Answer:
1 A
Explanation:
P = IV
120 W = I (120 V)
I = 1 A
g Which one of the following statements is false?The "ray model" of light can explain how images are formed by spherical mirrors. The "ray model" of light can explain the phenomenon known as a shadow. The "wave model" of light is necessary to explain phenomena such as diffraction and interference. The "ray model" of light can explain how images are formed by thin lenses. According to the "ray model", light is an electromagnetic wave that consists of oscillating electric and magnetic fields.
Answer
Correct option : According to the "ray model", light is an electromagnetic wave that consists of oscillating electric and magnetic fields.
Explanation : According to Ray model light is a simple transverse wave not a electromagnetic wave.
Which theory states that deviance results not only from what people do, but also from how others respond to those actions?
A.
strain theory
B.
labeling theory
C.
differential association theory
D.
control theory
E.
conflict theory
The theory states that deviance results not only from what people do but also from how others respond to those actions are labeling theory. Hence, the option B is correct.
What is labeling theory?This theory states that deviance and conformity result not so much from what people do but also from how others respond. It is called labeling theory. Eg: Skipping school, and underage drinking.
It also suggests that any deviance results in how society responds to certain behaviors. It defines the behavior of human beings influenced by other members of society.
It also notes that a person is made to act in a negative way by the manner in which society identifies him. If a person is identified as a criminal then he involves in the criminal activities.
Conflict theory refers to the theory linking deviance to social inequality. Anticipating the consequences of a person's behavior is control theory.
Hence, the correct option is B) labeling theory.
To learn more about the labeling theory:
https://brainly.com/question/31366596
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The space probe Deep Space 1 was launched on October 24th, 1998 and it used a type of engine called an ion propulsion drive. An ion propulsion drive generates only a weak force (or thrust), but can do so for long periods of time using only a small amount of fuel. Suppose the probe, which has a mass of 474 kg is travelling at an initial speed of 275 m/s. No forces act on it except the 5.60 x 10⁻² N thrust from the engine. This external force is directed PARALLEL to the displacement. The displacement has a magnitude of 2.42 x 10⁹ m. {PART A} Calculate the INITIAL kinetic energy of the probe [2 marks] {PART B} Find the work done BY THE ENGINE on the space probe [2 marks] {PART C} Calculate the FINAL KINETIC ENERGY of the probe (Hint W=∆E) [2 marks] {PART D} Determine the final speed of the probe, assuming that its mass remains constant [3 marks]
Answer:
Explanation:
mass of probe m = 474 Kg
initial speed u = 275 m /s
force acting on it F = 5.6 x 10⁻² N
displacement s = 2.42 x 10⁹ m
A )
initial kinetic energy = 1/2 m u² , m is mass of probe.
= .5 x 474 x 275²
= 17923125 J
B )
work done by engine
= force x displacement
= 5.6 x 10⁻² x 2.42 x 10⁹
= 13.55 x 10⁷ J
C ) Final kinetic energy
= Initial K E + work done by force on it
= 17923125 +13.55 x 10⁷
= 1.79 x 10⁷ + 13.55 x 10⁷
= 15.34 x 10⁷ J
D ) If v be its velocity
1/2 m v² = 15.34 x 10⁷
1/2 x 474 x v² = 15.34 x 10⁷
v² = 64.72 x 10⁴
v = 8.04 x 10² m /s
= 804 m /s
An experiment based at New Mexico’s Apache Point observatory uses a laser beam to measure the distance to the Moon with millimeter precision. The laser power is 120 GW, although it’s pulsed on for only 90 ps. The beam emerges from the laser with a diameter of 7.0 mm. It’s then beamed into a telescope aimed at the Moon. When the beam leaves the telescope, it has the telescope’s full 3.5-mm diameter. By the time it reaches the Moon, the beam has expanded to a diameter of 6.5 km.
a. Find the intensity of the beam as it leaves the laser. Express your answer with the appropriate units.
b. Find the intensity of the beam as it leaves the telescope. Express your answer with the appropriate units.
Answer:
Sorry but i dont know
Explanation:
The main component of all computer memory is
Answer: R.A.M
Explanation:
Which is a characteristic of diatoms?
A)whip-like tail
B)heterotrophic
C)multicellular
D)glass-ive cell wall
Answer:
b
Explanation:
Answer:
The correct answer is D, diatoms have glass-like cell walls.
A convex mirror of focal length 33 cm forms an image of a soda bottle at a distance of 19 cm behind the mirror.If the height of the image is 7.0 cm,where is the object located,and how tall is it? What is the magnification of the image? Is the image virtual or real? Is the image inverted or upright? Draw a ray diagram to confirm your results.
Answer:
Image distance = 44.8cm, Image height = 16.5cm, Magnification = 0.42
The image is a virtual and upright image.
Explanation:
The nature of image formed by an object placed in front of a convex mirror is always diminished, virtual and erect.
The focal length f and the image distance are always NEGATIVE beacause the image is formed behind the mirror.
Given f = -33.0cm, v = -19.0cm
using thr mirror formula to get the object distance u, we have;
[tex]\frac{1}{f}=\frac{1}{u} + \frac{1}{v}\\ \frac{1}{u}=\frac{1}{f} - \frac{1}{v}\\\frac{1}{u}=\frac{1}{-33} - \frac{1}{-19}\\\frac{1}{u}=\frac{-19+33}{627} \\\frac{1}{u}=\frac{14}{627} \\u=\frac{627}{14} \\u = 44.8cm[/tex]
To calculate the image height, we will use the magnification formula
M = [tex]\frac{image\ height}{object\ height}=\frac{image\ distance}{object\ distance} \\[/tex]
M = [tex]\frac{Hi}{HI}=\frac{v}{u}[/tex]
Given Hi = 7.0cm
v = 19.0cm
u = 44.8cm
HI = 7*44.8/19
HI = 16.5cm
The object height is 16.5cm
Magnification = v/u = 19.0/44.8 = 0.42
SInce the image is formed behind the mirror, the image is a VIRTUAL and UPRIGHT image
Speed is the rate of change of position expressed as____ Traveled per unit of time
Answer:
distance
Explanation:
Speed is the rate of change of position expressed as distance travelled per unit of time
Many more earthquakes occur along the San Andreas Fault in California than in other parts of the United States. Scientists claim that these earthquakes are caused by activity along a transform plate boundary.
Which evidence would best support this claim?
Subduction causes new crust to form.
Subduction causes old crust to be destroyed.
Plates slip past each other at the boundary.
Plates collide with each other at the boundary.
The correct answer is C. Plates slip past each other at the boundary.
Explanation
A geological fault is a fracture in the Earth's crust, in which the rocky blocks that compose the crust move. Faults often occur in transform plate boundaries in which the two plates displace laterally with respect to the other plate, this means plates slip past each other. Evidence of this type of fault is in the United States, where the cities of Los Angeles and San Francisco in California, United States, located on different plates of the San Andres fault move to different sides due to the transform plate boundary. In this way, the city of San Francisco moves to the East while Los Angeles moves to the West as the plates in which the cities are located slip past each other or move laterally. So, the correct answer is C. Plates slip past each other at the boundary.
Answer:
c
Explanation:
the medium of a mechanical wave is the matter through wich it travels? tru or false
Answer:
it can only travel through matter
Explanation:
the mechanical wave can be any type of matter
1. Deri had a large tank of oil (s-0.8) and was requested to determine the viscosity of that fluid. To assist with the process, she was given a 0.25-inch-diameter steel ball (sphere, s=8.0) to conduct the test. From the tests, she found that the terminal velocity of the sphere was 2.5 fpm. What is the viscosity of the oil? Remember, the volume of a sphere is (pi D3 /6). ANS. viscosity is 0.258 lb-s/ft2
Answer:
0.25916 lb-s/ft^2
Explanation:
Given:-
- The specific gravity of oil, SGo = 0.8
- The specific gravity of sphere, SGo = 8
- Terminal velocity of sphere, v = 2.5 fpm
- The diameter of sphere, D = 0.25 in
Find:-
What is the viscosity of the oil?
Solution:-
- Consider a sphere completely submerged into oil and travelling with terminal velocity ( v ).
- Develop a free body diagram for the sphere. There are forces acting on the sphere.
- The downward acting force is due to the weight of the sphere ( W ):
[tex]W = m_s*g[/tex]
Where,
The mass ( m_s ) of the sphere is given as:
[tex]m_s = S.G_s*p_w*V_s[/tex]
Where,
ρ_w : Density of water = 1.940 slugs/ft3
V_s: The volume of object ( sphere )
- The volume of sphere is expressed as a function of radius:
[tex]V_s = \frac{\pi *D^3}{6}[/tex]
Hence,
[tex]W = S.G_s*p_w*\frac{\pi*D^3 }{6}* g\\\\W = 8*1.940*\frac{\pi*(0.25/12)^3 }{6}*32\\\\W = 0.00235 lb[/tex]
- One of the upward acting force is the buoyant force ( Fb ) that is proportional to the volume of fluid displaced by the immersed object.
- The buoyant force ( Fb ) is given by:
[tex]F_b = S.G_o*p_w*V_s*g[/tex]
- Therefore the buoyant force ( Fb ) becomes:
[tex]F_b = 0.8*1.94*\frac{\pi*(0.25/12)^3 }{6} *32\\\\F_b = (4.73451*10^-^6)*(49.664)\\\\F_b = 0.00023 lb[/tex]
- The other upward acting force is the frictional drag ( F_d ) i.e the resistive frictional force acting on the contact points of the sphere and the fluid oil.
- From stokes formulations the drag force acting on a spherical object which is completely immersed in a fluid is given as:
[tex]F_d = 3*\pi*D*u*v[/tex]
Where,
μ: The viscosity of fluid
v : The velocity of object
Therefore,
[tex]F_d = 3*\pi*\frac{0.25}{12} *u*0.041666\\\\F_d = 0.00818*u\\[/tex]
- Apply Newton's second law of motion for the sphere travelling in the fluid:
[tex]F_n_e_t = m_s*a[/tex]
Where,
a: Acceleration of object = 0 ( Terminal velocity condition )
[tex]F_n_e_t = 0[/tex]
- Plug in the three forces acting on the metal sphere:
[tex]F_d + F_b - W = 0\\\\F_d = W - F_b\\\\0.00818*u = 0.00235 - 0.00023\\\\u = \frac{0.00212}{0.00818} = 0.25916 \frac{lb-s}{ft^2}[/tex]
Which of the following is correct? When a current carrying wire is in your left hand, thumb in the direction of the magnetic field lines, your fingers point in the direction of the current. When a current carrying wire is in your left hand, thumb in the direction of the current, your fingers point in the direction of the magnetic field lines. When a current carrying wire is in your right hand, thumb in the direction of the magnetic field lines, your fingers point in the direction of the current. When a current carrying wire is in your right hand, thumb in the direction of the current, your fingers point in the direction of the magnetic field lines.
Answer:
When a current carrying wire is in your left hand, thumb in the direction of the magnetic field lines, your fingers point in the direction of the current
Explanation:
This is in line with the left hand rule
B.) When the current flows up the wire, the magnetic field flows out on the left side of the wire and in on the right side of the wire.
A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed uniformly throughout the volume of the shell. What is the magnitude of the electric field at a distance r = 11.2 cm from the center of the shell? (ε0 = 8.85 × 10-12 C2/N ∙ m2) (Give your answer to the nearest 0.01 MN/C)
Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
[tex]\int EdS=\frac{Q_{in}}{\epsilon_o}[/tex] (1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
[tex]\int EdS=ES=E(4\pi r^2)[/tex] (2)
Qin is calculate by using the charge density:
[tex]Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho[/tex] (3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
[tex]\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}[/tex]
Next, you use the results of (3), (2) and (1):
[tex]E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})[/tex]
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
[tex]E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}[/tex]
hence, the electric field is 1580594.95 N/C
"You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 9.00 s after it was thrown. What is the speed of the rock just before it reaches the water 21.0 m below the point where the rock left your hand
Answer:
48.54 m/s
Explanation:
If the rock takes 9 seconds to reach your position after being thrown, it reaches its maximum height in 4.5 seconds.
The height the rock reaches above your position is ...
h = (1/2)gt^2 = (4.9 m/s^2)(4.5 s)^2 = 99.225 m
This height is an additional 21 m above the water, so the maximum height above the water is ...
99.225 m +21.0 m = 120.225 m
The velocity (v) achieved when falling from this distance is found from ...
v^2 = 2gh
v = √(2(9.8)(120.225)) = √2356.41 ≈ 48.543 . . . . m/s
The speed of the rock when it hits the water is about 48.54 m/s.
A turntable has a moment of inertia of 3.00 x 10-2 kgm2 and spins freely on a frictionless bearing at 25.0 rev/min. A 0.300 kg ball of putty is dropped vertically on the turntable and sticks at a point 0.10m from the center. The total moment of inertia of the system increases, and the turntable slows down. But by what factor does the angular momentum of the system change after the putty is dropped onto the turntable
Answer:
There will be no change in the angular momentum of the system.
Explanation:
Total angular momentum of the system will remain unchanged . We can apply law of conservation of momentum because no external torque is acting on the system . There is increase in the momentum of inertia due to dropping of ball of putty . In order to conserve angular momentum , the system decreases its angular velocity . Hence the final angular momentum remains unchanged .
A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 3.7 mm will experience only elastic deformation when a tensile load of 1890 N is applied. Calculate the maximum length of the specimen before deformation if the maximum allowable elongation is 0.45 mm.
Answer:
L= 276.4 mm
Explanation:
Given that
E= 180 GPa
d= 3.7 mm
F= 1890 N
ΔL= 0.45 mm
We know that ,elongation due to load F in a cylindrical bar is given as follows
[tex]\Delta L =\dfrac{FL}{AE}[/tex]
[tex]L=\dfrac{\Delta L\times AE}{F}[/tex]
Now by putting the values in the above equation we get
[tex]L=\dfrac{0.45\times 10^{-3}\times \dfrac{\pi}{4}\times (3.7\times 10^{-3})^2\times 108\times 10^9}{1890}\ m[/tex]
L=0.2764 m
L= 276.4 mm
Therefore the length of the specimen will be 276.4 mm
Visible light travels in air, which has an index of refraction of 1.0. It is incident on plastic which is covered by a coating (such that light hits the coating before the plastic). The coating has an index of refraction of 1.6 and is 0.50 microns thick. The ray reflected off the air-coating interface and the ray reflected off the coating-plastic interface experience interference. What frequencies could the light be
Answer:
λ = 0.64 10⁻⁶ m = 640 nm
Explanation:
The two reflected rays experience constructive interference, if we can see them, so we can one condition for interference, but let's see two phenomena that occur
* A ray when it is reflected by a surface of major spare part index has a phase change of 180º, these passes at the air-plastic interface
* A ray when passing a material with a refractive index changes its wavelength
λ= λ₀ / n
taking into account these facts the condition in constructive interference is
2 n t = (m + ½) λ
λ= 2 n t / (m + ½)
if we suppose that we have the first inference m = 0
λ = 2 1.6 0.5 10⁻⁶ (0 + ½)
λ = 3.2 10⁻⁶ m
this wavelength is in the infrared
suppose an interference of m = 1
λ = 1.6 10⁻⁶ / (1 + 1/2)
λ = 1.06 10-6 m
m = 2
λ = 1.6 10⁻⁶ / (2 + 1.5)
λ = 0.64 10⁻⁶ m
this wavelength corresponds to the visible range (640 nm) orange
A skydiver just after leaping out of an airplane is it isinertial frames of reference or not?
Answer:
He starts accelerating until he reaches terminal speed. (Or the maximum velocity attainable by an object)
Explanation:
When we apply the energy conversation principle to a cylinder rolling down an incline without sliding, we exclude the work done by friction because: A. there is no friction present B. the angular velocity of the center of mass about the point of contact is zero C. the coefficient of kinetic friction is zero D. the linear velocity of the point of contact (relative to the inclined surface) is zero E. the coefficient of static and kinetic friction are equal
Answer:
D. the linear velocity of the point of contact (relative to the inclined surface) is zero
Explanation:
The force of friction emerges only when there is relative velocity between two objects . In case of perfect rolling , there is no sliding so relative velocity between the surface and the point of contact is zero . In other words the velocity of point of contact becomes zero , even though , the whole body is in linear motion . It happens due point of contact having two velocities which are equal and opposite . One of the velocity is in forward direction and the other velocity which is due to rotation is in backward direction . So net velocity of point of contact becomes zero . Due to absence of sliding , displacement due to friction becomes zero . Hence work done by friction becomes zero.
Each propeller of the twin-screw ship develops a full-speed thrust of F = 285 kN. In maneuvering the ship, one propeller is turning full speed ahead and the other full speed in reverse. What thrust P must each tug exert on the ship to counteract the effect of the ship's propellers?
Answer:
tug_tug = 570 10³ l
Explanation:
In this problem, each propeller creates a force that makes the boat rotate, so the tugs have to create a die of equal magnitude rep from the opposite direction
∑ τ = 0
F1 la+ (-F1) (-l) = τ-tug
τ-tug = 2 f1 l
τ-tug = 2 28510³ l
tug_tug = 570 10³ l
where the is the distance from the propane axis to the point where the ship turns
This force may be less depending on where the tug is.
Can a moving object have gravitational energy?
Answer:
It is worth noting that the higher the gravitational energy of an object moving downwards, the lower the kinetic energy, and the lower the kinetic energy of an object moving upwards, the higher its gravitational energy.
Gravitational potential energy is acquired by an object when it has been moved against a gravitational field. For example, an object raised above the surface of the Earth will gain energy, which is released if the object is allowed to fall back to the ground.
a 1200 kg trailer is hitched to a 1400 kg car. the car and trailer are traveling at 72 km.h when the driver applies the brakes on both the car and the trailer. knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N respectively, determine (a) the distance traveled by the car and trailer before they come to a stop and (b) the horizontal component of the force exerted by the trailer hitch
Answer:
a) 8.67m
b) 1000N
Explanation:
(a) To find the distance you use the second Newton Law for both car and trailer, in order to calculate the dis-acceleration of the system:
[tex]F=ma\\\\a_=\frac{F}{m}=\frac{5000N+4000N}{1400kg+1200kg}=3.46\frac{m}{s^2}[/tex]
once you have this value, you use the the following kinematic equation to calculate the distance traveled by both car and trailer:
[tex]v^2=v_o^2-2ax\\\\x=\frac{-v^2+v_o^2}{2a}[/tex]
v: final velocity=0
vo: initial velocity = 72km/h = 60 m/s
by replacing the values of these parameters you obtain for x:
[tex]x=\frac{-0m/s+60m/s}{2(3.46m/s^2)}\\\\x=8.67m[/tex]
(b) The horizontal component of the force exerted by the trailer hitch is given by:
[tex]F_T=5000N-4000N=1000N[/tex]
When a star explodes as a supernova, the expanding gas can pass through the interstellar medium triggering...
A.) an increase in variable star activity
B.) formation of new stars
C.) life to develop
D.) additional supernovas
B. Formation of new stars
1) Calculate the equivalent resistance of 2 resistors wired in parallel. The value of the resistors are
4 ohms and 8 ohms.
Explanation:
[tex]r = \: ( { \frac{4 \times 8}{4 + 8} } \\ r= \: ohm[/tex]2.67
Answer:2 2/3 ohms
Explanation:
equivalent resistance=R
r1=4 ohms
r2=8 ohms
For parallel connection
1/R=1/(r1) + 1/(r2)
1/R=1/4 + 1/8
1/R=(2x1+1x1)/8
1/R=(2+1)/8
1/R=3/8
Cross multiply
1x8=3xR
8=3R
Divide both sides by 3
8/3=3R/3
8/3=R
R=8/3 or 2 2/3 ohms
Dual Nature of Light
Assignment
Active
Explaining the Nature of Light
Why do scientists believe that light is made of streams of
particles?
Exercise should challenge your body and be at a greater intensity than your usual bif daily activity. Discuss
Answer:
A fan pushes hot air out of a vent and into a room. The hot air displaces cold air in the room, causing the cold air to move closer to the floor.
The hot air displacing the cold air is an example of transfer by
Explanation:
Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 210 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 191 degrees in a room at 64 degrees, determine when the coffee reaches a temperature of 156 degrees.
Answer:
Explanation:
The problem is based on Newton's law of cooling .
According to Newton's law
dQ / dt = k ( T - T₀ ) ,
dT / dt = k' ( T - T₀ ) ; dT / dt is rate of fall of temperature.
T is average temperature of hot body , T₀ is temperature of surrounding .
In the first case rate of fall of temperature = (210 - 191) / 2.5
= 7.6 degree / s
average temperature T = (210 + 191) /2
= 200.5
Putting in the equation
7.6 = k' ( 200.5 - 64 )
k' = 7.6 / 136.5
= .055677
In the second case :---
In the second case, rate of fall of temperature = (191 - 156) / t
= 35 / t , t is time required.
average temperature T = (156 + 191) /2
= 173.5
Putting in the equation
35 / t = .05567 ( 173.5 - 64 )
t = 5.74 minute .
A person drives north 6 blocks, then turns west, and drives 6 blocks. The driver then turns south and drives 6 blocks. How could the driver have made the distance shorter while maintaining the same displacement?
Answer:
Considering that there is no obstructions, he could go west from the start.
Explanation:
2. Air at a temperature of 20 ºC passes through a pipe with a constant velocity of 40 m/s. The pipe goes through a heat exchanger in which hot gases outside the pipe heat up the air to 820 ºC. It then enters a turbine with a velocity of 40 m/s and expands till the temperature falls to 620 ºC. The air stream loses 4.3 kW heat in the turbine. If the air flow rate is 2.5 kg/s, calculate (a) How much heat is transferred to the air in the heat exchanger. (b) The power output of the turbine.
Answer:
a) Q = 1436 kW
b) P ≈ 776 kW
Explanation:
Let's begin by listing out the given parameters:
T1 = 20 °C, u = 40 m/s, T2 = 820 °C, P = 4.3 kW, m = 2.5 kg/s, T3 = 510 °C, V1 = 40 m/s,
V2 = 40 m/s, V3 = 55 m/s, ṁ = 2.5 kg/s
To solve the question, we make this assumption that the size of the pipe is constant
a) No change in velocity implies that heat added is isochoric
Q = m * C * ΔT
Cv of air at 300 K(≈20 °C) = 0.718
Q = 2.5 * 0.718 * (820 − 20)
Q = 1436 kW
b) P = ṁ * Cp * ΔT + ṁ * (V2² - V3²) ÷ 2000] - Ql
V2² - V3² = 55² - 40² = 1425
ΔT = T2 - T3 = 820 - 510 = 310 °C
Cp of air at 300 K(≠20 °C) = 1.005 kJ/kgK
Ql = 4.3 kW
P = 2.5 * (1.005 * 310) + 2.5 * (1425 ÷ 2000) - 4.3
P = 778.875 + 1.78125 - 4.3 = 776.35625
P ≈ 776 kW
A landscaper is shopping for landscaping materials. She wants to use materials through which water flows easily.
Which materials should she choose? Select three options.
clay
gravel
granite
rocks with cracks
loosely packed soil
Next
Mark this and retum
Save and Exit
Submit
Answer:
He needs clay gravel and rocks with cracks
