Answer:
The magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N
Explanation:
Given;
speed of the electron, v = 8.0 × 10⁶ m/s
magnetic field strength, B = 2.5 T
angle of inclination of the field, θ = 60°
The magnetic force experienced by the electron in the magnetic field is given as;
F = qvBsinθ
where;
q is charge of electron = 1.6 x 10⁻¹⁹ C
B is strength of magnetic field
v is speed of the electron
Substitute the given values and solve for F
F = (1.6 x 10⁻¹⁹)( 8.0 × 10⁶)(2.5)sin60
F = 2.77 x 10⁻¹² N
Thus, the magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N
The magnitude of the magnetic force on the electron is approximately 2.06 x 10^-13 N.
Explanation:In this scenario, the magnitude of the magnetic force on the electron can be calculated using the equation F = qvBsinθ, where q is the charge of the electron (+1.6 x 10-19 C), v is its velocity (8.0 x 106 m/s), B is the magnetic field (2.5 T), and θ is the angle between the velocity and the magnetic field (60°). Plugging in the values, we get:
F = (1.6 x 10-19 C)(8.0 x 106 m/s)(2.5 T)sin(60°)
Using the trigonometric identity sin(60°) = sqrt(3)/2, we can simplify the equation and calculate the magnitude of the magnetic force to be approximately 2.06 x 10-13 N.
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A man driving a car traveling at 30m/s slams on the brakes and accelerates in the negative direction at 4.75 m/s2, how far does the car travel before it stops?
Answer:
94.73 meters.
Explanation:
Using SUVAT.
s = ?
u = 30
v = 0
a = -4.75
t = ?
[tex]v^{2}=u^2+2as\\0=30^2+2(-4.75)s\\\\900=9.5s\\s = 94.73[/tex]
The car will travel approximately 94.74 meters before it stops.
To answer this question, we can use the equations of motion.
When the car stops, its final velocity will be 0 m/s.
We can use the equation v^2 = u^2 + 2as,
where v is the final velocity,
u is the initial velocity,
a is the acceleration, and
s is the distance.
Plugging in the values, we have (0)^2 = (30)^2 + 2(-4.75)s.
Solving for s, we get s = 900/9.5, which is approximately 94.74 meters.
Therefore, the car will travel approximately 94.74 meters before it stops.
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A helium balloon containing 5m³ of gas at a pressure of 30kPa is released into the air. Assuming that the temperature is constant calculate the volume the balloon would have when the pressure inside the balloon has fallen to 5kPa. Answer in m³.
Answer:
30 m³
Explanation:
Parameters given:
Initial volume of helium, V1 = 5 m³
Initial pressure in balloon, P1 = 30 kPa
Final pressure, P2 = 5 kPa
To find the volume of the balloon at that volume, we apply Boyle's law.
It states that at constant temperature, the pressure of a gas is inversely proportional to the volume of the gas.
Mathematically:
P = k / V
Where k = constant of proportionality
This implies that:
P * V = k
This means that if the pressure or volume of the gas changes at the same temperature, the product of the pressure and volume would be the same:
Hence:
P1 * V1 = P2 * V2
Hence, to find the final volume:
30 * 5 = 5 * V2
=> V2 = (30 * 5) / 5
V2 = 30 m³
The volume of the gas when the pressure is 5 kPa is 30 m³.
Which statement best describes what happens when a sound source is moving? (1 point)
Grupo de opciones de respuesta
As the sound source approaches the observer, the pitch decreases and continues to do so as it passes the observer.
As the sound approaches the observer, the pitch increases; as it passes the observer, the pitch decreases.
As the sound source approaches the observer, the pitch decreases; as it passes the observer, the pitch increases.
As the sound source approaches the observer, the pitch increases and continues to do so as it passes the observer.
Answer:
As the sound approaches the observer, the pitch increases; as it passes the observer, the pitch decreases.
Explanation:
The crests come closer when it approaching, and go farther when the source passes the observer.
How do the magnetic poles of earth serve to protect life on earth?
Answer:
The block solar winds and other harmful things
Explanation:
The Earth's magnetic field serves to deflect most of the solar wind, whose charged particles would otherwise strip away the ozone layer that protects the Earth from harmful ultraviolet radiation.
Answer:
Explanation:The Earth's magnetic field serves to deflect most of the solar wind, whose charged particles would otherwise strip away the ozone layer that protects the Earth from harmful ultraviolet radiation. One stripping mechanism is for gas to be caught in bubbles of magnetic field, which are ripped off by solar winds.
Suppose you want to make a nested function call (i.e. a call to a function from inside of another function) using a jal rather than a call for performance reasons. How would the push and pop pseudo-ops be proprely ordered along with the jal so that the previous return address isn't lost?
a) pop $ra
jal nested_function_label
nop
push $ra
b) push $ra
jal nested_function_label
nop
pop $ra
c) push $ra
pop $ra
jal nested_function_label
nop
d) jal nested_function_label
nop
pop $ra
push $ra
Answer:
As we need to use a nested loop in our function,hence push $ra
pop $ra
jal nested_function_label
nop is the correct option.
Quasar spectra often show many absorption lines that all appear to be as a result of the same electron transition (such as level 1 to level 2 in hydrogen) but that fall at different wavelengths in the spectrum. Why do we think this is the case?
Answer: Because of different redshift of cloud.
Explanation:
We are seeing absorption lines from clouds of gas that lie between us and the quasar, and therefore each cloud has a different redshift.
A quasar's spectrum is hugely redshifted. And most astronomers think this large redshift tells us about the distance to the quasar.
Describe how the properties of water contribute to two of the following. Transpiration thermoregulation in endotherms plasma membrane structure
Answer:
Explanation:
The contributions of water properties in Thermoregulation in endotherms and in the plasma membrane structure.
A) Thermoregulation in endotherms: Water contributes to thermoregulation by its property of High specific heat. Water's high specific heat acts as a heat buffer, it is much harder to cool down/heat up, therefore, it will make warmer than the cold environment around it or colder than the hot environment around it. Thermoregulation works in humansas well. Recall Homeostasis which is the body's way of regulating our temperature.
B) Plasma membrane structure:
The property of water, polarity, contributes to the plasma membrane because it creates the arrangement of phospholipids making it a semi-permeable membrane.
Recall that the Plasma Membrane has a hydrophilic end on the outside that is charged and likes water and a hydrophobic end on the inside that has no charge and doesn't like water.
The membrane that separates the interior of the cell from the external environment is known as the plasma membrane, sometimes known as the cell membrane, and it is present in all cells. A cell wall is affixed to the plasma membrane on the exterior of bacterial and plant cells.
What is transpiration?Like all other living things, plants need an excretory system to remove extra water from their bodies. Transpiration is the term for this process of removing extra water from the body of the plant. Typically, it is the evaporation of water from the leaf surface.
Now, according to the question :
(A) Thermoregulation in endotherms :
Water helps regulate body temperature since it has a high specific heat. Due to its high specific heat, which acts as a heat buffer, water will either make an area warmer than the surrounding cold environment or colder than the surrounding hot environment. Humans are capable of thermoregulation as well. Remember that the body uses homeostasis to control our body temperature.
(B) Plasma membrane structure:
Because it forms the arrangement of phospholipids that makes the plasma membrane a semi-permeable membrane, water's polarity contributes to the plasma membrane.
Remember that the Plasma Membrane has an exterior hydrophilic end that is charged and prefers water, and an interior hydrophobic end that is uncharged and dislikes water.
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When different resistors are connected in parallel across an ideal battery, we can be certain that...
Answer:
Explained Below.
Explanation:
Whenever the different resistors are fixed parallel along with an ideal battery , we can be certain that it carries the same potential difference across each of them.
And potential difference is nothing but the energy that charge moves with. It can also be denoted as (p.d.) and potential difference is been calculated as volts that is been denoted as (V).
he graph shows all of the stable isotopes of elements according to the numbers of protons and neutrons in their nuclei.
A graph titled zone of nuclear stability with number of protons from 0 to 80 on the x-axis with number of neutrons on the y-axis from 0 to 140. There are two lines, one with a slope of 1 from (0, 0). The other line has points of (20, 24), (40, 52) and (60, 120) labeled stability zone.
Which statements are supported by the graph? Check all that apply.
A. For large atoms, more neutrons than protons are needed to be stable.
B. Nuclei that have 90 or greater protons are always radioactive.
C. Atoms must have equal numbers of protons and neutrons to be stable.
D. Atoms that have less than 10 protons do not need neutrons to be stable.
Answer:
A and B i think
Explanation:
(A) For large atoms, more neutrons than protons are needed to be stable.
(B) Nuclei that have 90 or greater protons are always radioactive.
Stability of nucleus of an atom
A stable nucleus of an atom must have nutron-to-proton ratio of at least one. This implies that the nucleus must have more neutrons than protons.
From the graph we can conclude the following about stability of nucleus of an atom;
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A dog has a mass of 60kg and an acceleration of 2m/s/s. What is the force of the dog?
Answer:
120 N
Explanation:
F=ma therefore 60kg times 2m/s^2 is 120 N
A source of emf is connected by wires to a resistor, and electrons flow in the circuit. The wire diameter is the same throughout the circuit. Compared to the drift speed of the electrons before entering the source of emf, the drift speed of the electrons after leaving the source of emf is __________.A. faster. B. slower. C. the same. D. either A or B depending on circumstances. E. any of A, B, or C depending on circumstances.
Answer:
C. the same.
Explanation:
The drift speed of electrons in a circuit is the same all through the circuit.
The drift speed of the electrons remains the same before and after they pass through the source of emf because the current is constant throughout the circuit assuming a steady state situation. The emf source does not affect the drift speed despite functioning like a pump that maintains the potential difference.
Explanation:The drift speed of the electrons after leaving the source of emf is the same as before entering it. This is so because, in a steady-state situation where the current is constant throughout a circuit, the rate at which electrons enter any given section of the circuit must equal the rate at which they exit. The drift speed is the average speed of the electrons as they move through the conductor, and it remains constant throughout the entirety of the circuit assuming the diameter of the wire is consistent.
It's important to understand that the emf source acts like a pump, helping move the electrons through the circuit and maintaining a potential difference. However, this doesn't result in an increase in the electron drift speed, as the overall velocity of the electron stays steady. Even in the presence of an electric field, the phenomenon known as drift velocity remains unaffected due to the near-random movements of the free electrons caused by their collisions with atoms and other free electrons.
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Consider a loop of wire whose plane is horizontal and that carries a current in the clockwise direction when viewed from above. If we were to represent the current loop as a bar magnet or magnetic dipole, in what direction would the north pole be pointing?
Answer:
Downwards into the plane
Explanation:
Solution:-
- This is a conceptual application of hand rule. We will place our palm fingers open vertical to a plane surface. Then curl our fingers in and naturally point the thumb.
- The direction of curl of fingers denotes the direction of of current flow in the coil. Which in our case is "clockwise direction". We will orient/invert our right hand palm in such a way that we curl our fingers in clockwise fashion. Then stick the thumb out to give us the direction of magnetic field or North pole end. In our case the the thumb points downwards into the plane denoting that the magnetic field within the loop is also acting downwards into the plane.
- The bar magnet would be placed in such a way that North pole is pointing downward into the plane in the direction of magnetic field and end up at south pole pointing up out of the plane.
The speed of sound in air is 320 ms-1 and in water it is 1600 ms-1. It takes 2.5 s for sound to reach a certain distance from the source placed in air. a. Find the distance. b. How much time will it take for sound to travel the same distance when the source is in water?
Answer:
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
Explanation:
Given:
Speed of sound in air = 320 m/s
Speed of sound in water = 1600 m/s
Time taken to reach certain distance in air = 2.5 sec
a.
We have to find the distance traveled by sound in air.
Distance = Product of speed and time.
⇒ [tex]Distance = Speed\times time\ taken[/tex]
⇒ [tex]Distance = 320\times 2.5[/tex]
⇒ [tex]Distance = 800[/tex] meters.
b.
Now we have to find how much time the sound will take to travel in water.
⇒ Time = Ratio of distance and speed
⇒ [tex]Time =\frac{distance}{speed}[/tex]
⇒ [tex]Time =\frac{800}{1600}[/tex] ...distance = 800 m and speed = 1600 m/s
⇒ [tex]Time =\frac{1}{2}[/tex]
⇒ [tex]Time =0.5[/tex] seconds.
Distance covered by the sound in air is 800 meter and the time taken by the sound in water for the same distance is 0.5 seconds.
The photoelectric effect can be thought of as a simple function: light is an input, and electrons, which produce electricity, are the output. The model in this interactive is missing a few parts to harness this electricity effectively, what are they?
Answer:
(1) Conductive medium (2) Load.
Explanation:
Electricity is essentially Flow of electrons in conductive medium such as wire and to harness it there needs to be a load in the circuitry to use up the energy of the moving electrons through the wire.
this is how solar cells work, in them there are Silicon P and N Junctions connected in series and parallel ( quite alot of them) forming a large plate , which then can be connected to a load through wires to harness electricity produced.
How much time would it take to exert 700W of power while doing 700J of work?
Answer:
The answer to your question is time = 1 s
Explanation:
Data
time = ?
Power = 700 W
Work = 700 J
Power is defined as the work done per unit of time.
Formula
Power = Work / Time
-Solve for time
Time = Work/Power
-Substitution
Time = 700 / 700
-Result
Time = 1 s
Conclusion
In 1 second and with 700 J there will be produced 700 watts
The function x = (5.2 m) cos[(5πrad/s)t + π/5 rad] gives the simple harmonic motion of a body. At t = 5.3 s, what are the (a) displacement, (b) velocity, (c) acceleration, and (d) phase of the motion? Also, what are the (e) frequency and (f) period of the motion?
Answer:
(a) Displacement = - 3.0576 m
(b) Velocity [tex]=-66.48[/tex] m/s
(c)Acceleration = -753.39 m²/s
(d)The phase motion is 26.7 [tex]\pi[/tex].
(e)Frequency =2.5 Hz.
(f)Time period =0.4 s
Explanation:
Given function is
[tex]x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
(a)
The displacement includes the parameter t, so,at time t=5.3 s
[tex]x|_{t=5.3}= (5.2 m)cos[ (5\pi \ rad/s)5.3+ \frac\pi5][/tex]
[tex]= (5.2 m)cos[ 26.5\pi+ \frac\pi5][/tex]
=(5.2)(-0.588)m
= - 3.0576 m
(b)
[tex]x= (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
To find the velocity of simple harmonic motion, we need to find out the first order derivative of the function.
[tex]v=\frac{dx}{dt}[/tex]
[tex]=\frac{d}{dt} (5.2 m)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= (5.2 m)(-5\pi)sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
Now we can plug our value t=5.3 into the above equation
[tex]v= -26\pi sin[ (5\pi \ rad/s)5.3\ s+ \frac\pi5][/tex]
[tex]=-66.48[/tex] m/s
(c)
To find the acceleration of simple harmonic motion, we need to find out the second order derivative of the function.
[tex]v= -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]a=\frac{d^2x}{dt^2}[/tex]
[tex]=\frac{dv}{dt}[/tex]
[tex]=\frac{d}{dt}( -26\pi sin[ (5\pi \ rad/s)t+ \frac\pi5])[/tex]
[tex]= -26\pi (5\pi)cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
[tex]= -130\pi^2cos[ (5\pi \ rad/s)t+ \frac\pi5][/tex]
Now we can plug our value t=5.3 into the above equation
[tex]a= -130\pi^2cos[ (5\pi \ rad/s)5.3 \ s+ \frac\pi5][/tex]
= -753.39 m²/s
(d)
The general equation of SHM is
[tex]x=x_mcos(\omega t+\phi)[/tex]
[tex]x_m[/tex] is amplitude of the displacement, [tex](\omega t+\phi)[/tex] is phase of motion, [tex]\phi[/tex] is phase constant.
So,
[tex](\omega t+\phi)=5\pi t+\frac\pi5[/tex]
Now plugging t=5.3s
[tex](\omega t+\phi)=5\pi \times 5.3+\frac\pi5[/tex]
=26.7 [tex]\pi[/tex]
The phase motion is 26.7 [tex]\pi[/tex].
The angular frequency [tex]\omega = 5\pi[/tex]
(e)
The relation between angular frequency and frequency is
[tex]\omega =2\pi f[/tex]
[tex]\therefore f=\frac{\omega}{2\pi}[/tex]
[tex]=\frac{5\pi}{2\pi}[/tex]
[tex]=\frac52[/tex]
= 2.5 Hz
Frequency =2.5 Hz.
(f)
The relation between frequency and time period is
[tex]T=\frac1 f[/tex]
[tex]=\frac1{2.5}[/tex]
=0.4 s
Time period =0.4 s
Susan's 12.0 kg baby brother Paul sits on a mat. Susan pulls the man across the floor using a rope that is angled 30 degree above the floor. The tension is a constant 29.0 N and the coefficient of friction is 0.210. Use work and energy to find Paul's speed after being pulled 2.90 m.
Answer:
Paul speed after being pulled 2.9 m is 2.68m/s.
Explanation:
The work energy theorem, change in kinetic energy of the object from initial position to the final position is equal to the work done on the object ie when the force is applied on the object the object changes its position and work is done on the object.
According to the law of conservation of energy ,
ΔE = W, eqn 1
where ΔE is the change in object energy
W is the all the work done on the object.
Work done is written as W =F dcosθ
Where F is the force,
d is the distance,
θ is the angle between the force and displacement vector.
From the figure given below,
The work of friction is given by W₁ = F₁ d cos180°
The work of pulling force is given by W₂ =F₂ dcos 30°
Change in object energy ΔE = mv²/2.
Applying Newton first law along Y axis,
Fsin30° + N =mg
Normal force N =mg - Fsin30°
Frictional Force F₁ =μN =μ(mg - Fsin30°)
Substituting in eqn 1
mv²/2 = F₂ dcos 30°+ μ(mg - Fsin30°)d cos180°
=[tex]\frac{\sqrt{3} }{2}[/tex] F₂ d - μ(mg - [tex]\frac{F}{2}[/tex])d
v² = [tex]\sqrt{3}[/tex] [tex]\frac{F}{m}[/tex]d - 2μgd +
here m = 12 kg,
d = 2.9 m.
μ = 0.21
F = 29 N
Sub all values,
v² = 7.2
v = 2.68m/s
Paul speed after being pulled 2.9 m is 2.68m/s.
1. What will happen to the brightness of the light bulb if the switch in this circuit is suddenly closed?
Answer:
There is no closed-loop path for the current to flow through the circuit.When the switch is closed,the light bulb operates since the current flows through the circuit.The bulb glows at its full brightness since its receives its full 120 volts and has the design current flow.
Explanation:
A student waiting on a platform for an incoming train hears a train whistle blow some place down the track but she is not certain from which direction it came. If the student notices that the frequency of the train whistle slowly decreases. What can the student conclude about the speed and direction of the train blowing the whistle
Answer:
Considering the two cases of when the train is moving towards and away from the stationary observer. The observed frequency of sound waves from the train can decrease when
1) The speed of the train is slowly reducing and its direction is towards the stationary observer.
2) The speed of the train is slowly increasing and its direction is away from the observer for the sound's observed frequency to keep decreasing.
Explanation:
This phenomenon is due to Doppler's Effect.
Doppler's Effect explains how relative frequency of a sound source varies with the velocity of the source or the observer.
Generally, the mathematical expression for Doppler's Effect is given below
f' = f [(v + v₀)/(v - vₛ)]
where
f' = observed frequency
f = actual frequency
v = velocity of sound waves
v₀ = velocity of observer
vₛ = velocity of sound source
When the train is moving towards the stationary observer,
f' = observed frequency of the sound wave = f₁
f = actual frequency of the sound wave = f
v = velocity of sound waves = v
v₀ = velocity of observer = 0 m/s
vₛ = velocity of sound source = vₛ
f' = f [(v + v₀)/(v - vₛ)]
f₁ = f [(v + 0)/(v - vₛ)]
f₁ = fv/(v - vₛ) (eqn 1)
Observed frequency is obviously higher than the frequency of the train.
But if the train keeps reducing its speed (vₛ) towards from the stationary observer, the observed frequency will decrease.
When the train is moving away from the stationary observer,
f' = observed frequency of the sound wave = f₂
f = actual frequency of the sound wave = f
v = velocity of sound waves = v
v₀ = velocity of observer = 0 m/s
vₛ = velocity of sound source = - vₛ (train is moving away from the observer, hence, the negative sign)
f' = f [(v + v₀)/(v - vₛ)]
f₂ = f [(v + 0)/(v - (-vₛ)]
f₂ = fv/(v + vₛ) (eqn 2)
From the expression, it is clear that the observed frequency is smaller than the frequency of the sound waves and it keeps decreasing as the speed of the train (vₛ) increases as the train moves away from the stationary observer.
Hope this Helps!!!
Answer:
The student can conclude that the speed of the train is increasing while the direction of the train is away from her or him
Explanation:
Here we have the Doppler effect given by
For an approaching wave
[tex]F_{obs} =\left [ \frac{v}{v - v_{source}} \right ] f_{source[/tex]
For a receding wave we have
[tex]F_{obs} =\left [ \frac{v}{v + v_{source}} \right ] f_{source[/tex]
Therefore, comparing the denominators of the two equation, it is seen that the frequency heard by the observer is decreases for a receding wave.
That is the student can conclude that the train is moving away from her or him or that the speed of the train is increasing while the direction of the train is away from her or him.
Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyond Mars when furthest from the Sun (aphelion), but with an average distance (semi-major axis) of 1 AU. How long will it take to complete an orbit and where will it spend most of its time
Answer:
16.63min
Explanation:
The question is about the period of the comet in its orbit.
To find the period you can use one of the Kepler's law:
[tex]T^2=\frac{4\pi}{GM}r^3[/tex]
T: period
G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2
r: average distance = 1UA = 1.5*10^11m
M: mass of the sun = 1.99*10^30 kg
By replacing you obtain:
[tex]T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min[/tex]
the comet takes around 16.63min
If a wave has a speed of 100 m/s and a wavelength of 20 meters, what is the frequency?
Answer:
Frequency 5hz
Explanation:
Final answer:
To calculate the frequency of a wave with a speed of 100 m/s and a wavelength of 20 meters, divide the speed by the wavelength, resulting in a frequency of 5 Hz.
Explanation:
To find the frequency of a wave when you know its speed and wavelength, you can use the formula: speed = (wavelength) x (frequency). Rearranging this formula to solve for frequency gives us frequency = speed / wavelength. In this case, with a wave speed of 100 m/s and a wavelength of 20 meters, the calculation would be frequency = 100 m/s / 20 m.
Therefore, the frequency of the wave is 5 Hz. This means the wave completes 5 cycles per second.
A 1-kg rock is suspended from the tip of a horizontal meterstick at the 0-cm mark so that the meterstick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark. From this information, the mass of the meterstick is _________.
A) 1/4 kg.B) 1/2 kg.C) 3/4 kg.D) 1 kg.E) none of the above
Explanation:
Given that,
Mass if the rock, m = 1 kg
It is suspended from the tip of a horizontal meter stick at the 0-cm mark so that the meter stick barely balances like a seesaw when its fulcrum is at the 12.5-cm mark.
We need to find the mass of the meter stick. The force acting by the stone is
F = 1 × 9.8 = 9.8 N
Let W be the weight of the meter stick. If the net torque is zero on the stick then the stick does not move and it remains in equilibrium condition. So, taking torque about the pivot.
[tex]9.8\times 12.5=W\times (50-12.5)\\\\W=\dfrac{9.8\times 12.5}{37.5}[/tex]
W = 3.266 N
The mass of the meters stick is :
[tex]m=\dfrac{W}{g}\\\\m=\dfrac{3.266}{9.81}\\\\m=0.333\ kg[/tex]
So, the mass of the meter stick is 0.333 kg.
The mass of the meterstick is calculated using the principle of torque balance. The meterstick must have a mass that creates an equal opposite torque to the one created by the 1-kg rock. The calculation shows that the meterstick has a mass of 1/3 kg, which is not listed in the options, so the answer is (E) none of the above.
Explanation:The mass of the meterstick can be found using the principle of moments, also known as the principle of leverage, which states that for an object to be in rotational equilibrium, the sum of the clockwise moments about the pivot (fulcrum) must equal the sum of the counterclockwise moments. In this case, the 1-kg rock is hanging from the 0-cm mark and the meterstick balances when the fulcrum is at the 12.5-cm mark. The torque created by the 1-kg mass is given by its mass multiplied by its distance from the fulcrum, or 1 kg × 12.5 cm.
For the meterstick to balance, its center of mass must be located directly above the fulcrum. Since the meterstick is uniform, its center of mass is in the middle, at the 50-cm mark. This means that the mass of the meterstick acts 50 cm - 12.5 cm = 37.5 cm away from the fulcrum. Let's denote the mass of the meterstick as 'M'. The counterbalance torque provided by the meterstick is given by M kg × 37.5 cm.
Setting the torques equal for a balanced system:
M kg × 37.5 cm = 1 kg × 12.5 cm, which simplifies to M = 1/3 kg. Since 1/3 kg is equivalent to approximately 0.33 kg, the correct answer is (E) none of the above, as none of the provided options matches this result.
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what is the difference between condensation and conduction
Answer:
Convection is the heat transfer due to the bulk movement of molecules within fluids such as gases and liquids, including molten rock. Condensation is the process by which water vapor in the air is changed into liquid water. Condensation is crucial to the water cycle because it is responsible for the formation of clouds. These clouds may produce precipitation, which is the primary route for water to return to the Earth's surface within the water cycle.
Explanation:
Their a difference....... A huge One
Condensation is the transition from gas to liquid, whereas conduction is a form of heat transfer through direct contact between objects.
The difference between condensation and conduction lies in their definition and the processes they describe. Condensation is a phase change in which a substance transitions from a gaseous state to a liquid state, often as a result of cooling. It is not directly related to heat transfer but is a thermodynamic process. On the other hand, conduction is a mode of heat transfer where thermal energy is transmitted through stationary matter due to physical contact, as when heat from an electric burner moves to the bottom of a pan.
Examples of conduction include the warming of a glass when holding it in a warm hand or the transfer of heat from a stove's burner to a pan. Convection is another mode of heat transfer involving the movement of fluids, which plays an important role in weather systems and heating systems. In contrast, conduction relies on direct contact between objects for heat to be transferred.
donna is attempting to make an electromagnet to pick up old nails in and around her driveway. which of these materals would be best to use the core in order to make the strongest electromagnet to pick up the most nails
Answer:
D. Iron nail
Explanation:
Recently, astronomers have observed stars and other objects that orbit the center of the Milky Way Galaxy farther out than our Sun, but move around faster than we do. How do astronomers think such an observation can be explained?
Answer:
That scenario can be explained by the idea of the contribution of dark matter on that point.
Explanation:
It can be explained through the idea of dark matter, this one was born to explain why stars (or any object) that were farther for the supermassive black hole in the center of the Milky Way galaxy didn't decrease it rotational velocity as it was expected according to equation 1.
[tex]v = \sqrt{\frac{G M}{r}}[/tex] (1)
Where v is the rotational velocity, G is the gravitational constant, M is the mass of the supermassive black hole, and r is the orbital radius.
Notice, that If the distance increases the orbital speed decreases (inversely proportional).
Final answer:
Astronomers explain the high velocities of stars orbiting the Milky Way's center outside our Sun's orbit by the gravitational influence of dark matter. This halo of invisible matter adds mass to the galaxy, affecting the orbital speeds in accordance with Kepler's third law.
Explanation:
Astronomers have observed that stars and other objects orbiting the center of the Milky Way Galaxy farther out than our Sun move around faster than we do, which is an unexpected observation according to Kepler's third law. Typically, as per Kepler's third law, we would expect objects that are farther from the center of mass to orbit more slowly. However, the presence of invisible matter, which we now understand as dark matter, affects the gravitational pull and thus the orbital velocities of these outer objects, leading them to move at unexpectedly high speeds.
The Milky Way is surrounded by a halo of this invisible matter, which does not emit or reflect light, making it undetectable through traditional means. Instead, its presence is inferred from its gravitational effects on the motions of stars and gas in the galaxy. The observations mentioned, where objects beyond the luminous part of the Milky Way are moving faster than expected, provide evidence for the existence of dark matter, as this additional mass can account for the higher orbital velocities.
Three different planet-star systems, which are far apart from one another, are shown above. The masses of the planets are much less than the masses of the stars.
In System A , Planet A of mass Mp orbits Star A of mass Ms in a circular orbit of radius R .
In System B , Planet B of mass 4Mp orbits Star B of mass Ms in a circular orbit of radius R .
In System C , Planet C of mass Mp orbits Star C of mass 4Ms in a circular orbit of radius R .
(a) The gravitational force exerted on Planet A by Star A has a magnitude of F0 . Determine the magnitudes of the gravitational forces exerted in System B and System C .
___ Magnitude of gravitational force exerted on Planet B by Star B
___ Magnitude of gravitational force exerted on Planet C by Star C
(b) How do the tangential speeds of planets B and C compare to that of Planet A ? In a clear, coherent paragraph-length response that may also contain equations and/or drawings, provide claims about
why the tangential speed of Planet B is either greater than, less than, or the same as that of Planet A , and
why the tangential speed of Planet C is either greater than, less than, or the same as that of Planet A .
a) 4F0
b) Speed of planet B is the same as speed of planet A
Speed of planet C is twice the speed of planet A
Explanation:
a)
The magnitude of the gravitational force between two objects is given by the formula
[tex]F=G\frac{m_1 m_2}{r^2}[/tex]
where
G is the gravitational constant
m1, m2 are the masses of the 2 objects
r is the separation between the objects
For the system planet A - Star A, we have:
[tex]m_1=M_p\\m_2 = M_s\\r=R[/tex]
So the force is
[tex]F_A=G\frac{M_p M_s}{R^2}=F_0[/tex]
For the system planet B - Star B, we have:
[tex]m_1 = 4 M_p\\m_2 = M_s\\r=R[/tex]
So the force is
[tex]F=G\frac{4M_p M_s}{R^2}=4F_0[/tex]
So, the magnitude of the gravitational force exerted on planet B by star B is 4F0.
For the system planet C - Star C, we have:
[tex]m_1 = M_p\\m_2 = 4M_s\\r=R[/tex]
So the force is
[tex]F=G\frac{M_p (4M_s)}{R^2}=4F_0[/tex]
So, the magnitude of the gravitational force exerted on planet C by star C is 4F0.
b)
The gravitational force on the planet orbiting around the star is equal to the centripetal force, therefore we can write:
[tex]G\frac{mM}{r^2}=m\frac{v^2}{r}[/tex]
where
m is the mass of the planet
M is the mass of the star
v is the tangential speed
We can re-arrange the equation solving for v, and we find an expression for the speed:
[tex]v=\sqrt{\frac{GM}{r}}[/tex]
For System A,
[tex]M=M_s\\r=R[/tex]
So the tangential speed is
[tex]v_A=\sqrt{\frac{GM_s}{R}}[/tex]
For system B,
[tex]M=M_s\\r=R[/tex]
So the tangential speed is
[tex]v_B=\sqrt{\frac{GM_s}{R}}=v_A[/tex]
So, the speed of planet B is the same as planet A.
For system C,
[tex]M=4M_s\\r=R[/tex]
So the tangential speed is
[tex]v_C=\sqrt{\frac{G(4M_s)}{R}}=2(\sqrt{\frac{GM_s}{R}})=2v_A[/tex]
So, the speed of planet C is twice the speed of planet A.
What are the similarities and differences between novae and supernovae?
Answer:
Explanation:
A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.
A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.
From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae
The speed of propagation equals the frequency times the wavelength.
O
True
O
False
Answer:
true
Explanation:
because I say it's true
Answer:help me with my physics please
Explanation:
An ordinary egg can be approximated as a 5.5-cm diameter sphere. The egg is initially at a uniform temperature of 8°C and is dropped into boiling water at 97°C. Taking the properties of the egg to be r 1020 kg/m3 and cp 3.32kJ/kg · °C, determine (a) how much heat is transferred to the egg by the time the average temperature of the egg rises to70°C and (b) the amount of entropy generation associated with this heat transfer process.
Answer:
a) [tex]Q_{in} = 13.742\,kW[/tex], b) [tex]\Delta S = 370.15\,\frac{kJ}{K}[/tex]
Explanation:
a) The heat transfered to the egg is computed by the First Law of Thermodynamics:
[tex]Q_{in} +U_{sys,1} - U_{sys,2} = 0[/tex]
[tex]Q_{in} = U_{sys,2} - U_{sys,1}[/tex]
[tex]Q_{in} = \rho_{egg}\cdot \left(\frac{4\pi}{3}\cdot r^{3}\right)\cdot c \cdot (T_{2}-T_{1})[/tex]
[tex]Q_{in} = \left(1020\,\frac{kg}{m^{3}}\right)\cdot \left(\frac{4\pi}{3}\right)\cdot (0.025\,m)^{3}\cdot \left(3.32\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (70\,^{\textdegree}C - 8\,^{\textdegree}C)[/tex]
[tex]Q_{in} = 13.742\,kW[/tex]
b) The amount of entropy generation is determined by the Second Law of Thermodynamics:
[tex]\Delta S = \frac{Q_{in}}{T_{in}}[/tex]
[tex]\Delta S = \frac{13.742\,kJ}{370.15\,K}[/tex]
[tex]\Delta S = 370.15\,\frac{kJ}{K}[/tex]
The heat transferred to the egg is approximately 18.24 kJ. The entropy generation associated with this heat transfer is about 5.68 kJ/K.
To determine how much heat is transferred to the egg, we use the following formula for heat transfer in a substance:
Q = m × cp × (T_final - T_initial)
Given data:
Initial temperature, T_initial: 8°CFinal temperature, T_final: 70°CDensity, ρ: 1020 kg/m³Specific heat, cp: 3.32 kJ/kg·°CDiameter of the egg: 5.5 cm (radius is 2.75 cm or 0.0275 m)The volume of the egg as a sphere is:
V = (4/3) × π × r³ = (4/3) × π × (0.0275)³ ≈ 8.72 × 10⁻⁵ m³
The mass of the egg is then:
m = ρ × V = 1020 kg/m³ × 8.72 × 10⁻⁵ m³ ≈ 0.0889 kg
Now, calculate the heat transferred:
Q = m × cp × (T_final - T_initial) = 0.0889 kg × 3.32 kJ/kg·°C × (70°C - 8°C) ≈ 18.24 kJ
(b) Calculating Entropy Generation:
Using the formula for entropy change, we have:
ΔS = m × cp × ln(T_final/T_initial)
Temperatures in Kelvin:
T_initial = 8°C + 273.15 = 281.15 KT_final = 70°C + 273.15 = 343.15 KThen:
ΔS = 0.0889 kg × 3.32 kJ/kg·°C × ln(343.15 / 281.15) ≈ 5.68 kJ/K
Hence, the amount of entropy generated during the process is approximately 5.68 kJ/K.
Technician A says a pull to the right during braking could be caused by a defective metering valve. Technician B says a pull to the left could be caused by a defective proportioning valve. Which technician is correct
Answer:
The correct option is;
Neither A nor B
Explanation:
The common cause of a vehicle pulling to the right or to the left during braking is due to a contaminated breaking surface or a faulty caliper. Other causes include tire size variation or worn out suspension components.
Other causes include;
Leaking break fluid
Piston frozen in wheel cylinder or caliper
Adjuster of rear break
Tire fault.
As soon as it is observed that the vehicle pulls on one side when braking, the vehicle should should be taking for checks by a qualified mechanic as soon as possible.
Final answer:
Both Technician A and B are incorrect, as a pull during braking is more likely due to uneven brake pad wear, sticking calipers, or collapsed brake lines rather than a defective metering or proportioning valve.
Explanation:
The question relates to whether a pull to the right during braking could be caused by a defective metering valve, and whether a pull to the left could be caused by a defective proportioning valve. Neither of these assertions is typically accurate.
A metering valve is designed to ensure that the rear brakes engage just after the front brakes, to prevent rear wheel lockup during early brake application. A proportioning valve adjusts the pressure between the front and rear brakes to prevent the rear wheels from locking up during heavy braking. Neither valve is designed to balance the braking force from side to side.
A pull to one side during braking is more likely to be caused by uneven brake pad wear, sticking calipers, or a collapsed brake line, rather than issues with the metering or proportioning valves. Therefore, both Technician A and Technician B are incorrect regarding the cause of a pull during braking.