planets shine more stedily because they are closer to earth and so appear as pinpoints but as tiny disks in our sky.
Two forces are exerted on an object in the vertical direction: a 20 N force downward and a 10 N force upward. The mass of the object is 25 kg. (1) What are some possibilities about the motion of this object? (2) Represent the motion of the object with a force diagram and a motion diagram.
Answer:
They are equal.
Explanation:
FIND THE DISTANCE BETWEEN TWO GIVEN POINTS.
1. S (5, -1) and T (5, 7)
Answer:
8 units
Explanation:
Flying insects such as bees may accumulate a small positive electric charge as they fly. In one experiment, the mean electric charge of 50 bees was measured to be +(30±5)pC+(30±5)pC per bee. Researchers also observed the electrical properties of a plant consisting of a flower atop a long stem. The charge on the stem was measured as a positively charged bee approached, landed, and flew away. Plants are normally electrically neutral, so the measured net electric charge on the stem was zero when the bee was very far away. As the bee approached the flower, a small net positive charge was detected in the stem, even before the bee landed. Once the bee landed, the whole plant became positively charged, and this positive charge remained on the plant after the bee flew away. By creating artificial flowers with various charge values, experimenters found that bees can distinguish between charged and uncharged flowers and may use the positive electric charge left by a previous bee as a cue indicating whether a plant has already been visited (in which case, little pollen may remain). What is the best explanation for the observation that the electric charge on the stem became positive as the charged bee approached (before it landed)?
(a) Because air is a good conductor, the positive charge on the bee’s surface flowed through the air from bee to plant.
(b) Because the earth is a reservoir of large amounts of charge, positive ions were drawn up the stem from the ground toward the charged bee.
(c) The plant became electrically polarized as the charged bee approached.
(d) Bees that had visited the plant earlier deposited a positive charge on the stem.
Answer:
a) True
Explanation:
There are several possible explanations for this positive charge
* The explanation of the small positive charge in the plant when the bee approaches is like a defense system of the plants,
to prevent the bees from taking the pollen, but the flowers need the bees to transport the pollen for fertilization, so this possibility is not correct
* The air is conductive so the bee indexes a charge in the nearby air, this charge must be negative and this charge induced in the air induces a charge on the flower that must be positive.
When reviewing the different statements we have
a) True, it agrees with the second explanation of the phenomenon
b) False. The earth is a deposit of negative charge
c) false. If this is the case the charge should be negative
d) False. This residual charge from the other bees is quickly neutralized by the charge from the Earth.
Answer:
Explanation:
.
on both sides.
F
10cm
2cm
(2 marks)
(a)
a
State the type of the lens in the box and explain your answer.
Answer:
please put pic of the questions
Consider the balloon and air inside the flask to be a closed system. Use the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.
Consider the balloon and air inside the flask to be in a closed system. Using the First Law of Thermodynamics to explain what happened to the balloon as heat was added by the environment.
Balloon will burst, due to more heat and in another flask air gets heated.
What is the first law of thermodynamics ?"The first law of thermodynamics states that energy can neither be created nor destroyed, only altered in form. For any system, energy transfer is associated with mass crossing the control boundary, external work, or heat transfer across the boundary. These produce a change of stored energy within the control volume."
What is heat ?"Heat is the energy that is transferred from one body to another as the result of a difference in temperature. If two bodies at different temperatures are brought together, energy is transferred i.e, heat flows—from the hotter body to the colder."
Know more about heat here
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What is the density of a 36 g object with a volume of 15 cm3? (Density: D = )
0.42 g/cm3
0.54 g/cm3
2.4 g/cm3
5.4 g/cm3
Answer:
density = mass/volume
so . . .
density = (36 g)/(15 cm³) = 2.4 g/cm³
Explanation:
HELP PLEASE DUE IN 3 MINUTES!!!! 60 POINTS
Wind wears down rocks by blowing _________ them.
ice on
sand against
gently on
rocks toward
Answer:
i think rocks towards is correct answer
Answer:
It is gently on
Explanation:
I'm am not a 100 percent sure but try this o took that before
A beam of light, incident on a flat water surface, reflects from the mirror-like surface so that the angle of incidence equals the angle of reflection. The water has waves. Would individual light beams obey the law of reflection in this case?
Answer:
a protractor
Explanation:
because protractors measure angles
Which of the following would MOST likely slow Earth's tectonic activity?
O A. Earth's crust becomes cooler.
O B. Earth's mantle becomes warmer.
O C. Earth's mantle becomes cooler.
O D. Earth's outer core becomes warmer.
The Answer to your question is:
A.
The Earth's tectonic activity occurs by the movement of the fourteen main tectonic plates of the planet, which move over the mantle continuously, and result in the formation of mountains, earthquakes, tsunamis, volcanic activities, etc.
The decrease in the Earth's tectonic activity would therefore occur if the Earth's mantle became cooler, as the tectonic plates move over the magma, which is a paste formed by silicate of iron and magnesium and whose temperature reaches 600° and 1,200° Celsius , driven by forces from inside the planet.
Therefore the letter C is correct, as with the Earth's mantle cooler, the magma would become more solid which would decrease the Earth's tectonic activity.
Learn more here:
https://brainly.com/question/17220703
Consider a uniformly charged sphere of total charge Q and radius R centered at the origin. We want to find the electric field inside the sphere (r
Answer:
Hello your question is incomplete attached below is the complete question
answer :
Total charge enclosed within the sphere : [tex]\frac{q_{r1} }{4\pi e_{0}R^3 } . r[/tex]
Total charge enclosed outside the sphere : [tex]\frac{q}{4\pi e_{0}r^2 } .r[/tex]
Explanation:
Given data:
Total charge of a uniformly charged sphere = Q
radius = R
first step : find the electric field inside and outside the uniformly charged sphere
2nd step : determine the total charge enclosed within and outside the sphere
make a sketch of the uniformly charged sphere
Attached below is a detailed solution
A factory emits pollutants at a rate of 25 g/s. The factory is located between two mountain ranges resulting in an effective valley width of 5000 m. The height of the ceiling due to atmospheric conditions is 1000 m. During the fall and spring, a breeze that travels at a velocity of 4 m/s flows through the valley. Fortunately, the breeze coming in contains no pollutants. Based on the information given, the steady state concentration of pollutants in the valley, in micrograms per cubic meter is.
Answer:
[tex]1.25\ \mu\text{g/m}^3[/tex]
Explanation:
v = Velocity of the breeze = 4 m/s
w = Width of the valley = 5000 m
h = Height of the valley = 1000 m
Volumetric flow rate is given by
[tex]\dot{V}=vwh\\\Rightarrow \dot{V}=4\times 5000\times 1000\\\Rightarrow \dot{V}=2\times10^{7}\ \text{m}^3/\text{s}[/tex]
[tex]\dot{m}[/tex] = Mass flow rate of pollutant = 25 g/s = [tex]25\times 10^6\ \mu\text{g/s}[/tex]
Concentration is given by
[tex]C=\dfrac{\dot{m}}{\dot{V}}\\\Rightarrow C=\dfrac{25\times 10^6}{2\times 10^7}\\\Rightarrow C=1.25\ \mu\text{g/m}^3[/tex]
The steady state concentration of pollutants in the valley, is [tex]1.25\ \mu\text{g/m}^3[/tex].
Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
Answer:
Hello your question is poorly written below is the complete question
Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?
answer :
a) 231.48 days
b) n = 3.125 * 10^15
Explanation:
Battery moved 10,000 coulombs
current rate = 0.5 mA
A) Determine how long the clock run on the battery. use the relation below
q = i * t ----- ( 1 )
q = charge , i = current , t = time
10000 = 0.5 * 10^-3 * t
hence t = 2 * 10^7 secs
hence the time = 231.48 days
B) Determine how many electrons per second flowed
q = n*e ------ ( 2 )
n = number of electrons
e = 1.6 * 10^-19
q = 0.5 * 10^-3 coulomb ( charge flowing per electron )
back to equation 2
n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )
hence : n = 3.125 * 10^15
what is transactive memory
Here is a copy+pasted definition...
Transactive memory is a psychological hypothesis first proposed by Daniel Wegner in 1985 as a response to earlier theories of "group mind" such as groupthink.[1] A transactive memory system is a mechanism through which groups collectively encode, store, and retrieve knowledge. Transactive memory was initially studied in couples and families where individuals had close relationships but was later extended to teams, larger groups, and organizations to explain how they develop a "group mind",[1] a memory system that is more complex and potentially more effective than that of any of its individual constituents. A transactive memory system includes memory stored in each individual, the interactions between memory within the individuals, as well as the processes that update this memory. Transactive memory, then, is the shared store of knowledge.
In simple terms...
A transactive memory is a psychological hypothesis. It is a system in the brain which allows human's to encode, store, or get information.
For example...
let's say I learn 5 colors when I'm younger. My "transactive memory" will store it and will allow my brain to "retrieve" the information to allow me to remember colors I previously "stored".
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Two long parallel wires 20 cm apart carry currents of 5.0 A and 8.0 A in the between the two wires where the magnetic field is zero?
a. yes, midway between the wires
b. yes, 12 cm from the 5-A wire
c. yes, 7.7 cm from the 5-A wire
d. no
Answer:
c. yes, 7.7 cm from the 5-A wire
Explanation:
Given;
distance between the two wires, r = 20 cm = 0.2 m
first current, I₁ = 5.0 A
second current, I₂ = 8.0 A
The magnetic field due to the two wires occurs in different directions and it can be zero at this region.
Let x be the distance from 5 A wire where the magnetic field is zero.
[tex]B = \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{I_1}{x} -\frac{I_2}{r-x} ] = 0\\\\ \frac{\mu_o}{2\pi} [\frac{5}{x} -\frac{8}{0.2-x} ] = 0\\\\\frac{5}{x} -\frac{8}{0.2-x} = 0\\\\\frac{5}{x} = \frac{8}{0.2-x}\\\\5(0.2-x) = 8x\\\\1 -5x = 8x\\\\1 = 5x \ + \ 8x\\\\1 = 13x\\\\x = \frac{1}{13} \\\\x = 0.077 \ m\\\\x = 7.7 \ cm[/tex]
Therefore, the correct option is c. yes, 7.7 cm from the 5-A wire
An object has a kinetic energy of 14 J and a mass of 17 kg , how fast is the object
moving?
In anticipation of a long 10o upgrade, a bus driver accelerates at a constant rate of 5 ft/s^2 while still on a level section of a highway. Knowing that the speed of the bus was 80 mph as it begins to go up the hill and that the driver does not change the setting on his throttle or shift gears, determine the distance traveled (in miles) by the bus up the hill when its speed decreased to 50 mph.
Answer:
The distance (in miles) by the bus up the hill when its speed decreased to 50 mph is approximately 1.353 miles
Explanation:
The parameters of the motion of the driver are;
The upgrade of the road, θ = 10°
The rate of constant acceleration of the bus driver = 5 ft./s²
The speed of the bus as it begins to go up the hill, v₁ = 80 mph = 117.3228 ft./s
The speed of the driver at a point on the hill, v₂ = 50 mph ≈ 73.32677 ft./s
The acceleration due to gravity, g ≈ 32.1740 ft./s²
Therefore, we have;
The acceleration due to gravity down the incline plane, gₓ = g·sinθ
∴ gₓ = g·sin(θ) ≈ 32.1740 ft./s² × sin(10°) ≈ 5.587 ft/s²
The net acceleration of the bus, on the incline plane, [tex]a_{Net}[/tex] = gₓ - a = 5.587 ft./s² -5 ft./s² = 0.587 ft./s²
The vertical component of the velocity, [tex]v_y[/tex] = v × sin(θ)
∴ [tex]v_y[/tex] = 117.3228 ft./s × sin(10°) ≈ 20.37289 ft./s
vₓ = 117.3228 ft./s × cos(10°) ≈ 115.5404 ft./s
The velocity of the car, v₂, on the inclined plane is given as follows;
v₂ = v₁ - [tex]a_{Net}[/tex] × t
∴ t = (v₁ - v₂)/[tex]a_{Net}[/tex] = (117.3228 ft./s - 73.32677 ft./s)/(0.587 ft./s²) ≈ 74.95 s
The distance covered, 's', is given as follows;
s = v₁·t - 1/2·[tex]a_{Net}[/tex]·t²
∴ s = 117.3228 × 74.95 - 1/2 × 0.587 × 74.95² ≈ 7144.6069 ft.
The distance travelled up the hill, s ≈ 7144.6069 ft. ≈ 1.3531452 miles ≈ 1.353 miles
2. Using a giant screw, a crew does 650 J of work to drill a hole into a rock.
The screw does 65 J of work. What is the efficiency of the screw? Show your
work. Hellpppp
Answer:
42,250
Explanation:
It goes inside=
Displacemt
It does work=
Work done
To find efficiency of jule we do=
Dicplacement × Work done
650 × 65
42,250
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What is the cost, in dollars, of heating the hot tub, assuming 75.0% efficiency to account for heat transfer to the surroundings?
Answer:
$ 5.93
Explanation:
If we are to put into consideration the following factors even though it was not all stated in the question.
What is the cost of heating a hot tub containing 1500 kg of water from 10.0°C to 40.0°C, assuming 75.0% efficiency to take heat loss to surroundings into account? The cost of electricity is ¢8.50/kW·h.
Q = mc∆T .........1
Where m is the mass of water = 1500kg
C is the specific heat capacity of water = 4.184
∆T = temperature change
= 40° - 10°
= 30°
Now, substitute the values into equation 1
= 1500×4.184×30
= 188280
Q in kw/s
= 188280/3600
= 52.3kw/hr
Efficiency = 75%
= 75/100
= 0.75
Hence we have
52.3/0.75
= 69.73kw/hr
C = Q × cost per kw/hr
C = 69.73× $8.5/100
C = 5.92705
C =$ 5.93
Hence, the cost of heating the hot tub is $ 5.93
walking dancing and even some household chores are?
Answer:
Actions
Explanation:
if u mean what are these called then it's actions
Answer: Regular Physical Activity
Explanation:
You are walking down a straight path in a park and notice there is another person walking some distance ahead of you. The distance between the two of you remains the same, so you deduce that you are walking at the same speed of 1.17 m/s. Suddenly, you notice a wallet on the ground. You pick it up and realize it belongs to the person in front of you. To catch up, you start running at a speed of 2.75 m/s. It takes you 10.5 s to catch up and deliver the lost wallet. How far ahead of you was this person when you started running
Answer:
16.6 m
Explanation:
Let d be the distance the other person is ahead of you. Since the other person is walking at a speed, v = 1.17 m/s, after picking the wallet, the other person moves a distance , vt in time, t = 10.5 s, the total distance covered by you till catch up is D = d + vt.
Also, you moves with a speed of v' = 2.75 m/s in time t = 10.5 s as you pick up the wallet, you covers a distance d' = v't at catch up.
At catch up, D = d'
d + vt = v't
d = v't - vt
d = (v' - v)t
Substituting the values of the variables into d, we have
d = (2.75 m/s - 1.17 m/s)10.5 s
d = (1.58 m/s)10.5 s
d = 16.59 m
d ≅ 16.6 m
So, the other person was 16.6 m ahead of you when you started running.
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Anna weighs 132 lb. Determine her mass in kilograms using the conversion 1 kg equal 2.2 lb. Use this mass to answer this question.
calculate Anna's weight on the Moon. (G = 1.6 m / S2) must include a unit with your answer
1). If 2.2 lb actually = 1 kg, then
(132lb) x (1 kg/2.2lb) = 60 kg.
(Sadly, 2.2 lb doesn't " = " 1 kg. The explanation, unfortunately, is beyond the scope of this discussion.)
2). Weight = (mass) x (gravity)
Moon weight = (60 kg)x(1.6 m/s^2)
Moon weight = 96 Newtons
a. If we increase the wind velocity, the maximum vertical dispersal height and rate of diffusion will decrease____.
b. If we increase the humidity, the maximum vertical dispersal height will ___________increase____ after 24 hours.
c. If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will ____________________.
Answer:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Explanation:
a.If we increase the wind velocity, the maximum vertical dispersal height will decrease, while the rate of diffusion will increase
b.If we increase the humidity, the maximum vertical dispersal height will increase after 24 hours.
c.If we increase the lapse rate, the maximum vertical dispersal height of the pollutants will increase
Answer:
a decrease
b increase
c increase
Stored energy due to vertical position is known as
Elastic Potential energy
Vibrational energy
Kinetic energy
O Gravitational Potential energy
1
2
3
4
5
Answer: gravitational potential energy
Explanation:
8+10÷5(5×4+2)=?
it ıs said that this question was very diffıcult
can you slove?
Answer:
220 bastanyan sagot ko yawa
A simple pendulum is used to measure gravity using the following theoretical equation,TT=2ππ�LL/gg ,where L is the length of the pendulum, g is gravity, andT is the period of pendulum.Twenty measurements of T give a mean of 1.823 seconds and a standard deviation of 0.0671 s. The device used to measure time has a resolution of 0.02 s. The pendulum length is measured once to be 0.823 m (with a scale having a resolution of 0.001 m). Determine the value of g and its uncertainty (assume 90% confidence where necessary). You may use any method of uncertainty propagation that we covered in class.
Answer:
g ±Δg = (9.8 ± 0.2) m / s²
Explanation:
For the calculation of the acceleration of gravity they indicate the equation of the simple pendulum to use
T = [tex]2\pi \sqrt{ \frac{L}{g} }[/tex]
T² = [tex]4\pi ^2 \frac{L}{g}[/tex]4pi2 L / g
g = [tex]4\pi ^2 \frac{L}{T^2}[/tex]
They indicate the average time of 20 measurements 1,823 s, each with an oscillation
let's calculate the magnitude
g = [tex]4\pi ^2 \frac{0.823}{1.823^2}[/tex]4 pi2 0.823 / 1.823 2
g = 9.7766 m / s²
now let's look for the uncertainty of gravity, as it was obtained from an equation we can use the following error propagation
for the period
T = t / n
ΔT = [tex]\frac{dT}{dt}[/tex] Δt + [tex]\frac{dT}{dn}[/tex] ΔDn
In general, the number of oscillations is small, so we can assume that there are no errors, in this case the number of oscillations of n = 1, consequently
ΔT = Δt / n
ΔT = Δt
now let's look for the uncertainty of g
Δg = [tex]\frac{dg}{dL}[/tex] ΔL + [tex]\frac{dg}{dT}[/tex] ΔT
Δg = [tex]4\pi ^2 \frac{1}{T2}[/tex] ΔL + 4π²L (-2 T⁻³) ΔT
a more manageable way is with the relative error
[tex]\frac{\Delta g}{g} = \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}[/tex]
we substitute
Δg = g ( \frac{\Delta L }{L} + \frac{1}{2} \frac{\Delta T}{T}DL / L + ½ Dt / T)
the error in time give us the stanndard deviation
let's calculate
Δg = 9.7766 ([tex]\frac{0.001}{0.823} + \frac{1}{2} \ \frac{0.671}{1.823}[/tex])
Δg = 9.7766 (0.001215 + 0.0184)
Δg = 0.19 m / s²
the absolute uncertainty must be true to a significant figure
Δg = 0.2 m / s2
therefore the correct result is
g ±Δg = (9.8 ± 0.2) m / s²
A cylindrical metal rod has a resistance . If both its length and its diameter are quadrupled, its new resistance will be:________.
A. 16R
B. R/4
C. R
D. 4R
Answer:
R' = R/4
Explanation:
The resistance of a metal rod is R. It is given by the relation as follows :
[tex]R=\rho\dfrac{l}{A}[/tex]
Where
l is the length and A is the area of cross-section
[tex]A=\pi r^2=\pi (\dfrac{d}{2})^2[/tex]
If both its length and its diameter are quadrupled, it means,
l' = 4l
and d'= 4d
It means,
[tex]A'=\pi (\dfrac{4d}{2})^2[/tex]
Let new resistance be R'. So,
[tex]R'=\rho\dfrac{l'}{A'}\\\\R'=\rho\dfrac{4l}{\pi (\dfrac{4d}{2})^2}\\\\=\rho \dfrac{4l}{\pi \dfrac{16d^2}{2}}\\\\=\dfrac{4}{16}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\=\dfrac{1}{4}\times \dfrac{\rho l}{\pi \dfrac{d^2}{2}}\\\\R'=\dfrac{R}{4}[/tex]
So, the correct option is (B) "R/4".
(It made me choose physics as the category, but the class is Astronomy)
The Earth's Escape velocity, the velocity needed to escape the Earth's
gravitational field, is
Name the nutrients required for the body
Answer:
Explanation:
Water.
Carbohydrates.
Protein-Amino acids.
Fat.
Vitamins.
Minerals.
Omega-3 fatty acids.
A rocket, with a mass of 5100 kg, has an engine that provides a net upward force of 8.0 x 10^5 N. It starts from rest and reaches a maximum speed of 900 m/s. How long does it take to reach that maximum velocity?
Answer:
5.7375 seconds
Explanation:
The computation of the time required to reach that maximum velocity is shown below:
Given that
Mass = m = 5100 kg
Net upward force F = 8 × times 10^5 N
Initial speed = V_i = 0
Maximum speed = V = 900 m.s
Based on the above information
Impluse J = m(V - V_i)
= 5100 (900 - 0)
= 459 × 10^4 kg m.s
As we know that
J = FT
So
T = J ÷ F
= (459 × 10^4) ÷ (8 × 10^5)
= 5.7375 seconds
How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
This question is incomplete, the complete question is;
A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?
Answer:
the required heat energy is 16 J
Explanation:
Given the data in the question;
Lets consider the ideal gas equation;
PV = nRT
from the image, we calculate initial pressure;
Pi = ( 2000N/M × 0.06m) / 0.0008 m²
Pi = 15 × 10⁴ Pa
next we find Initial velocity
Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²
now we find the number of moles
n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K
N = 6.6 × 10⁻³ mol
next we calculate the final temperature;
Pf = ( 2000N/m × 0.08) / 0.0008 m²
Pf = 2 × 10⁵ Pa
Calculate the final Volume
Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³
we also determine the final temperature
[tex]T_{f}[/tex] = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK
[tex]T_{f}[/tex] = 466.8 K
so change in temperature ΔT
ΔT = 466.8 K - 300K = 166.8 K
we then calculate the change in thermal energy
ΔU = nCΔT
ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K
ΔU = 13.761 J
C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic
now we calculate the work done;
W = 1/2 × K( x[tex]_{i\\}[/tex]² - x[tex]_{f\\}[/tex]² )
W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )
= - 2.8 J
and we then calculate the heat energy using the following expression;
Q = ΔU - W
we substitute
Q = 13.761 - (- 2.8 J)
Q = 13.761 + 2.8 J)
Q = 16 J
Therefore, the required heat energy is 16 J