Answer:
60 kg
Explanation:
An astronaut weighs 588 N on the earth
He also weighs 98 N on the moon
Therefore the the mass of the astronaut on the moon can be calculated as follows
= 588/98
= 6 × 10
= 60 kg
Hence the mass of the astronaut on the moon is 60 kg
The mass of the astronaut on the moon will be equal to 60 kg whereas the gravitational acceleration on the moon is 1.63 [tex]\rm m/s^2[/tex].
Given information:
An astronaut weighing 588 N on the earth notices that he weighs only 98 N on the moon.
The mass of any object in the whole universe doesn't change but the gravitational force or weight can change. This is because the gravitational acceleration can change.
The mass of the object in the earth or anywhere will be,
[tex]w=mg\\588=m\times 9.8\\m=60\rm \; kg[/tex]
So, the astronaut has 60 kg mass which is constant.
The gravitational acceleration on the moon will be,
[tex]98=60a\\a=1.63\rm\; m/s^2[/tex]
Therefore, the mass of the astronaut on the moon will be equal to 60 kg whereas the gravitational acceleration on the moon is 1.63 [tex]\rm m/s^2[/tex].
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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...
explain what happent to the pressure exerted by an object when the area over which it is exerted:
a) increase
b) decrease
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
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]
I don't quite understand. Can you help please?
-5 1/3 - 2 1/3 in math
Answer:
3
Explanation:
5 1/3 - 2 1/3 = 3
Broken down: 1/3 - 1/3 = 0 5 - 2 = 2
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
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
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.
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
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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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
Tension force ........... throughout a string that changes direction over a pulley.
(Fill in the blank)
Answer:
"is constant"
Which two types of waves require matter in order to travel A:light waves B:sound waves C:electromagnetic waves D:Water waves
Answer:
Sound Waves
Light Waves
Explanation:
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
Which electromagnetic wave is used in dental scans?
Answer:
X-rays are most often used to examine bones and teeth.
Hope this helps!
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.
what is 25.61 cm to Gm?
Explanation:
25.61 cm * 1 Gm
100000000000 cm = 2.561e-10 Gm
1 Gm = 10^9 m =10^11 cm
then 10^11 cm = 1Gm
25.61 cm = 25.61 /10^11
=25.61 * 10^-11 Gm
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.
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
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:
PHYSICS HELP URGENT!!! While standing at the edge of the roof of a building, a man throws a stone upward with an initial speed of 7.59 m/s. The stone subsequently falls to the ground, which is 18.9 m below the point where the stone leaves his hand. At what speed does the stone impact the ground? Ignore air resistance and use =9.81 m/s2 for the acceleration due to gravity. I need help finding out the impact speed, please.
Answer:
v = 20.69 m/s
Explanation:
Given that,
Initial speed of a stone, u = 7.59 m/s
he stone subsequently falls to the ground, which is 18.9 m below the point where the stone leaves his hand, h = 18.9 m
We need to find the speed of the stone impact the ground. Let the speed be v. Using the equation of kinematics to find v. So,
[tex]v^2-u^2=2as[/tex]
Here, a = g
So,
[tex]v^2=2gs+u^2\\\\v=\sqrt{2gs+u^2} \\\\v=\sqrt{2\times 9.81\times 18.9+(7.59)^2} \\\\v=20.69\ m/s[/tex]
So, the speed of the stone impact the ground is 20.69 m/s.
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
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.
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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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
which one is not a derived unit?
1. Hertz
2. mol
3. Watt
4. Newton
Answer:
mol
Explanation:
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
You do 32 joules of work using a pair of scissors. The scissors do 25 joules of
work cutting a piece of fabric. What is the efficiency of the scissors?
Answer:
Efficiency = 65%
Explanation:
The formula of Efficiency applied to any circumstance is:
Efficiency = Useful Energy / Energy applied
Then replacing the values given its:
Efficiency = 25 J / 32 J
Efficiency = 0.65
0.65 written as percentage is 65%, then:
Efficiency = 65%
As you do 32 joules of work using a pair of scissors and the scissors do 25 joules of work cutting a piece of fabric, the efficiency of the scissors is 78.125%.
What is efficiency?
Efficiency is the proportion of work done by a machine or throughout a process to the overall amount of energy or heat used.
The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.
Mathematically,
efficiency of a machine = (work output/work input)×100%
Given parameters:
Input work to the pair of scissors= 32 joules.
Output work from the pair of scissors= 25 joules.
Hence,
The efficiency of a machine = (work output/work input)×100%
= ( 25 joule/32joule)×100%
= 78.125%
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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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