The sun is at the focus of elliptical planetary orbits, according to the first law. The radius vector from the sun to a planet sweeps equal regions in equal amounts of time, according to the second law.
According to Kepler's Third Law,
T^2 is directly proportional to r^3
(T1/T2)^2 = (r1/r2)^3
Time taken by earth to complete 1 revolution, T1 = 1 year
Time taken by asteroid to complete 1 revolution T2
distance of earth r1 = r
distance of asteroid r2 = 4.55r
So,
1/T2 = r/4.55r
T2 = 4.55 years
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Which cloud types would most likely indicate that a thunderstorm is on the way?
cirrus clouds
dull, gray, stratus clouds
cumulus clouds that are small and round
cumulus clouds that are tall with a flat top
Answer:
The aswer is D
Explanation:
Answer:
d
Explanation:
cumulus clouds that are tall with a flat top
What is the work done by the 200.-N tension shown if it is used to drag the 150-N crate 25 m across the floor at a constant speed?
Answer:
0 J
Explanation:
Work equals force times distance, but the force is zero because the crate being dragged will have zero acceleration. Force equals mass times acceleration and since acceleration is zero, force has to equal zero as well. Since the force is zero, the work required also has to be zero.
Which is true regarding a child standing up for their own rights?
Answer:
hey mate......looks like the question is incomplete
a stationary object experiences two forces as shown in the diagram below
Answer: the answer is B
Explanation: 80 is not the same as 150 so it will go the way 150 units of force is pulling.
Sparks occur when the electric field in air exceeds 3 x 106 N/C. This is because free electrons normally present in air are accelerated to such high speeds that their kinetic energy will overcome the potential energy holding other electrons to atoms. When those electrons rearrange themselves after such a collision, a flash of light is emitted. Let us suppose that the work done on an electron must give it an energy of 3 x 10-19 J to cause this ionization. How far does an electron involved in making in a spark travel through the air before it collides with an atom
Answer:
h = 5.38 10¹⁶ m
Explanation:
Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy
starting point. Just before ionization
Em₀ = U = qE
final point. Right after ionization
Em_f = K = ½ m v²
Energy is conserved
Em₀ = Em_f
q E = ½ m v²
v² = 2qE / m
Now we can use the relationship between net work and kinetic energy
W_net = ΔK
net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J
W - Ф = K_f - K₀
we assume that the electron converts all its initial initial kinetic energy to be zero
W -Ф = ½ m v² - 0
W = ½ m v² +Ф
we substitute
W = 1/2 m 2qE/m + E
W = qE +Ф
W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ + 3 10⁻¹⁹
W = 4.8 10⁻¹³ J
When the electron is in air, its kinetic energy can be transformed into gravitational potential energy
As the electron is in the air, all work is transformed into scientific energy
W = K
starting point Em₀ = K = W
end point Em_F = U = m g h
energy conservation Em₀ = Em_f
W = m g h
h = [tex]\frac{W}{mg}[/tex]
let's calculate
h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]
h = 5.38 10¹⁶ m
Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.
What is an electric field?An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.
Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.
Energy before ionization;
[tex]\rm Em_0 = U = qE[/tex]
Energy after ionization;
[tex]Em_f = K = \frac{1}{2} mv^2[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]
The relationship between net work and kinetic energy;
[tex]W_{net} = \triangle K[/tex]
The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.
[tex]W - \phi = K_f - K_0[/tex]
[tex]W = K_f+ \phi[/tex]
[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]
EMF at starting point;
[tex]\rm Em_0 = K = W[/tex]
EMF at the endpoint;
[tex]\rm Em_F = U = m g h[/tex]
From the law of conservation of energy principle;
[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]
Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.
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A 0.545-kg ball is hung vertically from a spring. The spring stretches by 3.56 cm from its natural length when the ball is hanging at equilibrium. A child comes along and pulls the ball down an additional 5cm, then lets go. How long (in seconds) will it take the ball to swing up and down exactly 4 times, making 4 complete oscillations before again hitting its lowest position
Answer:
t = 9.52 s
Explanation:
This is an oscillatory motion exercise, in which the angular velocity is
w = [tex]\sqrt{ \frac{k}{m} }[/tex]
Let's use hooke's law to find the spring constant, let's write the equilibrium equation
F_e - W = 0
F_e = W
k x = m g
k = [tex]\frac{m g}{x}[/tex]
k = 0.545 9.8 /0.0356
k = 150 N / m
now the angular velocity is related to the period
W = 2π / T
we substitute
4π² T² = k /m
T = 4pi² [tex]\sqrt{ \frac{m}{k} }[/tex]
we substitute
T = 4 pi² [tex]\sqrt{ \frac{0.545}{150} }[/tex]
T = 2.38 s
therefore for the spring to oscillate 4 complete periods the time is
t = 4 T
t = 4 2.38
t = 9.52 s
The force of Earths gravity keeps earth in orbit true or false
Answer:
True
Explanation:
The force of gravity keeps all of the planets in orbit around the sun
True. The force of gravity keeps all of the planets in orbit around the sun.
What is Gravity?
The force that pulls items toward the center of a planet or other entity is called gravity. All of the planets are kept in orbit around the sun by gravity.
Gravity applies to everything that has mass. Gravity is stronger for objects with higher mass. Along with distance, gravity weakens as well. Therefore, the gravitational pull of two things becomes stronger the closer they are to one another.
The mass of the Earth is what creates gravity. The combined gravitational force of all of its mass acts on the mass in your body.
Therefore, True. The force of gravity keeps all of the planets in orbit around the sun.
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PLEASE HELP! I'LL GIVE BRAINLEST
Answer:
1.62 m/s²
Explanation:
A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.3 kg raven flying horizontally at 5.0 m/s flies into this door at its center and bounces back at 2.0 m/s in the opposite direction.
Required:
a. What is the angular speed of the gate just after it is struck by the unfortunate raven?
b. During the collision, why is the angular momentum conserved but not the linear momentum?
Answer:
its a. and jusing that youl
Calculate the induced electric field (in V/m) in a 52-turn coil with a diameter of 17 cm that is placed in a spatially uniform magnetic field of magnitude 0.45 T so that the face of the coil and the magnetic field are perpendicular. This magnetic field is reduced to zero in 0.10 seconds. Assume that the magnetic field is cylindrically symmetric with respect to the central axis of the coil. (Enter the magnitude.)
Answer:
the induced electric field is 9.95 V/m
Explanation:
Given the data in the question;
Number of turns N = 52
Diameter of coil D = 17 cm = 0.17 m
Radius r = D/2 = 0.17/2 = 0.085 m
Now,
cross-section area A of the coil = πr²
A = π × ( 0.085 m )²
A = 0.0227 m²
Also given that;
Initial magnetic field B₁ = 0.45 T
Final magnetic field B₂ = 0
∴ change in magnetic field ΔB = B₁ - B₂ = 0.45 T - 0 = 0.45 T
Time taken dT = 0.10 seconds
Now, we know that;
Induced emf ∈ = N[tex](\frac{d\eta }{dt} )[/tex]
where η = BAcosθ
We know that, magnetic field is cylindrically symmetric, coil is also perpendicular to magnetic field.
Hence, the angle between B & A is 0°
∴ θ = 0°
Induced emf ε = N[tex](\frac{d }{dt} )BAcos\theta[/tex]
we substitute
ε = N[tex](\frac{d }{dt} )[/tex] (BAcos0°)
A is constant and cos0° = 1
so
ε = NA[tex](\frac{dB }{dt} )[/tex]
We now substitute in our values;
ε = 52 × 0.0227 m² × [tex](\frac{0.45T }{0.10s} )[/tex]
ε = 5.3118 V
we know that, from the relation between electric and emf
ε = ∫∈.dl or ε = ∈∫dl { for coil; ∫dl = πD }
so we have;
ε = ∈πD
solve for ∈
∈ = ε/πD
we substitute
∈ = 5.3118 V / ( π × 0.17 m )
∈ = 9.95 V/m
Therefore, the induced electric field is 9.95 V/m
3 CO2
How many atoms of Oxygen (O) are there?
Answer:
6 oxygen
Explanation:
3 multiply O2 and 6 Oxygen
A textbook of mass 2.10 kg rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.100 m , to a hanging book with mass 3.10 kg . The system is released from rest, and the books are observed to move a distance 1.29 m over a time interval of 0.850 s . Part A What is the tension in the part of the cord attached to the textbook
Answer:
the tension in the part of the cord attached to the textbook is 7.4989 N
Explanation:
Given the data in the question;
As illustrated in the image below;
first we determine the value of the acceleration,
along vertical direction; we use the second equation of motion;
y = ut + [tex]\frac{1}{2}[/tex]a[tex]_y[/tex]t²
we substitute;
0 m/s for u, 1.29 m for y, 0.850 s for t,
1.29 = 0×0.850 + [tex]\frac{1}{2}[/tex]×a[tex]_y[/tex]×(0.850)²
1.29 = 0.36125a[tex]_y[/tex]
a[tex]_y[/tex] = 1.29 / 0.36125
a[tex]_y[/tex] = 3.5709 m/s²
Now when the text book is moving with acceleration , the dynamic equation will be;
T₁ = m₁a[tex]_y[/tex]
where m₁ is the mass of the text book ( 2.10 kg )
a[tex]_y[/tex] is the vertical acceleration ( 3.5709 m/s² )
so we substitute
T₁ = 2.10 × 3.5709
T₁ = 7.4989 N
Therefore, the tension in the part of the cord attached to the textbook is 7.4989 N
Boxes A and B are in contact on a horizontal, frictionless surface. Box A has mass 21.0 kg and box B has mass 8.0 kg. A horizontal force of 100N is exerted on box A. What is the magnitude of the force that box A exerts on box B?
Answer:
2.75 N
Explanation:
Given that,
Box A has a mass 21.0 kg and box B has a mass 8.0 kg.
A horizontal force of 100N is exerted on box A.
Let a be the acceleration of the system. Using second law of motion,
[tex]F=(m_A+m_B)a\\\\a=\dfrac{F}{(m_A+m_B)}\\\\a=\dfrac{10}{(21+8)}\\\\a=0.344\ m/s^2[/tex]
Now applying Newton's second law to box B. So,
[tex]F_A=m_Ba\\\\=8\times 0.344\\\\=2.75\ N[/tex]
So, 2.75 N is the force that box A exerts on box B.
The key concept here is that when two objects interact, the forces they exert on each other are equal in magnitude but opposite in direction. The magnitude of the force that box A exerts on box B is also 100N.
Newton's third law of motion states that for every action, there is an equal and opposite reaction. It is one of the three fundamental principles described by Sir Isaac Newton in his Laws of Motion.
According to Newton's third law, when an object exerts a force on another object, the second object simultaneously exerts a force of equal magnitude but in the opposite direction on the first object. In simpler terms, if object A applies a force on object B, then object B applies an equal force in the opposite direction on object A.
Since boxes A and B are in contact on a frictionless surface, the force exerted on box A will be transmitted to box B. According to Newton's third law of motion, the magnitude of the force that box A exerts on box B will be equal in magnitude but opposite in direction to the force exerted on box A.
Therefore, the magnitude of the force that box A exerts on box B is also 100N.
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A 25.0kg girl pushes a 50.0kg boy so that he accelerates at 4.00m/s2. What is the force of the boy on the girl? A. 200N B. 100N C. 12.5 D. 400N
Answer:
a
Explanation:
so the answer is 200N
and I hope it is correct
what must be the mass of a rock if a boy applies a 64N force and causes it to accelerate at 4.51m/s2
first of all the formula of force is F=ma,so we are searching for m,so we can divide a on both sides F/a=m, after this substitute the values given above 64N/4.51=14.2°Kg
A dump truck contains a load of soil. Which action will leave the dump truck's
inertia unchanged?
A. Dump out some of the soil.
B. Add gas to its fuel tank.
C. Add more soil.
D. Increase the force applied by the engine.
Answer:
D
Explanation:
This will not change the weight and therefore not change the inertia
A dump truck contains a load of soil. The action that will leave the dump truck's inertia unchanged is that increase the force applied by the engine. Hence, option D is correct.
What is inertia?A body's ability to resist being propelled into motion or, if already moving, to modify the direction or magnitude of its velocity is known as inertia.
An object's lethargy is a passive quality that prevents it from doing anything other than obstructing active forces and torques. The only reason a moving body continues to move is the lack of a force that might slow it down, alter its trajectory, or accelerate it. This is not due to inertia.
A body's inertia moment about a certain axis and its mass, which determine how resistant it is to the application of forces to that axis, respectively, are two statistical measures of inertia.
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What do we call the small changes that
could result in large future changes?
A. the "butterfly effect"
B. the "snowflake effect"
C. the "ripple effect"
D. the "trickle-down effect"
Answer:
The "butterfly Effect"
Explanation:
The "butterfly effect" will probably have big changes in the future.
At the end of the passage, Sarah says to Emma, “You’re not the only one with tricks up your sleeves.” Explain what Sarah means by this. Use information from the passage to support your answer.
Answer:
Sarah plans to use trickery and cunning to get back at Emma. Sarah thinks she is smarter than Emma.
Explanation:
i think this is right i dont know tho
True or false? Pls help
False.
Tripling the height will triple the potential energy.
Speed has nothing to do with potential energy.
49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner, as the drawing shows. The centripetal acceleration measured at corner A is n times as great as that measured at corner B. What is the ratio L1/L2 of the lengths of the sides of the rectangle when n
Answer:
[tex]\frac{L_1}{L_2} = \sqrt{(n^2 - 1)}[/tex]
Explanation:
For this interesting problem, we use the definition of centripetal acceleration
a = v² / r
angular and linear velocity are related
v = w r
we substitute
a = w² r
the rectangular body rotates at an angular velocity w
We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A
the distance OB = L₂
the distance AB = L₁
the sides of the rectangle
It is indicated that the acceleration in in A and B are related
[tex]a_A = n \ a_B[/tex]
we substitute the value of the acceleration
w² r_A = n r_B
the distance from the each corner is
r_B = L₂
r_A = [tex]\sqrt{L_1^2 + L_2^2}[/tex]
we substitute
\sqrt{L_1^2 + L_2^2} = n L₂
L₁² + L₂² = n² L₂²
L₁² = (n²-1) L₂²
an arrow is shot horizontally from the top of a tower at a speed of 15m/s and hits the ground with a speed of 25m/s. calculate the height of the tower
The height of the tower is 20.41 m.
To determine the height of the tower, we need to understand the concept of the energy conservation principle since the speed and acceleration due to gravity are involved in the system.
What is the energy conservation principle?The principle of energy conservation lets us know that in an isolated system, energy can neither be created nor destroyed.
It can be expressed using the formula:[tex]\mathbf{mgh = \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2}[/tex]
From the parameters given:The initial speed [tex]v_1[/tex] = 15 m/sThe final speed [tex]v_2[/tex] = 25 m/sBy applying the energy conservation principle, we have:
[tex]\mathbf{gh +\dfrac{1}{2}v_1^2 = \dfrac{1}{2}v_2^2}[/tex]
[tex]\mathbf{h = \dfrac{v_2^2- v_1^2 }{2 \times g}}[/tex]
[tex]\mathbf{h = \dfrac{25^2-15^2 }{2 \times 9.8}}[/tex]
h = 20.41 m
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The work done is a vector quantity and SI base unit is J
Answer:
Is this your question? Also I think work done is a scalar quantity.
Explanation:
Please help due today
Answer:
8
Explanation:
(8√2)² = x² + x²
8² × √2² = 2x²
64 × 2 = 2x²
128 = 2x²
64 = x²
x = 8
give me brainliest please
what do solar winds and the earths magnetic field create
Answer:
bc earth rotates
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Explanation:
Answer:
The interaction between the solar wind and Earth's magnetic field, and the influence of the underlying atmosphere and ionosphere, creates various regions of fields, plasmas, and currents inside the magnetosphere such as the plasmasphere, the ring current, and radiation belts.
Explanation:
Time Vs Position of Battery Operated Car what type of relationship is shown in the graph?
Calculate the acceleration of a car if the force on the car is 450 Newtons and the mass is 1300 kilograms.
[tex] \Large {\underline { \sf {Required \; Solution :}}}[/tex]
We have ―
Force, F = 450 NMass of the car, m = 1300 kgWe have been asked to calculate the acceleration of the car.
[tex]\qquad \implies\boxed{\red{\sf{ F = ma }}}\\[/tex]
F denotes Forcem denotes massa denotes acceleration[tex] \quad \twoheadrightarrow\sf { 450 = 1300a} \\ [/tex]
[tex] \quad \twoheadrightarrow\sf {\cancel{ \dfrac{450}{1300}} = 1300a} \\ [/tex]
[tex]\quad\twoheadrightarrow\boxed{\red{\sf{0.346 \; ms^{-2} = a }}}\\[/tex]
Therefore, acceleration of the car is 0.346 m/s².
The aircraft wing from problem 6 experiences temperature extremes that span 210 degrees Celsius. The component for the wing will have a length of exactly 3 meters. Testing indicates that the aircraft wing will remain stable only if the component never expands to a length larger than 3.017 meters. If the component is made from the metal alloy in question, will it meet this requirement. An unknown metal alloy is being tested to discover its thermal properties to see if it is suitable for use as a component in an aircraft wing. The alloy is formed into a bar measuring 1 meter in length, and is then heated from its starting temp. of 30°C to a final temperature of 100°C. The length of the heated bar is measured to be exactly 1.002 meters in length.
Required:
What is the coefficient of thermal expansion of the alloy?
Answer:
α = 2,857 10⁻⁵ ºC⁻¹
Explanation:
The thermal expansion of materials is described by the expression
ΔL = α Lo ΔT
α = [tex]\frac{\Delta L}{L_o \ \Delta T}[/tex]
in the case of the bar the expansion is
ΔL = L_f - L₀
ΔL= 1.002 -1
ΔL = 0.002 m
the temperature variation is
ΔT = 100 - 30
ΔT = 70º C
we calculate
α = 0.002 / 1 70
α = 2,857 10⁻⁵ ºC⁻¹
If a weather service map has a circle that
is shaded completely in, what does that
mean about the cloud cover in that area?
A. There is 100% cloud cover in that area.
B. There is 0% cloud cover in that area.
C. There is a good chance of rain.
D. There are sunny skies.
Answer:
A. There is 100% cloud cover in that area.
Explanation:
Cloud cover is recorded on weather charts by shading in parts of the circle.
If there are no clouds, the circle is left white and if the sky is completely covered in cloud, the circle is shaded completely in which means 100% cloud cover in that area.
A certain type of laser emits light that has a frequency of 4.9 x 1014 Hz. The light, however, occurs as a series of short pulses, each lasting for a time of 2.9 x 10-11 s. The light enters a pool of water. The frequency of the light remains the same, but the speed of light slows down to 2.3 x 108 m/s. In the water, how many wavelengths are in one pulse
Answer:
N = 1.42 × 10⁴ cycles
Explanation:
Given that:
frequency f = 4.9 × 10¹⁴ Hz
Time = 2.9 × 10⁻¹¹ s
Speed = 2.3 × 10⁸ m/s
Recall that:
wavelength [tex]\lambda = \dfrac{c}{f} \\ \\[/tex]
Horizontal distance [tex]\Delta x = ct[/tex]
Number of wavelengths [tex](N) = \dfrac{\Delta x}{\lambda}[/tex]
[tex]N = \dfrac{ct}{c/f} \\ \\ N= ft[/tex]
N = (4.9 × 10¹⁴ cycles/s) (2.9 × 10⁻¹¹ s)
N = 14210
N = 1.42 × 10⁴ cycles
Plz help
What factors determine
how the speed of the marbles changes in a
collision?
Answer:
Force,friction,inertia and momentum
Explanation:
The speed that the marble is moving at can be determined by the amount of force used when pushed or pulled and what kind of surface it's on.Momentum is also a factor because of the mass of the marbles.