The shape of the Earth's orbit around the Sun is not a perfect circle, but rather an ellipse or oval shape. This means that the distance between the Earth and the Sun varies throughout the year.
When the Earth is closest to the Sun, it is at a point in its orbit called perihelion, while when it is farthest from the Sun, it is at a point called aphelion. The magnitude of the gravitational force between the Earth and the Sun is directly related to their distance from each other. When the Earth is at perihelion, it is closer to the Sun and therefore experiences a stronger gravitational force than when it is at aphelion. This means that the Earth moves faster when it is at perihelion and slower when it is at aphelion. The force of gravity between two objects depends on their masses and the distance between them. Since the mass of the Earth is much smaller than the mass of the Sun, the gravitational force that the Sun exerts on the Earth is much greater than the force that the Earth exerts on the Sun. This force keeps the Earth in its elliptical orbit around the Sun. In summary, the oval shape of the Earth's orbit around the Sun affects the distance between the two objects, which in turn affects the magnitude of the gravitational force between them. This force is what keeps the Earth in its orbit around the Sun.
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when you heat a sample of a gas, what happens to the particles that make up that gas?
Answer:
the particles that heat up the gas become more active (vibrating around)
Water (density = 1 ´ 103 kg/m3) flows at 15 m/s through a pipe with radius 0. 040 m. The pipe goes up to the second floor of the building, 3. 0 m higher, and the pressure remains unchanged. What is the speed of the water flow in the pipe on the second floor?
The is that the velocity of water flow in the pipe on the second floor is 13.6 m/s.
To calculate the velocity of water flow on the second floor, we can use Bernoulli’s principle which states that the pressure and speed of a fluid are inversely proportional to each other. Pressure + (1/2) * Density * Velocity^2 + Density * g * Height
= Constant (where g = acceleration due to gravity)In this problem, since the pressure remains unchanged, we can eliminate the first term on both sides of the equation for both positions. Using Bernoulli’s principle for position 1 and 2 as:
(1/2) * Density * Velocity1^2 + Density * g * H1 = ConstantPosition 2:
(1/2) * Density * Velocity2^2 + Density * g * H2
= ConstantConstant
= (1/2) * Density * Velocity1^2 + Density * g *
H1Taking the difference of the above equations we get:1/2 * ρ * (V2^2 - V1^2)
= -ρghV2^2 - V1^2
= -2ghV2^2 = V1^2 - 2ghV2
= √(V1^2 - 2gh)
The negative sign is ignored as it represents the direction of flow. Putting the values we get:
V2 = √(15^2 - 2 × 9.81 × 3)
= 13.6 m/s
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If 1.32 mol of H2 gas is added holding the volume and temperature constant, the average kinetic energy of the total system will
- increase
- not enough information to answer the question
- decrease
- remain the same
The average kinetic energy of a gas is directly proportional to the temperature of the gas. Since the volume and temperature are being held constant and the amount of gas is increasing, the average kinetic energy of the total system will remain the same. Thus, the correct option is "remain the same".
Explanation:In the kinetic molecular theory, the average kinetic energy of the gas molecules is directly proportional to the temperature of the gas. The average kinetic energy of a gas can be calculated using the equation KE = (3/2) kT, where k is the Boltzmann constant and T is the temperature of the gas.
Since the volume and temperature of the gas are being held constant, the only factor that is changing is the amount of gas. If 1.32 mol of H2 gas is added, the number of gas molecules will increase, but the temperature will remain the same. Therefore, the average kinetic energy of the total system will remain the same.
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If you double the current through a resistor, by what factor does the power dissipated by the resistor change?.
If you double the current through a resistor, the power dissipated by the resistor change by factor of 4.
What is electric current?The stream of positive charges which flow from the positive terminal to the negative terminal of the battery attached in a circuit.
Power dissipated through a resistor is related to current as
P =I²R
When the current is doubled, the new power will be
P' = (2I)²R
P' = 4 I²R
P' = 4P
Thus, the power change by the factor of 4.
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The initial mass of the beaker and water was 0.080 kg.
The final mass of the beaker and water was 0.071 kg.
The energy transferred by the immersion heater as the water boiled was 25 200 J.
Calculate the specific latent heat of vaporisation of water given by the student’s data. Give the unit.
Q = 25200 J.By entering these values into to the formula, we obtain the following results: 25200 J = (-0.009 kg) x L,If we solve for L, we obtain:L = 2800000 J/kg, L = -25200 J/(-0.009 kg), etc.The student's results indicate that the specific vaporization latent heat of water is 2800000 J/kg, the with unit joules per kg (J/kg).
What is the latent heat of vaporization?A substance's vaporization latent heat is one of its physical properties. The amount of heat required to change one mole of water at full boil under mean atmospheric pressure is its definition.It is expressed as kg/mol or kJ/kg.
What does vaporization's latent heat mean?Instead, it makes the water turn from a to a gas, or evaporate.One kilogram of water needs an additional 970 Btu of heat to evaporate into one pounds of vapor (2256.2 kJ/kg).Since a thermometer cannot detect it, it is known as latent heat.
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A star has a mass of 5.9M
sun. Hydrogen gas accounts for 68.9%
. of the star's mass. What is the mawx of all the hydrogen in this star? mass of hydrogen:
The mass of all the hydrogen in this star is approximately 4.0661 times the mass of the Sun.
The mass of all the hydrogen in this star can be calculated by multiplying the mass of the star (5.9M sun) by the fraction of the star's mass accounted for by hydrogen gas (68.9%).
The mass of hydrogen in a star is significant as it determines the star's energy production through nuclear fusion. Hydrogen fusion reactions occur in the star's core, releasing immense amounts of energy in the form of light and heat. This energy sustains the star's luminosity and enables it to radiate heat and light into space. Additionally, hydrogen is the primary fuel source for stars, and its abundance directly influences the star's lifespan, size, and overall evolution.
To calculate the mass of hydrogen, we can use the following formula:
Mass of hydrogen = Mass of the star * Fraction of mass accounted for by hydrogen gas
Substituting the given values:
Mass of hydrogen = 5.9M sun * 0.689 = 4.0661M sun
Therefore, the mass of all the hydrogen in this star is approximately 4.0661 times the mass of the Sun.
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If 400 g of a (80 g/mol) and 1700 g of b (85 g/mol) are mixed, what is the vapor pressure of b?
Finally, the vapor pressure of substance b can be determined using Raoult's law: vapor pressure of b = mole fraction of b * vapor pressure of pure b.
However, we need additional information such as the vapor pressure of pure b to calculate the vapor pressure of b accurately.
To determine the vapor pressure of substance b when 400 g of substance a (with a molar mass of 80 g/mol) and 1700 g of substance b (with a molar mass of 85 g/mol) are mixed, we need to consider the mole fractions of the two substances.
First, let's calculate the number of moles for each substance.
For substance a:
moles of a = mass of a / molar mass of a
moles of a = 400 g / 80 g/mol
moles of a = 5 mol
For substance b:
moles of b = mass of b / molar mass of b
moles of b = 1700 g / 85 g/mol
moles of b = 20 mol
Next, we need to calculate the total number of moles in the mixture:
total moles = moles of a + moles of b
total moles = 5 mol + 20 mol
total moles = 25 mol
Now, let's calculate the mole fraction of substance b:
mole fraction of b = moles of b / total moles
mole fraction of b = 20 mol / 25 mol
mole fraction of b = 0.8
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If the distance to a given star were increased by a factor of four, by what factor would its apparent. brightness change? Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. 1 a 1/4 b 1/8 с 1/16 d 1/32
If the distance to a given star were increased by a factor of four, its apparent brightness would decrease by a factor of 16.
The apparent brightness of a star is directly proportional to the inverse square of the distance from Earth. The inverse square law states that the apparent brightness of an object decreases as the square of the distance from the observer increases. So, if the distance to a given star were increased by a factor of four, its apparent brightness would decrease by a factor of 4^2 or 16.
Therefore, the answer is option (c) 1/16.
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What is the acceleration of a rocket that speeds up from 50 m/s to 1000 m/s in 3 seconds?
Answer:
= 316.67 m/s^2
Explanation:
Acceleration formula =
A = \(\frac{V_{f} - V_{i} }{t}\)
\(A = \frac{1000 - 50}{3}\)
= \(\frac{950}{3}\)
= 316.67 m/s^2
if a slab is rotating about its center of mass g, its angular momentum about any arbitrary point p is __________ its angular momentum computed about g (i.e., i_gω).
If a slab is rotating about its center of mass G, its angular momentum about any arbitrary point P is equal to its angular momentum computed about G (i.e., I_Gω).
To clarify this, let's break it down step-by-step:
1. The slab is rotating about its center of mass G.
2. Angular momentum (L) is calculated using the formula L = Iω, where I is the moment of inertia and ω is the angular velocity.
3. When calculating angular momentum about G, we use I_G (the moment of inertia about G) in the formula.
4. To find the angular momentum about any arbitrary point P, we will still use the same formula L = Iω, but with the same I_Gω value computed about G, as the rotation is still happening around the center of mass G.
So, the angular momentum about any arbitrary point P is equal to its angular momentum computed about G (I_Gω).
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ASAP Question 3 of 10
In the diagram shown below, the blue light ray represents a ray of light
incident on a boundary between media. The diagram represents:
D
O A. indention
B. diffraction
C. reflection.
O D. refraction
Answer:
the answer is refraction, btw could u
say which grades question this is
How long does it take an automobile traveling 66.7 km/h to become even with a car that is traveling in another lane at 52.7 km/h if the cars' front bumpers are initially 119 m apart?
Answer:
The time taken is \(t = 32.5 \ s\)
Explanation:
From the question we are told that
The speed of first car is \(v_1 = 66.7 \ km/h = 18.3 \ m/s\)
The speed of second car is \(v_2 = 52.7 \ km/h = 14.64 \ m/s\)
The initial distance of separation is \(d = 119 \ m\)
The distance covered by first car is mathematically represented as
\(d_t = d_i + d_f\)
Here \(d_i\) is the initial distance which is 0 m/s
and \(d_f\) is the final distance covered which is evaluated as \(d_f = v_1 * t\)
So
\(d_t = 0 \ m/s + (v_1 * t )\)
\(d_t = 0 \ m/s + (18.3 * t )\)
The distance covered by second car is mathematically represented as
\(d_t = d_i + d_f\)
Here \(d_i\) is the initial distance which is 119 m
and \(d_f\) is the final distance covered which is evaluated as \(d_f = v_2* t\)
\(d_t = 119 + 14.64 * t\)
Given that the two car are now in the same position we have that
\(119 + 14.64 * t = 0 + (18.3 * t )\)
\(t = 32.5 \ s\)
Which color of flames would represent the HOTTEST temperature emitted from a star? Red Yellow Orange Blue
The color of flames that represents the hottest temperature emitted from a star is (d) Blue.
Blue flames are associated with the highest temperatures. In the context of stars, the color of a star's light is related to its surface temperature. The hottest stars, such as blue supergiants or blue-white stars, emit a significant amount of blue light due to their high temperatures.
As the temperature decreases, stars emit light in different colors, progressing from blue to white, yellow, orange, and finally red, with cooler stars emitting predominantly red light.
Therefore, if we consider the color of flames as an analogy, blue flames would correspond to the hottest temperature emitted from a star.
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whats an astral projection
Answer:Esotericism
Explanation:
it’s something that’s in intentional out of body experience
Answer:
Astral projection is a term used in esotericism to describe an intentional out-of-body experience that assumes the existence of a soul or consciousness called an "astral body" that is separate from the physical body and capable of travelling outside it throughout the universe.
Explanation:
A 2.0 kg 2.0kg2, point, 0, start text, k, g, end text cart moving right at 5.0 m s 5.0 s m 5, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction on a frictionless track collides with a 3.0 kg 3.0kg3, point, 0, start text, k, g, end text cart initially at rest. The 2.0 kg 2.0kg2, point, 0, start text, k, g, end text cart has a final speed of 1.0 m s 1.0 s m 1, point, 0, start fraction, start text, m, end text, divided by, start text, s, end text, end fraction to the left. What is the final speed of the 3.0 kg 3.0kg3, point, 0, start text, k, g, end text cart?
The final speed of the 3.0 kg cart is 4 m/s.
What is conservation of momentum principle?When two bodies of different masses move together each other and have head on collision, they travel to same or different direction after collision.
The external force is not acting here, so the initial momentum is equal to the final momentum. For elastic collision, final velocities is different for both the bodies.
m₁u₁ +m₂u₂ = m₁v₁ +m₂v₂
A 2.0 kg cart moving right at 5.0 m/s on a frictionless track collides with a 3.0 kg cart initially at rest. The 2.0 kg cart has a final speed of 1.0 m/s to the left.
Substitute the values for m₁ = 2kg, m₂ =3kg, u₁ =5 m/s and u₂ =0 m/s, then the final velocity will be
3x0 +2x5 = -2x1 + 3v₂
v₂ = 4 m/s
Thus, the the final velocity 3.0 kg cart is 4 m/s.
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I NEED THE RIGHT ANSWER ASAP NO LINKS !!!
This is a Science question
Answer:
12+7 and 12+7and 12-7AND -12+7
A plane travels 2400 km in 3 hours. Calculate its average speed
Explanation:
Speed =distance travelled/time taken
A 1.50 kg mass acclerates accross a smooth table at 15.0m/s. What is the net force?
How do you find net force?
I think its 20 points, and if there is more than one answer, will mark brainliest!
Answer:
We follow here the basics of problem-solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion. Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.
Explanation:
Which of the following is an example of a difference between planets in the solar system?
A planet can have a surface made of either liquid or gas.
Some planets have thick atmospheres, while others have thin atmospheres.
A planet must have a moon, but it can differ in terms of its size.
All planets have rings, but some are very hard to see with your eyes.
Answer:
A planet can have a surface made of either liquid or gas.
Hope it helped brainiest plz and thank you.
Explanation:
Which of the following is an example of a difference between planets in the solar system?
"A planet can have a surface made of either liquid or gas."
that's the answer:A planet can have a surface made of either liquid or gas.
Answer:
A i know its correct please give brainliest and does anydoby here play sky: children of light
Explanation:
during weathering, the surface area of the rock exposed will ________. because of this, the rate of weathering will ________.
Answer:During weathering, the surface area of the rock exposed will increase. This is because the outer layers of the rock will gradually erode or break down, exposing more surface area of the underlying rock.
As the surface area of the rock exposed increases, the rate of weathering will also increase. This is because there is more surface area available for weathering agents, such as water, wind, or chemicals, to react with the rock. The greater the surface area exposed, the greater the contact between the rock and the weathering agents, which leads to faster and more effective weathering.
Therefore, increasing the surface area of the rock by weathering accelerates the process of weathering.
Explanation:
An Atwood machine consists of 40-kg object A and 43-kg object B. Suppose
object A is released from rest. Determine the magnitude of the acceleration for
both objects. Assume the rope and pulley are massless.
Help
The magnitude of the acceleration for both objects is 0.36 m/s².
Acceleration is the name we supply to any system wherein the velocity modifications. for the reason, that velocity is a velocity and a path, there are only methods in order to accelerate: trade your velocity or exchange your path or exchange each.
Acceleration is the fee of alternating of the velocity of an object with recognition to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by means of the orientation of the internet pressure acting on that object.
Uniform acceleration, non-uniform acceleration, and average acceleration are the three varieties of multiplied motions. The term uniform acceleration refers to a motion in which an object travels in an immediate line with an increase in velocity at the same intervals of time.
calculation:-
mass 1 = 43 kg
F₁ = 430 N
mass 2= 40 kg
F₂ = 400 N
Tension = (F₁m₂ + F₂m₁)/ m₁ + m₂
= (430 × 40 + 400 × 43) / 83
=34400 /83
= 414.46 N
a = (F₁ - F₂)/ toatl mass
= 430 N - 400N / 83
= 0.36 m/s²
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if 80 btu of heat energy are added to 2 lb of water at 408f, what will be the final temperature of the water?
If 80 BTU of heat energy are added to 2 lb of water at 40°F, the final temperature of the water will be 80°F.
One BTU (British Thermal Unit) is the amount of heat required to raise the temperature of one pound of water by one degree Fahrenheit.
So, if we have 80 BTU of heat energy and 2 lb of water, we can use the following formula to find the final temperature:
Final temperature = Initial temperature + (BTU / lb of water)
Final temperature = 40°F + (80 BTU / 2 lb)
Final temperature = 40°F + 40°F
Final temperature = 80°F
Therefore, the final temperature of the water will be 80°F.
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an infinite plane of charge has surface charge density 5.6 mu or micro cc/m2. how far apart are the equipotential surfaces whose potentials differ by 100 v?
Consequently, it is 3.16×10⁻⁴m away from the plate.
By equipotential, what do you mean?Points of Equipotential: Equipotential points are those in an electric field that are all of the same electric potential. A line or curve connecting these places is referred to as a equipotential line. An equipotential surface is one where such points are present.
What does the electrical term "equipotential" mean?Equipotential bonding is simply a electrical connection that maintains a range of exposed and auxiliary conducting components' potentials near one another. The earthed imaginary horizontal zone is formed by attaching exposed conductive components and unrelated conducting components to one another.
Charge density of infinite sheet is given as:
σ = 5.6μC/m²
Now the electric field due to this charged sheet will be constant and it is given as:
E = σ/2ε₀
E = 5.6×10⁻⁶/2×8.85×10⁻¹²
E = 3.16 ×10⁵N/C
Now the relation between electric field intensity and potential difference is given as:
ΔV= E.d
ΔV=100Volts
E = 3.16 ×10⁵N/C
Now we have:
d=ΔVE
d=100/3.16 ×10⁵
d=3.16×10⁻⁴m
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light cannot _ around objects
Answer:
mark me Brainliest
Explanation:
light cannot travel around objects
Answer:
Translucent objects allow light to pass through them but they distort it.
Explanation:
Example: frosted or stained glass. Opaque objects do not allow light to pass through them. Example: brick walls, human beings.
How much heat is lost by 2.0 grams of water if the temperature drops from 31 °C to 29 °C? The specific heat of water is 4.184 J/(g • °C) *
1 point
a) 4.0 J
b) - 7.5 J
c)- 8.4 J
d) - 16.7 J
Given :
Mass of water, m = 2 grams.
The temperature of water drops from 31 °C to 29 °C .
The specific heat of water is 4.184 J/(g • °C).
To Find :
Amount of heat lost in this process.
Solution :
We know, heat lost is given by :
\(Heat\ lost,H = ms( T_f - T_i)\\\\H = 2\times 4.184 \times ( 31 - 29 )\ J\\\\H = 16.736\ J\)
Therefore, amount of heat lost in this process is 16.736 J.
when a force is exerted on an object it moves at a certain velocity if a force is applied immediately after the object is moving how does the speed change g
When a force is applied immediately after an object is moving, the speed of the object can either increase or decrease depending on the direction and magnitude of the applied force.
The motion of an object is determined by the net force acting on it. If the applied force acts in the same direction as the object's velocity, it will accelerate the object, causing its speed to increase. This is known as a positive acceleration.
On the other hand, if the applied force acts in the opposite direction to the object's velocity, it will decelerate the object, causing its speed to decrease. This is known as a negative acceleration or deceleration.
If the applied force is equal in magnitude but opposite in direction to the force that was initially propelling the object, it will result in the object coming to a stop. This is because the two forces cancel each other out, resulting in a net force of zero and no change in velocity.
It's important to note that the mass of the object also plays a role in how its speed changes. Objects with larger masses require more force to produce the same acceleration as objects with smaller masses.
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what if your outgoing,fun,kind,and a bit loud and ppl still don't what to be your friend? What am I doing wrong?
Answer:
Maybe you're hanging out around the wrong crowd. The people you are probably choosing to be around are more quiet then you.
Explanation:
Las ondas tienen elementos de posición y temporales uno de ellos se define como, La distancia recorrida por una onda en un período se llama Opciones: Longitud de onda Frecuencia Período Velocidad
Answer:
Velocidad.
Explanation:
Físicamente hablando, la velocidad es la distancia recorrida por unidad de tiempo. En el caso de las ondas se trata de la distancia recorrida en un período de tiempo.
Por ende, la opción correcta es: Velocidad.
A washing machine heats 10kg of water in each wash cycle. How much energy is saved by washing at 30'c instead of 50'c if the starting temperature of the cold water is 16'c? The specific heat capacity of water is 4200 J/Kg'c
Answer:
\(8.4 \times 10^{5}\; \rm J\), assuming that there's no heat exchange between the washing machine and the environment.
Explanation:
Let \(m\) denote the mass of water and \(c\) the specific heat capacity of water. The energy required to raise the temperature of that much water by \(\Delta T\) would be:
\(Q = c \cdot m \cdot \Delta T\).
Washing at \(30\; \rm ^{\circ} C\) would require a temperature change of \(\Delta T = 30\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 14\; \rm K\).
Washing at \(50\; \rm ^{\circ} C\) would require a temperature change of \(\Delta T = 50\; \rm ^{\circ} C - 16\; ^{\circ} \rm C = 34\; \rm K\).
In both situations, \(c = 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1}\) while \(m = 10\; \rm kg\).
Calculate the energy required in either situation:
Washing at \(30\; \rm ^{\circ} C\):
\(\begin{aligned}& Q({30\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 14\; \rm K \\ &= 588000 \times 10^{5}\; \rm J\end{aligned}\).
Washing at \(50\; ^{\circ} {\rm C}\):
\(\begin{aligned}& Q({50\; ^{\circ} {\rm C}}) \\ &= c \cdot m \cdot \Delta T \\ &= 4.2 \times 10^{3}\; \rm J \cdot kg \cdot K^{-1} \times 10\; \rm kg \times 34\; \rm K \\ &= 1428000 \; \rm J\end{aligned}\).
\(1428000\; \rm J - 588000\; \rm J = 8.4 \times 10^{5}\; \rm J\).
A baseball player hits a 140 g baseball with a force of 2800 N. What is the
acceleration of the ball as it leaves the bat?
A. 0.050 m/s2
B. 20 m/s2
C. 390 m/s2
D. 20,000 m/s2