You plan a trip to the moon. You will need to leave the earth with enough speed to make it to the moon. The following is important information that will help you:Mass of earth=5.9742*10^24kg,Mass of the moon=7.36*10^22kg,G=6.67428*10^-11 N-m^2/kg^2,Radius of earth=6.3781*10^6 meter.Radius of the moon=1.7374*10^6 meter.Distance from earth to moon=3.844*10^8 meter.1) So, on you first attempt you leave the surface of the earth at v=5054m/s. How far from the center of the earth will you get?
Answers
We are given that an object leaves the earth at a speed of 5054 m/s. We want to determine how far from the center of the earth the object travels. To do that we will do an energy balance. We will have into account that the total energy at the surface of the earth is the kinetic energy of the object plus the gravitational potential energy, and when the object stops the energy is equal to the gravitational potential energy:
\(\begin{gathered} E_1=E_k+E_{g1} \\ E_2=E_{g1} \end{gathered}\)Where:
\(\begin{gathered} E_1=\text{ energy at the surface of the earth} \\ E_k=\text{ kinetic energy} \\ E_g=\text{ }gravitational\text{ potential energy} \\ E_2=\text{ energy when the object stops} \end{gathered}\)The kinetic energy is given by:
\(E_k=\frac{1}{2}mv^2\)Where:
\(\begin{gathered} m=\text{ mass} \\ v=\text{ velocity} \end{gathered}\)The gravitational potential energy is given by:
\(E_g=-G\frac{Mm}{R}\)Where:
\(\begin{gathered} G=\text{ gravitational contant} \\ M=\text{ mass of th earth} \\ m=\text{ mass of the object} \\ R=\text{ distance to the center of the earth at the given point} \end{gathered}\)Substituting the values we get:
\(\begin{gathered} E_1=\frac{1}{2}mv^2-G\frac{Mm}{R_1} \\ \\ E_2=-G\frac{Mm}{R_2} \end{gathered}\)In this case, R1 is the radius of the earth since the object leaves from the surface of the earth and R2 the distance to the center of the earth from the object stops as shown in the diagram above.
Now we can set them equal due to conservation of energy:
\(\frac{1}{2}mv^2-G\frac{Mm}{R_1}=-G\frac{Mm}{R_2}\)Now we may cancel out the "m":
\(\frac{1}{2}v^2-G\frac{M}{R_1}=-G\frac{M}{R_2}\)We wish to determine the distance to the center of the earth when the object stops, therefore, we solve for "R2". We do this first by multiplying both sides by R2:
\(R_2(\frac{1}{2}v^2-G\frac{M}{R_1})=-GM\)Now we divide both sides by the factor multiplying R2:
\(R_2=-\frac{GM}{(\frac{1}{2}v^2-G\frac{M}{R_1})}\)Now we substitute the known values:
\(R_2=-\frac{(6.67428\times10^{-11}\frac{Nm^2}{\operatorname{kg}^2})(5.9742\times10^{24}\operatorname{kg})}{(\frac{1}{2}(5054\frac{m}{s})^2-((6.67428\times10^{-11}\frac{Nm^2}{\operatorname{kg}^2})\frac{5.9742\times10^{24}\operatorname{kg}}{(6.3781\times10^6_{}m)})}\)Solving the operations we get:
\(R_2=8.06\times10^6m\)Therefore, the distance from the earth is 8.06x10^6 meters.
Related Questions
What is the net force net
on an airplane window of area 1800 cm2
if the pressure inside the cabin is 0.95 atm
and the pressure outside is 0.76 atm
?
Answers
The net force on the airplane window of area 1800 cm² is 3469.47 Pa.m² .
Given:
Pressure inside the cabin: 0.95 atm
Pressure outside the cabin: 0.76 atm
Area of the airplane window: 1800 cm²
Now to find the net force on the airplane window, we can calculate the pressure difference between the inside and outside of the cabin and to calculate the pressure difference, we subtract the outside pressure from the inside pressure.
Pressure Difference = Pressure inside - Pressure outside
Pressure Difference = 0.95 atm - 0.76 atm
Pressure Difference = 0.19 atm
The area of the airplane window is given as 1800 cm². To simplify calculations in SI unit we convert the area to square meters:
Area in m² = (Area in cm²) / 10,000
Area in m² = 1800 cm² / 10,000
Area in m² = 0.18 m²
As we know,
Net Force = Pressure Difference * Area
Net Force = 0.19 atm * 0.18 m²
Net Force = 0.0342 atm·m²
To convert the net force to pascals (Pa), we use 1 atm = 101325 Pa. Multiplying the net force by 101325 Pa, we get
Net Force = 3469.47 Pa·m²
Therefore, the net force on the airplane window is approximately 3469.47 pascals times square meters (Pa.m²).
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https://brainly.com/question/143618794. In a humidifier, air at a dry bulb temperature of 40 oC and relative humidity of 10% is humidified to a relative humidity of 40%. Determine the amount of moisture added in the humidifier per kg of dry air.
Answers
Answer:
0.0141 kg/kg.
Explanation:
So, we are given the following parameters or information from the problem above:
The relative humidity = 10% was humidified to 40%, the temperature = 40°C.
From the chart, at temperature = 40°C and relative humidity = 10%, the weight of dry air = a1 = 0.0046 kg/kg, volume = 0.89 m³/kg.
Also, from the chart at temperature = 40°C and relative humidity = 40%, the weight of dry air = a2 = 0.00187 kg/kg, volume = 0.913 m³/kg.
Therefore, the amount of moisture added in the humidifier per kg of dry air = 0.0046 kg/kg - 0.00187 kg/kg = 0.0141 kg/kg.
1st block
Worksheet: Metric Prefixes
A. Circle the larger unit:
1.millimeter, centimeter
2. kilogram, megagram
3. microsecond, millisecond
4.dL, mL
5.mg, kg
6. Mm,mm
7. S, cs
8. M, mm
9. U s, ks
Answers
Solution:
1. millimeter, centimeter
1 centimeter = 10 millimeter
So, centimeter is the larger unit here.
2. kilogram, megagram
1 megagram = 1000 kilogram
So, megagram is the larger unit here.
3. microsecond, millisecond
1 millisecond = 1000 microsecond
So, millisecond is the larger unit here.
4.dL, mL
1 dL = 100 mL
So, dL is the larger unit here.
5.mg, kg
1 mg = 0.000001 kg
So, kg is the larger unit here.
6. Mm,mm
1 mm = 1 mm
Both are same
7. S, cs
1 second = 100 centisecond
So, S is the larger unit.
8. M, mm
1 M = 1000 mm
So, M is the larger unit here.
9. U s, ks
Us is larger.
What is an example of total internal reflection at work?
A.
A ray of light has the same intensity both entering and exiting a fiber optic cable.
B.
A ray of light entering a glass cube gets refracted.
C.
A ray of light in air hits a shiny surface and bounces off.
D.
A ray of light entering a ruby gets refracted.
Answers
\(\huge\underline{\underline{\boxed{{ \mathbb{SOLUTION:}}}}}\)
\(\leadsto\) Total internal reflection occurs when a ray of light is traveling through a medium and, when that medium changes, it does not refract into the second medium but instead it reflects back into the first medium.
By Snell's law, we know that
\(\longrightarrow \sf{n_1 sin \emptyset_1 = n_2 sin \emptyset_2}\)
If this happens, then \(\sf{n_2 \angle n_1}\), and the ray does not lose intensity due to refraction.
\(\huge\underline{\underline{\boxed{\mathbb{ANSWER:}}}}\)
\(\large \bm{ A.}\) A ray of light has the same intensity both entering and exiting a fiber optic cable.
An object traveling at 1.5 rad
accelerates at 0.75d for 12
S
seconds. What is the object's final
velocity?
Answers
The object's final velocity, given the data is 10.5 rad/s
What is acceleration?This is defined as the rate of change of velocity which time. It is expressed as
a = (v – u) / t
Where
a is the acceleration v is the final velocity u is the initial velocity t is the time How to determine the final velocityThe following data were obtained from the question
Initial velocity (u) = 1.5 rad/sAcceleration (a) = 0.75 rad/s²Time (t) = 12 sFinal velocity (v) = ?The final velocity can be obtained as follow:
a = (v – u) / t
0.75 = (v – 1.5) / 12
Cross multiply
v – 1.5 = 0.75 × 12
v – 1.5 = 9
Collect like terms
v = 9 + 1.5
v = 10.5 rad/s
Thus, the final velocity of the object is 10.5 rad/s
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Water flows at a speed of 15 m/s through a pipe that has a radius of 0.40 m. The water then flows through a smaller pipe at
a speed of 45 m/s. What is the radius of the smaller pipe?
Answers
Answer:
Explanation: The volume flow rate of the water remains constant as it flows from the larger pipe to the smaller pipe. Therefore, we can use the equation:
A1v1 = A2v2
where A1 and v1 are the cross-sectional area and velocity of the larger pipe, and A2 and v2 are the corresponding values for the smaller pipe.
The cross-sectional area of a pipe is given by the formula:
A = πr^2
where r is the radius of the pipe.
Substituting the given values, we get:
π(0.4)^2(15) = π(r^2)(45)
Simplifying and solving for r, we get:
r = 0.13 m
Therefore, the radius of the smaller pipe is 0.13 m.
Question 2 The pulley on a machine is 230 mm diameter. It is to be driven at 183 rev/min. A main shaft to drive the machine has a pulley of diameter 140 mm. What is the speed of the running shaft driving the machine? (10)
Answers
The speed of the running shaft driving the machine, given that the main shaft to drive the machine has a pulley of diameter 140 mm is 111.4 rev/min
How do i determine the speed of the running shaft?First, we shall list out the given parameters from the question. Details below:
Speed of the main shaft (S₁) = 183 rev/minDiameter of the main shaft (D₁) = 140 mmDiameter of the second pulley (D₂) = 230 mmSpeed of the running shaft i.e second pulley (S₂) = ?The speed of the running shaft can be obtain as shown below:
S₁D₁ = S₂D₂
183 × 140 = S₂ × 230
183 × 140 = S₂ × 230
25620 = S₂ × 230
Divide both sides by 230
S₂ = 25620 / 230
S₂ = 111.4 rev/min
Thus, we can conclude that the speed of the running shaft driving the machine is 111.4 rev/min
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A 1,100 kg car is traveling at 20 m/s as it begins going down a 40 m hill. What is its total mechanical energy?(1 point)
A 651,200 J
B 211,200 J
C 431,200 J
D 220,000 J
Answers
Answer: 651,200 J
Explanation:
The total mechanical energy of the car is determined as 651,200 J.
Option A is the correct answer.
What is the total mechanical energy of the car?
The total mechanical energy of the car is the sum of the kinetic energy and potential energy of the car.
Mathematically, the total mechanical energy of the car is given as;
M = K.E + P.E
where;
K.E is the kinetic energy of the carP.E is the potential energy of the carM = ¹/₂mv² + mgh
where;
m is the mass of the carv is the velocity of the carh is the height of the hillg is acceleration due to gravityThe given parameters include;
m is the mass of the car = 1,100 kgv is the velocity of the car = 20 m/sh is the height of the hill = 40 mg is acceleration due to gravity = 9.8 m/s²M = ¹/₂(1100)(20²) + (1100)(9.8)40)
M = 651,200 J
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Question 1
Describe the path light takes as it travels through air and into glass
Question 2
Explain the brightness of light using the wave model of light
Answers
Answer:
here is answer!
Explanation:
Question 1:
Light bends when it transitions from air to glass due to differences in refractive indices. It follows an incident path in air, refracts at the air-glass boundary, and continues through the glass as a transmitted ray. Total internal reflection may occur if the angle of incidence is large enough.
Question 2:
Brightness in the wave model of light is determined by the amplitude and intensity of the light waves. Higher amplitudes and intensities correspond to brighter light. When multiple light waves overlap, their amplitudes add up, resulting in increased brightness
The path light bends as it travels through air and into glass.
What is light?
The light is the ray form of energy obtained from the Sun.
The path of light will look like bended when it has crossed the interface of two medium.
As. the light ray falls on the surface of glass travelling through air, the ray appears to bend after refraction.
Thus, the path of light bends as it travels through air and into glass.
———————
Question 2 :
According to the wave theory of light, the energy of radiation depends only on the intensity or wave amplitude (brightness), not the frequency (what type of light; e.g red light or green light; visible light or gamma) According to the particle theory of light states the energy of radiation depend only on the frequency
In an application, Germanium is
made p-type material by adding
Indium. The rate of adding is one
indium atom
per
3 Germanium
atom. Assume donor density to be
zero and ni = 6.2 Cubic
metre at room temperature.
Determine the value
of
the
aceeptor atom density
Answers
Answer:
produce electronics
Explanation:
The uses of Germanium are recorded beneath: Germanium's principle use is to deliver strong state hardware, semiconductors and fiber optic frameworks. As a phosphor in fluorescent lights.
Students measure how much work it takes to lift an object two meters. They use an inclined plane that is 3 meters long, an inclined plane that is 4 meters long, and a compound pulley. Which statement is correct?
A. All three simple machines will require the same amount of work
B. The 4-meter plane will require the least amount of work
C. The 3-meter plane will require the least amount of work
D. The compound pulley will require the least amount of work
Answers
With only one applied force and one load force, a simple machine may perform work.
What is true about all simple machines?With only one applied force and one load force, a simple machine may perform work.The work performed by the load equals the work performed by the applied force when friction losses are taken into account.Inclinable planes, levers, and pulleys are the three simplest simple machines.While simple machines may increase or decrease the forces that can be applied to them, they have no effect on the overall amount of work required to complete the activity.There are three ways simple machines facilitate work: by extending the distance over which force is applied, by reversing the direction in which force is applied, or by boosting the force and speed of the energy applied.To learn more about applied force refer
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D. 5.0 g of water at 25°C is dropped onto a large block of ice at 0 °C. The water cools to 0 °C and
some of the ice melts. Assume that all the energy lost by the water is gained by the ice. What is
the mass of ice that melts? The specific heat capacity of water is 4.2J/(g°C). The specific latent
heat of fusion of ice is 340J/g.
Answers
Answer:
remeber me u wher rude to me so ur points are mine now
Explanation:
Calculate the net force on the particle q1.
Answers
Answer:
-12.1
Explanation:
i’m almost sure this is it, i’m checking my old answers
if not let me know and i’ll give you some more answers
A body dropped over a fixed rough inclined plane of inclination 45 from height h. If after collision velocity of body becomes horizontal
then co-efficient of restitution if co-efficient of friction is 1/2
Answers
As per the given scenario, in this case, the coefficient of friction () is half and the coefficient of restitution (e) is zero.
Identify the body's starting velocity:
We may use the equation of motion to get the body's initial velocity (u)
\(v^2 = u^2 + 2as\)
\(0 = u^2 + 2(-9.8)m/s^2 * h\)
\(u^2 = 19.6h\)
u = √(19.6h)
Determine the coefficient of restitution (e):The body's initial velocity (u) and initial relative velocity (u_rel) are the same.
The body's horizontal velocity immediately following the collision, which is zero, is the final relative velocity (v_rel).
\(e = v_{rel }/ u_{rel}\)
e = 0 / u_rel = 0 / u
Now, one can investigate the forces affecting the body: When a body is on an inclined plane.
There are two main forces at work on it: the frictional force that prevents the body from moving and the gravitational force that pulls it downward (mg).
The gravitational force has two components that act perpendicular to and parallel to the inclined plane, respectively: m*g*cos(45°) and m*g*sin(45°).
Determine the conditions for the body to stop:
μ * N = m * g * sin(45°)
μ * (m * g * cos(45°)) = m * g * sin(45°)
μ * cos(45°) = sin(45°)
(1/2) * cos(45°) = sin(45°)
Simplifying further, we have:
√2 / 4 = √2 / 2
Thus, the body will come to rest following the collision if the equation is valid.
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The position of a particle is r⃗ (t)=(3.0t2iˆ+5.0jˆ−6.0tkˆ)m. (a) Determine its velocity and acceleration as functions of time. (b) What are its velocity and acceleration at time t = 0?
Answers
Explanation:
It is given that,
The position of a particle is given by :
\(r(t)=(3t^2i+5j-6tk)\ m\)
(a) Velocity of a particle is given by :
\(v=\dfrac{dr(t)}{dt}\)
Putting values,
\(v=\dfrac{d}{dt}(3t^2+5-6t)\\\\v=(6ti-6k)\ m/s\)
The acceleration of the particle is given by :
\(a=\dfrac{dv}{dt}\\\\a=\dfrac{d}{dt}(6t-6)\\\\a=6i\ m/s^2\)
(b) At t = 0,
Velocity, v = 6k m/s
Acceleration, a = 6i m/s²
PLEASE HELP! Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an angle of 63.0° with the
horizontal. What is the height of the building?
A. 174 m
B. 110 m
C. 98 m
D. 50 m
Answers
Answer:
The height of building should be 98.13 m plus the height of Daniel. Since the 63° was measured from his eye level.
Explanation:
To move the Center of Gravity (CG) forward, you could do which of the following?
Add clay to the nose.
Increase the density of the nose material.
Decrease the volume of the horizontal and vertical stabilizer material.
All of the above.
its not Increase the density of the nose material bc i tried that so help pleasee
Answers
Answer:
I think it's no. 3
Decrease the volume of the horizontal and vertical stabilizer material.
Explanation:
Lemme know if it's correct
Answer:
all of the above
Explanation:
A motorcycle stoop is at a traffic light, when the light turns green, the motorcycle accelerates to a speed of 78 km/h over a distance of 50 m. What is the average acceleration of the motorcycle over this distance?
Answers
The average acceleration of the motorcycle over the given distance is approximately 9.39 m/s².
To calculate the average acceleration of the motorcycle, we can use the formula:
Average acceleration = (final velocity - initial velocity) / time
First, let's convert the final velocity from km/h to m/s since the distance is given in meters. We know that 1 km/h is equal to 0.2778 m/s.
Converting the final velocity:
Final velocity = 78 km/h * 0.2778 m/s = 21.67 m/s
Since the motorcycle starts from rest (initial velocity is zero), the formula becomes:
Average acceleration = (21.67 m/s - 0 m/s) / time
To find the time taken to reach this velocity, we need to use the formula for average speed:
Average speed = total distance/time
Rearranging the formula:
time = total distance / average speed
Plugging in the values:
time = 50 m / 21.67 m/s ≈ 2.31 seconds
Now we can calculate the average acceleration:
Average acceleration = (21.67 m/s - 0 m/s) / 2.31 s ≈ 9.39 m/s²
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A pitching machine is programmed to pitch baseballs horizontally at a speed of 134 km/h. The machine is mounted on a truck and aimed forward. As the truck drives toward you at a speed of 85 km/h, the machine shoots a ball toward you. A pickup truck moves to the left at a constant velocity. A pitching machine sits in the bed of the pickup truck. The pitching machine launches a baseball to the right with a different constant velocity. A man with a baseball mitt stands at rest some distance to the right of the truck. For each of the object pairings listed, determine the correct relative speed. The speed of the pitching machine relative to the truck The speed of the pitched ball relative to the truck The speed of the pitching machine relative to you The speed of the pitched ball relative to you
Answers
Answer: 134 = 143 = 151 = 166 = 176
Hope this helps!!
Sorry if it's incorrect!!
:'(
A carousel is (more or less) a disk of mass, 15,000 kg, with a radius of 6.14. What torque must be applied to create an angular acceleration of 0.0500 rad/s^2?round to 3 significant figures
(Plssss help me im suffering from severe brainrot)
Answers
To calculate the torque required to create an angular acceleration, we can use the formula:
Torque = Moment of Inertia × Angular Acceleration
The moment of inertia of a disk can be calculated using the formula:
Moment of Inertia = (1/2) × Mass × Radius^2
Given:
Mass = 15,000 kg
Radius = 6.14 m
Angular Acceleration = 0.0500 rad/s^2
First, calculate the moment of inertia:
Moment of Inertia = (1/2) × Mass × Radius^2
Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2
Next, calculate the torque:
Torque = Moment of Inertia × Angular Acceleration
Torque = Moment of Inertia × 0.0500 rad/s^2
Now, let's plug in the values and calculate:
Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2
Moment of Inertia ≈ 283,594.13 kg·m^2
Torque = 283,594.13 kg·m^2 × 0.0500 rad/s^2
Torque ≈ 14,179.71 N·m
Rounding to three significant figures, the torque required to create an angular acceleration of 0.0500 rad/s^2 is approximately 14,180 N·m.
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a 4 kg block is moving at 12 m/s on a horizontal frictionless surface. a constant force is applied such that the block slows with an acceleration of 3 m/s^2. how much work must this force do to stop the block?
a. -576 J
b. -360 J
c. -288 J
d. 360 J
e. 576 J
walk me through this please. NO LINKS.
Answers
Answer:
Is there a picture?
Explanation:
Electric field is always perpendicular to the equipotential surface.
a. True
b. False
Answers
Answer:
a: true.
Explanation:
We can define an equipotential surface as a surface where the potential at any point of the surface is constant.
For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.
Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.
Now we want to see if the electric field is always perpendicular to these equipotential surfaces.
You can see that in the two previous examples this is true, but let's see for a general case.
Now suppose that you have a given field, and you have a test charge in one equipotential surface.
So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.
But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.
Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.
The correct option is a.
write down the value of
920 kg in g
Answers
Answer:
920000
Explanation:
Each kg contains 1,000 grams
identify the types of motion in each activity.1.walking a long a hallway. 2.motion of the blades of the fan. 3.earths rotation 4.ball moving on the ground. 5.soldiers marching.
Answers
Answer:
Explanatation
1 is just walking
2 spinging
3 roatating
4 rolling
5 stomping there feet
Those should be right but if im wrong then just someone eles the brainly
A thin half ring with a radius of R = 10 cm is uniformly charged with a linear density of = 1 Mikrokulon/m and located in a vacuum. Determine the force F of interaction between the half ring and a point charge q = 20 nC located at the center of curvature. (don't use chatgpt and add an explain please )
Answers
Explanation:
To determine the force F of interaction between the half ring and the point charge, we can use the principle of superposition, which states that the total force on a point charge due to a collection of other charges is the vector sum of the individual forces that each of those charges would exert on the point charge if it were the only charge present.
First, we need to find the electric field at the center of curvature due to the charged half ring. The electric field at a point on the axis of a uniformly charged ring is given by:
E = kqz / (z^2 + R^2)^(3/2)
where k is Coulomb's constant, q is the linear charge density, z is the distance from the center of the ring to the point on the axis, and R is the radius of the ring.
At the center of curvature of the half ring, z = R, so the electric field is:
E = kq / (2R)
Next, we can use the electric field to find the force on the point charge q:
F = qE
Substituting the given values, we get:
F = (20 x 10^-9 C) x (9 x 10^9 N·m^2/C^2) x (1/20 cm)
F = 9 x 10^-3 N
Therefore, the force of interaction between the half ring and the point charge is 9 x 10^-3 N.
This force can also be interpreted as the force required to hold the point charge at the center of curvature against the electric field due to the charged half ring. It is an attractive force because the point charge is opposite in sign to the charged half ring.
Answer:
5.65N.
Explanation:
Solution Given:
radius of R = 10 cm=10/100=0.1 m
linear density λ = 1 Mikrokulon/m= 10^-6 Coulomb/m
force F=?
q1 = 20nC=20*10^-9 C
we have
Coulomb's law, which states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, this can be expressed as:
F = k*(q1*q2)/r^2
since q1 is located at the center of curvature of the half ring , so the half ring is uniformly charged with a linear density of λ= 1 μC/m.
again
equation becomes.
F=k*(q1*λL)/r^2
Since the half ring is a semicircle,
we have L=πr
F=k*(q1*λ*πr)/r^2
substituting value
F=9 x 10^9 N*m^2/C^2 *(20*10^-9 C* 10^-6 m^3*π*0.1 m)/(0.1m^2)
F=5.65 N
Therefore, the force of interaction between the half ring and the point charge is 5.65N.
What is the bond between oxygen and hydrogen in nitric acid
Answers
Answer:
The spectra are consistent with a structure in which the nitric acid forms a near-linear, 1.78 Å hydrogen bond to the oxygen of the water.
Explanation:
Hope this helps you
A car travels a distance of 140 km at 70.0 km/hr. It then travels an additional distance of 60.0 km at 40.0 km/hr. The average speed is
Answers
Answer:
57.1 km/hr
Explanation:
To find the average speed you take the total distance divided by the total elapsed time.
So, the total distance is 140 + 60 = 200
the total elapsed time is found by taking 140/70=2 and 60/40=1.5
2+1.5=3.5
The plug the numbers into the equation,
200/3.5=57.1
1. A material emits electrons when the electromagnetic radiation with a frequency equal to or less than the material's threshold frequency strikes the material.
True
False
2. In a particle model to show energy transfer, which of the following is correct?
a. Energy jumps are best described of electrons in packages known as photons
b. In energy transfers, electrons can absorb or emit energy
c. Photons carry energy from one location or object to another
d. All of the above are correct of particle models
Answers
1) The photoelectric effect is the phenomena by which, the metals release electrons when they are exposed to electromagnetic radiation with the suitable frequency. The photoelectrons are the electrons that are released in this process.
So, the statement is true.
2) In a particle model, energy transfer can be done through many ways such as:
The energy jumps between the particles in the form of photons.
Electrons can absorb or emit energy during energy transfer.
The energy is transferred between different objects in the form of photons.
So, all of them are true.
To learn more about photons, click:
https://brainly.com/question/20912241
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how many times can the earth revolve around the sun in the time it takes Jupiter to revolve around the sun once?????????
Answers
Answer:
Jupiter fully revolves around the sun in about 11.9 Earth years, where as Earth fully revolves around it in a year.
Explanation:
Is it possible to stand backwards on a flight of stairs?
Answers
Answer:
yes
Explanation:
if your destination is on the top floor, but you are facing towards the bottom floor, you are facing the wrong way. It's all relative. Honestly, I hate this question, because then it turns into, is it possible to go backwards anywhere. Backwards and forwards, right and left, it's all relative to which way you are facing and where you want to go. If something is on the left and you turn 180 degrees, then that thing is on the right. If you are going forward towards your house and you turn around, you are now going backwards, relative to your house. But if your perspective is relative to that tree in the opposite direction of your house, you are now facing the right way. See, it's all relative to where you are going.