The ratio of the mass of copper to water will be 10.8
The quantity of heat energy needed to raise a body's temperature per unit of mass is known as specific heat. Other names for specific heat include mass specific heat and specific heat capacity.
The quantity of heat in joules needed to elevate one gram of a material by one Kelvin is known as specific heat in SI units (symbol: c). The standard way to report specific heat is in joules (J).
We know that
The product of mass of copper and mass of water is equal to the product of the specific heat of copper and Water
Therefore ,
[tex]m_wc_w = m_cc_c[/tex]
We know that
Specific heat of water is 4.18 J
Specific heat of copper is 0.39 J
Therefore,
[tex]m_c*(0.39)=m_w*(4.18)\\\frac{m_c}{m_w} = \frac{4.18}{0.39}\\ = 10.8[/tex]
Hence The ratio of the mass of copper to water will be 10.8
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At what height from the surface of the earth does the value of acceleration due to gravity be 2.45m/s square where the radius of the earth is 6400 km.
Answer: 6,277,647m
Explanation:
Radius of Earth = 6400km
To calculate the gravitational acceleration of a planet, we use the following formula:
g = mG/r^2
Gravitational acceleration is equal to the mass of the planet multiplied by the gravitational constant all divided by the radius of the planet squared.
We already know what the gravitational acceleration will be, 2.45m/s^2.
So, 2.45m/s^2 = mG/r^2
the mass of the earth is equal to 5.9*10^24.
And the gravitational constant is equal to 6.67408 * 10^-11.
We don't know the radius though.
2.45m/s^2 = 5.9*10^24 * 6.67408 * 10^-11 divided by r^2
2.45m/s^2 = 3.93 * 10^14 divide by r^2
Now, we can cross-multiply.
2.45m/s^2 * r^2 = 3.93 * 10^14
divide r^2 from both sides.
r^2 = 3.93 * 10^14 divided by 2.45m/s^2
r^2 = 1.6*10^14.
Now, take the square root of both sides.
r = 12,677,647 meters from the center of the Earth.
To calculate the height from the surface of the Earth, we need to subtract r by the Earth's radius.
That will be 12,677,647-6.400.000m = 6,277,647m from the surface of Earth.
A generator produces 30 pulsesin 5 second. its frequency is?
Answer:
6 pulse per second
Explanation:
if,
5 secs = 30 pulse
1 sec = 30/5
1 sec = 6
The force between two interacting charges is 9.0 × 10-5 newtons. They are kept 1 meter apart. If the magnitude of one charge is 1.0 × 10-6 coulombs, what is the magnitude of the other charge?
Pluto users ??
The magnitude of other charge will be 1 × 10⁻² coulomb
The formula of electrostatic force is
Electrostatic force = K q1 q1 / r²
where k is the coulomb's constant whose value is 9 × 10⁹
q1 and a2 are the magnitude of charges
and r is the distance between them
magnitude of the force given to us is 9.0 × 10⁻⁵ newtons
magnitude of one charge = 1.0 × 10⁻⁶ coulomb
Force = K q1 q2 / r²
9.0 × 10⁻⁵ = ( ( 9 × 10⁹ ) × ( 1.0 × 10⁻⁶ ) × q2 ) / 1
9.0 × 10⁻⁵ = 9 × 10³ × q2
10⁻² = q2
Charge on q2 is 1 × 10⁻² coulomb
So the magnitude of the second charge is came out to be 1 × 10⁻² coulomb after applying the formula of electrostatic force.
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A jumbo jet must reach a speed of 360 km/h on the runway for takeoff. What is the lowest constant acceleration needed for takeoff from a 1.80 km runway
The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
To find the answer, we need to know about the Newton's equation of motion.
What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aSV= final velocity, U= initial velocity, S= distance and a= accelerationWhat's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/sSo, 100²-0= 2×a×1800=> 10000= 3600a
=> a= 10000/3600 = 2.8 m/s²
Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
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You have a string with a mass of 0.0135 kg. You stretch the string with a force of 8.29 N, giving it a length of 1.83 m. Then you vibrate the string transversely at precisely the frequency that corresponds to its fourth normal mode, that is, at its fourth harmonic. What is the wavelength of the standing wave you create in the string
The wavelength of the standing wave at fourth harmonic is; λ = 0.985 m and the frequency of the wave at the calculated wavelength is; f = 36.84 Hz
Given Conditions:
mass of string; m = 0.0133 kg
Force on the string; F = 8.89 N
Length of string; L = 1.97 m
1. To find the wavelength at the fourth normal node.
At the fourth harmonic, there will be 2 nodes.
Thus, the wavelength will be;
λ = L/2
λ = 1.97/2
λ = 0.985 m
2. To find the velocity of the wave from the formula;
v = √(F/(m/L)
Plugging in the relevant values gives;
v = √(8.89/(0.0133/1.97)
v = 36.2876 m/s
Now, formula for frequency here is;
f = v/λ
f = 36.2876/0.985
f = 36.84 Hz
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A current magnetic relay is often used to assist in the starting of the compressor, but most new appliances use a(n) ____________________ device.
A current magnetic relay is often used to assist in the starting of the compressor, but most new appliances use positive temperature coefficient (PTC) device.
What is a compressor?A compressor can be defined as a mechanical device that is designed and developed to provide power to refrigerators, especially by increasing the pressure on air or other applicable gases.
According to heating, ventilation, and air conditioning (HVAC), a current magnetic relay is mostly used to assist in the starting of the compressor, but many new electrical appliances use positive temperature coefficient (PTC) device.
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A rectangular box of lines and an extra vertical line near the right end of the box has circles with X's in them on the right side, on the extra line, and on the top side between the extra line and a stack of vertical lines near the left end. The top circle is A. The right side circle is C. The circle on the extra line is B. The bottom side has 2 small circles separated by a short segment of the side near the left end. The top side has a stack of short horizontal lines labeled A in its middle, which are from left to right short, very short, short, very short.
Use the diagram to answer each question with yes or no.
If you removed bulb A, would bulbs B and C continue to shine?
If you removed bulb B, would bulbs A and C continue to shine?
Light bulbs 1, 2, and 4 will continue to shine when Linh switches off bulb 3.
What exactly is a circuit?The battery-powered electrical circuit where a voltage differential is generated across a capacitor, a resistor, or an inductor.
As soon as Linh cuts the middle branch, the bulbs in positions 1, 2, and 4 will function as a resistor in a series circuit. Except for resistor number 3, the current through the other resistors will be constant.
In conclusion, if Linh switches off Bulb 3, only Bulbs 1, 2, and 4 will still be on.
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Answer:
1. no
2. yes
Explanation:
on edge 2022
Absorbed solar radiation undergoes irreversible degradations, transferring from reservoir to another and ends up as what?
Answer:
Absorbed solar radiation undergoes irreversible degradations, transferring from reservoir to another converting from light energy to radiation energy.
Explanation:
The solar energy cycle functions in the following form:
Considering that 100% of energy is directed at earth.When 100% light is incident on the Earth's surface about 6-7% of it is reflected back by the atmosphere into space.20% is reflected back by the clouds.And about 4% is reflected back by Earth's surface.16% of solar energy is absorbed by the atmosphere, 3% is absorbed by clouds and 51% is absorbed by land and ocean.From the absorbed energy, 64% is radiated back to space from clouds and atmosphere and 6% is directly radiated back to space from earth.From the radiated energy 15% is absorbed by the atmosphere and 23% is carried by the clouds and atmosphere through the latent heat in water vapor.Therefore, absorbed solar radiation undergoes irreversible degradations, transferring from reservoir to another converting from light energy to radiation energy.
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A 1500 kg weather rocket accelerates upward at 10.0 m/s . It explodes 2.00 s after liftoff and breaks into two fragments, one twice as massive as the other. Photos reveal that the lighter fragment traveled straight up and reached a maximum height of 530 m. What were the speed and direction of the heavier fragment just after the explosion
20 m/s is the speed of the heavier fragment just after the explosion.
Newton's second law of motion states that the resultant force applied to an object is directly proportional to the mass and acceleration of the object.
F = ma where, F = Force (Newton)
m= mass
a = acceleration
Given:
mass of rocket = M = 1500 kg
acceleration of rocket = a = 10 m/s²
elapsed time = t = 2.00 s
mass of lighter fragment = m₁ = m = 500 kg
mass of heavier fragment = m₂ = 2m = 1000 kg
maximum height of lighter fragment = h = 530 m
Let's calculate the final speed of the rocket just before the explosion:
v = u + at
v = 0 + 10(2)
v = 20 m/s
Then, we will calculate the height of the rocket just before the explosion:
[tex]h' = ut + \frac{1}{2}at^{2}[/tex]
[tex]h' = 0 + \frac{1}{2} (10)(2.00)^{2}[/tex]
[tex]h' =20m[/tex]
The initial speed of lighter fragment just after the explosion:
[tex]v^{2}[/tex]₁ [tex]= u^{2}[/tex]₁ - [tex]2g[/tex]Δ[tex]h[/tex]
[tex]v^{2}[/tex]₁ [tex]= u^{2}[/tex]₁ - [tex]2g[/tex] [tex](h - h')[/tex]
[tex]0^{2} = u^{2}[/tex]₁[tex]- 2(9.8) (530-20)[/tex]
[tex]u^{2}[/tex]₁[tex]=9996[/tex]
[tex]u[/tex]₁ [tex]=\sqrt{9996} m/s[/tex]
Using Conservation of Momentum Law :
[tex]M v=m[/tex]₁ [tex]u[/tex]₁ [tex]+ m[/tex]₂[tex]u[/tex]₂
[tex]1500 (20)= 500(\sqrt{9996} ) + 1000u[/tex]₂
[tex]u[/tex]₂ ≅ [tex]- 20 m/s[/tex]
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Two plane waves of the same frequency and with vibrations in the z-direction are given by c (y, t) = (2 cm) cos a p 4 cm y - 20 s t + pb c(x, t) = (4 cm) cos a p 3 cm x - 20 s t + pb
The resultant wave at point (5, 2) is Ψ = 5.99 cos [ 7.15 - (20/s) t]
What are the plane waves:
A plane wave is a special case of wave or field: whose value, at any moment, is constant through any plane that is perpendicular to a fixed direction in space. plane waves are free-space modes.Here,
Two plane waves are given:
c (5, t) = 4 cos [(8π/3) - (20/s) t]
c (2, t) = 2 cos [(3π/2) - (20/s) t]
now, the waves as imaginary exponentials,
separating the spatial parts, and then adding them together
we get The resultant:
Ψ = [ 4 sin (8π/3) + 2 sin (3π/2) ]^2 + [ 4 cos(8/3 π) + 2 cos(3/2π) ]^2
Ψ = 5.99 tan(a) = 0.747/ 5.95
a = 7.15
Ψ = 5.99 cos [ 7.15 - (20/s) t]
hence,
The resultant wave is Ψ = 5.99 cos [ 7.15 - (20/s) t]
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Your question is incomplete, but most probably the full question was:
Two plane waves with the same frequency and with vibrations (measured by Psi) in the z-direction are given by c (x, t) = (4cm.) cos [pi/3cm. x - 20/s t + pi] c (y, t) = (2cm.) cos[pi/4cm. y - 20/s t + pi]
Express the waves as imaginary exponentials, separate the spatial parts, and add them together using a phasor diagram to find the resultant at the point x = 5cm. y = 2cm
. A pilot diverted a fighter jet 20.0 km [S 38.0° E] to avoid thunderstorm, and then flew 60.0 km [S 55.0° W] to get back on track. If the diversion lasted a total of 2.00 minutes, what was the average velocity of the jet during the diversion?
The average velocity of the jet during the diversion is 518.75 m/s.
Displacement of the pilot
The displacement of the pilot from the starting position is calculated as follows;
d² = 20² + 60² - 2(20)(60) x cos(θ)
where;
θ is the angle between 20 km and 60 km leg of triangle formed from the motion.θ = (90 - 38) + (90 - 55) = 87⁰
d² = 20² + 60² - 2(20)(60) x cos(87)
d² = 3874.39
d = 62.25 km
Average velocity of the pilotV = total displacement/total time
where;
time = 2mins = 0.033 hourV = (62.25 km)/(0.033 hr)
V = 1,867.5 km/hr = 518.75 m/s
Thus, the average velocity of the jet during the diversion is 518.75 m/s.
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what are effects of high gravity on blood pressure, blood circulatory system, size of lungs, bone fragility and density
Answer:
Explanation:
On return to Earth, gravity once again “pulls” the blood and fluids into the abdomen and legs. The loss of blood volume, combined with atrophy of the heart and blood vessels that can occur in space, reduces the ability to regulate a drop in blood pressure that happens when we stand on Earth.
A large truck is ahead of you and is turning right onto a street with two lanes in each direction. The truck:
May have to swing wide to complete the right turn
In any vehicle which makes any turn, the rear wheels follow a shorter path as compared to the front wheels. While the truck being larger than any usual vehicle, the distance between the wheels are even more, which makes it harder to complete the turn by going in its own lane or the other two right lanes.
So, the truck will try to go wide in left and then complete the right turn as by doing this the distance between the rear and front will get compensated by the width of the lanes making it easier for the truck to complete its turn.
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[Your question is incomplete, but most probably your full question was
May not turn without engaging its hazard lights
May complete its turn in either of the two lanes
Must stay in the right lane at all times while turning
May have to swing wide to complete the right turn]
How wide is the central diffraction peak on a screen 2.40 m behind a 0.0328- mm -wide slit illuminated by 588- nm light
The width of the central diffraction peak is 0.0754 m.
What is central diffraction:
It is distance between the 1st order minima from the center of the screen on both sides of the center.
The first diffraction minimum from the center is given as:
y = L*λ / awhere
y is the location of first minimum diffraction.
Distance of screen from slit (L) = 2.40 m
Width of the slit (a) = 0.0328 mm = 0.0328 × 10⁻³ m
Wavelength of light (λ) = 588 nm = 588 × 10⁻⁹ m
Width of central diffraction peak is,
W = 2*y = (2* L*λ ) / aNow, plug in the values given and solve for width 'W'.
This gives,
W = 2 * 2.40 * 588 × 10⁻⁹ / ( 0.0328 × 10⁻³ )
W = 0.0860 m
Therefore,
The width of the central diffraction peak is 0.0754 m.
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Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is |v with arrow| = 160 km/s and the orbital period of each is 14.3 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 1030 kg.)
The mass M of the star is mathematically given as
M=24.269kg
What is the mass M of the star?Question Parameters
Orbital period P= 14.3*24*60*60 s
P= 1235520s
Speed V=160000
Generally, the equation for the Radius is mathematically given as
R =[tex]\frac{ V*t}{ 2*pi}[/tex]
Therefore
R =[tex]\frac{ ( 160000 )*(1235520 s)}{ 2*3.142}[/tex]
R=[tex]3.14581795*10^{10}[/tex]
Where
M*v^2/R = GM^2/(2R)^2
Therefore
M=[tex]=4(v^2)R/G\\\\ =4(240^2)R/G\\\\ = \frac{4*(160000)^2)*3.14581795*10^{10}}{6.67*10^{-11}}[/tex]
M=4.82956159*10^{31}kg
Considering, that the mass of our Sun is 1.99*10^{30} kg.
The mass M of the star is
M[tex]=\frac{4.82956159*10^{31}}{1.99*10^{30}}[/tex]
M=24.269
In conclusion, the mass M of the star is
M=24.269
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Hello people ~
Antoine and Adriane are each at the top of a Ferris wheel. After one revolution, each will again be at the top. Which sinusoid (sine or cosine) should you graph to model the ride of your chosen rider? Why? Assume that each Ferris wheel starts at time 't = O' seconds.
Answer:
graph of cosine function
Explanation:
f(x) = sin(x)
The curve of the sine function crosses the y-axis at the origin, heads up to y = 1 at x = π/2 and down to y = -1 at x = 3π/2.
Domain: (-∞, ∞)
Range: [-1, 1]
f(x) = cos(x)
The curve of the cosine function crosses the y-axis at (0, 1) which is its maximum, heads down to y = -1 at x = π then heads up again.
Domain: (-∞, ∞)
Range: [-1, 1]
Given information:
Antoine and Adriane begin their journey at the top of the Ferris wheel.After one revolution, they will again be at the top of the wheel.If the wheel starts when time t = 0 seconds, and they begin their journey at the top of the wheel (the maximum point), the most appropriate graph to model their ride is the graph of the cosine function since it is at its highest point when x = 0.
Sally and ner pet puagie go everywnere
together. One afternoon Sally decides to go for
a walk. She walks 8.0 x 10² m [N] in 5.0
minutes, then walks 0.30 km [E] in 1.5 minutes.
(8 marks)
d₁ = 300 m [E]
d₂ = 800 m [N]
a. Calculate the total distance she walks. (Be
careful with the units.)
b. Her budgie is lazy and always flies along the
shortest path possible when they go on these
trips. Find the magnitude and direction of the
budgies path.
c. Determine Sally's average speed for the
journey.
d. Determine the average velocity of Sally's
budgie.
a)The total distance she walks will be 1100 m.
b)The magnitude and direction of the budgie's path will be 854.4 m at 69.44°.
c)Sally's average speed for the journey will be 2.82 m/sec
d)The average velocity of Sally's budgie willl be 2.19 m/sec North of East.
What is displacement?Displacement is defined as the shortest distance between the two points.
Distance travelled in north direction,d₁ = 8.0 x 10² m
Time elapsed travelled in north direction,t₁ = 5.0 minutes
Distance travelled in east direction,d₂ = 0.30 km = 3.0 × 10² m
Time elapsed travelled in east direction,t₂= 1.5 minutes
The total distance she walks,d =? m
The magnitude and direction of the budgies path will be, D
Sally's average speed for the journey, [tex]\rm S_{avg}[/tex]
The average velocity of Sally's budgie,[tex]\rm V_{avg}[/tex]
a)
The total distance she walks is found as;
d=d₁+d₂
d=800 m +300 m
d = 1100 m
b)
The magnitude and direction of the budgies path will be;
[tex]\rm D = \sqrt{d_1 ^2 + d_2^2 } \\\\ D = \sqrt{800^2 +300^2 } \\\\ D = 854.4 \ m[/tex]
The direction of the displacement is;
[tex]\rm \theta = tan^{-1} \frac{800}{300} \\\\ \theta = tan^{-1}(2.6) \\\\ \theta = 69.44 ^ 0[/tex]
c)journey is found as;
Sally's average speed for the
[tex]\rm S_{avg} = \frac{d}{t} \\\\ S_{avg} =\frac{100}{300+900} \\\\ S_{avg} =2.82 \ m/sec[/tex]
d)
The average velocity of Sally's budgie is found as;
[tex]\rm V_{avg} = \frac{854.4}{300+ 900 } \\\\ V_{avg} = 2.19 \ m/sec[/tex] North of east
Hence, the total distance she walks, the magnitude and direction of the budgie's path, Sally's average speed, and the average velocity of Sally's budgie will be 1100 m,854.4 m at 69.44°,2.82 m/sec, and 2.19 m/sec North of East.
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Most exceptions to the trend of decreasing radius moving to the right within a period occur in the __________.
Answer:
Most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.
What are the characteristics of d-block elements?
In a period as we advance from left to right, the number of valence electrons in the same shell increases due to which the effective nuclear charge increases and thus the atomic size decreases.In d-block atomic radius initially decreases then remains constant and increases towards the end.As one moves from Sc (scandium) to Zn (zinc), the effective nuclear charge increases by a factor of 1, this is because:The number of electrons are low in the inner shell.The shielding power of d-orbital is low.Inter electronic repulsions will be operating at a value less than the nuclear charge, which will result in decrease in atomic radii.As the number of electrons in the inner orbital increases the outer electrons repel one another which enables them to push away.Although d-orbital has less shielding power, the number of electrons present in it are high. Hence, the electron-electron repulsive force becomes dominant, this results in an increase in the atomic radii.Therefore, most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.
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To improve the acoustics in auditorium, a sound reflector with mass of 200 kg is suspended by a chain from the ceiling. what is its weight? what force (magnitude and direction) does the chain exert on it? what is the tension in the chain? assume that the mass of the chain itself is negligible.
Answer:
W = 1960 N
T = 1960 N in the upward direction
Explanation:
Let's apply Newton's second law to this exercise with acceleration equal to zero, translational equilibrium
Weight is
W = m g
W = 200 9.8
W = 1960 N
In the attachment we can see a diagram
T -W = 0
T = W
T = 1960 N
in the upward direction
credit - somebody else
What is the acceleration of a 600,000 kg freight train, if each of itsthree engines can provide 100,000 n of force?
The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².
How to calculate acceleration?The acceleration of a freight train can be calculated using the following formula:
Force = mass × acceleration
According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:
100,000 = 600,000 × a
100,000 = 600,000a
a = 0.167m/s²
Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².
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Suppose a 60-kg boy and a 41-kg girl use a massless rope in a tug-of-war on an icy, resistance-free surface. If the acceleration of the girl toward the boy is 3.0 m/s2, find the magnitude of the acceleration of the boy toward the girl.
A.
2.05 m/s2
B.
2 m/s2
C.
2.1 m/s2
Answer:
Approximately [tex]2.05\; {\rm m\cdot s^{-2}}[/tex].
Explanation:
The net force on the girl would be:
[tex]\begin{aligned}m(\text{girl}) \, a(\text{girl}) &= 41\; {\rm kg} \times 3.0\; {\rm m\cdot s^{-2}} \\ &= 123.0\; {\rm N} \end{aligned}[/tex].
Under the assumptions, the net force on this girl would be equal to the tension force in the rope. All other forces on the girl would be balanced.
In other words, the tension force that the rope exerted on the girl would be [tex]123.0\; {\rm N}[/tex]. The girl would exert a reaction force on the rope at the same magnitude ([tex]123.0\; {\rm N}\![/tex]) in the opposite direction. This force would translate to a [tex]123.0\; {\rm N}\!\![/tex] force on the boy towards the girl.
Under similar assumptions, the net force on the boy would also be [tex]123.0\; {\rm N}[/tex]. Since the mass of the boy is [tex]m(\text{boy}) = 60\; {\rm kg}[/tex], the acceleration of the boy would be:
[tex]\begin{aligned}a(\text{boy}) &= \frac{(\text{net force})}{m(\text{boy})} \\ &= \frac{123.0\; {\rm N}}{60\; {\rm kg}} \\ &= 2.05\; {\rm m\cdot s^{-2}}\end{aligned}[/tex].
In example 18. 4 of the text, the deflection angle of the laser beam as it exits the prism is 22. 6º. If the prism had been made of glass instead of polystyrene plastic, what would the deflection angle have been?.
The deflection angle is 37.29º if the prism was made of polystyrene plastic.
What is a laser beam?
Laser beam is a light beam propagating dominantly in one direction.
It is a beam of radiation produced from a laser.
Here given that,
refraction index of glass, n1= 1.52
refraction index of polystyrene plastic, n2 = 1.59
deflection angle, B = 22.6º
For the second surface,
B = 45º - 22.6º
B = 22.40º
Now from the formula of Snell's law:
n1 sinФ = n2 sin B
sin B / sinФ = n1 / n2
sinФ = (sin 22.40º) * ( 1/ 1.59)
Ф = 37.29º
Hence,
The deflection angle is 37.29º if the prism was made of polystyrene plastic.
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A coffee filter of mass 1.5 grams dropped from a height of 3 m reaches the ground with a speed of 0.7 m/s. How much kinetic energy Kair did the air molecules gain from the falling coffee filter? Start from the Energy Principle, and choose as the system the coffee filter, the Earth, and the air.
The kinetic energy gained by the air molecules is 0.0437 J
Given:
Mass of a coffee filter, m = 1.5 g
Height from which it is dropped, h = 3 m
Speed at ground, v = 0.7 m/s
Initially, the coffee filter has potential energy. It is given by :
[tex]P =mgh[/tex]
P = 1.5 × 10⁻³ kg × 9.8 m/s² × 3m
P = 0.0441 J
Finally, it will have kinetic energy. It is given by :
[tex]E= \frac{1}{2} mv^{2}[/tex]
[tex]E= \frac{1}{2}[/tex]×[tex]1.4[/tex] × 10⁻³ × (0.7)²
E = 0.000343 J
The kinetic energy Kair did the air molecules gain from the falling coffee filter is :
E = 0.000343 - 0.0441
= 0.0437 J
So, the kinetic energy Kair did the air molecules gain from the falling coffee filter is 0.0437 J
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A projectile is shot upward from the surface of Earth with an initial velocity of 120 meters per second. What is its velocity after 5 seconds
After 5 seconds the velocity of the projectile is 71m/s
Given that Initial velocity is 120m/s
Time t = 5s
We need to find the velocity after 5 sec
Projectile motion is the movement of an item hurled or projected into the air, which is only subject to the acceleration of gravity. The trajectory of the object is referred to as its trajectory.
The horizontal velocity of a projectile is constant,
There is a vertical acceleration caused by gravity and its value is 9.8 m/s
Now the velocity for 5 seconds
[tex]v(t)=v_0-gt\\[/tex]
[tex]v(t)=120-(9.8)(5)\\\\v(t) =71m/s[/tex]
Hence the velocity after 5 seconds is 71m/s
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Which part of the heating curve corresponds with the boiling of liquid water?
Ο Α. Α
OB. B
O C. D
OD. C
Answer:
The answer is B
Explanation:
The section C of the heating curve represents the liquid phase, which is where liquid water is heated. Hence option D is correct.
What is heating?The body or particle is considered to be heating up as the temperature rises.
100°C is the boiling point of water at 1 atm of pressure.
The term "enthalpy of vaporisation" refers to the amount of heat energy needed to change liquid water into steam.
Enthalpy of vaporisation has a greater value than enthalpy of fusion.
The amount of power needed to melt ice into liquid water is known as the enthalpy of fusion.
As a result, section C illustrates the heating of liquid water.
Hence option D is correct.
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The box resting on the inclined plane above has a mass of 20kg. The incline sits at a 30o angle. Find the friction force between the box and the incline if the box does not slide down the incline.
The friction force between the box and the incline if the box does not slide down the incline will be 0.577
The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.
Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle
We have to find the friction force between the box and the incline if the box does not slide down the incline
Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:
F₁ = F₂
μmgcosΘ = mgsinΘ
μ = (mgsinΘ)/(mgcosΘ)
μ = tanΘ
μ = 0.577
Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577
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:The resistance created by the box's friction with the incline The slope will be 0.577 if the box doesn't descend.
Friction is the force that prevents solid surfaces, fluid layers, and material constituents from sliding past one another. Different types of friction exist. Dry friction opposes the relative lateral motion of two touching solid surfaces.
Given that the 20-kg box lying on the inclined plane above and the incline's 30 degree inclination
If the box does not slide down the incline, we must determine the friction force between them.
Given that the gravitational force F2 down the gradient must be equal to or greater than the frictional force F1,
F₁ = F₂
If mgcosx = mgsinx
equals (mgsinx)/(mgcosx)
μ = tanx
μ = 0.577
Therefore, if the box does not slide down the incline, the friction force between the box and the incline will be 0.577.
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give the phases into which matter can be classified, and an example of each
Explanation:
there are 4 phases in which matter could be classified.
1. Solid: an example of this would be the metal copper.
2. Liquid: an example of this would be H2O or water.
3. Gas: an example of this would be oxygen gas or O2
4. Plasma: an example of this would be lightning.
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The single-pole, single-throw switch is normally wired ? between the source and the load to turn devices on and off.
In series.
Single-pole and single-throw switch:
A switch with only one input and one output is referred to as a Single Pole Single Throw (SPST) switch. This indicates that it has a single output terminal and a single input terminal.
A single pole, one throw switch functions as an on/off switch in circuits. The circuit is turned on when the switch is closed. The circuit is shut off when the switch is open.
Thus, SPST switches are relatively basic in design.
Circuit for a single-pole, single-throw (SPST) switch
Types:
According to the application, it can be divided into three categories, including:
Simple SPST(ON)-OFF, Push-to-close, SPST MomentaryON-(OFF), Push-to-Open, SPST MomentaryInches Switch SPSTLearn more about terminal here:
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Which of the following is NOT considered electromagnetic radiation? x-rays radio waves cosmic rays ultraviolet light
Answer:
Cosmic ray is not considered electromagnetic radiation.
Explanation:
Electromagnetic radiation refers to
It is Radiation that has both electric and magnetic fields and travels in waves.Electromagnetic radiation can vary in strength from low energy to high energy.X rays are the rays produced when a negatively charged electrode is heated by electricity and electrons are released, thereby producing energy. It is a type of radiation called EM waves.
Radio wave are wave from the portion of the electromagnetic spectrum at lower frequencies than microwaves. It is an EM wave.
Ultraviolet wave Invisible rays that are part of the energy that comes from the sun. It is an example of EM wave.
Cosmic ray is a high-speed particle either an atomic nucleus or an electron that travels through space.
Cosmic ray is not an electromagnetic wave.
Hence
Cosmic ray is not considered electromagnetic radiation
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A 60.0-kg skater begins a spin with an angular speed of 6.0 rad/s. By changing the position of her arms, the skater decreases her moment of inertia to one-half its initial value. What is the skater's final angular speed
The skater's final angular speed is equal to 12 rad/s.
When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.
The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.
In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.
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