Answer:
Note that:
Particles in a:
gas are well separated with no regular arrangement.
liquid are close together with no regular arrangement.
solid are tightly packed, usually in a regular pattern.
Explanation:
10 points
Calculate the force of attraction between the moon and the earth, if their masses and distance apart are 10^22Kg, 10^24kg and 6.4x10^6m. Take g = 6.67x10^-11Nm^2Kg^-2
a) 0.63x10^24N
b) 1.63x10^22N
c) 2.63x10^20N
d) 3.63x10^18N
Answer:
Explanation:
You didn't fill in the proper masses which is why you never got an answer to this. But that's ok...I got you. I happen to know what they are! We will use the universal law of gravitation and the gravitational constant to solve this.
[tex]F_g=\frac{Gm_1m_2}{r^2}[/tex] and filling in:
[tex]F_g=\frac{(6.67*10^{-11})(5.98*10^{24})(7.36*10^{22})}{(3.84*10^8)^2}[/tex] The denominator is the radius of the earth plus the radius of the moon plus the distance between their surfaces, just FYI.
That gives us that
[tex]F_g=1.99*10^{20}N[/tex] Not sure what your choices entail, but I'd have to say, taking into consideration that maybe your problem didn't figure in the distance between the surfaces, you'd be at choice B.
There is a bell at the top of a tower that is 45m high. The bell weighs 190N. The bell has ___________________ energy. Calculate it.
I NEED THE ANSWER PLEASE
Answer:
250mn
Explanation:
[tex]what \: is \: matter \: \: \: {?} [/tex]
Answer:
matter is any substance that has mass and takes up space by having volume.
Explanation:
Matter is anything that has a mass and it takes up space or has a volume .
matters are made up many atoms or molecules .
hope it is helpful to you
stay safe, happy and healthy
a bell rings at a frequency of 75hz on a warm 25 degree evening. calculate the...
a. speed of the sound wave at this temperature.
b. wavelength of the sound wave
Answer:
Explanation:
We need 2 different equations for this problem: first the velocity of sound equation, then the frequency of the sound equation.
The velocity of sound is found in:
v = 331.5 + .606T
We need to find that first in order to fill it into the frequency equation which is
[tex]f=\frac{v}{\lambda}[/tex] where v is the velocity we will find the part a, f is frequency and lambda is the wavelength. Starting with the velocity of the sound:
v = 331.5 + .606(25) and
v = 331.5 + 15 and rounding correctly using the rules for sig fig when adding:
v = 347 m/s
Filling that into the frequency equation:
[tex]75=\frac{347}{\lambda}[/tex] and
[tex]\lambda=\frac{347}{75}[/tex] so
[tex]\lambda=4.6m[/tex]
As Courtney switches on the TV set to watch her favorite cartoon, the electron beam in the TV tube is steered across the screen by the field between two charged plates. If the electron experiences a force of 3.0 * 10^6 N, how large is the field between the deflection plates?
Answer:
Explanation:
Force= (q1q2)/(4/\Ęr2)
3×10^6= (1.602×10^-19)^2/(r^2)
r^2=(2.27×10^-33)/(3×10^6)
r^2=8.55×10^-45
r= 9.25×10^-23
If you were testing materials to see whether they conduct electricity using a circuit and bulbs, what variable would you be changing?
A. The strength of the battery
B. The number of bulbs
C. The material
D. The time it takes the electricity to complete the circuit
Answer: C. The material
Explanation:
In order to check whether a material conducts electricity, the first thing to do would be to design a circuit. Connect a positive wire to the batteries and then another wire from the battery to the bulb and then a wire from the bulb as well.
Then bring different materials to test for conductivity by connecting the wires to the material. If the bulb lights then it is a conductor and if it doesn't the material is not a conductor. The material is therefore the changing variable.
1 Ten (10) ml aqueous solutions of drug A (10% w/v) and drug B (25% w/v) are stored in two identical test tubes under identical storage conditions at 37°C for 3 months. If both drugs degrade by first-order, which drug will retain the highest percentage of initial concentration?
Answer:
YOUR answer is given below:
Explanation:
Drug B will retain the highest percentage of initial concentration.
First order degradation means that both solution will degrade by same amount in same time because first order is directly proportional relationship between given both aqueous solution.
Given that Ten (10) ml aqueous solutions of drug A (10% w/v) and drug B (25% w/v) are stored in two identical test tubes.
Since, initially concentration of Drug B is more. Therefore, at all instant concentration of Drug will be highest.
Learn more:
brainly.com/question/17029235
6.05 Infections and Health
Plese don't report me i am trying to get the assigment done
Introduction
Contaminated water is a health risk, and not all populations have access to clean water. You are going to use household items to design, test and revise a water filtration system. New technologies undergo testing and revision before they are released to the public. This lab will simulate how the scientific process is used to solve problems.
Purpose
The Purpose of your design is to filter dirt, sand and other sediment from “swamp” water. You do not need to make it drinkable during this design. You will begin with a water sample that has visible dirt, mud or other contaminants. You will design a way to filter the water and evaluate your design. You will not be able to test for microorganisms or to filter them out during this design process. Do not test the water by drinking it.
Important: Do NOT drink the water in your experiment.
Materials
“Swamp” water (You can get this from an outside source like a hole in the ground, puddle, lake, river or swamp. Or, you may add dirt and sediment to tap water.) The starting sample should have visible dirt or sediment.
You may use any household materials for your water filtration system. Here are some suggestions:
Cheesecloth
OId t-shirt
Coffee filters
Pantyhose
Cotton balls
Sock
Funnel
Water bottle
Data for Design One (6 points)
Describe the appearance and smell of the “swamp” water before and after filtration. Use complete sentences and detailed examples.
Skilled Scientist level from Rubric: Student gives thorough details about the appearance and smell of the water before and after the filtration for Design One and Design Two.
Water before filtration
Water after filtration
Description of appearance
Description of smell or other physical characteristics
Rate the visible cleanliness of the water on a scale of 1 to 10 with 10 being the cleanest.
Feedback ( 5 points)
Consulat a peer or a family member about ways your design could be improved. Describe the feedback and provide the name of the reviewer. These are only suggestions that you can use to plan for Design Two. Use complete sentences and detailed examples.
Skilled Scientist level from Rubric: Student thoroughly describes the feedback received from peer/family member, and it is clear a meaningful exchange about ways to improve the design took place.
List the name of peer/family member and relationship:
Describe the feedback given by your peer/family member. What changes do they suggest you make to your design in order to clean the water better?
Planning for DesignTwo (8 points)
Use this space to plan out how to improve your water filter. Based on the feedback you received, what part are you trying to improve? Why did you make the changes that you did? Use complete sentences and detailed examples. You may include sketches or diagrams.
Skilled Scientist level from Rubric: Student thoroughly explains the changes for the design and gives ample details about any improvements that were made.
List materials used:
Describe how Design Two will work:
What part of your design are you trying to improve?
Why did you make the changes you did?
Data for Design Two (6 points)
Use the data and feedback from your first design to make changes to your water filtration system. Test your new design. Record the appearance and smell of the “swamp” water before and after filtration. Use complete sentences and detailed examples.
Skilled Scientist level from Rubric: Student gives thorough details about the appearance and smell of the water before and after the filtration for Design One and Design Two.
Water before filtration
Water after filtration
Description of appearance
Description of smell or other physical characteristics
Rate the visible cleanliness of the water on a scale of 1 to 10 with 10 being the cleanest.
Conclusion (10 points)
Use your data from Design One and Design Two to answer the following questions. Use complete sentences and detailed examples.
Skilled Scientist level from Rubric: Student answers all questions using ample details and showing critical thinking skills to reflect on the design process.
Summarize the appearance and smell of the “swamp” water before and after the filtering process using Design One.
2. Describe the changes you made to your water filtration system.
3. What feedback helped you make changes to your water filtration system?
4. Summarize the appearance and smell of the “swamp” water before and after the filtering process using Design Two.
5. What did you learn about the process of planning, testing and revising a design?
The process of filtration of swamp water using a filtering material such as neat clothes will make the water clearer and remove any bad smell.
What is filtration?Filtration is a separation technique used to separate insoluble solids from a liquid.
Filtration is employed in water purification.
Water purification is done in water purification plants.
A locally designed purification process for swamp water will involve using materials such as:
A T-shirt Funnel Water bottleThe T-shirt is used filter out mud and sediments.
Repeated filtrations will make the water clearer and remove any bad smell.
Learn more about filtration at: https://brainly.com/question/552187
Waves break on a beach due to:
reducing depth of water near beach
increasing wavelength near beach
turbulence near the shoreline
Answer:
increasing wavelength near beach
Explanation:
When wavelength increases, frequency of these waves decreases and the waves suddenly stop.
This is also called damped oscillation.
A carpenter applies a force of 60N horizontally to push a plane 40 cm along a piece of wood, how much work does she do?
Answer:
W = 24 J
Explanation:
Given that,
Applied force, F = 60 N
Distance moved, d = 40 cm = 0.4 m
We need to find the work done by the carpenter. We know that,
Work done, W = Fd
Put all the values,
W = 60 N × 0.4 m
= 24 J
Hence, the required work done is equal to 24 J.
a string 2m long used to whirl a 200gm stone in horizontal circle at a speed of 2m/s . find the tension in the string
Answer:
Explanation:
First off, we need the mass in kg and it's in g so we have to convert it. Then we can do the problem. 200 g = .200 kg. Moving on.
The equation used to find the tension in the string is the same one we use to find the centripetal force, because the tension is what is upplying the centripetal force needed to keep the stone moving in a circular manner. The formula for that is
[tex]F_c=T=\frac{mv^2}{r}[/tex] where ma is the mass in kg, v is the velocity in m/s, and r is the radius of the circle about which the stone spins in meters.
Filling in:
[tex]T=\frac{(.200)(2)^2}{2}[/tex] which gives us
T = .4 N
A cyclist is taking part in the Tour de France, which is a bicycle race that takes place every year.
a Two forces acting on the cyclist are weight and reaction. Name two other forces acting on him
as he cycles along.
If you have a cube that is 2x2x2 and weighs 5.29 g. How do you find the density?
Answer:
0.66
Explanation:
Follow the Density Formul d = m/v
so
d = 5.29/8
= 0.66 Density
Just correct me if I'm wrong, I mean like literally :> Thank you! ^^
Vector B has components Bx = 6.18 and By = 4.43. Find the magnitude and direction counter clockwise from the positive x-axis.
Answer:
Explanation:
The formula for the magnitude of this resultant vector, which is in Q1, btw, is:
[tex]B_{mag}=\sqrt{6.18^2+4.43^2}[/tex] which gives us, to 3 sig fig's:
7.60 m. Since the resultant vector is in Q1, we don't need to add anything to the angle once we find it to 3 sig fig:
[tex]tan^{-1}(\frac{4.43}{6.18})[/tex] gives us an angle of 35.6 degrees
How is an ammeter connected in a circuit to measure current flowing through it?
Answer:
It is connected in series with the circuit
Explanation:
This is because to measure the current in the circuit, the current in the circuit has to flow through the ammeter. As such, the ammeter must be connected in series with the circuit so as to measure the current flowing through the circuit.
So, to measure the current flowing through a circuit with an ammeter, the ammeter must be connected in series with the circuit.
1. Do you think that hand signals are important in officiating games? why?
Yeah, hand signals are important in officiating games. The hand signals are given by the referee who first signals the fault and then indicates which team has won the point. A point is indicated by one finger at the side of the court to indicate the winner of the rally.
the volume of vessel is 6 litres. convert it into ml
9) What is the temperature of 5 moles of nitrogen at 1 atm in a 2 liter container?*
- 4.88 K
- 0.0328 K
- 0.41 K.
- 0.0164 K
Answer:
4.88 K.
Explanation:
From the question given above, the following data were obtained:
Number of mole (n) = 5 moles
Pressure (P) = 1 atm
Volume (V) = 2 L
Gas constant (R) = 0.082 atm.L/Kmol
Temperature (T) =?
The temperature of the gas can be obtained by using the ideal gas equation as illustrated below:
PV = nRT
1 × 2 = 5 × 0.082 × T
2 = 0.41 × T
Divide both side by 0.41
T = 2 / 0.41
T = 4.88 K
Therefore, the temperature of the gas is 4.88 K.
A skier of weight 700 N is pointed down a ski hill that has a slope angle of 25° above horizontal.
What is the component of his weight pulling him down the slope.
O 634N
O 326N
O 296N
O 700N
Answer:
O 296N
Explanation:
F_y = F*sin(x)
F_y = 700N*sin(25°)
F_y = 295.83N or 296N
A sphere of diameter 3.0cm is mounted into a thin uniform wire of diameter 0.2mm calculate the length of the wire in meters
Answer:
Length of the wire in meters = 0.000628 meter (Approx.)
Explanation:
Given:
Diameter of wire = 0.2 mm
Find:
Length of the wire in meters
Computation:
Radius of wire = 0.2 / 2
Radius of wire = 0.1
Length of the wire = Circumference of circle
Circumference of circle = 2πr
Length of the wire = 2πr
Length of the wire = 2(3.14)(0.1)
Length of the wire = 0.628 mm (Approx.)
1 meter = 1,000 mm
So,
Length of the wire in meters = 0.000628 meter (Approx.)
a ball of mass 0.2 kg is dropped from a height of 20m on impact to the ground it loses in 30 joule of energy calculate the height it reaches on rebound
Answer:
new PE at top=original PE - energy lost
mgh=.2(9.8)20-30
Explanation:
mgh=.2(9.8)20-30 = 9.2
(b) During one day, 250 kg of water is pumped through
the solar panel. The température of this
water rises from 16°C to 38°C.
The water absorbs 25% of the energy incident on the solar panel. The
specific heat capacity
of water is 4200J/(kg °C).
Calculate the energy incident on the solar panel during that day.
Please explain how to find the incident energy
Answer: The energy incident on the solar panel during that day is [tex]9.24 \times 10^{7} J[/tex].
Explanation:
Given: Mass = 250 kg
Initial temperature = [tex]16^{o}C[/tex]
Final temperature = [tex]38^{o}C[/tex]
Specific heat capacity = 4200 [tex]J/kg^{o}C[/tex]
Formula used to calculate the energy is as follows.
[tex]q = m \times C \times (T_{2} - T_{1})[/tex]
where,
q = heat energy
m = mass of substance
C = specific heat capacity
[tex]T_{1}[/tex] = initial temperature
[tex]T_{2}[/tex] = final temperature
Substitute the values into above formula as follows.
[tex]q = 250 kg \times 4200 J/kg^{o}C \times (38 - 16)^{o}C\\= 250 kg \times 4200 J/kg^{o}C \times 22^{o}C[/tex]
As it is given that water absorbs 25% of the energy incident on the solar panel. Hence, energy incident on the solar panel can be calculated as follows.
[tex]\frac{25}{100} \times q = 250 kg \times 4200 J/kg^{o}C \times 22^{o}C\\q = 9.24 \times 10^{7} J[/tex]
Thus, we can conclude that the energy incident on the solar panel during that day is [tex]9.24 \times 10^{7} J[/tex].
what are scalar quantities and list their examples
Explanation:
these are quantities that have magnitude but no direction
Examples include : speed, mass, volume, density, everything, time
what is electricity?write any two defects of a simple cell.
Answer:
Electricity is the set of physical phenomena associated with the presence and motion of matter that has a property of electric charge. Electricity is related to magnetism, both being part of the phenomenon of electromagnetism, as described by Maxwell's equations.
A car with a mass of 500 kg is moving at a speed of 12 m/s. How much kinetic energy does it have?
someone please help me PLEASEEEE <3
Answer:
36000
Explanation:
1/2mv²=0.5×500×12²=36000j
In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.71 km mark at a time of 25.0 min . If he accelerated for 60 s and then maintained his increased speed for the remainder of the race, calculate his acceleration over the 60 s interval. Assume his instantaneous speed at the 7.71 km mark was the same as his overall average speed up to that time.
Answer:
0.18 m/s²
Explanation:
The total time taken to cover a 10 km race is 27 min, 43.6 seconds.
At 25 min, Pre was at 7.71 km mark. Therefore the average speed = 7.71 km / 25 min = 7710 m / (25 * 60) s = 5.14 m/s
The distance remaining = 10 km - 7.71 km = 2.29 km = 2290 m
The remaining time = 27 min, 43.6 seconds - 25 min = 2 min 43.6 second = 163.6 seconds
She accelerates for 60 seconds, therefore the distance covered (S) during the acceleration (a) is:
S₁ = 5.14(60) + 0.5a(60)² = 308.4 + 1800a
She maintains the speed for the remaining distance (S₂). The remaining time = 163.6 seconds - 60 seconds = 103.6 seconds. The final speed after the acceleration = (5.14 + 60a) m/s
S₂ = (5.14 + 60a)* 103.6 = 532.5 + 6216a
S₁ + S₂ = 2290 m
(308.4 + 1800a) + (532.5 + 6216a) = 2290
8016a + 840.9 = 2290
8016a = 1449.1
a = 0.18 m/s²
how does energy change ? explain with example . how does bulb generate ? write with it's part.
Answer:
According to the law of conservation of energy, energy is neither created nor destroyed but changes from one form to another
The energy in the light bulb comes from the movement of electrons through the resistance of the coil of the bulb, therefore, the energy in the bulb is changed from electric energy to heat and light energy as follows;
The friction between the moving electrons and the resistive filament of the light bulb results in the slowing down of the rapidly moving electrons
The part change in the (kinetic) energy of the electrons moving through the filament of the bulb as they slow down is observed as sensible heat which causes the bulb to be hot
Another part of the change in energy is given seen as the light given off by the bulb filament in an incandescent bulb as the filament becomes very hot from the movement of the electrons which causes the bulb to glow (produce light), due to the incandescent characteristics of the filament
Explanation:
A container of gas is at a pressure of 3.7 x 10^5 Pa. How much work is done by the gas if its volume expands by 1.6 m^3 ?
Answer:
592000 J
Explanation:
We'll begin by converting 3.7×10⁵ Pa to Kg/ms². This can be obtained as follow:
1 Pa = 1 Kg/ms²
Therefore,
3.7×10⁵ Pa = 3.7×10⁵ Kg/ms²
Next, we shall determine the workdone.
Workdone is given by the following equation:
Workdone (Wd) = pressure (P) × change in volume (ΔV)
Wd = PΔV
With the above formula, the work done can be obtained as follow:
Pressure (P) = 3.7×10⁵ Kg/ms²
Change in volume (ΔV) = 1.6 m³
Workdone (Wd) =?
Wd = PΔV
Wd = 3.7×10⁵ × 1.6
Wd = 592000 Kgm²/s²
Finally, we shall convert 592000 Kgm²/s² to Joule (J). This can be obtained as follow:
1 Kgm²/s² = 1 J
Therefore,
592000 Kgm²/s² = 592000 J
Therefore, the Workdone is 592000 J.
A gas in a sealed container has a pressure of 50 kPa at 27°C. What will the pressure of the gas be if the temperature rises to 87°C?
Answer:
the final pressure of the gas is 60 kPa.
Explanation:
Given;
initial pressure of the gas, P₁ = 50 kPa = 50,000 Pa
initial temperature of the gas, T₁ = 27⁰ C = 27 + 273 = 300 k
final temperature of the gas, T₂ = 87⁰ C = 87 + 273 = 360 K
Let the final pressure of the gas = P₂
Apply pressure law;
[tex]\frac{P_1}{T_1} = \frac{P_2}{T_2} \\\\P_2 = \frac{P_1T_2}{T_1} = \frac{50,000 \times 360}{300} = 60,000 \ Pa = 60 \ kPa[/tex]
Therefore, the final pressure of the gas is 60 kPa.
A certain capacitor, in series with a resistor, is being charged. At the end of 25 ms its charge is half the final value. The time constant for the process is about:
Answer:
τ = 36 10⁻³ s
Explanation:
The charge of a circuit Rc is given by the expression
q = q₀ (1 - [tex]e^{-t/ \tau }[/tex])
q / q₀ = 1 - e^{-t/ \tau }
in this exercise they indicate that
q / q₀ = ½
for a time t = 25 10⁻³ s
we substitute
½ = 1 - [tex]e^{- 25 \ 10^{-3} / \tau}[/tex]
e^{- 25 \ 10^{-3} / \tau} = 1 -½
- 25 10⁻³ /τ = ln 0.5
-25 10⁻³ / ln 0.5 = τ
τ = 36 10⁻³ s
