The acceleration of the mass when it is nudged ever so slightly on the incline: a = (g sin(∅)) - (μ g cos(∅))
If the minimum mass on the incline is nudged ever so slightly, it will experience a force due to gravity pulling it down the incline. This force can be broken down into two components: one parallel to the incline and one perpendicular to the incline.
Since the mass is on an incline, it experiences a normal force that is perpendicular to the incline.
The perpendicular component, also known as the weight component perpendicular to the incline, is given by mg cos(∅).
The force of friction can be represented as the product of the coefficient of friction (μ) and the normal force (mg cos(∅)).
Therefore, the net force along the incline is;
(mg sin(∅)) - (μ mg cos(∅)).
To calculate the acceleration, we use Newton's second law, which states that the net force acting on an object is equal to its mass (m) multiplied by its acceleration (a).
(mg sin(∅)) - (μ mg cos(∅)) = ma
Simplifying:
a = (g sin(∅)) - (μ g cos(∅))
This is the expression for the acceleration of the mass when it is nudged ever so slightly on the incline. The acceleration will depend on the angle of the incline (theta), the coefficient of friction (μ), and the acceleration due to gravity (g).
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The audible frequency range for a person is 30HZ to 16500HZ. Determine the largest wavelength of sound in air the person can detect (speed=340m/s)
Answer:
We have the formula,
V=λ×v
where,
V is the velocity of sound =330ms
−1
v is the frequency
λ is the wavelength
We have a range of frequencies from 20Hz to 20,000Hz
Hence for frequency v=20Hz, we have
330=20×λ
Therefore, λ=
20
330
m
Explanation:
Which of the following phrases best describes the term scientific model?
OA. A physical copy of a scientific object, system, or process
B. The application of scientific knowledge to make predictions about
an object, system, or process
C. An experiment in which variables are controlled
OD. A simplified representation used to explain or make predictions
about something
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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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.
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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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
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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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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Two concentric, coplanar, circular loops of wire of different diameter carry currents in the same direction. Describe the nature of the force exerted on the inner loop by the outer loop and on the outer loop by the inner loop.
Answer:
Both will be attractive in nature.
Explanation:
In the given case, the direction of the magnetic field is same in both loops as the direction of the current is same in both loops. When two parallel straight wire carrying current in the same direction are brought close to each other, the force between them is attractive in nature. In the same way, when two coplanar, circular and concentric loops of wire are carrying current in the same direction, the force between them is attractive in nature. It can be checked by using right hand thumb rule.
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why does the production of electricity by friction always yield equal amounts of positive and negative charge
Friction yields equal amounts of positive and negative charge as positive and negative charges occur in pairs.
What is frictional electricity:
When two charged particles are rubbed together, transfer of charge takes place. This leads to production of electricity, as one object loses electrons and gains positive charge while the other object gains electrons and becomes negatively charged.
All objects contain equal amounts of positive and negative charge. Thus, when friction is caused between them, it yields equal amounts of positive and negative charge which is known as static electricity.
One of the prime example of static electricity is rubbing silk cloth over glass rod. Here, the glass rod loses electrons while silk gains electrons. thus, the glass rod becomes positively charged and silk becomes negatively charged.
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A radio telescope detects an incoming radio wave which has a power of 1.00 x 10-24 W. If this signal creates a potential of 1.00 nV in the radio receiver, how many electrons move through the receiver circuit per second
Answer:
Therefore, the number of electrons moving through the receiver circuit per second is 6250 electrons/s.
Explanation:
Given:
Power of radio wave, P = 1 × 10⁻²⁴ W
Potential of signal, V = 1 × 10⁻⁹ V
Solution:
Consider the number of electrons flowing through the receiver circuit per second be 'n'.
The current flowing through the circuit will be:
I = P/V
= (1 × 10⁻²⁴ W)/(1 × 10⁻⁹ V)
= 1 × 10⁻¹⁵ A
Then, the number of electrons moving through the receiver circuit can be calculated as:
ne = I
where, e is charge on an electron
I is current flowing through the circuit
Applying values in above equation we get:
n(1.6×10⁻¹⁹ C) = (1 × 10⁻¹⁵ A)
n = (1 × 10⁻¹⁵ A)/(1.6×10⁻¹⁹ C)
n = 6250 electrons/s
Therefore, the number of electrons moving through the receiver circuit per second is 6250 electrons/s.
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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]
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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How can we change direction of light? Name this phenomenon. any indian didis and bhayas to help (ik i said australian any will work for me)goys please help im in grade 6 and my teacher didnt give us the answer to a question in notebook:(
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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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
Mature wind waves of one wavelength that form orderly undulations of the ocean surface and often precede a major storm system is known as
Mature wind waves of one wavelength that form orderly undulations of the ocean surface and often precede a major storm system is known as swells.
Meaning of mature wind and waves:
The waves in a fully developed sea outrun the storm that creates them, lengthening and reducing in height in the process.
There are 3 types of wind generated waves:
Capillary wavesGravity wavesSwellswe can say that,
Mature wind waves are known as swell, when they left the formation region.
They have traveled away from where they were raised by the wind, and have to a greater or lesser extent dispersed.
Hence,
Mature wind waves of one wavelength is known as swell.
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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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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
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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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.
66. Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water
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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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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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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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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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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A 31.4-kg wheel with radius 1.21 m is rotating at 283 rev/min. It must be brought to a stop in 14.8 s. Find the required average power. Assume the wheel to be a thin hoop.
The average power required to stop the wheel is 2795 Joule.
To find the answer, we need to know about the linear velocity, acceleration and force on the wheel.
What is the angular frequency of the rotating wheel?Mathematically, angular frequency= 2×π×frequencySo, angular frequency= 2×π× 283 rev/min= 2×π×(283/60) rev/s
= 30 rad/s
What's the expression of velocity from angular frequency?Mathematically, angular frequency= velocity/radiusSo, velocity= angular frequency × radius Here, radius of the wheel= 1.21mSo, velocity= 30×1.21 m = 36.3 m/s
What will be the acceleration of the wheel, if the final velocity is zero, initial velocity 36.3m/s and time is 14.8 s?Mathematically, acceleration= changeing velocity/timeHere, changing velocity= 36.3m/s and time = 14.8 sSo, acceleration= 36.3/14.8 = 2.45 m/s²What's the force experienced by the wheel?The force on the wheel= mass× acceleration
= 31.4 × 2.45 = 77 N
What's the average power of the wheel?Mathematically, power= work done / time = force×velocityPower= 77 × 36.3 = 2795 J.Thus, we can conclude that the average power of the wheel is 2795 J.
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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].
What happens if you move a bar magnet back and forth along the axis of the
coiled wire shown below?
c. A current is induced in the coiled wire, which lights the light bulb.
What is electromagnetic induction?If we kept the bar magnet stationary and moved the coil back and forth within the magnetic field an electric current would be induced in the coil.
Then by either moving the wire or changing the magnetic field we can induce a voltage and current within the coil and this process is known as Electromagnetic Induction and is the basic principle of operation of transformers, motors and generators.
When the magnet shown below is moved “towards” the coil, the pointer or needle of the Galvanometer, which is basically a very sensitive center zeroed moving-coil ammeter, will deflect away from its center position in one direction only.
When the magnet stops moving and is held stationary with regards to the coil the needle of the galvanometer returns back to zero as there is no physical movement of the magnetic field.
Therefore ,
If you move a bar magnet back and forth along the axis of the coiled wire shown below then a current is induced in the coiled wire, which lights the light bulb.
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If 710- nm and 660- nm light passes through two slits 0.65 mm apart, how far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away
0.23 mm far apart are the second-order fringes for these two wavelengths on a screen 1.5 m away.
Given wavelengths 710nm and 660nm,0.65mm apart two slits, and a screen 1.5m away.Position of n the order fringe = n λ D / d
for n = 2
position = 2 λ D / d
λ = 710 nm , D = 1.5m
d = .65 x 10⁻³
position 1 = 2 x 710 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3276.92 x 10⁻⁶ m
= 3.276x 10⁻³ m
= 3.276mm .
For λ = 660 nm
position = 2 λ D / d
λ = 660 nm , D = 1.5 m
d = .65 x 10⁻³
position 2 = 2 x 660 x 10⁻⁹ x 1.5 / .65 x 10⁻³
= 3046.15 x 10⁻⁶ m
= 3.046 x 10⁻³ m
= 3.046 mm .
Difference between their position
= 3.276mm ₋ 3.046 mm
= 0.23 mm .
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