The action force in this scenario is earth pushes back on the exhaust gases.
Explanation:The action force in this scenario is option C: earth pushes back on the exhaust gases.
Newton's third law states that for every action, there is an equal and opposite reaction. In this case, the action force is the exhaust gases pushing downward on the earth as the rocket launches. The reaction force is the earth pushing back on the exhaust gases with an equal force in the opposite direction.
So, while the rocket is traveling upward, the action force is the exhaust gases pushing downward on the earth and the reaction force is the earth pushing back on the exhaust gases.
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What is the energy of the spring-mass system when the mass first passes through the equilibrium position? (you may wish to include a logical test to help you find when this occurs)?
A ball is thrown at an angle of 45° to the ground. if the ball lands 87 m away, what was the initial speed of the ball? (round your answer to the nearest whole number. use g â 9.8 m/s2.)
The period of a wave is found to be 50 seconds. What is the frequency of the wave? 50 Hz 0.8 Hz 0.02 Hz not enough information given
The period of a wave is found to be 50 seconds. What is the frequency of the wave?
0.02 Hz
There is a 12 v potential difference between the positive and negative ends of the jumper cables, which are a short distance apart. an electron at the negative end ready to jump to the positive end has a certain amount of potential energy. on what quantities does this electrical potential energy depend? view available hint(s)
A force of 35 n acts on an object which has a mass of 5.4 kg. what acceleration (in m/s2) is produced by the force
A grindstone, initially at rest, is given a constant angular acceleration so that it makes 20.0 rev in the first 8.00 s. what is its angular acceleration?
The angular acceleration of a grindstone that makes 20 revolutions in 8 seconds from rest, convert the revolutions to radians and use the rotational kinematic equation, resulting in an angular acceleration of approximately 3.93 radians/s².
The angular acceleration of a grindstone that makes 20 revolutions in 8 seconds, starting from rest.
To find the angular acceleration, we can use the kinematic equation for rotational motion:
θ = ω₀t + 1/2αt²
where:
θ is the angular displacement in radians,
ω₀ is the initial angular velocity (which is 0 because it starts from rest),
t is the time in seconds,
and α is the angular acceleration in radians per second squared.
First, we convert the revolutions to radians:
20 revolutions * 2π radians/revolution = 40π radians
We then have:
40π radians = 0 + 1/2α(8²)
Solving for α gives:
α = (40π radians) / (1/2 * 64 s²)
α = 80π / 64 radians/s²
α ≈ 3.93 radians/s²
Therefore, the angular acceleration of the grindstone is approximately 3.93 radians/s².
If electrons of energy 12.8 ev are incident on a gas of hydrogen atoms in their ground state, what are the energies of the photons that are emitted by the excited gas?
When 12.8 eV energy electrons excite hydrogen atoms, the atoms are elevated to an energy level just below ionization. To determine the specific energy state, the energy formula for hydrogen atoms is used, then the energies of emitted photons are found by calculating the energy difference between the excited and final states.
Explanation:If electrons with an energy of 12.8 eV are incident on hydrogen atoms in their ground state, we first need to determine the energy level to which the hydrogen atoms are excited. The ground state energy of hydrogen is -13.6 eV. If 12.8 eV is provided, the electron reaches an energy level just below 0 eV (ionization energy), since 12.8 eV is not sufficient to completely ionize the atom (13.6 eV needed).
To find out which energy level the electron will jump to, we use the formula for the total energy of an electron in a hydrogen atom, En = (- 13.6 eV/n²), and solve for n.
Once the electron is excited, it will eventually drop back to a lower energy state, emitting a photon in the process.
The energy of the emitted photon can be calculated using the difference in energy levels, ΔE = E1 - Ef.
You are at home in your air conditioned garage. You are planning a family road trip from New York to Florida. You are lnfating the tires of your suv. You fill each car-tire with about 10 Liters of air. The temperature inside your air conditioned garage is about 10 degrees Celsius. Your family packs the car, and you begin your trip to Florida. The temperature in New York is about 21 degrees Celsius, and the happen temperature in Florida is 32 degrees Celsius. Assuming the pressure stays constant from the garage all the way to Florida, what could to your car tires during your road trip.
Write a paragraph that makes a claim about what happen to the car tires
A centrifuge in a medical laboratory rotates at an angular speed of 3,650 rev/min. when switched off, it rotates through 48.0 revolutions before coming to rest. find the constant angular acceleration (in rad/s2) of the centrifuge.
A centrifuge in a medical laboratory rotates at an angular speed of 3,650 rev/min, the constant angular acceleration of the centrifuge is approximately -1522.71 rad/s².
We can apply the following formula to determine the centrifuge's constant angular acceleration:
Δθ = ω₀t + (1/2)αt²
Δθ = ω₀t + (1/2)αt²
Since the final angular speed is 0, the formula becomes:
Δθ = ω₀t
48 rev * (2π rad/rev) = 381.92 rad/s * t
Simplifying:
96π rad = 381.92 rad/s * t
Dividing both sides by 381.92 rad/s:
t ≈ 0.251 s
Now, we can calculate the angular acceleration (α) using the rearranged formula:
α = (0 - ω₀) / t
α = (0 - 381.92 rad/s) / 0.251 s
α = -1522.71 rad/s²
Thus, the constant angular acceleration of the centrifuge is approximately -1522.71 rad/s².
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A heat engine takes in 840 kJ per cycle from a heat reservoir. Which is not a possible value of the engine's heat output per cycle?
for a bohr model,how many protons,electrons,and neutron does Sodium has
When populations of two species work together to obtain resources, they are
Answer:
cooperating.
Explanation:
:)
An experimenter finds that no photoelectrons are emitted from a particular metal unless the wavelength of light is less than 295 nm. her experiment will require photoelectrons of maximum kinetic energy 2.4 ev. what frequency light should be used to illuminate the metal?
The frequency of light is approximately [tex]\( 1.016 \times 10^{15} \)[/tex] Hz.
To determine the frequency of light that should be used to illuminate the metal in order to produce photoelectrons with a maximum kinetic energy of 2.4 eV, we can use the relationship between energy, frequency, and wavelength in the context of the photoelectric effect.
The energy of a photon (light particle) is given by Planck's equation:
[tex]\[ E = h \cdot f \][/tex]
Where:
- [tex]\( E \)[/tex] is the energy of the photon in joules (J)
-[tex]\( h \)[/tex] is Planck's constant, approximately [tex]\( 6.626 \times 10^{-34} \)[/tex] J·s
- [tex]\( f \)[/tex] is the frequency of the light in hertz (Hz)
We are given the maximum kinetic energy of the photoelectrons as 2.4 eV. To convert this to joules, we use the conversion factor [tex]\( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \)[/tex]
[tex]\[ E_{\text{max}} = 2.4 \, \text{eV} \times (1.602 \times 10^{-19} \, \text{J/eV}) \]\\\[ E_{\text{max}} = 3.845 \times 10^{-19} \, \text{J} \][/tex]
Now, since the photoelectrons are emitted only when the wavelength of light is less than 295 nm (nanometers), we can use the speed of light equation to relate frequency and wavelength:
[tex]\[ c = f \cdot \lambda \][/tex]
Where:
- [tex]\( c \)[/tex] is the speed of light, approximately [tex]\( 3.00 \times 10^8 \)[/tex] m/s
- [tex]\( f \)[/tex] is the frequency of the light in hertz (Hz)
- [tex]\( \lambda \)[/tex] is the wavelength of the light in meters (m)
First, we convert the wavelength limit of 295 nm to meters:
[tex]\[ \lambda_{\text{max}} = 295 \, \text{nm} \times (1 \, \text{m} / 10^9 \, \text{nm}) \][/tex]
[tex]\[ \lambda_{\text{max}} = 2.95 \times 10^{-7} \, \text{m} \][/tex]
Now, we can rearrange the speed of light equation to solve for frequency:
[tex]\[ f = \frac{c}{\lambda_{\text{max}}} \][/tex]
[tex]\[ f = \frac{3.00 \times 10^8 \, \text{m/s}}{2.95 \times 10^{-7} \, \text{m}} \][/tex]
[tex]\[ f = 1.016 \times 10^{15} Hz[/tex]
Therefore, the frequency of light that should be used to illuminate the metal and produce photoelectrons with a maximum kinetic energy of 2.4 eV is approximately [tex]\( 1.016 \times 10^{15} \)[/tex] Hz.
A child goes down a playground slide with an acceleration of 1.12 m/s2 .find the coefficient of kinetic friction between the child and the slide if the slide is inclined at an angle of 27.0 ∘ below the horizontal.
The coefficient of kinetic friction is obtained by considering the forces acting on the child as they slide. These are the force of gravity (mg sin θ) and the force of kinetic friction (μk mg cos θ). Setting these forces equal gives us an equation that can be solved to find the coefficient of kinetic friction, μk.
Explanation:To figure out the coefficient of kinetic friction, we need to consider the forces acting on the child as they slide down the playground. There are two key forces to consider, the force caused by gravity which is pulling the child down the slide, and the force of kinetic friction which is trying to slow the child down.
The force of gravity working to pull the child down the slide, or the component of the weight down the slope is given as mg sin θ. Here m is the mass of the child, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of the slide. The force of kinetic friction, which opposes the motion, is expressed as μk mg cos θ, where μk is the coefficient of kinetic friction we're trying to find.
Since the child is sliding down the slide with an acceleration of 1.12 m/s², the force due to gravity exceeds the force of friction. Setting up the equation μk mg cos θ = mg sin θ - ma, where a is the acceleration of 1.12 m/s², and solving for μk gives us the coefficient of kinetic friction between the child and the slide.
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A sinusoidal wave travels with speed 250 m/s . its wavelength is 3.5 m . part a what is its frequency
At the moment the acceleration of the blue ball is 9.40 m/s2, what is the ratio of the magnitudes of the force of air resistance to the force of gravity acting on the blue ball? use 9.81 m/s2 for the standard value of g at the location of the experiment.
The ratio of the magnitudes of the force of air resistance to the force of gravity acting on the ball is approximately 1:24.
The student is asking about the ratio of the force of air resistance to the force of gravity acting on a blue ball, when the ball has an acceleration of 9.40 m/s2. To calculate this ratio, we can use Newton's second law, F = ma, where F is the force, m is the mass of the ball, and a is the acceleration of the ball due to the net force acting on it. If we denote the force of air resistance as Fair and the force of gravity as Fgrav, we can write Fnet = Fgrav - Fair, since the air resistance works in the opposite direction of gravity.
Given that the acceleration due to gravity is approximately 9.81 m/s2, the force due to gravity (weight) can be expressed as Fgrav = mg. Here, the acceleration is less than g due to air resistance, thus we can write Fnet = ma = mg - Fair. Since the acceleration of the ball is given as 9.40 m/s2, we can now express Fair as mg - ma. Taking the ratio of Fair to mg (the weight), we get:
(mg - ma) / mg = (g - a) / g = (9.81 - 9.40) / 9.81.
Therefore, the ratio of the magnitudes of the force of air resistance to the force of gravity is equal to 0.0418, which simplifies to approximately 1:24.
"a metal sphere of radius 5.00 cm is initially uncharged. how many electrons would have to be placed on the sphere to produce an electric field of magnitude 1.53 ✕ 105 n/c at a point 8.64 cm from the center of the sphere?"
a 13 kg sled is moving at a speed of 3.0 m/s. at which of the following speeds will the sled have tace as mucb as the kinetic
A proton moves perpendicularly to a uniform magnetic field b with a speed of 3.7 × 107 m/s and experiences an acceleration of 5 × 1013 m/s 2 in the positive x direction when its velocity is in the positive z direction. the mass of a proton is 1.673 × 10−27 kg. find the magnitude of the field. answer in units of t.
A charge of 50 µC is pushed by a force of 25 µN a distance of 15 m in an electric field. What is the electric potential difference? Recall that Fe = qE.
The electric potential difference is 7.5 V.
Electric potential difference between two points is defined as the work done in moving unit positive charge between the two points. Electric potential difference can also be defined as the work done W per unit charge q in an electric field.
[tex]\Delta V=\frac{W}{q}[/tex]......(1)
Work done by a force is the product of the force F and the displacement s made in the direction of the force. Assuming that the displacement is parallel to the line of action of the force,
[tex]W=F.s[/tex]......(2)
Rewrite equation (1) using equation (2).
[tex]\Delta V=\frac{Fs}{q}[/tex]......(3)
Substitute the given values of force, displacement and charge in the equation (3).
[tex]\Delta V=\frac{Fs}{q}\\ =\frac{(25*10^-^6N)(15m)}{(50*10^-^6C)} \\ =7.5V[/tex]
The potential difference between the two points is 7.5 V
A man drags a 12.0 kg bag of mulch at a constant speed applying a 39.5 N force at 41°. What is the normal force acting on the bag?
A 0.32 μc particle moves with a speed of 18 m/s through a region where the magnetic field has a strength of 0.95 t . you may want to review (pages 773 - 777) . part a at what angle to the field is the particle moving if the force exerted on it is 4.8×10−6n? express your answer using two significant figures. θ = ∘ request answer part b at what angle to the field is the particle moving if the force exerted on it is 3.0×10−6n? express your answer using two significant figures.
The force on a particle in a magnetic field can be determined using F = qvBsin(\theta). By rearranging the equation to solve for \theta, we can find the angle the particle's velocity makes with the magnetic field for any given force, charge, velocity, and field strength.
Explanation:The situation described is a classic example of Lorenz force experienced by a charged particle moving through a magnetic field. The force exerted on a charged particle when it moves through a magnetic field can be calculated with the following equation:
F = qvBsin(\theta)
where:
F is the magnetic force,q is the charge of the particle,v is the velocity of the particle,B is the magnetic field strength, and\theta is the angle between the velocity vector and the magnetic field.To find the angle \(\theta\), we can rearrange the equation:
\theta = arcsin(\frac{F}{q v B})
For part A:
Using the information given, where F = 4.8 \times 10^{-6} N, q = 0.32 \times 10^{-6} C, v = 18 m/s, and B = 0.95 T, we can calculate the angle.
\(\theta = arcsin(\frac{4.8 \times 10^{-6}}{0.32 \times 10^{-6} \times 18 \times 0.95}) = arcsin(\frac{4.8}{0.32 \times 18 \times 0.95})\)
This returns an angle \(\theta\), which will be in degrees once evaluated using a calculator.
For part B, you would apply the same approach but with F = 3.0 \times 10^{-6} N, to find the different angle.
A rock is thrown vertically upward from ground level at time t = 0. at t = t, it passes the top of the tower and ât later (i.e., when t = t + ât) it reaches the maximum height. what is the height of the tower? leave g as g â do not substitute with numbers
"The height of the tower is given by the equation:
[tex]\[ h = \frac{1}{2} g (t + \hat{t})^2 - \frac{1}{2} g t^2 \][/tex]
This equation represents the difference in height between the maximum height the rock reaches and the height it reaches at time \( t \) when it passes the top of the tower. Here, \( g \) is the acceleration due to gravity, \( t \) is the time it takes for the rock to reach the top of the tower, and \( \hat{t} \) is the additional time it takes for the rock to reach the maximum height after passing the top of the tower.
To find the height of the tower, we can simplify the equation:
[tex]\[ h = \frac{1}{2} g ((t + \hat{t})^2 - t^2) \][/tex]
[tex]\[ h = \frac{1}{2} g (t^2 + 2t\hat{t} + \hat{t}^2 - t^2) \][/tex]
[tex]\[ h = \frac{1}{2} g (2t\hat{t} + \hat{t}^2) \][/tex]
[tex]\[ h = g t\hat{t} + \frac{1}{2} g \hat{t}^2 \][/tex]
Since [tex]\[ u = g \hat{t} \][/tex]is the time it takes for the rock to reach the maximum height from the top of the tower, and at the maximum height the velocity of the rock is zero, we can use the kinematic equation for the final velocity \( v \) at the maximum height:
[tex]\[ v = u + g t \][/tex]
Here, \( u \) is the initial velocity (which is the velocity of the rock at the top of the tower), \( g \) is the acceleration due to gravity, and \( t \) is the time taken to reach the maximum height, which is \( \hat{t} \). At the maximum height, \( v = 0 \), so:
[tex]\[ 0 = u - g \hat{t} \][/tex]
[tex]\[ u = g \hat{t} \][/tex]
Now, we can substitute \( u \) back into the height equation:
[tex]\[ h = g t\hat{t} + \frac{1}{2} g \hat{t}^2 \][/tex]
[tex]\[ h = g t\left(\frac{u}{g}\right) + \frac{1}{2} g \left(\frac{u}{g}\right)^2 \][/tex]
[tex]\[ h = u t + \frac{1}{2} \frac{u^2}{g} \][/tex]
Since \( u t \) represents the distance the rock would have traveled in time \( t \) if it continued at a constant velocity \( u \), and \( \frac{1}{2} \frac{u^2}{g} \) represents the total height the rock would reach if it continued to move upward with initial velocity \( u \) and was only acted upon by gravity, the term \( u t \) is actually the height of the tower. Thus, the height of the tower is:
[tex]\[ h = u t \][/tex]
This is the final answer, and it shows that the height of the tower is the product of the initial velocity of the rock at the top of the tower and the time it takes to reach the top of the tower. The term[tex]\( \frac{1}{2} \frac{u^2}{g} \)[/tex]is not needed for the height of the tower, as it represents the additional height the rock reaches after passing the top of the tower until it reaches the maximum height."
After harvesting their main crops, such as corn and wheat, many farmers plant a second crop called a cover crop. This cover crop grows during the fall and winter and is then plowed over in the spring. What is the main purpose of a cover crop?
A.
to get rid of excess seed rather than storing during the winter
B.
to provide food and shelter for birds and animals during the winter
C.
to prevent water runoff into other parts of the farmers' field
D.
to prevent erosion of the topsoil by wind and water
Answer:
Option (D)
Explanation:
A cover crop is usually defined as a special type of crop that is cultivated in order to increase the productivity of the soil, rather than focusing on the productivity of the crop. These are very commonly used and it helps in the decreasing amount of weeds, controlling pests and diseases, increasing the fertility of the soil, reduction of soil erosion. It also enhances the biodiversity.
Thus, the main purpose of the cover crop is to prevent the erosion of the topsoil by the agents such as water and wind.
Hence, the correct answer is option (D).
A water bed weighs 1025 N, and is 1.5 m wide by 2.5 m long. How much pressure does the water bed exert if the entire lower surface of the bed makes contact with the floor?
a.
3.6 x 10^2 Pa
b.
2.7 x 10^4 Pa
c.
1.2 x 10^3 Pa
d.
2.7 x 10^2 Pa
Which labels correctly identify the layers most closely associated with gamma rays and visible light? Z: Gamma rays X: Visible light X: Gamma rays Z: Visible light Z: Gamma rays Y: Visible light Y: Gamma rays Z: Visible light
Answer:
A on edge
Explanation:
Five metal samples, with equal masses, are heated to 200oC. Each solid is dropped into a beaker containing 200 ml 15oC water. Which metal will cool the fastest?
A) aluminum
B) copper
C) gold
D) platinum
Answer:
aluminum...
Explanation:
You are asked to design a horizontal curve with a 40-degree central angle (' = 40) for a two-lane road with 11-ft lanes. the design speed is 70 mi/h and superelevation is limited to 0.06 f
a. True
b. Falset. give the radius, degree of curvature, and length of curve that you would recommend. 3.43 for the h
To design a horizontal curve for a two-lane road, you can calculate the radius, degree of curvature, and length of the curve using formulas.
Explanation:To design a horizontal curve, we need to calculate the radius, degree of curvature, and length of the curve. First, we can calculate the radius using the formula: R = (V2) / (15 * f), where R is the radius in feet, V is the design speed in miles per hour, and f is the superelevation. Plugging in the values, we get: R = (702) / (15 * 0.06) = 43,333.33 feet. The degree of curvature can be calculated using the formula: (180 * L) / (π * R), where L is the length of the curve in feet. Since the central angle is given as 40 degrees, the degree of curvature is also 40 degrees. Finally, we can calculate the length of the curve using the formula: L = (π * R * d) / 180, where d is the degree of curvature. Plugging in the values, we get: L = (π * 43,333.33 * 40) / 180 = 9,632.87 feet.
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The Doppler effect is a shift in the _______________ of an oscillation caused by the motion of the source of the oscillation, and occurs at speeds below the speed of sound. a) amplitude b) frequency c) speed
what is air resistance?
Air resistance is the pushing of air against an object that is moving. Both the air and the object rub together and this slows the down the object and making it use more energy to reach a required speed. If an object moves faster, the greater the air resistance it encounters.