The ball stayed 2.525 seconds in the air.
How to find how long the ball stayed in the air?To know how long the ball stayed in the air, we will use the vertical motion model with Initial Velocity .
[tex]h(t) = -16t^2 + v*t + h0[/tex]
Where h(t) = height above the ground after t seconds
v = 40ft/s initial velocity
h0 = 4 ft (initial object height)
t = motion time
[tex]h(t) = -16t^2 + 40*t + 4[/tex]
find t such that h(t) = 3ft
[tex]3 = -16t^2 + 40*t + 4[/tex]
[tex]-16t^2 + 40*t + 4 = 3[/tex]
t = 2.525 seconds
Therefore, the ball stayed 2.525 seconds in the air.
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A .35 kg block at -27.5 ºC is added to .217 kg of water at 25.0 ºC. They come to equilibrium at 16.4 ºC. What is the specific heat of the block?
Answer:
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To solve this problem, we can use the formula for heat transfer:
q = mcΔT
where q is the heat transferred, m is the mass of the object, c is its specific heat capacity, and ΔT is the change in temperature.
We know that the mass of the block is 0.35 kg and that its initial temperature is -27.5 ºC. We also know that the mass of water is 0.217 kg and that its initial temperature is 25.0 ºC.
When they come to equilibrium at 16.4 ºC, we can calculate how much heat was transferred from the water to the block:
q = mcΔT q = (0.217 kg)(4186 J/kg ºC)(25.0 ºC - 16.4 ºC) q = 1825 J
This amount of heat was transferred from the water to the block, so we can set it equal to the amount of heat absorbed by the block:
q = mcΔT 1825 J = (0.35 kg)c(16.4 ºC - (-27.5 ºC)) 1825 J = (0.35 kg)c(43.9 ºC) c = 148 J/kg ºC
Therefore, the specific heat capacity of the block is 148 J/kg ºC.
Explanation:
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a wave has a frequency of 40 hertz and a wavelength of 2 meters . what is the wave speed ?
Answer:
[tex]80\; {\rm m\cdot s^{-1}}[/tex].
Explanation:
The frequency [tex]f[/tex] of a wave is the number of cycles completed in unit time ([tex]1\; {\rm s}[/tex] in this example.) In this question, [tex]f = 40\; {\rm s^{-1}}[/tex] ([tex]1\; {\rm Hz} = 1\; {\rm s^{-1}}[/tex]) means that the wave would complete [tex]40[/tex] cycles in every [tex]1\; {\rm s}[/tex].
The wavelength [tex]\lambda[/tex] of a wave is the distance the wave travels in each cycle. It is given that [tex]\lambda = 2\; {\rm m}[/tex].
The goal is to find the wave speed, which is the distance that this wave travels in unit time ([tex]1\; {\rm s}[/tex].)
In this question, it is given that [tex]\lambda = 2\; {\rm m}[/tex] and [tex]f = 40\; {\rm s^{-1}}[/tex]. Thus, this wave would travel a total of [tex]40\, (2\; {\rm m}) = 80\; {\rm m}[/tex] for the [tex]40[/tex] cycles completed in each unit time of [tex]1\; {\rm s}[/tex] ([tex]\lambda = 2\; {\rm m}[/tex] for each cycle.) The speed of this wave would be [tex]80\; {\rm m\cdot s^{-1}}[/tex].
Formally, the speed [tex]v[/tex] of this wave can be found by multiplying the wavelength [tex]\lambda[/tex] of this wave by its frequency [tex]f[/tex]:
[tex]\begin{aligned}v &= \lambda\, f \\ &= (2\; {\rm m})\, (40\; {\rm s^{-1}) \\ &= 80\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
The following graph shows the kinetic energy of a roller coaster car as it passes through a loop.
Roller Coaster Car's Kinetic Energy
O B.
Kinetic Energy (kilojoules)
OC.
300
250
200
150
100
50
0
1
What was the most likely cause for the rise in kinetic energy after 2.5 seconds?
A. The roller coaster was pulled with decreasing gravitational force.
The roller coaster began slowing down.
The roller coaster started gaining elevation.
O D. The roller coaster began speeding up.
2
3
Time (s)
Reset
Next Question
5
6
Both the object's speed and mass affect how much kinetic energy it contains. Motional energy is produced while the roller coaster descends. The roller coaster's bottom of the track position is where the most kinetic energy is produced. Kinetic energy changes to potential energy when it starts to rise.
Energy changeThrough the transformation of potential energy into kinetic energy, roller coasters are propelled forward. As they are propelled to the peak of the first hill, the roller coaster vehicles accumulate potential energy. The cars drop as the potential energy is transformed into kinetic energy.Kinetic energy is produced by converting potential energy. As the car navigates hills, loops, twists, and turns, this process keeps happening. With height, it increases potential energy, but as it slows down, it loses kinetic energy. Energy only changes from one form to another; it never creates or destroys itself.For more information on kinetic energy of roller coaster kindly visit to
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30 POINTS!!!! NO CHATGPT OR ANY BOTS_
As you sit in a fishing boat, you notice that 12 waves pass the boat every 45 s
. If the distance from one crest to the next is 9.0 m
, what is the speed of these waves?
Express your answer to two significant figures and include the appropriate units.
The speed of the waves can be expressed to two significant figures as 0.2 m/s. The unit for this expression is meters per second (m/s).
What is wave crest?A wave crest is the highest point of a wave. It is the top of the wave, where the wave is moving most up and away from the equilibrium position. It is the point of highest amplitude (height) of the wave and is followed by a wave trough, which is the lowest point of the wave.
The speed of the waves can be calculated using the formula speed = distance over time.
We know the distance between wave crests is 9.0 m and the time it takes for 12 waves to pass the boat is 45 s. Therefore, the speed of the waves can be calculated as:
Speed = 9.0 m / 45 s
Speed = 0.2 m/s
The speed of the waves can be expressed to two significant figures as 0.2 m/s. The unit for this expression is meters per second (m/s).
This calculation shows that the speed of the waves passing the boat is 0.2 m/s. This speed can be further broken down into how many meters the waves travel in one second if necessary.
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A Carnot Engine operates between two heat reservoirs. The cold reservoir is maintained at 20.0 °C. What temperature must the hot reservoir be at in order for the efficiency of the engine to be 20.0 %?
A heat engine with a 65.0% Carnot efficiency is currently being developed. Between a reservoir that is 25.00C and one that is 3750C, a heat engine is operational.
What is the formula for Carnot efficiency ?The equation is: Carnot efficiency is equal to 1 - Tc/Th, wherein Tc is the cycle's cold end temperature and Th is its hot end temperature. In other words, efficiency is equal to one minus the difference between the hot and cold temperatures.
Explanation: The cold reservoir's temperature is TL=20C=20+273=293K. T L = 20 ∘ C = 20 + 273 = 293 K .
A Carnot cycle running between both of these two reservoirs has a thermal efficiency of = 1 TC/TH. This value exceeds the value of the Otto cycle, which is operating between similar reservoirs by a large margin.
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On the water surface, there are two sources of oscillating waves of the same phase located at A and B, emitting two coherent waves of wavelength λ. Let Δ be the line perpendicular to AB at B. On Δ there are 16 interference maxima, the distance between the two closest and farthest interference maxima is 2.71 cm and 229.55 cm, respectively. . Which of the following is the length of line segment AB closest to?
Two or greater sources are said to be coherent if they emit waves that have the identical wavelength (or frequency) and amplitude and which maintain a steady phase difference.
Do two coherent sources have equal wavelength?If two sources have the identical wavelength, frequency, and segment difference, they are said to be coherent. Therefore, we can conclude that coherent sources have the identical wavelength.
Two microwave coherent factor sources emitting waves of wavelenths λare positioned at 5λdistance apart. The interference is being observed on a flat non-reflecting surface alongside a line passing through on sources ,in a course perpendicular to the line joining the two sources
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https://brainly.com/question/12495315#SPJ1If the sun were more massive, what would happen to Earth’s gravity with the sun?
A. decrease
B. would be infinite
C. would be 0
D. increase
Answer: d. increase
Explanation:
If the sun were more massive, the gravitational force between the sun and Earth would increase. This means that Earth's gravity with the sun would also increase. Therefore, the correct answer is (D) increase.
The gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So, if the mass of one of the objects increases, the gravitational force between them will also increase. In this case, if the mass of the sun were to increase, the gravitational force between the sun and Earth would become stronger, and hence, Earth's gravity with the sun would also increase.
A 0.80kg block of carbon (solid) is dropped into 1.4kg of water. If the carbon starts at -20C, the water starts at 92C, and they have equal final temperatures, what is the final temperature of the system?
The system's final temperature is roughly 16.7°C.
What is a system's final temperature?You may determine your substance's final heat by multiplying the temperature change by the initial temperature. Your water's final temperature would be 24 + 6, or 30 degrees Celsius, for instance, if it started off at 24 degrees Celsius.
The following is the formula for energy conservation:
Q1 + Q2 = 0
Q = mcΔT
Q1 + Q2 = 0
568.8
Simplifying and solving for
6394.4 - 106768 = 0
= 16.7°C
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Would you expect the smallest guitar string to produce waves in the glass of water at a higher or lower frequency?
Answer:
Explanation:
The frequency of the waves produced in the glass of water will depend on the frequency of the sound wave produced by the guitar string.
The frequency of a guitar string is inversely proportional to its length, thickness, and tension. Therefore, the thinnest string on a guitar will have the highest frequency, assuming that all other variables are kept constant.
Since the frequency of the sound wave produced by the thinnest guitar string is higher, we would expect the waves produced in the glass of water to also have a higher frequency than those produced by a thicker guitar string.
A researcher investigated whether job applicants with popular (i.e. common) names are viewed more favorably than equally qualified applicants with less popular (i.e. uncommon) names. Participants in one group read resumes of job applicants with popular (i.e. common) names, while participants in the other group read the same resumes of the same job applicants but with unpopular (i.e. uncommon) names. The results showed that the differences in the evaluations of the applicants by the two groups were not significant at the .001 level
The researcher did not find strong evidence to support the idea that job applicants with popular names are viewed more favorably than equally qualified applicants with less popular names.
What factors plan an important role in the hiring process for a job?It sounds like the researcher conducted an experiment to investigate whether job applicants with popular names are viewed more favorably than equally qualified applicants with less popular names.
Based on the information provided, the researcher found that the differences in the evaluations of the applicants by the two groups were not significant at the .001 level.
The factors that play an important role in the hiring process for a job:
(1) Qualifications and experience: Employers typically look for candidates who possess the necessary qualifications and experience for the job. This includes education, training, certifications, and work experience.
(2) Skills and abilities: Employers also consider a candidate's skills and abilities related to the job. These may include technical, interpersonal, communication, and problem-solving skills.
(3) Personal characteristics: Personal characteristics, such as motivation, work ethic, and adaptability, can also play a role in the hiring process. Employers may look for candidates who demonstrate a positive attitude, a willingness to learn, and the ability to work well with others.
(4) Fit with company culture: Companies may also consider whether a candidate fits with their company culture, values, and mission. This can include factors such as teamwork, creativity, and innovation.
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A 27 g block of ice is cooled to −65 ◦C. It is added to 525 g of water in an 80 g copper calorimeter at a temperature of 25◦C. Find the final temperature. The specific
heat of copper is 387 J/kg ◦C and of ice is 2090 J/kg ◦C . The latent heat of fusion of
water is 3.33 × 105 J/kg and its specific heat is 4186 J/kg ◦C . Answer in units of ◦C.
The final temperature after adding the ice to the water and calorimeter will be approximately 8.37 ◦C.
What is Temperature?
Temperature is a measure of the average kinetic energy of the particles in a substance or system. It is a scalar quantity that indicates how hot or cold an object or medium is. Temperature is commonly measured using various scales, such as Celsius (°C), Fahrenheit (°F), and Kelvin (K), which represent different reference points and units of measurement.
Since energy is conserved, we can set Q_ice equal to Q_water+calorimeter:
m_ice * c_ice * ΔT_ice = (m_water + m_calorimeter) * c_water+calorimeter * ΔT_water+calorimeter
27 g * 2090 J/kg ◦C * (T_f + 65) = (525 g + 80 g) * (4186 J/kg ◦C + 387 J/kg ◦C) * (T_f - 25)
Simplifying and solving for T_f:
27 * 2090 * (T_f + 65) = 605 * (T_f - 25)
56130 T_f + 361350 = 605 T_f - 15125
56130 T_f - 605 T_f = -15125 - 361350
-44,970 T_f = -376475
T_f = (-376475) / (-44,970)
T_f ≈ 8.37 ◦C
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In the diagram, q₁, q2, and q3 are in a straight line.
Each of these particles has a charge of
-2.35 x 10-6 C. Particles q₁ and q2 are separated
by 0.100 m and particles q2 and q3 are separated
by 0.100 m. What is the net force on particle q₁?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
The net force on particle q₁ is 9.22 × 10^-13 N, and it points to the left.
How do we calculate?The net force on particle q₁ due to particles q2 and q3 can be found using Coulomb's law.
Coulomb's law states that the force between two charged particles is given as
F= k * (q₁ * q₂) / r^2
Since q₁ and q₂ have the same charge, the force between them is repulsive, i.e., it points to the left. Using Coulomb's law, we can find the magnitude of this force:
F₁₂ = k * (q₁ * q₂) / r₁₂^2
F₁₂ = (9 × 10^9 Nm^2/C^2) * (-2.35 × 10^-6 C)^2 / (0.100 m)^2
F₁₂ = -4.61 × 10^-13 N
Here, the force between q₁ and q₂ points to the left, and its magnitude is 4.61 × 10^-13 N.
The force between q₂ and q₃ also points to the left, and its magnitude is given as
F₂₃ = k * (q₂ * q₃) / r₂₃^2
F₂₃ = (9 × 10^9 Nm^2/C^2) * (-2.35 × 10^-6 C)^2 / (0.100 m)^2
F₂₃ = -4.61 × 10^-13 N
Here, the force between q₂ and q₃ also points to the left, and its magnitude is 4.61 × 10^-13 N.
F_net = -F₁₂ - F₂₃
F_net = -(-4.61 × 10^-13 N) - (-4.61 × 10^-13 N)
F_net = 9.22 × 10^-13 N
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franchising why is it the best option for you as an entrepreneur
Answer:
ttrockstars
Explanation:
it's math you to be an expert at math thank you
A model rocket blast off and moves upward with an acceleration of 12m/s2 until it reaches a height of 26m, at which point its engine shuts off and it continues its flight in free fall.
a) What is the maximum height attained by the rocket?
b) What is the speed of the rocket just before it hits the ground?
c) What is the total duration of the rocket's flight?
A U-tube is open to the atmosphere at both ends. Water is poured into the tube until the water rises part-way along the straight sides, and then some oil with a density of is poured into one end. This causes the water surface on that side of the tube to go down by and the surface on the other side to go up by the same amount. How much higher is the top surface of the oil on that side of the tube compared with the surface of the water on the other side of the tube?
The top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.
Describe principle of hydrostatics?The principle of hydrostatics, also known as Pascal's principle, states that when an external pressure is applied to a fluid in a container, that pressure is transmitted uniformly in all directions within the fluid, regardless of the shape or volume of the container. In other words, the pressure applied to a confined fluid will be distributed evenly throughout the fluid and will not change in magnitude at any point within the fluid. This principle is important in a number of applications, such as hydraulic systems, which use fluids to transmit force and pressure from one point to another. It is also used to explain how liquids exert pressure on the walls of their container and how objects can float or sink in fluids.
We can use the principles of hydrostatics to solve this problem. Let's call the height difference between the two water surfaces h. We can assume that the oil completely covers the water on one side of the tube and does not mix with it, so the oil and water form two separate liquid columns with a common interface. Let's call the height difference between the oil and water surfaces on the same side of the tube H.
The pressure at any given point in a fluid depends only on the depth of that point below the surface of the fluid and the density of the fluid. Since the two water columns are at the same height, they experience the same pressure from the atmosphere. Similarly, the two oil columns experience the same pressure from the atmosphere.
Now consider a point on the interface between the oil and water on the same side of the tube. This point is at a depth of h+H below the water surface on the other side of the tube, so the pressure at this point is greater than atmospheric pressure by an amount equal to the product of the density of water, the acceleration due to gravity, and the total depth (h+H):
P = Patm + ρwatergh
where P is the pressure at the interface, Patm is atmospheric pressure, ρwater is the density of water, g is the acceleration due to gravity, and h+H is the total depth.
Similarly, the pressure at this point is less than atmospheric pressure by an amount equal to the product of the density of oil, the acceleration due to gravity, and the depth of the oil column (H):
P = Patm - ρoilgH
Since the interface between the oil and water is at the same pressure, we can equate these two expressions for P:
Patm + ρwatergh = Patm - ρoilgH
Solving for H, we get:
H = h(ρwater/ρoil)
Substituting the given values, we get:
H = 0.6h
Therefore, the top surface of the oil on that side of the tube is 0.6 times higher than the surface of the water on the other side of the tube.
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5. Two equal charges are situated in a vacuum 10.0cm apart, if they repel each other with a force of 0.5N, calculate the value of the charge on each. [4π)¹ = 9.0 x 10⁹ I
The value of the charge on each particle is [tex]1.05 x 10^-8 C[/tex].
What is Coulomb's law?Coulomb's law is a fundamental principle of electrostatics that describes the interaction between electric charges. It states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We can use Coulomb's law to solve this problem. Mathematically,
[tex]F = k(q1q2)/r^2[/tex]
where F is the force of attraction or repulsion between the two charged particles,[tex]q1[/tex] and [tex]q2[/tex] are the magnitudes of the charges on the two particles, r is the distance between them, and k is Coulomb's constant, which has a value of [tex]9.0 x 10^9 Nm^2/C^2.[/tex]
In this problem, we know that the charges are equal and the distance between them is 10.0 cm. We also know that the force between them is 0.5 N. Therefore,
[tex]0.5 N = k(q^2)/(0.1 m)^2[/tex]
Solving for q, we get:
[tex]q = \sqrt{[(0.5 N)(0.1 m)^2/k]}[/tex]
[tex]q = \sqrt{(0.5 N)(0.01 m)/(9.0 x 10^9 Nm^2/C^2)}[/tex]
[tex]q = 1.05 x 10^-8 C[/tex]
Therefore, the value of the charge on each particle is [tex]1.05 x 10^-8 C.[/tex]
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Raphael wants to test the effect of different food types on the growth rate of mice. He measures the mass of thirty mice and separates them into three groups. Each group is given a different type of feed. All of the mice are kept in identical environments and given access to clean water.
After three months, Raphael measures the mass of the mice again. The results of Raphael's experiment are shown below.
Food Type Average Growth (g)
oat grains 1.5 g
cereal flakes 0.3 g
sunflower seeds 2.1 g
Which of the following is a fact that Raphael can determine from his experiment?
A.
Mice do not like the taste of cereal flakes.
B.
Sunflower seeds are the best type of food to feed pet mice.
C.
Mice that ate sunflower seeds gained an average of 2.1 grams.
D.
Bigger mice are more desirable as pets than smaller mice.
Mice that ate sunflower seeds gained an average of 2.1 grams that Raphael can determine from his experiment. Each group is given a different type of feed.
What is grams ?Grams (g) is a unit of measurement for mass in the International System of Units (SI). It is the base unit of mass in the SI, and is defined as being equal to the mass of a physical prototype, which is kept at the International Bureau of Weights and Measures. In practical terms, 1 gram is equal to 0.0352739619 ounces, or 0.00220462262 pounds. Grams are often used to measure the weight of food, medicines, and other small objects.
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If the speed of a wave is 400 cm/s with a frequency of 80 Hz, what is the wavelength for this wave?
32,000 cm
32,000 m
5 cm
5m
Within the living area of the colony, what atmospheric gases must be present on Venus?
Humans would need a breathable environment like that on Earth in the living section of a colony on Venus in order to survive. Nitrogen, oxygen, and trace amounts of other gases, such as carbon dioxide, make up the majority of the atmosphere on Earth.
What gases are present in Venus' atmosphere?The clouds are made of sulfuric acid, and the atmosphere is primarily carbon dioxide, the same gas that causes the greenhouse effect on Venus and Earth. And the heated, high-pressure carbon dioxide acts corrosively at the surface.
What gases are found in Mars' and Venus' atmospheres?For instance, compared to Earth, which has 99% nitrogen and oxygen in its atmosphere, Venus and Mars both contain more than 98% carbon dioxide and nitrogen.
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A 25 kg child plays on a swing having support ropes that are 2.20 m long. A friend pulls her back until the ropes are ăÿÿfrom the vertical and releases her from rest. (a) What is the potential energy for the child just as she is released compared with the potential energy at the bottom of the swing? (b) How fast will she be moving at the bottom of the swing? (c) How much work does the tension in the ropes do as the child swings from the initial position to the bottom?
Answer:
A) P.E = 138.44 J
B) The velocity of swing at bottom, v = 3.33 m/s
C) The work done, W = -138.44 J
Explanation:
Given,
The mass of the child, m = 25 Kg
The length of the swing rope, L = 2.2 m
The angle of the swing to the vertical position, ∅ = 42°
A) The potential energy at the initial position ∅ = 42° is given by the relation
P.E = mgh joule
Considering h = 0 for the vertical position
The h at ∅ = 42° is h = L (1 - cos∅)
P.E = mgL (1 - cos∅)
Substituting the given values in the above equation
P.E = 25 x 9.8 x 2.2 (1 - cos42°)
= 138.44 J
The potential energy for the child just as she is released, compared to the potential energy at the bottom of the swing is, P.E = 138.44 J
B) The velocity of the swing at the bottom.
At bottom of the swing the P.E is completely transformed into the K.E
∴ K.E = P.E
1/2 mv² = 138.44
1/2 x 25 x v² 138.44
v² = 11.0752
v = 3.33 m/s
The velocity of the swing at the bottom is, v = 3.33 m/s
C) The work done by the tension in the rope from initial position to the bottom
Tension on string, T = Force acting on the swing, F
=
= - 2.2 x 25 x 9.8 [cos0 - cos 42°]
= - 138.44 J
The negative sign in the in energy is that the work done is towards the gravitational force of attraction.
The work done by the tension in the ropes as the child swings from the initial position to the bottom of the swing, W = - 138.44 J
A large piston in a hydraulic lift has an area of 100 cm2. The force needed to a small piston with an area of 15 cm2 to lift a 1800 kg car is _ kg
The force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).
The hydraulic lift works based on Pascal's principle, which states that the pressure applied to a confined fluid is transmitted equally in all directions throughout the fluid.
Assuming there is no loss of energy due to friction or other factors, the force exerted on the small piston will be equal to the force exerted on the large piston. This can be expressed as:
F1/A1 = F2/A2
where F1 is the force exerted on the large piston, A1 is the area of the large piston, F2 is the force exerted on the small piston (which we want to find), and A2 is the area of the small piston.
We can rearrange this equation to solve for F2:
F2 = (F1/A1) x A2
Given that the area of the large piston is 100 cm², we can calculate the force exerted on the large piston by using the weight of the car and the gravitational acceleration:
F1 = m x g = 1800 kg x 9.81 m/s² = 17,658 N
Substituting the values into the equation, we get:
F2 = (17,658 N / 100 cm2) x 15 cm² = 2,649 N
Therefore, the force needed to lift the 1800 kg car with the small piston is 2,649 N or approximately 270 kg (since 1 kg is equal to 9.81 N).
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How long does it take for radiation from a cesuim-133 atom to complete 1.5 million cycles
A cesium-133 atom's radiation goes through 1.5 million cycles in around 0.1633 microseconds (or 163.3 nanoseconds).
What frequency does one kind of radiation that cesium-133 emits have?9,192,631,770 hertz (cycles per second) is the frequency of the microwave spectral line that the isotope cesium-133 emits. The basic unit of time is provided by this. Cesium clocks have an accuracy and stability of 1 second in 1.4 million years.
The radiation emitted by cesium-133 has a frequency of 9,192,631,770 cycles per second, or 9.192631770 109 Hz.
The following formula may be used to determine how long 1.5 million radiation cycles take to complete:
Time is equal to the frequency of cycles.
Plugging in the numbers, we get:
time = 1.5 million / 9.192631770 × 10^9 Hz
time = 1.632995101 × 10^-7 seconds
So it takes approximately 0.1633 microseconds (or 163.3 nanoseconds) for radiation from a cesium-133 atom to complete 1.5 million cycles.
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The nearest neighboring star to the Sun is about 4 light-years away. If a planet happened to be orbiting this star at an orbital radius equal to that of the Earth-Sun distance, what minimum diameter would an Earth-based telescope's aperture have to be in order to obtain an image that resolved this star-planet system? Assume the light emitted by the star and planet has a wavelength of 550 nm
. The Earth-Sun distance is 149.6×106km
, and 1ly=9.461×1015m
.
To resolve the star-planet system at a distance of 4 light-years, a telescope on Earth would need an aperture with a minimum diameter of 55.88 mm.
What does microscopy's Rayleigh criterion mean?In optical microscopy, the Rayleigh criterion is frequently used to estimate the resolution of the microscope. The resolution limit imposed by this criterion has long been regarded as a roadblock to using an optical microscope to study biological phenomena at the nanoscale.
We can use the Rayleigh criterion,
θ = 1.22 λ / D
θ = angular resolution
λ = wavelength of light
D = diameter of the telescope's aperture
θ = arctan (r / d)
r = radius of the planet's orbit
d = distance to the star
Now, we use the given values,
r = 149.6×106 km = 149.6×109 m
d = 4 × 9.461×1015 m = 3.7844×1016 m
λ = 550 nm = 550×10-9 m
θ = arctan (r / d)
=arctan (149.6×109 / 3.7844×1016) = 0.000012 radians
we can use the Rayleigh criterion,
θ = 1.22 λ / D
D = 1.22 λ / θ
D = 1.22 × 550×10-9 / 0.000012
D = 55.88 mm
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A rock climber stands on top of a 59 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 1.7 m/s . Include value and units.
a) How long after the release of the first stone does the second stone hit the water?
b) What was the initial speed of the second stone?
c) What is the speed of the first stone as it hits the water?
d) What is the speed of the second stone as it hits the water?
a) The time after the release of the first stone that the second stone hits the water is 2.0 s.
b) 15.7 m/s is the initial speed of the second stone.
c) The speed of the first stone as it hits the water is 15.7 m/s.
d) The speed of the second stone as it hits the water is 28.2 m/s.
What is velocity?Velocity is a vector quantity that measures both the speed and direction of an object's motion. It is equal to the rate of change of an object's position with respect to time. Velocity is usually represented by the symbol v and is measured in meters per second (m/s).
a) The time between first and second stone's release is 1.0 s. Since the time of release of first stone and the time of splash of both stones are same, the time between the release of second stone and the splash of both stones is 1.0 s.
Thus, the time after the release of the first stone that the second stone hits the water is 2.0 s.
b) The initial speed of the second stone can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
c) The speed of the first stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (1.7)² + 2(9.8) * 59
v = 15.7 m/s
d) The speed of the second stone as it hits the water can be calculated using the equation of motion,
v² = u² + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration due to gravity (9.8 m/s²), and s is the displacement.
Substituting the values,
v² = (15.7)² + 2(9.8) * 59
v = 28.2 m/s
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A locust jumps at an angle of 55.0° and lands 0.750 m from where it jumped.
Aproximately 1.09 m/s was the locust's first speed.
What role do vectors have in mechanics?In engineering mechanics, vectors are used to express values with both a magnitude and a direction. For analysis, vector representations of a variety of engineering quantities—including forces, displacements, velocities, and accelerations—are required.
Δy = vsin(θ)t - 0.5gt²
0 = v*sin(55°)t - 0.5(-9.81 m/s²)*t²
t = 2vsin(55°)/g
Now, we can use the horizontal motion of the locust to find the initial velocity v. The horizontal distance traveled by the locust is given by:
Δx = v*cos(55°)*t
Substituting the expression for t that we just found:
0.750 m = vcos(55°)2vsin(55°)/g
Solving for v:
v = √(0.750 mg/(2sin(55°)*cos(55°)))
v ≈ 1.09 m/s
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Which are different forms of an element that have different numbers of neutrons?
ions
isotopes
compounds
molecules
.
.
Answer:A
Explanation:
Isotopes are members of a family of an element that all have the same number of protons but different numbers of neutrons. The number of protons in a nucleus determines the element's atomic number on the Periodic Table.
The length of the river span of a bridge is 2799.0 ft. The total length of the bridge is 6998ft. Convert the length of the river span of the bridge to meters.
According to the question the length of the river span of the bridge in meters is 853.3232 m.
What is Length?Length is a physical quantity that measures the distance between two points. It is one of the fundamental units in the International System of Units (SI). It is usually measured in meters, although it can also be measured in other units such as centimeters, kilometers, feet, yards, miles, and so on.
The length of the river span of the bridge is 2799.0 ft. To convert this length to meters, we need to use a conversion factor. There are 0.3048 meters in one foot, so the conversion factor we will use is 1 ft
= 0.3048 m.
To convert 2799.0 ft to meters, we multiply by the conversion factor:
2799.0 ft * 0.3048 m/ft
= 853.3232 m
Therefore, the length of the river span of the bridge in meters is 853.3232 m.
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5. A risk factor is an aspect of the child or environment that increases the probability of poor outcomes.
Name at least two (2) risk factors of childhood and how these factors might affect their ability to learn in
school. (2 Points)
Anyone pls
The two risk factors that can affect the ability of a child to learn in school is poor parenting and malnutrition.
What is a risk factor?A risk factor can be defined as any predisposing factor that can expose an individual to harm.
A risk factor that affects a child is an aspect of the child or environment that increases the probability of poor outcomes.
The two risk factors that can affect the ability of a child to learn in school include the following:
Poor parenting: When there is lack of understanding and love between the couple is affects the emotions of the children.Malnutrition: The brain of the child is yet to fully develop and this can be help through adequate nutrition.Learn more about nutrition here:
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1. A kid jumps straight up at 7.20 m/s. How long will he remain in the air?
The time takes the kid to remain in the air is 0.735 s.
What is time?Time is the duration of an events. The s.i unit of time is seconds.
To calculate how long the kid will be in the air, we use the formula below.
Formula:
t = (v-u)/g.................................... Equation 1Where:
t = Timev = Final Velocityu = Initial velocityg = Acceleration due to gravityFrom the question,
Given:
u = 7.20 m/sv = 0 m/sg = -9.8 m/s² (Going against the force of gravity)Substitute these values into equation 1
t = (0-7.20)/-9.8t = -7.20/-9.80t = 0.735 secondsHence, the time is 0.735 s.
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For every baryon in the Universe, there are about 109 photons. The ratio of photons to baryons has been
constant since a few seconds after the big bang. This is a crucial number that sets the stage for much of
the future evolution of the Universe. If the number were just a little different, the Universe would be a
very different place, and life could possibly not exist. In this question we will use the photon-to-baryon
ratio to work out the redshift at which the Universe becomes dominated by matter, instead of by
radiation.
Assume that most of the photons in the present Universe are cosmic microwave radiation photons that
are a relic of the big bang. (It turns out that this is not a bad assumption). For simplicity, also assume
that all the photons have the energy corresponding to the wavelength of the peak of a 2.73K black-body
radiation curve. At approximately what redshift will the energy density in radiation be equal to the
energy density in matter?
The Universe became dominated by matter instead of radiation at a redshift of around 3300.
To determine at what redshift the Universe became dominated by matter, we need to find the redshift at which the energy density of matter becomes equal to the energy density of radiation.
Let's start with the energy density of radiation, which can be calculated using the Stefan-Boltzmann law:
$[tex]u_{rad} = \frac{4\sigma}{c}T^4$[/tex]
where $\sigma$ is the Stefan-Boltzmann constant, $c$ is the speed of light, and $T$ is the temperature of the radiation. Since we are assuming that the cosmic microwave radiation is a black-body radiation, we can use the temperature of 2.73 K, which corresponds to the peak of the radiation curve:
[tex]$u_{rad} = \frac{4\sigma}{c}(2.73K)^4 \approx 0.261 \text{ eV/cm}^3$[/tex]
Next, let's calculate the energy density of matter. We know that the number density of baryons is [tex]$n_b \approx \frac{1}{10^9}n_{\gamma}$, where $n_{\gamma}$[/tex] is the number density of photons. Since we are assuming that the photon-to-baryon ratio is constant, we can write:
[tex]$\frac{\rho_b}{\rho_{\gamma}} = \frac{m_b n_b}{\frac{4}{3}\sigma T^4} = \frac{3m_b}{4\sigma T^3 n_{\gamma}} \approx \frac{3m_b}{4\sigma T^3}\frac{1}{n_{\gamma}}$[/tex]
where $m_b$ is the mass of a baryon. Substituting the values, we get:
[tex]$\frac{\rho_b}{\rho_{\gamma}} \approx 4.15 \times 10^{-10}$[/tex]
Since the total energy density of the Universe is given by:
[tex]$\rho_{tot} = \rho_b + \rho_{\gamma}$[/tex]
we can write:
[tex]$\frac{\rho_b}{\rho_{tot}} = \frac{\rho_b}{\rho_b + \rho_{\gamma}} \approx \frac{\rho_b}{\rho_{\gamma}} = 4.15 \times 10^{-10}$[/tex]
At the redshift $z$, the energy density of radiation will be diluted by a factor of $[tex](1+z)^4[/tex]$, while the energy density of matter will be diluted by a factor of $[tex](1+z)^3[/tex]$. Thus, at some redshift $z$, we will have:
$ [tex]\frac{\rho_b}{\rho_{tot}} = \frac{\rho_b}{\rho_b + \rho_{\gamma}} = \frac{1}{1+z}\frac{3m_b}{4\sigma T^3 n_{\gamma}}[/tex] $
Setting this equal to the value we calculated above, we can solve for $z$:
$ [tex]\frac{1}{1+z}\frac{3m_b}{4\sigma T^3 n_{\gamma}} \approx 4.15 \times 10^{-10}[/tex] $
$ [tex]1+z \approx \frac{3m_b}{4\sigma T^3 n_{\gamma}}\frac{1}{4.15 \times 10^{-10}}[/tex] $
$ [tex]z \approx 3300[/tex] $
Therefore, the Universe became dominated by matter instead of radiation at a redshift of around 3300.
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