Answer:
From plants and animals that die and fall to forests floor
Explanation:
Hope it helps
Mark brainliest
Which ray diagram demonstrates the phenomenon of absorption?
An illustration with a vector pointed right going through an opening in a boundary and splitting into 3 vectors. One up and to the right, one straight and one down to the right.
An illustration with a vector pointed right going through an opening in a boundary and turning into a transverse wave on the other side.
An illustration with a vector striking a boundary at an angle and a second vector coming off the boundary at the exact same angle.
Answer:
B
Explanation:
on edge
The illustration with a vector pointed right going through an opening in a boundary and turning into a transverse wave on the other side demonstrates the phenomenon of absorption, so, option B is correct.
What is absorption?Absorption, in wave motion, is the process by which a wave's energy is transferred to matter when the wave travels through it. The energy of an electromagnetic, acoustic, or other wave is related to the square of its amplitude, which is the maximum displacement or movement of a point on the wave.
The amplitude of a wave continuously diminishes as it travels through a substance. The medium is described as being transparent to a specific type of radiation if just a tiny portion of the energy is absorbed, whereas it is described as opaque if all the energy is lost.
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Which statement best explains how wave one compares with wave 2
A. wave one has louder sound
B. wave one has softer sound
C. wave one had higher pitch
D. wave one has a lower pitch
Answer:
D. wave one has a lower pitch
Explanation: because The sound an object makes changes depending on how fast it is vibrating. When an object vibrates quickly, high-pitched sounds are heard. Low-pitched sounds come from things that vibrate more slowly.
A ball is thrown vertically upward from the top of a 100 foot tower, with an initial velocity of 10 ft/sec. Its position function is s(t)=−16t2+10t+100.
a. What is its velocity in ft/sec when t = 2 seconds? (Solve by using instantaneous rate.)
b. Determine the equation of a line, in slope-intercept form, that passes through the points (5, 6) and (10, 2).
Answer;
-54ft/s
y = -4/5 x + 10
Explanation
Given the position of an object expressed by the function
s(t)=−16t²+10t+100
Velocity is the change in position with respect to time
v(t) = ds(t)/dt
v(t) = -32t + 10
When t = 2
v(2) = -32(2)+10
v(2) = -64+10
v(2) = -54
Hence the velocity of the object is -54ft/s
b) The standard equation of a line in point slope form is expressed as;
y = MX+c
M is the slope
c is the y-intercept
Given the coordinate (5, 6) and (10, 2)
M = 2-6/10-5
M = -4/5
Get the y-intercept
Substitute m = -4/5 and any point say (5,6) into the expression y = mx+c
6 = -4/5 (5) + c
6 = -4+c
c = 6+4
c = 10
Get the required equation
Recall that: y = mx+c
y = -4/5 x + 10
Hence the equation of a line, in slope-intercept form is y = -4/5 x + 10
William Gilbert used a terella in an attempt to explain _____.
why like poles of a magnet repel
why unlike poles of a magnet attract
why compasses point toward the north and south poles of the earth
the properties of lodestone
Answer:
why compasses point towards the north and south poles of the earth.
Explanation:
hope it helps!!
A 87 kg man has a total mechanical energy of 1780 J .If he is swinging downward and is currently 1.4 m above the ground, what is his speed? Use g = 10 m/s^2
Answer:
6.4m/s
Explanation:
The total mechanical energy of the man is 1780J.
This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.
Also, mass = 87kg
Kinetic energy = [tex]\frac{1}{2}[/tex] m v²
m is the mass
v is the velocity
1780 = [tex]\frac{1}{2}[/tex] x 87 x v²
v² = 40.9
v = 6.4m/s
The speed of sound in air is 340 m/s. A sound wave that is traveling through air has a wavelength of 17 m. What is the frequency of this sound? *
Answer:
Assuming the speed of sound in air is 345 m/s, determine the wavelength of this infrasonic sound wave. Let λ = wavelength. Use v = f • λ where v = 345 m/s and f = 10 Hz.
Explanation:nothing is my explanation lol
I WILL GIVE YOU BRAINLIEST! PLEASE HELP
A daydreaming soccer player takes a 0.47 kg ball to the face experiencing an impact force of 1060.9 N. If the ball hit the
player's face with a speed of 14.5 m/s and bounces off in the opposite direction with the same speed, calculate the time of
impact. Note: The time of impact will be a fraction of a second so answer with at least 5 decimal places
Answer:
0.00547s
Explanation:
Step one:
given data
mass= 0.4kg
force= 1060.9N
velocity = 14.5m/s
Frome
Ft= mv
substitute
t= mv/F
t=0.4*14.5/1060.9
t=5.8/1060.9
t=0.00547s
True or False do Eclipses, tides, season, and moon phases ALL have to do with the positions of the Earth, Sun, and Mars.
Explain the relationship between the current output of the power supply and the current through each component in the parallel circuit.
Explanation:
Current output at the battery will be current of entire circuit, while the current through each bulb in the parallel circuit is the total current circuit.
So, current output through power supply is i and current through each component be [tex]i_1, i_2 , i_3[/tex] considering only three component.
Then in a parallel circuit
[tex]i = i_1+i_2+i_3[/tex]
This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.
(i) Including the water in its internal cavity, the squid has a total mass of 6.50 kg.
(ii) The mass of the water in its cavity is 1.75 kg.
(iii) In order to escape its predators, the squid needs to achieve an escape speed of 2.5 m/s.
Answer:
6.79 m/s
Explanation:
By applying the principle of conservation of momentum.
The total momentum = MV - mv = 0 (since the squid is beginning at rest)
the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg
= 4.75 kg
speed of the squid (V) = 2.5 m/s
mass of the water expelled (m) = 1.75 kg
speed of the water (v) = ???
∴
4.75 × 2.5 = 1.75 × v
[tex]v = \dfrac{4.75 \times 2.5}{1.75 }[/tex]
v = 6.79 m/s
A U-tube manometer is used to measure the pressure at the stagnation point of a model in a wind tunnel. One side of the manometer goes to an orifice at the stagnation point; the other side is open to the atmosphere ( Fig. P1.24 ). If there is a difference of 3.0 cm in the mercury levels in the two tubes, what is the pressure difference in N>m2 .
Answer:
ΔP = 3.98 10³ Pa
Explanation:
We use that the pressure is given
P = ρ g y
where ρ is the density of mercury (ρ = 13534 kg / m³) and y the height of the column
In this case, to measure the pressure, a line is drawn at the lowest point of the mercury column, on the right side of the U-shaped tube, for this point the pressure on both sides of the tube is the same
ΔP = P_l{eft} - P_{right}
ΔP = (P_a + P) - P_a
done P is the gauge pressure on the left side and P_a is the atmospheric pressure
ΔP = ρ g h
let's calculate
ΔP = 13534 9.8 0.03
ΔP = 3.98 10³ Pa
remember that [Pa] = [N / m²]
A 100 V battery is connected across a
resistor and causes 5 mA current to flow.
Determine the resistance of the resistor.
If the voltage is now reduced to 25 V,
what will be the new value of the current
flowing?
1 mA = 0.001 A
Therefore, 5 mA = 0.001 * 5
=0.005 A
Resistance = voltage / current
= 100 / 0.005
= 20000 ohms
Current = voltage / resistance
= 25 / 20000
= 0.00125 A (or) 1.25 mA
A flywheel of mass 182 kg has an effective radius of 0.62 m (assume the mass is concentrated along a circumference located at the effective radius of the flywheel).
(a) How much work is done to bring this wheel from rest to a speed of 120 rev/min in a time. interval of 30.0 s?
(b) What is the applied torque on the fly-wheel (assumed constant)?
Answer:
A)5524J,
B) 29.2Nm
Explanation:
This question can be treated using work- energy theorem
Work= change in Kinectic energy
W= Δ KE
Work= difference between the final Kinectic energy and intial Kinectic energy.
We know that
Kinectic energy= 1/2 mv^2 .............eqn(1)
This can be written in term of angular velocity, as
KE= 1/2 I
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.69 times a second. A tack is stuck in the tire at a distance of 0.331 m from the rotation axis. Noting that for every rotation the tack travels one circumference, find the tack's tangential speed.
Answer:
the tack's tangential speed is 5.59 m/s
Explanation:
Given that;
R = 0.331 m
wheel rotates 2.69 times a second which means, the wheel complete 2.69 revolutions in a second, so
ω = 2.69 rev/s × 2π/1s = 16.9 rad/s
using the relation of angular speed with tangential speed
tangential speed v of the tack is expressed as;
v = R × ω
so we substitute
v = 0.331 m × 16.9 rad/s
v = 5.59 m/s
Therefore, the tack's tangential speed is 5.59 m/s
What is the correct organization of living things, from smallest to largest?
Cells - Tissues - Organs - Organ Systems - Organism
Organs - Tissues - Cells - Organ Systems - Organism
Cells - Organs - Tissues - Organism - Organ Systems
Cells - Organism - Tissues - Organ Systems - Organs
A certain force gives object m1 an acceleration of 12.0 m/s2. The same force gives object m2 an acceleration of 3.30 m/s2. What acceleration would the force give to an object whose mass is (a) the difference between m1 and m2 and (b) the sum ofm1 andm2
Answer:
a) a = 4,552 m / s², b) a = 2,588 m / s²
Explanation:
Newton's second law is
F = ma
a = F / m
in this case the force remains constant
indicate us
* for a mass m₁
a₁ = F/m₁
a₁ = 12, m/ s²
* for a mass m₂
a₂= 3.3 m / s²
a) acceleration
m = m₂-m₁
we substitute
a = [tex]\frac{F}{m_2 - m_1}[/tex]
1 / a = [tex]\frac{m_2}{F} - \frac{m_1}{F}[/tex]
let's calculate
[tex]\frac{1}{a}[/tex] = [tex]\frac{1}{3.3} - \frac{1}{12}[/tex]
[tex]\frac{1}{a}[/tex] = 0.21969
a = 4,552 m / s²
b) m = m₂ + m₁
a = F / (m₂ + m₁)
[tex]\frac{1}{a} = \frac{m_2}{F} + \frac{m_1}{F}[/tex]
we substitute
[tex]\frac{1}{a} = \frac{1}{3.3} + \frac{1}{12}[/tex]
a = 2,588 m / s²
Why is damage from sound waves is an issue on the launchpad but not in the air
(I would have done more points for answering but I'm almost out sry. just pls answer and help.)
The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
Closeness to the Sound Source: When a rocket is fired on the launchpad, it creates a tremendous amount of noise in proximity to the nearby equipment and structures.
When a rocket is launched, concentrated sound waves are created that can seriously harm neighboring structures, especially if such structures are not built to handle such strong vibrations. In contrast, once a rocket is in the air, the sound waves spread out and become less forceful as they travel through the atmosphere, decreasing the possibility that they may cause harm.
Reflection and Amplification: The launchpad environment can serve as an echo chamber for sound waves because of its huge, solid structures.
Hence, The sound wave does not damage the air because no external factors such as reflection, amplification, and vibrations are present. However, in the launch pad factors such as reflection, amplification, and vibrations are present which damages the sound wave.
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Which action will leave the dump trucks inertia unchanged?? PLEASE ANSWER FAST!!!
A. add gas
B. increase force applied to engine
Answer:
B.
Explanation:
Can a single atom be considered a molecule?
A:only if the atom is found in water
B:no, it takes two or more atoms bonded to create a molecule
C:only if it is an oxygen atom floating in the air
D:yes, all atoms are made up of many different molecules
what is the result of seafloor spreading?
Answer:
Seafloor spreading occurs at divergent plate boundaries. As tectonic plates slowly move away from each other, heat from the mantle's convection currents makes the crust more plastic and less dense. The less-dense material rises, often forming a mountain or elevated area of the seafloor. Eventually, the crust cracks.
Explanation:
eventually the crust cracks.
How many miles per day can you walk at a MODERATE Intensity level and your heart rate is 170?
Answer:
Not enough detail as it is very defendant on the person and a bunch of factors in health, but overall your heart rate shouldn't reach 170 as an adult walking at a moderate intensity level, that would be closer to extreme intensity.
Explanation:
According to Newton’s second law of motion, when an object is acted on by an unbalanced force, how will that object respond?
It will stop moving.
It will accelerate.
It will decrease speed.
It will increase velocity.
It will accelerate. A force of magnitude F exerted on an object of mass m applies an acceleration a according to
F = m a
Remember that force and accleration are vector quantities, so the object's acceleration would point in the same direction as the applied force. The other choices describe some possible outcomes, but
• it the object starts in motion, it can only stop moving if the force opposes the motion and acts in the opposite direction. The object would eventually stop, but only for an instant before changing direction and starting to move again. Consider a ball being thrown directly upward, reaching its highest point, then falling again.
• its speed (which is a scalar quantity) would only decrease if the object starts in motion and is slowed down. But if it starts at rest, its speed can only increase. Consider a ball being dropped from some height and gaining speed as it falls.
• its velocity will certainly change, but can only increase if the object is at rest, or if it's already in motion and the force points in the same direction.
Answer:
It will accelerate.
Explanation:
i go to K12
Introduction: The specific heat capacity of a substance is the amount of energy needed to change the temperature of that substance by 1 °C. Specific heat capacity can be calculated using the following equation:
q = mc deltaT
In the equation q represents the amount of heat energy gained or lost in joules), m is the mass of the substance (in grams), c is the specific heat capacity of the substance (in J/g °C), and AT is the temperature change of the substance in °C).
Goal: Calculate the specific heat capacities of copper, granite, lead, and ice.
Solve: When you mix two substances, the heat gained by one substance is equal to the heat lost by the other substance. Suppose you place 125 g of aluminum in a calorimeter with 1,000 g of water. The water changes temperature by 2 °C and the aluminum changes temperature by -74.95 °C.
A. Water has a known specific heat capacity of 4.184 J/g °C. Use the specific heat equation to find out how much heat energy the water gained (q).
B. Assume that the heat energy gained by the water is equal to the heat energy lost by the aluminum. Use the specific heat equation to solve for the specific heat of aluminum. Aluminum's accepted specific heat value is 0.900 J/g °C. Use this value to check your work.
Answer:
A) 8,368 J
B) ) 0.893 J/gºC
Explanation:
A)
The heat gained by the water can be obtained solving the following equation:[tex]q_{g} = c_{w} * m * \Delta T (1)[/tex]
where cw = specific heat of water = 4.184 J/gºCm= mass of water = 1,000 gΔT = 2ºC Replacing these values in (1) we get:[tex]q_{g} = c_{w} * m * \Delta T = 4.184 J/gºC*1,000 g* 2ºC = 8,368 J (2)[/tex]
B)
Assuming that the heat energy gained by the water is equal to the one lost by the aluminum, we can use the same equation, taking into account that the energy is lost by the aluminum, so the sign is negative: -8,368 J.Replacing by the mass of aluminum (125 g), and the change in temperature (-74.95ºC), in (1), we can solve for the specific heat of aluminum, as follows:[tex]q_{l} = c_{Al} * m_{Al} * \Delta T (3)[/tex]
⇒ [tex]-8,368 J = c_{Al}* 125 g * (-74.95ºC) (4)[/tex]
[tex]c_{Al} = \frac{-8,368J}{125g*(-74.95ºC} = 0.893 J/gºC (5)[/tex]
which is pretty close to the Aluminum's accepted specific heat value of 0.900 J/gºC.
WILL MARK BRAINLIEST Why do we see sedimentary rocks more often than igneous and metamorphic rocks?
O Because they are extrusive
O Because they are intrusive
O because they are the rocks found in the ocean
O because they are the rocks that line the surface of our planet
Answer:
because they are the rocks that line the surface of our planet
Explanation:
We see sedimentary rocks more than other rock types because they are the rocks that line the surface of our planet.
Sedimentary rocks typically form the earth cover due to the way they are formed.
These rocks are produced by the weathering, transportation and deposition of sediments within a basin. In this basin, the sediment is lithified and converted to sedimentary rocks. These processes are driven by the external heat engineTherefore, it is confined to the surface. Igneous and metamorphic rock's processes are confined to the subsurface.
A 50kg refrigerator is being moved across a kitchen floor with an applied force of
300N. There is a known friction force of 50N acting against the motion of the
refrigerator.
What was the acceleration of the refrigerator? [ Select ]
m/s2
Answer:
5m/s²
Explanation:
Given parameters:
Mass of refrigerator = 50kg
Applied force = 300N
Frictional force = 50N
Unknown:
Acceleration of the refrigerator = ?
Solution:
To solve this problem:
Net force = m x a
m is the mass
a is the acceleration
Net force = Applied force - Frictional force
300 - 50 = 50 x a
250 = 50 x a
a = 5m/s²
Which of the following does NOT describe a major function of the reproductive system?
a. produce sex cells
b. produce hormones
c. produce waste products
d. produce offspring
1. The volume of a given mass of gas is 20cm when its
pressure is 400mmHg. Calculate its pressure when the
volume becomes 80cm'at constant temperature,
Answer:
Explanation:
The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.
100 mmHg
Givens
V1 = 20 cm^3
V2 = 80 cm^3
P1 = 400 mmHg
P2 = ?
Formula
V1 * P1 = V2 * P2
Solution
20 * 400 = 80 * P2 Divide by 80
20 * 400/80 = P2
P2 = 8000 / 80
P2 = 100 mmHg
PLEASEEEEEE HELPPPPPPPPPPP
Describe how a battery works.
Answer:
A battery works because of the electrical energy inside of the battery, which is connected through a wire, which transmits energy through the metal inside of the wire to power to object that the wire is connected to.
Explanation:
Please mark as Brainliest!!!How do I solve this problem please help
Answer:
som,eythiung did cant wait
Explanation:
ya so i texted maria and she said i con c()me over so letyeyey yeah betty letty
PAINN I MISS JUOICE WORLDD PAINNNN
Heather drives her Super-Beetle around a turn on a circular track which has a radius of 200 m. The Super-Beetle has a mass of 1500 kg and the coefficient of static friction between the road and tires is 0.6.
a. What is the force of static friction the road can apply batore the car starts to selon (use Ft= uFn).
b. What is the maximum speed the car can travel before it would start to slide?
Answer:
a) The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.
Explanation:
a) Let suppose that the car is on a horizontal ground and travels at constant speed. The vehicle experiments a centripetal acceleration due to friction, which can be seen in the Free Body Diagram (please see image attached for further details). By Newton's Laws, we construct the following equations of equilibrium:
[tex]\Sigma F_{x} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R}[/tex] (1)
[tex]\Sigma F_{y} = N -m\cdot g = 0[/tex] (2)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]N[/tex] - Normal force from ground to the car, measured in newtons.
[tex]v[/tex] - Maximum speed of the car, measured in meters per second.
[tex]R[/tex] - Radius of the circular track, measured in meters.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
By applying (2) in (1):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex] (3)
The force of static friction the road can apply in the car ([tex]f[/tex]), measured in newtons, is: ([tex]\mu_{s} = 0.6[/tex], [tex]m = 1500\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex])
[tex]f = \mu_{s}\cdot m \cdot g[/tex]
[tex]f = (0.6)\cdot (1500\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)[/tex]
[tex]f = 8826.3\,N[/tex]
The force of static friction the road can apply before the car starts to move is 8826.3 newtons.
b) Then, we calculate the maximum speed of the car by (3):
[tex]\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}[/tex]
[tex]\mu_{s}\cdot g = \frac{v^{2}}{R}[/tex]
[tex]v = \sqrt{\mu_{s}\cdot g\cdot R}[/tex]
If we know that [tex]\mu_{s} = 0.6[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]R = 200\,m[/tex], then the maximum speed of the car can travel before it would start to slide is:
[tex]v =\sqrt{(0.6)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (200\,m)}[/tex]
[tex]v \approx 34.305\,\frac{m}{s}[/tex]
The maximum speed that a car can travel before it would start to slide is approximately 34.305 meters per second.