Answer:
1.Divergent boundaries (crust moves apart, magma fills in) Convergent boundaries (magma fills when one plate goes beneath another) Hot spots (a large magma plume rises from mantle)
2.Some mountains are made of solid rock, like the Rocky Mountains or the Swiss Alps. But, some mountains are actually volcanoes.
3.an area under the rocky outer layer of Earth, called the crust, where magma is hotter than surrounding magma.
4.hydrogen sulfide, carbon monoxide, hydrogen chloride, hydrogen fluoride, and other minor gas species.
5.Extinct volcanoes are not expected to erupt in the future. Inside an active volcano is a chamber in which molten rock, called magma, collects. Pressure builds up inside the magma chamber, causing the magma to move through channels in the rock and escape onto the planet's surface.
Explanation:
hope its help:)
Which is the correct ranking of the speed of light in a substance from fastest to *
slowest
Diamond, Glass, Water, Helium
Helium, Water, Glass, Diamond
Water, Helium, Diamond, Glass
Sugar, Diamond, Glass, Helium
Answer:
Helium, Water, Glass, Diamond
Explanation:
light travels slowest though a diamond
If the period of oscillation of a simple pendulum is 4s, find its length. If the velocity of the bob
at the mean position is 40cms−1
, find its amplitude. (take gravity = 9.81ms−2
Answer:
Explanation:
Because we assume the pendulum is a "mathematical pendulum" (neglecting the moment of inertia of the bob), we can find:
[tex]T=2\pi\sqrt{\frac{L}{g}} \rightarrow 4=2\pi\sqrt{\frac{L}{9.81}} \rightarrow \frac{4}{\pi^{2}}=\frac{L}{9.81} \rightarrow L \approx 3.97 m[/tex]
By using the [tex]y=A\sin(\omega t) \rightarrow v = \frac{dy}{dt}=\omega A \cos\omega t = \omega\sqrt{A^{2}-y^{2}}[/tex]
The mean position is the position when y = 0, so:
[tex]\omega = \frac{2\pi}{T}=\frac{2\pi}{4}=0.5\pi[/tex] rad/s
and [tex]v = \omega A \rightarrow A=\frac{40}{0.5\pi}=\frac{80}{\pi}[/tex] in centimeters (cm).
A ball which is dropped from the top of building strikes the ground in 2.6 sec. Assume air resistance can be ignored. The height of the building is
approximately
Compare and contrast heat and temperature. Cite examples from the readings to support your comparison. Also cite examples from what you may already know.
(THIS NEEDS TO BE LONG I WILL GIVE 100 POINTS!!!!)
Answer:
Heat is a form of energy. It is measured in Joules. ... The temperature of an object relates to both the kinetic energy of its particles and the number of particles. Temperature is the average kinetic energy of the particles in a substance.
Explanation:
The two basic properties of all matter are ______ and ______.
options for first blank
a. length B. volume
options for second blank
a. density B. mass
Answer:
volume , mass
Explanation:
the 3 basic properties of a matter is volume, mass and shape.
1 If electromagnetic radiation acted like particles in the double-slit experiment, what would be observed?
a The screen would remain dark because no radiation would reach the screen.
b One bright band would appear in the center of the screen.
c A series of light and dark bands would appear on the screen.
d Two bright bands would appear on the screen in line with the slits.
2 Which statement about the interference behavior of electromagnetic radiation seen in the double-slit test experiment is true?
a Waves that make up the radiation collide with each other so that they add together or cancel each other out.
b Particles that make up the radiation collide with each other and scatter randomly.
c Particles that make up the radiation collide with each other so that they add together or cancel each other out.
d Waves that make up the radiation do not interact with each other.
3 Which statement about the observed results of the double-slit experiment is true?
a Waves that are out of phase constructively interfere to create bright bands.
b Waves that are in phase destructively interfere to create bright bands.
c Waves that are out of phase constructively interfere to create bright bands.
d Waves that are in phase constructively interfere to create bright bands.
4 Which statement about the observed results of the double-slit experiment is true?
a Waves that are in phase constructively interfere to form dark bands.
b Waves that are out of phase constructively interfere to form dark bands.
c Waves that are in phase destructively interfere to form dark bands.
d Waves that are out of phase destructively interfere to form dark bands.
5 A scientist decreases the wavelength of the light used in a double-slit experiment and keeps every other aspect the same. What will be true about the new interference pattern seen on the screen compared to the original interference pattern?
a The spacing between the dark fringes will increase.
b The spacing between the bright fringes will increase.
c The spacing between the bright fringes will decrease.
d The spacing between the dark fringes will remain the same.
6 Consider the two-slit interference experiment. Electromagnetic radiation passes through the two slits that are a distance of 0.0170 nm apart. A fourth-order bright fringe forms at an angle of 8.0 degrees relative to the incident beam. What is the wavelength of the light?
a 789 nm
b 420 nm
c 581 nm
d 591 nm
Answer:
1. Two bright bands would appear on the screen in line with the slits.
2. Waves that make up the radiation collide with each other so that they add together or cancel each other out.
3. Waves that are in phase constructively interfere to create bright bands.
4. Waves that are out of phase destructively interfere to form dark bands.
5. The spacing between the bright fringes will decrease.
6. 581 nm
Explanation:
QUESTION 9
What determines the evolutionary path a star takes through its life?
O a. Diameter
O b. Who made it
O c. Color
O d. Mass
Answer:
Mass
Explanation:
In the core of the red giant, helium fuses into carbon. All stars evolve the same way up to the red giant phase. The amount of mass a star has determines which of the following life cycle paths it will take from there. The life cycle of a low mass star (left oval) and a high mass star (right oval).May 7, 2015
Problem 19: Oil with a density of 892 kg/m is flowing smoothly through a pipe, as
shown. In the lower portion, the oil is flowing at vi = 1.84 m/s, and the pressure gauge
indicates P, = 237 kPa. In the upper portion of the pipe, oil is flowing at V2 = 3.61 m/s at a
height that is 8.63 m above the lower portion.
The pressure of the oil with the given properties flowing in the horizontal pipe at the upper portion of the pipe is 157 kPa.
Pressure in the upper portion of the pipe
The pressure in the upper portion of the pipe is calculated by applying Bernoulli's equation,
P₁ + ¹/₂ρv₁² + ρgh₁ = P₂ + ¹/₂ρv₂² + ρgh₂
Given;
P₁ = 237 kPav₁ = 1.84 m/sh₁ = 0v₂ = 3.61 m/sh₂ = 8.63 mρ = 892 kg/m³P₂ = ?Susbtsitute the given parameters and solve the for the pressure in the upper portion of the pipe.
237,000 + ¹/₂(892)(1.84)² + (892)(9.8)(0) = P₂ + ¹/₂(892)(3.61)² + (892)(9.8)(8.63)
238,509.9776 = P₂ + 81,252.325
P₂ = 238,509.9776 - 81,252.325
P₂ = 157,257.65 Pa
P₂ ≅ 157 kPa
Thus, the pressure of the oil with the given properties flowing in the horizontal pipe at the upper portion of the pipe is 157 kPa.
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A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figure below. The string goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is k = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the string does not stretch and does not slip on the pulley, and that the mass of the string is negligible. Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown.
The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
Angular Speed of the pulley
The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;
K.E = P.E
[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]
[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]
Substitute the given parameters and solve for the angular speed;
[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]
Linear speed of the blockThe linear speed of the block after travelling 0.7 m;
v = ωR₂
v = 35.39 x 0.03
v = 1.1 m/s
Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
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Two cars collide at an intersection. One car has a mass of 1300 kg and is
moving 12 m/s to the north, while the other has a mass of 1400 kg and is
moving 11 m/s to the south. What is their combined momentum?
O A. 31,000 kg m/s south
O B. 200 kg m/s north
O C. 31,000 kg m/s north
O D. 200 kg m/s south
Answer:B
Explanation:sorry they removed my answer for some reason.
Answer: B
Explanation: Sorry these brainly trolls deleted my first one
In the figure below (Figure 1), the upper ball is released from rest, collides with the stationary lower ball, and sticks to it. The strings are both 50.0 cm long. The upper ball has a mass of 2.20 kg and it is initially 10.0 cm higher than the lower ball, which has a mass of 2.70 kg. Find the frequency of the motion after the collision. Find the maximum angular displacement of the motion after the collision.
(a) The frequency of the motion after the collision is 0.71 Hz.
(b) The maximum angular displacement of the motion after the collision is 16.3⁰.
Speed of the 2.2 kg ball when it collides with 2.7 kg ballThe speed of the 2.2 kg ball which was initially 10 cm higher that 2.7 kg ball is calculated as follows;
K.E = P.E
¹/₂mv² = mgh
v² = 2gh
v = √2gh
v = √(2 x 9.8 x 0.1)
v = 1.4 m/s
Final speed of both balls after collisionThe final speed of both balls after the collision is determined from the principle of conservation of linear momentum.
Pi = Pf
m₁v₁ + m₂v₂ = vf(m₁ + m₂)
2.2(1.4) + 2.7(0) = vf(2.2 + 2.7)
3.08 = 4.9vf
vf = 3.08/4.9
vf = 0.63 m/s
Maximum displacement of the balls after the collisionP.E = K.E
[tex]mgh_f = \frac{1}{2} mv_f^2\\\\h_f = \frac{v_f^2}{2g} \\\\h_f = \frac{(0.63)^2}{2(9.8)} \\\\h_f = 0.02 \ m[/tex]
Maximum angular displacementThe maximum angular displacement of the balls after the collision is calculated as follows;
[tex]cos \theta = \frac{L - h_f}{L} \\\\cos\theta = \frac{0.5 - 0.02}{0.5} \\\\cos\theta = 0.96\\\\\theta = cos^{-1}(0.96)\\\\\theta = 16.3 \ ^0[/tex]
Frequency of the motion[tex]f = \frac{1}{2\pi} \sqrt{\frac{g}{L} } \\\\f = \frac{1}{2\pi } \sqrt{\frac{9.8}{0.5} } \\\\f = 0.71 \ Hz[/tex]
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how to make sure a snow leopard does not escape?
build a fence.....................
The diagram shows a heating curve for water.
Which statement best describes what is happening at
segment W-X?
Heating Curve for Water
O Water is absorbing heat.
O Water is releasing heat.
O Aliquid is changing to a gas.
O A gas is changing to a liquid.
z
110
100
90
x
х
Y
80
Temperature (Cº)
70
60
50
40
w
Heat (J)
Answer:
a water is heating up
Explanation:
the water's temperature is inceasing
A person sitting on a pier observes incoming waves that have a sinusoidal form with a distance of 2.5 m between the crests. If a wave laps against the pier every 5.0 s, what are (a) the frequency and (b) the speed of the wave?
Remember to identity all of your data, write the equation, and show your work. Additionally, be sure you have answered both a and b of this question.
Answer:The focal length of a lens is determined by the curve of the lens and the material that the lens is made from.
Explanation:
i just took the test
The frequency is 0.25Hz and the speed of the wave is 0.4m/s.
What is frequency and speed of the wave?Wave frequency is the number of waves that pass a fixed point in a given amount of time. This equation shows how the three factors are related: Speed = Wavelength x Wave Frequency.
To find the frequency and speed of the wave.λ=1.6m,T=4s
f=1/T
f=1/4s
f=0.25Hz
The frequency of the wave is 0.25Hz
speed=fλ
speed=(0.25×1.6)m/s
speed=0.4m/s.
The speed of the wave is 0.4m/s.
The frequency is 0.25Hz and the speed of the wave is 0.4m/s.
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17. How long will it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s? (A) 0.6s (B) 1.2 s (C) 2.4 s (D) 3.6 s (E) 4.8 s
Answer:
(B)
Explanation:
Time = change of velocity ÷ acceleration
= (6-0) ÷ 5
= 1.2
1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
To calculate the time it takes for an object to change its velocity from 0 to 6 m/s, we can use the formula:
time = change in velocity / acceleration
Given that the change in velocity (Δv) is 6 m/s and the acceleration (a) is 5 m/s², we can plug these values into the formula:
time = 6 m/s / 5 m/s²
time = 1.2 seconds
Therefore, 1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
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If a snowboarder’s initial speed is 4 m/s and comes to rest when making it to the upper level. With a slightly greater initial speed of 5 m/s, the snowboarder is moving to the right on the upper level. His final speed in this case is 3 m/s. Suppose this situation is repeated on planet Epsilon, where gravity is less than gravity on earth.
A. Would the height of the hill on Epsilon cause a reduction in speed from 4 m/s to 0 greater than, less than, or equal to the height of the corresponding hill on earth? Explain.
B. Consider the hill on Epsilon discussed in part A. If the initial speed at the bottom of the hill is 5 m/s, will the final speed at the top of the hill be greater than, less than, or equal to 3 m/s? Explain.
(a) At a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
(b) If the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
Conservation of mechanical energy
The effect of height and gravity on speed on the given planet Epislon is determined by applying the principle of conservation of mechanical energy as shown below;
ΔK.E = ΔP.E
¹/₂m(v²- u²) = mg(hi - hf)
¹/₂(v²- u²) = g(0 - hf)
v² - u² = -2ghf
v² = u² - 2ghf
where;
v is the final velocity at upper levelu is the initial velocityhf is final heightg is acceleration due to gravitywhen u² = 2gh, then v² = 0,
when gravity reduces, u² > 2gh, and v² > 0
Thus, at a corresponding hill on Earth and a lesser gravity on planet Epslion, the height of the hill will cause a reduction in the initial speed of the snowboarder from 4 m/s to a value greater than zero (0).
Final speedv² = u² - 2ghf
where;
u is the initial speed = 5 m/sg is acceleration due to gravity and its less than 9.8 m/s²v is final speedhf is equal heightSince g on Epislon is less than 9.8 m/s² of Earth;
5² - 2ghf > 3 m/s
Thus, if the initial speed at the bottom of the hill is 5 m/s, the final speed at the top of the hill be greater than 3 m/s.
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what does a snow leopard eat?
Answer:
They eat Himalayan blue sheep, boar, deers, wild mountain goats, pikas etc.
Explanation:
Explain the use of each part of Vernier callipers?
A Vernier caliper has total of four jaws, with two upper jaws used for measuring the internal distances and two lower jaws for measuring the internal distances of objects. The two upper jaws are the smaller jaws of the vernier that are used to measure the internal distances between two parallel sides of an object or an internal diameter.
If you go into a store to buy a digital thermometer and there are 8 instruments of the same brand and model, at the same price, but they indicate different temperature values, explain which one you choose and why.
Answer: I would buy the thermometer with the most features because they all cost the same price so why not you getting more than what you are paying for than buying something the same but with less feature and things it can do for the same price.
Hope this helps :) If so brainliest would mean alot thanks!
Explanation:
What are the Applications of electronic polarization?
Answer:
it can be used for communication, industrial application and also for instrumentation application
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
a. What is surface area being vibrated by the time the sound reaches the listener (you may
have to look up the equation for the area of a sphere).
b. What is the intensity of the sound wave as it reaches the person listening?
c. What is the relative intensity of the sound as heard by the listener?
2. A police car siren has a frequency of 512 Hz. They are chasing a suspect at 65.0 m/s.
a. What is the speed of sound if the air temperature is 15.0o?
b. If the suspect is running away from the police car at 10.0 m/s, what is the frequency of
the sound he hears?
(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².
(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².
(1.c) The relative intensity of the sound as heard by the listener is 103 dB.
(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.
(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.
Surface area being vibratedThe surface area being vibrated by the time the sound reaches the listener is calculated as follows;
A = 4πr²
A = 4π x (20)²
A = 5,026.55 m²
Intensity of the soundThe intensity of the sound is calculated as follows;
I = P/A
I = (100) / (5,026.55)
I = 0.02 W/m²
Relative intensity of the sound[tex]B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB[/tex]
Speed of sound at the given temperature[tex]v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s[/tex]
Frequency of the soundThe frequency of the sound heard is determined by applying Doppler effect.
[tex]f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )[/tex]
where;
-v₀ is velocity of the observer moving away from the source-vs is the velocity of the source moving towards the observerfs is the source frequencyfo is the observed frequencyv is speed of sound[tex]f_0 = f_s(\frac{v-v_0}{v- v_s} )[/tex]
[tex]f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz[/tex]
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A car travel from A to B at a speed of 30km per hour the average speed of the car the whole journey is?
What is the difference between magnitude of displacement and magnitude of velocity
Which angle is the angle of refraction?
Explanation:
[tex]\longmapsto\: \red{ hello \: mate}[/tex]
OPTION (C) 3
is the angle of refraction
Answer:
Option C. 3
Explanation:
Refraction is the change in direction of light passing from one medium to another. Angle 1 and 2 are reflection due to the bouncing of light.
a person have a kinetic energy = 32, if the mass increase two times and the velocity decrease four times, What would be the kinetic energy?
Answer:
4 J
Explanation:
Original KE
KE = 1/2mv²⇒ 32 = 1/2mv²Mass increases x 2
KE ∝ mm x 2 ∝ KE x 232 x 2KE' = 64Velocity decreases x 4
KE' ∝ v²KE' x (1/4)² ∝ (1/4)²KE' x 1/16 = 64 x 1/16 = 4 Jhelp me solve the following question below in the attachment
Answer:
Explanation:
The point P is influenced by two electric fields, that are from Q1 and Q2. Because Q1 is a positive point charge, the E-fields is going to the right and the E-fields from Q2 is going to the left because of negative point charge. So:
[tex]E_{1}=\frac{kQ_{1}}{R_{1}^{2}}[/tex]
[tex]E_{2}=\frac{kQ_{2}}{R_{2}^{2}}[/tex]
Where R1 = 2 + 3 = 5 m, and R2 = 3 m. With k = [tex]k=9\times 10^{9}[/tex] in [tex]Nm^{2}/C^{2}[/tex], we can obtain:
[tex]E_{1}=(9\times 10^{9})\frac{500\times10^{-6}}{25}=180\times 10^{3}[/tex] N/C (+)
[tex]E_{2}=(9\times 10^{9})\frac{100\times 10^{-6}}{9}=100\times 10^{3}[/tex] N/C (-)
So: the Electric Field at P : [tex]E_{P}=E_{1}-E_{2}=(180-100)\times 10^{3}=8\times 10^{4}[/tex] N/C in magnitude
Calculate the electric field at x = 2 given the electric potential at this point is ϕ=3x3.
Hi there!
Recall the following relationship:
[tex]E = -\frac{dV}{dx}[/tex]
E = Electric field strength (V/m)
V = Potential Difference (V)
Take the derivative of the given electric potential equation using the power rule:
[tex]\frac{d}{dx}x^n = nx^{n-1}[/tex]
[tex]-\frac{dV}{dx} = -(3 \cdot 3 \cdot x^2) = -9x^2[/tex]
Evaluate at x = 2:
[tex]E(2) = -9(2^2) = \boxed{-36 \frac{V}{m}}[/tex]
**The magnitude of the field strength would simply be 36 V/m.
Why is gravitational force always towards the center?
Answer:
i beleave cuz of the Earth is spherical
Explanation:
If a lever was a meter long and the fulcrum was placed at 50mm, what would the mechanical advantage be?
Please show work if possible
Answer:
The mechanical advantage would depend on where the load force and the balancing force were placed on the stick
MA = (50 - F) / (L - 50)
Supoose the balancing force was placed at zero and the load at 75 cm
then MA = (50 - 0) / (75 - 50) = 50 / 25 = 2
or the balancing force was applied at 10 cm and the load force was applied at 60 cm then
MA = (50 - 10) / (60 - 50) = 40 / 10 = 4
Two particles each with a charge of 1μC are separated a distance of 0.4 m. A test charge is placed just above them but directly between them. (b) What is the direction of the resultant electric field on the test charge?
Answer:
Forces on the test (test is positive) charge
Σ Fx = 0 net result of x-compoment charges is zero
Σ Fy = 2 F sin θ where F is the force due to the distance of the test charge from one of the charges and sin θ is the component of that force that is perpendicular to the line separating the charges
The direction of the force is perpendicular to the line separating the two given charges