Answer:
C
Explanation:
Although this may seem surprising, water beneath the ground is commonplace. Usually groundwater travels slowly and silently beneath the surface, but in some locations it bubbles to the surface at springs. The products of erosion and deposition by groundwater were described in the Erosion and Deposition chapter.
Groundwater is the largest reservoir of liquid fresh water on Earth and is found in aquifers, porous rock and sediment with water in between. Water is attracted to the soil particles and capillary action, which describes how water moves through a porous media, moves water from wet soil to dry areas.
Aquifers are found at different depths. Some are just below the surface and some are found much deeper below the land surface. A region may have more than one aquifer beneath it and even most deserts are above aquifers. The source region for an aquifer beneath a desert is likely to be far from where the aquifer is located; for example, it may be in a mountain area.
The amount of water that is available to enter groundwater in a region is influenced by the local climate, the slope of the land, the type of rock found at the surface, the vegetation cover, land use in the area, and water retention, which is the amount of water that remains in the ground. More water goes into the ground where there is a lot of rain, flat land, porous rock, exposed soil, and where water is not already filling the soil and rock.
The residence time of water in a groundwater aquifer can be from minutes to thousands of years. Groundwater is often called “fossil water” because it has remained in the ground for so long, often since the end of the ice ages.
The type of water occurs as a liquid resource that is dispersed through
numerous holes is known as Ground water.
Ground water is formed from the liquid on the land being dispersed deep
down into the soil. Water is readily absorbed into the soil depending on the
type of soil. Sandy soil is known to have a higher degree of absorption.
The water absorbed is usually as a result of numerous holes, pores, fractures,and cavities in rocks and sediments and is also dependent on the climate and rate of precipitation in the area. This is why water is usually found in the soil when dug to certain levels.
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g Suppose you have this brilliant idea: A Ferris wheel has radial metallic spokes between the hub and the circular rim (of radius roughly 16 m). These spokes move in the magnetic field of the Earth (5e-05 T), so each spoke acts like a rotating bar in a magnetic field. The magnetic field points perpendicular to the plane of the Ferris wheel. You plan to use the emf generated by the rotation of the Ferris wheel to power the light-bulbs on the wheel. Suppose the period of rotation for the Ferris wheel is 90 seconds. What is the magnitude of the induced emf between the hub and the rim
Answer:
[tex]4.46\times 10^{-4}\ \text{V}[/tex]
Explanation:
B = Magnetic field = [tex]5\times 10^{-5}\ \text{T}[/tex]
r = Radius of rim = 16 m
t = Time = 90 seconds
A = Area of rim = [tex]\pi r^2[/tex]
EMF is given by
[tex]\varepsilon=\dfrac{BA}{t}\\\Rightarrow \varepsilon=\dfrac{5\times 10^{-5}\times \pi\times 16^2}{90}\\\Rightarrow \varepsilon=0.000446=4.46\times 10^{-4}\ \text{V}[/tex]
The magnitude of the induced emf between the hub and the rim is [tex]4.46\times 10^{-4}\ \text{V}[/tex].
what makes a funnel appear black
Answer:
Air
Explanation:
The mixing of cooler air in the lower troposphere with air flowing in a different direction in the middle troposphere causes the rotation on a horizontal axis, which, when deflected and tightened vertically by convective updrafts, forms a vertical rotation that can cause condensation to form a funnel cloud.
7. Copper can be coated on the surface of iron but not on silver? why
Answer:
Silver has greater electronegativity (pull on electrons) than copper, so it will be reduced rather than oxidized. Silver ions will plate out on copper metal. Sir the coating of iron with copper surface can be referred as IRON-COPPER PLATING.
Explanation:
Iron is used for the electroplating of so copper because iron falls above copper in the electrochemical series whereas silver will fall off below the copper in electrochemical series that makes it not reliable for the electroplating purpose
PLEASE HELP ME WITH THIS ONE QUESTION
The specific heat of aluminum is 899 J/kg*C. How much aluminum is changed by 1.00°C assuming 1000. J of heat is added to the piece of aluminum?
Answer:
Q = c * m * T where c = specific heat, m = mass, T = temperature change
m = Q / (c * T) = 1000 J / (899 J / kg deg C * 1 deg C)
m = 1000 / 899 kg = 1.11 kg
*
Check all of the things that can affect power
Time
Work
Distance
Force
Answer:
fhddmvxmvydlyghclhchc
An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s1045 rad/s ). If a particular disk is spun at 734.1 rad/s 734.1 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds0.569 seconds , what is the magnitude of the average angular acceleration of the disk
Answer:
[tex]1290.16\ \text{rad/s}^2[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 734.1 rad/s
[tex]\omega_f[/tex] = Final angular velocity = 0
t = Time = 0.569 seconds
[tex]\alpha[/tex] = Angular acceleration
From the kinematic equations of rotational motion we have
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{0-734.1}{0.569}\\\Rightarrow \alpha=-1290.16\ \text{rad/s}^2[/tex]
The magnitude of the average angular acceleration of the disk is [tex]1290.16\ \text{rad/s}^2[/tex].
A cyclist traveling at 5m/s uniformly accelerates up to 10 m/s in 2 seconds. Each tire of the bike has a 35 cm radius, and a small pebble is caught in the tread of one of them. (A) What is the angular acceleration of the pebble during those two seconds
Answer:
[tex]a=2.5\ m/s^2[/tex]
Explanation:
Given that,
Initial speed, u = 5 m/s
Final speed, v = 10 m/s
Time, t = 2 s
The radius of the tire of the bike, r = 35 cm
We need to find the angular acceleration of the pebble during those two seconds. It can be calculated as follows.
[tex]a=\dfrac{v-u}t{}\\\\a=\dfrac{10-5}{2}\\\\a=2.5\ m/s^2[/tex]
So, the required angular acceleration of the pebble is equal to [tex]2.5\ m/s^2[/tex].
A 60 kg swimmer at a water park enters a pool using a 2 m high slide. Find the velocity of the swimmer
at the bottom of the slide.
Answer:
the velocity of the swimmer at the bottom of the slide is 6.26 m/s
Explanation:
The computation of the velocity of the swimmer at the bottom of the slide is given below:
v = √2gh
= √2 × 9.8 × 2
= 6.26 m/s
Hence, the velocity of the swimmer at the bottom of the slide is 6.26 m/s
The velocity of the swimmer at the bottom of the slide will be 6.26 m/s.The pace of displacement change with reference to time is referred to as the velocity
What is velocity?The change of displacement with respect to time is defined as the velocity. Velocity is a vector quantity. it is a time-based component. Velocity at any angle is resolved to get its component of x and y-direction.
The velocity is found as;
[tex]\rm v= \sqrt{2gh} \\\\ \rm v= \sqrt{2\times 9.81 \times 2}\\\\ \rm v= 6.26 \ m/sec[/tex]
Hence velocity of the swimmer at the bottom of the slide will be 6.26 m/s.
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how many atmospheres is at depth of 100 meters of ocean water?
Explanation:
At 100m above sea level, the air pressure is approximately 990mbar, so the air pressure has decreased by approximately 10mbar.
There would be 10 atmospheric pressure at a depth of 100 meters of ocean water because approximately 10 meters depth of water is equivalent to 1 atmospheric pressure.
What is pressure?The total applied force per unit of area is known as the pressure.
The pressure depends both on externally applied force as well the area on which it is applied.
As given in the problem we have to find out the atmospheric pressure equivalent at a depth of 100 meters of ocean water,
10 meters depth of water = 1 atmospheric pressure
100-meter depth of the ocean water = 10 atmospheric pressure
Thus, there would be 10 atmospheric pressure at a depth of 100 meters of ocean water.
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3) if you have a convex (converging) lens (f#12 cm) that produced an image with a
magnification of -2.5, how far must the object be placed from the lens?
I
b) Draw a ray-tracing diagram of this situation below (label all points in cm)
Explanation:
step 1. since the magnification is negative this means the object is real and is farther away from the lens than the focal point
step 2. 1/f = 1/o + 1/i where f is the focal length, o is the object distance from the lens and i is the image distance from the lens
step 3. 1/12 = 1/o + 1/i, -i/o = -2.5 (these are our 2 equations or system
step 4. 1/12 = 1/o + 1/2.5o
step 5. o = 12 + 12/2.5 = 16.8cm
step 6. i = 2.5o = 2 5(16) = 42cm
5/6 When switched on, the grinding machine accelerates from rest to its operating speed of 3450 rev/min in 6 seconds. When switched off, it coasts to rest in 32 seconds. Determine the number of revolutions turned during both the startup and shutdown periods. Also determine the number of revolutions turned during the first half of each period. Assume uniform angular acceleration in both cases.
Answer:
Δθ₁ = 172.5 rev
Δθ₁h = 43.1 rev
Δθ₂ = 920 rev
Δθ₂h = 690 rev
Explanation:
Assuming uniform angular acceleration, we can use the following kinematic equation in order to find the total angle rotated during the acceleration process, from rest to its operating speed:[tex]\Delta \theta = \frac{1}{2} *\alpha *(\Delta t)^{2} (1)[/tex]
Now, we need first to find the value of the angular acceleration, that we can get from the following expression:[tex]\omega_{f1} = \omega_{o} + \alpha * \Delta t (2)[/tex]
Since the machine starts from rest, ω₀ = 0.We know the value of ωf₁ (the operating speed) in rev/min.Due to the time is expressed in seconds, it is suitable to convert rev/min to rev/sec, as follows:[tex]3450 \frac{rev}{min} * \frac{1 min}{60s} = 57.5 rev/sec (3)[/tex]
Replacing by the givens in (2):[tex]57.5 rev/sec = 0 + \alpha * 6 s (4)[/tex]
Solving for α:[tex]\alpha = \frac{\omega_{f1}}{\Delta t} = \frac{57.5 rev/sec}{6 sec} = 9.6 rev/sec2 (5)[/tex]
Replacing (5) and Δt in (1), we get:[tex]\Delta \theta_{1} = \frac{1}{2} *\alpha *(\Delta t)^{2} = \frac{1}{2} * 6.9 rev/sec2* 36 sec2 = 172.5 rev (6)[/tex]
in order to get the number of revolutions during the first half of this period, we need just to replace Δt in (6) by Δt/2, as follows:[tex]\Delta \theta_{1h} = \frac{1}{2} *\alpha *(\Delta t/2)^{2} = \frac{1}{2} * 6.9 rev/sec2* 9 sec2 = 43.2 rev (7)[/tex]
In order to get the number of revolutions rotated during the deceleration period, assuming constant deceleration, we can use the following kinematic equation:[tex]\Delta \theta = \omega_{o} * \Delta t + \frac{1}{2} *\alpha *(\Delta t)^{2} (8)[/tex]
First of all, we need to find the value of the angular acceleration during the second period.We can use again (2) replacing by the givens:ωf =0 (the machine finally comes to an stop) ω₀ = ωf₁ = 57.5 rev/secΔt = 32 s[tex]0 = 57.5 rev/sec + \alpha * 32 s (9)[/tex]
Solving for α in (9), we get:[tex]\alpha_{2} =- \frac{\omega_{f1}}{\Delta t} = \frac{-57.5 rev/sec}{32 sec} = -1.8 rev/sec2 (10)[/tex]
Now, we can replace the values of ω₀, Δt and α₂ in (8), as follows:[tex]\Delta \theta_{2} = (57.5 rev/sec*32) s -\frac{1}{2} * 1.8 rev/sec2\alpha *(32s)^{2} = 920 rev (11)[/tex]
In order to get finally the number of revolutions rotated during the first half of the second period, we need just to replace 32 s by 16 s, as follows:[tex]\Delta \theta_{2h} = (57.5 rev/sec*16 s) -\frac{1}{2} * 1.8 rev/sec2\alpha *(16s)^{2} = 690 rev (12)[/tex]effect of high pitch on humans
Answer:
High frequency sound causes two types of health effects: on the one hand objective health effects such as hearing loss (in case of protracted exposure) and on the other hand subjective effects which may already occur after a few minutes: headache, tinnitus, fatigue, dizziness and nausea.
3) A rather large fish is about to eat an unsuspecting small fish. The big fish has a mass of 5kg and is
swimming at 8 m/s, while the small fish has a mass of 1 kg and is swimming at -4 m/s. What is the
velocity of big fish after lunch?
Answer:
the velocity of the big fish after the launch is 6 m/s.
Explanation:
Given;
mass of the big fish, m₁ = 5 kg
velocity of the big fish, u₁ = 8 m/s
mass of the small fish, m₂ = 1 kg
velocity of the small fish, u₂ = -4 m/s
Let the final velocity of the big fish after launch = v
Apply the principle of conservation linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
5 x 8 + 1 x (-4) = v(5 + 1)
40 - 4 = 6v
36 = 6v
v = 36/6
v = 6 m/s.
Therefore, the velocity of the big fish after the launch is 6 m/s.
What conversion takes place in a motor?
A. An electric current into a magnetic field
B. Mechanical energy into electric energy
C. Electric energy into mechanical energy
D. A lower voltage into a higher voltage
Thank you!! I will mark brainliest!!
Pesticides sprayed on the plants in fields become When they settle and drip onto the soil .
A. Nutrients
B.Pollutants
C.Acid rain
D.Ammonia
Answer:nutrients
Explanation:
Answer:
pollutants is the right answer.
The circuit is working, and all three bulbs are lit. If a switch at B is opened, what will happen to the circuit? Photo included PLS ONLY ANSWER IF YOU KNOW 100%
Answer:
D. Bulb 2 will go out but bulbs 1 and 3 will remain lit.
Explanation:
If switch at B is opened then, Bulb 2 will go out but bulbs 1 and 3 will remain lit. This is because the positive and negative current flow still fully passes through the two closed switches at switch A and C which allows the bulbs to receive both currents and thus allowing it to turn on. When Switch B is opened it cuts off access for the negative charge to get to bulb 2 and without it, bulb 2 will not turn on.
Answer:all 3 bulbs will go out
Explanation:
Just did it and got it right
Which term does this explain?
This is a non-mathematical explanation of how nature works. It must be
supported by a large body of evidence.
Fact
Law
Theory
Hypothesis
Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is placed on the scalp, and a brief burst of current in the coil produces a rapidly changing magnetic field inside the brain. The induced emf can be sufficient to stimulate neuronal activity. One such device generates a magnetic field within the brain that rises from zero to 1.2 T in 100 ms. Determine the magnitude of the induced emf within a circle of tissue of radius 1.3 mm and that is perpendicular to the direction of the field.
poste en français s’il vous plaît
Define threshing?
Brainliest For the right answer
Answer:
Process used for separating grains from the stalks is known as threshing, In this process, stalks are beaten to free the grain seeds.
Answer:
Threshing is extraction of wheat germ from the stalk. In today's usage the combine tractor cuts and threshes the wheat at the same time. Imagine a big lawn mower with a rotating drum inside.
The drum turns and shakes the germ out of the wheat, the seeds falling through small holes onto a conveyor belt one way, the leftover grass dumping out the other way. The grain is poured into a truck driving beside the combine.
In old times, grain had to be beaten out of the grass on a Threshing Floor.
How do you find the period of a sound wave?
Answer:
Period refers to the time for something to happen and is measured in seconds/cycle. In this case, there are 11 seconds per 33 vibrational cycles. Thus the period is (11 s) / (33 cycles) = 0.33 seconds. We now know that the period is 3.2 seconds and that the frequency is 0.31 Hz.
Which disciplines were developed based on the Greek questioning of the elements?
A. art and music
B. chemistry and biology
C. tragedy and comedy
The motor of a washing machine rotates with a period of 28 ms. What is the angular speed, in units of rad/s?
Answer:
2π/[28 x (10^-3)]
Explanation:
Angular speed : ω=2π/T
T = 28ms = 28 x (10^-3) s
Angular speed = 2π/[28 x (10^-3)]
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I believe you are incorrect. A weathered mountain would appear more jagged.
I do believe with a lot of exposure to weather will make the mountain appear somewhat more jagged compared to a mountain that is less weathered.
If this is incorrect, please, don't refrain to tell me.
Fill in the graph for 50 points
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Answer:
Speed: 3, 4, 5, 6. Distance: 1, 2, 3, 4, 5
Explanation:
If a true bearing of a ship at sea is 227°, what is its direction angle?
A. 43°
B.313°
C. 223°
Which of the following relationships is correct?
2 points
1 N = 1 kg
1 N = 1 kg·m
1 N = 1 kg·m/s
1 N = 1 kg·m/s2
Can someone please write a summery about Yes day the movie from 2021 and it has to have something about Hygiene in it and this is for 100 points so can you please help me I will give you $100 and 100 points for a 2 page essay
Answer: This book is about a mom that is always saying no to her kids and is always pushing them to focus on what they are doing, but one day the mom or dad runs into a man that does what he calls "A Yes Day" with his kids and what this is, is for 24 hours you say yes to whatever you want to do and you aren't able to say no. So the mom tries yes day with her kids and they go through many crazy/extreme events. This has to deal with hygiene because the kids trash the house and get very dirty in a few parts of the movie. They always try to be as clean as possible, but the limits have been broken... This is why I think this movie has to deal with hygiene.
Answer:
I can't do 2 pages but I'll try
Explanation:
Yes day is a movie about parents and kids who usually fought, until someone suggested a yes day a day when parents can't say no. In the beginning of the movie, you'll first see that the parents used to say yes to everything, until they has kids they started saying no to everything. When someone suggested yes day they had to earn it, they did all the chores and they got a yes day. They made a list of 5 things they wanna do during yes day. i don't know what else to put for hygiene but yea.
A 20-cm-diameter disk emits light uniformly from its surface. 40 cm from this disk, along its axis, is a 16.0-cm-diameter opaque black disk; the faces of the two disks are parallel. 40 cm beyond the black disk is a white viewing screen. The lighted disk illuminates the screen, but there's a shadow in the center due to the black disk. What is the diameter of the completely dark part of this shadow
Answer:
132
Explanation:
if you round it is correct
If
[tex] \binom{13}{7} [/tex]
N has a half-life of about 10.0 min, how long will it take for 20 g of the isotope to decay to 1.9 g? [2 marks] 2. Radon-22
Answer:
the time it will take the element to decay to 1.9 g is 34.8 mins.
Explanation:
Given;
half life of Nitrogen, t = 10 min
initial mass of the element, N₀ = 20 g
final mass of the element, N = 1.9 g
The time taken for the element to decay to final mass is calculated as follows;
time (min) mass remaining
0 ----------------------------------20 g
10 mins ------------------------- 10 g
20 mins ------------------------- 5 g
30 mins -------------------------- 2.5 g
40 mins --------------------------- 1.25 g
Interpolate between 2.5 g and 1.25 to obtain the time for 1.9 g
30 min ------------------------- 2.5 g
x ----------------------------------- 1.9 g
40 min -------------------------- 1.25 g
[tex]\frac{30 - x}{40- 30} = \frac{2.5 - 1.9}{1.25 - 2.5} \\\\-1.25(30-x) = 6\\\\-37.5 + 1.25x = 6\\\\1.25x = 6+37.5\\\\1.25x = 43.5\\\\x = \frac{43.5}{1.25} \\\\x = 34.8 \ mins[/tex]
Therefore, the time it will take the element to decay to 1.9 g is 34.8 mins.