Answer:
B
Explanation:
Answer:
Its B)Cryosphere and hydrosphere! I took the test and got that question correct!
Explanation:
A bar of gold has a temperature of 22 Celsius, and a bar of aluminum has a temperature of 27 Celsius. Which statement explains why the bar of aluminum as a higher temperature
Answer:Type of metal
Explanation:Basically certain types of metal may heat and get cold faster maybe the alumm may collect more sun and heat more and stay that way
Due to the higher thermal conductivity of aluminium, it has higher temperature than gold.
What is meant by thermal conductivity ?The ability of a material to conduct or transport heat is referred to as thermal conductivity.
Here,
It is not the solid itself that moves in bulk as a result of thermal conductivity; rather, it is molecular agitation and contact.
A region of high temperature and high molecular energy is moved towards a region of lower temperature and lower molecular energy via a temperature gradient.
The density of aluminium is small. It is easily castable, machinable, and formable, and it has a high thermal conductivity and great corrosion resistance.
At 20 °C, gold has an electrical resistivity of 0.022 micro-ohm m and a thermal conductivity of 310 W m-1 K-1.
Aluminium conducts more heat than gold.
Hence,
Due to the higher thermal conductivity of aluminium, it has higher temperature than gold.
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which one is not a derived unit?
1. Hertz
2. mol
3. Watt
4. Newton
Answer:
mol
Explanation:
If a car travels 600m west in 25 seconds, what is its velocity?
Answer:
The answer is 24 m/sExplanation:
The velocity of the car can be found by using the formula
[tex]v = \frac{d}{t} \\ [/tex]
d is the distance
t is the time taken
From the question we have
[tex]v = \frac{600}{25} \\ [/tex]
We have the final answer as
24 m/sHope this helps you
Do 2. A cyclist starts from rest and accelerates along a straight path to a speed of
12.15 m/s in a time of 4.5 seconds.
What is the cyclist's acceleration to the nearest tenth?
O
54.7 m/s2
o
24.3 m/s?
2.7 m/s2
De
3.4 m/s?
The cyclist's acceleration to the nearest tenth is: C. 2.7 [tex]m/s^2[/tex].
Given the following data:
Initial velocity = 0 m/s (since the cyclist starts from rest).Final velocity = 12.5 m/s.Time = 4.5 seconds.To find the cyclist's acceleration to the nearest tenth:
Acceleration is calculated by subtracting the initial velocity from the final velocity and dividing by the time.
Mathematically, acceleration is given by the formula;
[tex]A = \frac{V - U}{t}[/tex]
Where:
A is the acceleration.V is the final velocity.U is the initial velocity.t is the time measured in seconds.Substituting the given parameters into the formula, we have;
[tex]A = \frac{12.5\; - \;0}{4.5} \\\\A = \frac{12.5}{4.5}[/tex]
Acceleration, A = 2.7 [tex]m/s^2[/tex]
Therefore, the cyclist's acceleration to the nearest tenth is 2.7 [tex]m/s^2[/tex].
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I WILL MARK YOU AS BRAINLIEST IF RIGHT
A 1250 kg car was traveling at 10 m/s toward a traffic signal when the light suddenly turned red. The driver slammed on his brakes, stopping the car in 3.3 seconds. What was the acceleration of the car ?
Answer:
a = - 3 m/s^2 (negative 3 meters per second squared)
Explanation:
We use the formula for acceleration as the change in velocities (from initial velocity "vi" to final velocity "vf") divided the time "t" it took. In our case:
a = (vf - vi) / t = (0 - 10) / 3.3 = -10 /3.33 = -3 m/s^2
what is the initial position
Answer:
Initial position of a body is the position of the body before accelerating or increasing its velocity the position changes and then that position is the final position.
hope it is helpful...
A 10-cm-thick aluminum plate (α = 97.1 × 10−6 m2/s) is being heated in liquid with temperature of 550°C. The aluminum plate has a uniform initial temperature of 25°C. If the surface temperature of the aluminum plate is approximately the liquid temperature, determine the temperature at the center plane of the aluminum plate after 15 s of heating. Solve this problem using the analytical one-term approximation method. The temperature at the center plane after 15 s of heating is
Answer:
356°C.
Explanation:
(1). The first step to the solution to this particular Question/problem is to determine the Biot number, and after that to check the equivalent value of the Biot number with plate constants.
That is, Biot number = (length × ∞)÷ thermal conductivity. Which gives us the answer as ∞. Therefore, the equivalent value of the ∞ on the plates constant = 1.2732 for A and 1.5708 for λ.
(2). The next thing to do is to determine the fourier number.
fourier number = [α = 97.1 × 10−6 m2/s × 15 s] ÷ (.05m)^2 = 0.5826.
(3). The next thing is to determine the temperature at the center plane after 15 s of heating.
The temperature at the center plane after 15 s of heating = 500°C [ 25°C - 500°C ] [1.2732] × e^(-1.5708)^2 ( 0.5826).
The temperature at the center plane after 15 s of heating = 356°C.
By applying the third law of motion, which force is greater when you push on a wall? Your force, the wall’s force, your force and the wall’s force are equal or the force saved before pushing the wall.
Hello!
By applying the third law of motion, your force and the wall's force are equal.
I hope this helps you! Have a great day!
- Mal
Answer:
they are equal
Explanation:
since it's an equal and opposite force exerted on both bodies, obviously there's no force greater than the other. however, perhaps you could say that since your mass is smaller, the impact of the force exerted on you is greater than that exerted on the wall since F=ma.
I forgot how to breath
Answer:
1-steal an inhaler 2-use it 3-your good
orrrr
1-swallow air(preferably air from space)
Answer:
then breath
Explanation:
A scientist just learned that she will not receive enough money to complete her year-long study about the
relationship between certain diseases and the foods that people eat.
How can she best overcome this limitation?
conduct smaller studies for more than a one-year period
study only a very small group of people
conduct a study about something else
use data from a similar study and adjust it to fit her study
Please answer I need help
When faced with limited funding to complete a year-long study on the relationship between certain diseases and people's diets she can best overcome this by conduct smaller studies for more than a one-year period, study only a very small group of people, conduct a study about something else, use data from a similar study and adjust it to fit her study.
The correct answer would be all of the above.
There are several strategies the scientist can consider to overcome this limitation. Each option has its own advantages and potential drawbacks, so the scientist should carefully evaluate which approach aligns best with her research goals and available resources.
1. Conduct smaller studies for more than a one-year period: Instead of one large-scale study, the scientist can break down the research into smaller, more manageable studies. This approach allows for incremental progress, and findings from each smaller study can contribute to the overall understanding of the topic. By conducting multiple studies over an extended period, the scientist can still gather valuable data and draw meaningful conclusions.
2. Study only a very small group of people: Focusing on a small group of participants can reduce costs and streamline data collection and analysis. While the sample size may be limited, the scientist can still gain insights into the relationship between diseases and diet within this specific group. However, generalizing the findings to a larger population may be challenging due to the limited sample size.
3. Conduct a study about something else: If funding limitations prevent the scientist from conducting the intended study, she could consider redirecting her research efforts towards a related but more feasible topic. This allows her to leverage her expertise and resources while still generating valuable scientific knowledge.
4. Use data from a similar study and adjust it to fit her study: The scientist could explore existing datasets or previous studies that are relevant to her research question. By analyzing and adapting this data to fit her study's context, she can gain insights without incurring the costs and time associated with primary data collection. However, it is crucial to ensure that the adjusted data aligns with the specific objectives and parameters of her study.
Ultimately, the scientist should carefully assess the feasibility, potential impact, and trade-offs associated with each option. It may also be beneficial to seek guidance from peers, mentors, or funding agencies to explore alternative funding sources or collaborative opportunities that could support her research goals.
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1- Write Gauss’ Law.
2- Consider an electric field line passing through a closed surface. What is the sign of the flux if:
a) The line passes from inside to outside?
b)The line passes from outside to inside?
Think of a conductor as having both positive and negative mobile charges. Consider a conductor in electrostatic equilibrium – that is, whose charges are stationary.
3- Given that all of the charges are stationary, what must the E-field be at any point inside of the conductor?
1)The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through an area is defined as the electric field multiplied by the area of the surface projected in a plane perpendicular to the field.
Find the required angular speed, ω, of an ultracentrifuge for the radial acceleration of a point 2.10 cm from the axis to equal 5.00×105 g (where g is the free-fall acceleration)
Answer:
ω = 15275.25 rad/s
Explanation:
Given that,
Radial acceleration of an ultracentrifuge is, [tex]a=5\times 10^5g[/tex]
Distance from the axis, r = 2.1 cm = 0.021 m
g is the free-fall acceleration such that g = 9.8 m/s²
We need to find the angular speed of an ultracentrifuge. The formula that is used to find the angular speed is given by formula as follows :
[tex]a=r\omega^2[/tex]
Putting all the values,
[tex]\omega=\sqrt{\dfrac{a}{r}} \\\\\omega=\sqrt{\dfrac{5\times 10^5\times 9.8}{0.021}} \\\\\omega=15275.25\ rad/s[/tex]
So, the required angular speed, ω, of an ultracentrifuge is 15275.25 rad/s.
The coefficient of friction for an object sliding across a rough floor is 0.7. A 125 N horizontal force causes the object to accelerate at 1.2 m/s2 . What is the objects weight
Answer:
152 N
Explanation:
From the question,
F-F' = ma................. Equation 1
Where F = Horizontal force, F' = Frictional force, m = mass of the object, a = accleration of the object.
But,
F' = mgμ............... Equation 2
Where g = acceleration due to gravity, μ = coefficient of friction.
make m the subject of the equation
Substitute equation 2 into equation 1
F-mgμ = ma.
make m the subject of the equation
m = F/(gμ+a).............. Equation 3
Given: F = 125 N, g = 9.8 m/s², μ = 0.7, a = 1.2 m/s²
Substitute these value into equation 3
m = 125/[(9.8×0.7)+1.2)
m = 125/8.06
m = 15.51 kg.
But,
W = mg.............. Equation 4
Where W = weight of the object.
W = 15.51(9.8)
W = 152 N
Two dogs are running a race of 100m. The green dog runs the 100m in 10s. The
red dog runs the first half of the race in 3s and then turns around and runs back
to the starting line in 5s. Describe what the velocity time graph would look like of
the race.
Answer: What type of dog is green and red?
Explanation:
Which electromagnetic wave is used in dental scans?
Answer:
X-rays are most often used to examine bones and teeth.
Hope this helps!
I don't quite understand. Can you help please?
Will give brainliest!!!!! Help asap
Answer:
Correct choice: 5,400 Km/h
Explanation:
Unit Conversions
We'll use the following conversions:
1 Km = 1,000 m
1 h = 3,600 s
Convert the speed of sound in water (1.5\cdot 10^3\ m/s) to Km/h:
[tex]\displaystyle 1.5\cdot 10^3\ m/s*\frac{ 3,600\ s/h}{ 1000 \ Km/m}[/tex]
[tex]=5,400\ Km/h[/tex]
Correct choice: 5,400 Km/h
A 4kg object is at rest. How much force is required to get the object to a velocity of 20m/s in 2 seconds? (Show work and include units)
Explanation:
Answer:-The Object was at rest. So, Initial Velocity is Zero.
[Initial Velocity]u = 0[Final Velocity]v = 20 m/s[Time]t = 2 seconds Mass = 4kgForce = ?We know that:-
[tex]\sf{Force = Mass \times \dfrac{(v-u)}{t}}[/tex]
Applying it, we get:-
[tex]\sf{Force = 4 \times \dfrac{(20-0)}{2}}[/tex]
[tex]\sf{Force = 2 \times \dfrac{(20)}{2}}[/tex]
[tex]\sf{Force = 2 \times 10}[/tex]
[tex]\sf{Force = 20 \ N \ (Newton)}[/tex]
Hope it helps :)
explain what happent to the pressure exerted by an object when the area over which it is exerted:
a) increase
b) decrease
A device for acclimating military pilots to the high accelerations they must experience consists of a horizontal beam that rotates horizontally about one end while the pilot is seated at the other end. In order to achieve a radial acceleration of 29.9 m/s2 with a beam of length 5.33 m , what rotation frequency is required
Answer:
Rotation frequency is 0.377 hertz.
Explanation:
After a careful reading of statement, we need to apply the concept of radial acceleration due to uniform circular motion, whose formula is:
[tex]a_{r} = \omega^{2}\cdot L[/tex] (Eq. 1)
Where:
[tex]a_{r}[/tex] - Radial acceleration, measured in meters per square second.
[tex]\omega[/tex] - Angular velocity, measured in radians per second.
[tex]L[/tex] - Length of the beam, measured in meters.
Now we clear the angular velocity within:
[tex]\omega = \sqrt{\frac{a_{r}}{L} }[/tex]
If [tex]a_{r} = 29.9\,\frac{m}{s^{2}}[/tex] and [tex]L = 5.33\,m[/tex], the angular velocity is:
[tex]\omega = \sqrt{\frac{29.9\,\frac{m}{s^{2}} }{5.33\,m} }[/tex]
[tex]\omega \approx 2.368\,\frac{rad}{s}[/tex]
The frequency is the number of revolutions done by device per second and can be found by using this expression:
[tex]f = \frac{\omega}{2\pi}[/tex] (Eq. 2)
Where [tex]f[/tex] is the frequency, measured in hertz.
If we know that [tex]\omega \approx 2.368\,\frac{rad}{s}[/tex], then rotation frequency is:
[tex]f = \frac{2.368\,\frac{rad}{s} }{2\pi}[/tex]
[tex]f = 0.377\,hz[/tex]
Rotation frequency is 0.377 hertz.
In 2.5 s, a car increases its speed from 60 km/h to 65 km/h while a bicycle goes from rest to 5 km/h. Which undergoes the
greater acceleration?
What is the acceleration of each? (Don't forget your units when reporting answers).
Answer:
Same, 2 km/h/s
Explanation:
Acceleration is change in velocity over time.
a = Δv / Δt
The car's acceleration is:
a = (65 km/h − 60 km/h) / 2.5 s
a = 2 km/h/s
The bicycle's acceleration is:
a = (5 km/h − 0 km/h) / 2.5 s
a = 2 km/h/s
.a stalled car is being pushed up a hill at constant velocity by three people. the net force on the car is ?
Answer:
I think its 3 because each person would provide 1 net force the information is very vague here sorry wish I could help more
A sports car of mass 1.00x103 kg can accelerate from rest to 30.0 m/s in 7.00 s. What is the average forward force on the car?
Answer:
Explanation:
Before you can find the force, you need to find the acceleration.
Givens
vi = 0
vf = 30 m/s
t = 7 seconds
Formula
a = (vf - vi) / t
Solution
a = (30 - 0)/7
a = 4.28 m/s
Now you can look at the Force
F = m * a
F = 1.00*10^3 * 4.28
F = 4.28 * 10^3 N
Which of the following graphs shows the motion of an object that starts to travel forward, stops for several seconds, then returns to its original position?
Answer:
Pls can you send the following graphs
Convert 56,340,040 meters into scientific notation?
Answer:
5.634004 × 10^7
Explanation:
the number 7 is for the numbers before the number 5
derive ideal gas equation for n mole of gas.
Answer:
Explanation:
Ideal gas equation- The volume (V) occupied by the n moles of any gas has pressure(P) and temperature (T) Kelvin,
the relationship for these variables PV=nRT where R gas constant is called the ideal gas law
Derivation of the Ideal Gas Equation
Let us consider the pressure exerted by the gas to be ‘p,’
The volume of the gas be – ‘v’
Temperature be – T
n – be the number of moles of gas
Universal gas constant – R
According to Boyle’s Law,
it constant n & T, the volume bears an inverse relation with the pressure exerted by a gas.
i.e. v∝1p ………………………………(i)
According to Charles’ Law,
When p & n are constant, the volume of a gas bears a direct relation with the Temperature.
i.e. v∝T ………………………………(ii)
According to Avogadro’s Law,
When p & T are constant, then the volume of a gas bears a direct relation with the number of moles of gas.
i.e. v∝n ………………………………(iii)
Combining all the three equations, we have-
v∝nTp
or pv=nRT
where R is the Universal gas constant, which has a value of 8.314 J/mol-K
Tension force ........... throughout a string that changes direction over a pulley.
(Fill in the blank)
Answer:
"is constant"
If such a particle is moving, with respect to the laboratory, with a speed of 0.950 c , what average lifetime is measured in the laboratory?
Complete Question
The positive [tex]muon (^+)[/tex], an unstable particle, lives on average [tex]2.20 * 10^{-6}\ s[/tex] (measured in its own frame of reference) before decaying.
If such a particle is moving, with respect to the laboratory, with a speed of 0.950 c , what average lifetime is measured in the laboratory?
Answer:
The value is [tex]\Delat t = 7.046 *10^{-6} \ s[/tex]
Explanation:
From the question we are told that
The the average live time of [tex]muon (^+)[/tex] is [tex]\Delta t_o = 2.20 *10^{-6} \ s[/tex]
The speed of of [tex]muon (^+)[/tex] in the laboratory is [tex]v = 0.950 c[/tex]
Generally the average life time of the positive [tex]muon (^+)[/tex] measured in the laboratory is mathematically represented as
[tex]\Delat t = \frac{\Delta t_o }{ \sqrt{1 - \frac{v^2}{c^2} } }[/tex]
[tex]\Delat t = \frac{2.20 *10^{-6}}{ \sqrt{1 - \frac{(0.950 c)^2}{c^2} } }[/tex]
[tex]\Delat t = \frac{2.20 *10^{-6}}{ \sqrt{1 - \frac{0.9025 c^2}{c^2} } }[/tex]
[tex]\Delat t = \frac{2.20 *10^{-6}}{ \sqrt{1 - 0.9025 } }[/tex]
[tex]\Delat t = \frac{2.20 *10^{-6}}{ \sqrt{ 0.0975 } }[/tex]
[tex]\Delat t = 7.046 *10^{-6} \ s[/tex]
From the concepts you have learned in this module, how are you going to assess
the Earth's condition to support life? Explain your answer.
1. Temperature
2. Water
3. Atmosphere
4. Energy
5. Nutrients
Answer:
Find the explanation below.
Explanation:
Earth is properly designed to support life. This is seen in the favorable temperature that supports life, the water cycle that recycles water for plant and animal life, the atmosphere, energy, and nutrients.
1. Temperature: The temperature which is regulated by the different weather conditions such as the rains, snows, dry seasons all help to maintain a stable condition for life.
2. Water: The water cycle through processes like evaporation, condensation, precipitation, helps to ensure that there is never a lack of water in the earth. The numerous water bodies like the seas, oceans, rivers, lakes, also provide a habitat for some living things. Water makes up 70% of the earth.
3. Atmosphere: The atmosphere is a mixture of gases in the right proportions that are necessary for life. Oxygen, Nitrogen, Carbon, etc are released and inhaled by man and other living things. They are also involved in so many biochemical reactions that help in metabolism and catabolism.
4. Energy: Energy generated from the sun and within the earth is stored in various forms and is always conserved. This energy is converted to different states such as the potential, chemical, kinetic, mechanical forms to get work done and to release heat.
5. Nutrients: Though cycles such as the carbon, nitrogen, oxygen, and phosphorous cycles, the earth maintains its stock of essential nutrients that help to sustain life.
An interdisciplinary approach encompassing climatology, oceanography, environmental science, and other fields of study is necessary to evaluate the Earth's capacity to support life.
Temperature: Monitoring and analyzing climate data from numerous sources, including weather stations, satellites, and ocean buoys, is necessary to determine the Earth's temperature. To understand how temperature patterns vary over time, scientists look at long-term trends, seasonal variations, and severe events. They forecast future temperature increases and their possible effects on life and ecosystems using global climate models.
Water: Monitoring freshwater availability, water quality, and water distribution throughout various regions are all part of the assessment of Earth's water resources. Studies of precipitation patterns, data on ice melting from polar regions, and measurements of water levels in lakes, rivers, and aquifers are all conducted by researchers. Testing for toxins, pollutants, and chemical compositions is part of evaluating water quality to make sure it adheres to acceptable standards for both ecological and human health.
Atmosphere: scientists measure and research a number of factors, such as greenhouse gases, air quality, and atmospheric pressure, in order to evaluate the Earth's atmosphere. Carbon dioxide (CO2), methane (CH4), and other greenhouse gases are measured at monitoring sites throughout the globe to better understand how they contribute to climate change. Pollutants like particle matter and ozone, which have an influence on both human health and ecosystems, are measured by air quality monitoring stations.
Energy: studying diverse energy sources and their effects on the environment and ecosystems is necessary to evaluate the amount of energy present on Earth. Scientists assess the usage of non-renewable energy sources like fossil fuels as well as renewable energy sources like solar, wind, hydro, and geothermal energy. To create sustainable energy plans that support life on Earth, they examine energy consumption trends, carbon emissions, and energy efficiency.
Nutrients: studying nutrient cycles and availability in soils, oceans, and terrestrial ecosystems is necessary for evaluating the availability of nutrients in the Earth's ecosystems. To determine the nutrient levels for agriculture and plant growth, researchers examine soil samples. In order to gauge the productivity and availability of nutrients for marine life, they also research marine ecosystems.
Hence, an interdisciplinary approach encompassing climatology, oceanography, environmental science, and other fields of study is necessary to evaluate the Earth's capacity to support life.
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A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled? A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if the slit width is doubled? It will be cut in half. It will double. It will become eight times as large. It will be cut to one-quarter its original size. It will become four times as large.
Answer:
It will be cut in half
Explanation:
The diffraction of a slit is given by the formula
a sin θ = m where
a = width of the slit,
λ = wavelength and
m = integer that determines the order of diffraction.
Next we divide both sides by a, we have
sin θ = m λ / a
Also, recall that
a’ = 2 a
Then we substitute in the previous equation
2asin θ' = m λ, if divide by 2a, we have
sin θ' = (m λ / 2a).
Now again, from the first equation, we said that sin θ = m λ / a, so we substitute
sin θ ’= sin θ / 2
Then we use trigonometry to find the width, we say
tan θ = y / L
Since the angle is small, we then have
tan θ = sin θ / cos θ
tan θ = sin θ, this then means that
sin θ = y / L
we will then substitute
y’ / L = y/L 1/2
y' = y / 2
this means that when the slit width is doubled the pattern width will then be halved