Answer:
The nose sends the information of the smell to the brain.
Explanation:
i think but I'm not really sure
Any electricity-charged object creates an electric field. Walking across the carpet in wool socks can create an electric charge. This observation is evidence for which characteristics of electrical forces?
Explanation:
they generate electricity by contact
Which of the following scenarios will generate electricity using electromagnetic induction?
A. a highly conductive meatal moving within magnetic field
B. a changing magnetic field surrounded by plastic wires
C. electricity Being applied to a magnet
Answer:
A generates electricity, the others do not
Explanation:
If a person is putting in 500 J of work in a 25 second period, how much power are they producing?
4. The period of the sound wave produced by a 440 Hz tuning fork is
Answer:
.002273 seconds
Explanation:
1/f = period
1/ 440 c/s = .002273 seconds
Astronomers often detect stars that are rotating extremely rapidly, known as neutron stars. These stars are believed to have formed in the inner core of a larger star that collapsed, due to its own gravitation, to a star of a very small radius and very high density. Before collapse suppose the core of such star is the size of our Sun (R=7 x 10km) with mass 2.0 times as great as the Sun, and is rotating at a speed of 1 revolution every 10 days. If it were to undergo gravitational collapse to a neutron star of radius 10 km, what would its rotational speed be? Assume the star is a uniform solid sphere at all times. (1 = MR2) a
Based on law of conservation of angular momentum, the rotational speed of the star is equal to 6,000 rev/s.
Given the following data:
Radius of Sun = 7 × 10⁵ km.Mass of star = 2 Mass of Sun (M = 2M).Radius of star = 10 km.Time = 10 days.How to calculate the rotational speed.First of all, we would determine the initial angular speed of the neutron star as follows:
[tex]\omega_i = \frac{1 \;Rev}{10 \;days} \\\\\omega_i = \frac{1 \;Rev}{10 \times 24 \times 60 \times 60}\\\\\omega_i = 1.157 \times 10^{-6}\;rev/s[/tex]
Mathematically, the moment of inertia of a uniform solid sphere is given by this formula:
[tex]I=\frac{2}{5} mr^2[/tex]
Where:
I is the moment of inertia.m is the mass.r is the radius.In order to determine the rotational speed of this neutron star, we would apply the law of conservation of angular momentum:
[tex]L_1 = L_2\\\\I_1\omega_1 = I_2\omega_2\\\\\frac{2}{5} m_1r_1^2 \omega_1 = \frac{2}{5} m_2r_2^2\omega_2\\\\m_1r_1^2 \omega_1 = m_2r_2^2\omega_2\\\\\omega_2 =\frac{r_1^2 \omega_1}{r_2^2}[/tex]
Substituting the given parameters into the formula, we have;
[tex]\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ (10 )^2}\\\\\\\omega_2 = \frac{ (7 \times 10^5 )^2 \times 1.157 \times 10^{-6}}{ 10 0}\\\\\omega_2 =\frac{566,930}{100}[/tex]
Final angular speed = 5,669 ≈ 6,000 rev/s.
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D. Calculate the electric force (F.) between an electron and
proton that are 5.29 x 10-11 meters apart.
Answer:
See below
Explanation:
F = C q1 q2 /r^2
8.988 x 10^9 * (1.60217 x 10^-19)^2 / ( 5.29 x 10^-11)^2 =
.00000008244 N
82.4 nano N
Select all the correct answers.
A worker is holding a filled gas cylinder still. Which two sentences are true about the energy of the filled gas cylinder?
it has no energy because it's being held still
It has gravitational potential energy because of its height.
its atoms and molecules have thermal energy
It has motion energy because it will fall if let go
its kinetic energy is being converted to potential energy
ghts reserved
it has no energy because its being held still
it has motion energy because it will fall if let go
A wave is sent back and forth along a rope 4 m long with a mass of 0.6 kg by exerting a force a force of 30 N. Calculate the linear mass density of the rope (in kg/m).
The linear mass density of the rope in the given motion of the wave is 0.15 kg/m.
Linear mass density of the ropeThe linear mass density of the rope in the given motion of the wave is determined by dividing the mass of the rope with the length of the entire rope.
The linear mass density of the rope is calculated as follows;
μ = m/L
μ = 0.6 kg / 4 m
μ = 0.15 kg/m
Thus, the linear mass density of the rope in the given motion of the wave is 0.15 kg/m.
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In a steam engine, steam in a cylinder expands against a piston, exerting 10 atm of external pressure. The volume of the cylinder increases by 10 L and simultaneously the steam cools, losing 3,000 kJ of energy to the surroundings. What is the change in energy of the steam
If steam in a cylinder expands against a piston, exerting 10 atm of external pressure, the change in energy of the steam is mathematically given as
dE= -3010 kJ
What is the change in energy of the steam?Generally, the equation for the workdone is mathematically given as
Work = - (pressure * Volume)
Therefore
w= - (10 x 10)
w= -100 L atm
In conclusion
dEnergy = q + work
dE = -3000 + -10.13
dE= -3010 kJ
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explain the fleming left-hand rule with the diagram and what will the direction of the induced current in the figure if the magnet is going towards the coil.
how to make sure a snow leopard does not escape?
build a fence.....................
How much pressure is created when you apply a
50.0N force on a syringe plunger that has a diameter of 0.4
A garage hoist lifts a truck up 2 m above the ground in 15seconds.Find the power delivered to the truck (given 1000kg as the mass of the truck and gas 9.81m/s
The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.
What is power?
Power can be deifned as the rate at which work is done.
To calculate the power delivered to the truck, we use the formula below.
Formula:
P = mgh/t.............. Equation 1Where:
P = Power delivered to the truckm = Mass of the truckh = Heightg = Acceleration due to gravityt = timeFrom the question,
Given:
m = 1000 kgh = 2 mt = 15 secondsg = 9.81 m/s²Subsitute these values into equation 1
P = (1000×2×9.81)/15P = 19620/15P = 1308 WHence, The power delivered to the truck of mass 1000 kg that was lifted by a garage hoist, 2 m high above the ground in 15 seconds is 1308 W.
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A car of mass 1200kg falls a vertical distance of 24m starting from rest what is the workdone by the force of gravity on the car?Use the work energy theorem to find the final velocity of the car just before it hits the water(Treat the as a point like object)
Answer:
PE (relative to water) = M g h
PE = 1200 kg * 9.8 m/s^2 * 24 m = 2.82E5 Joules
KE = PE when vehicle strikes water
1/2 M V^2 = 2.82E5
V = (2.82E5 * 2 / 1200)^1/2 = 21.7 m/s
Check:
M g h = 1/2 M V^2
V = (2 g h)^1/2 = (2 * 9.8 * 24)^1/2 = 21.7 m/s
give an example of the 4 steps of the scientific method
Answer:
1) asking a question about something you observe, 2) doing background research to learn what is already known about the topic, 3) constructing a hypothesis, 4) experimenting to test the hypothesis
Answer:
1) make an observation that describes a problem - see something you can fix or improve and try to describe the problem like which type of vegetables does rabbit like
2) create a hypothesis - What do you think will happen - If the rabbit eat the lettuce than .........
3) test the hypothesis - Do your experiment with the variables and follow the procedures
4) draw conclusions and refine the hypothesis - see if your hypothesis was correct - In conclusion my hypothesis was not correct because.......
What are the Applications of electronic polarization?
Answer:
it can be used for communication, industrial application and also for instrumentation application
Question 5 of 10
Which of the following is most likely to happen when energy is transferred to
an object?
O A. The force on the object will decrease.
O B. The object will begin to move.
O C. The object will stay exactly the same.
O D. The mass of the object will increase.
Answer:
B: The object will start moving
Explanation:
If energy is transferred the object will definitely change so it can't be a. If you add energy the object will have more force so it cant be c. The mass of an object can't increase just by giving an object energy so it cant be d
divide 48 into the ratio 2:4
why did the ball move the same distance when it was on the flatbed truck that accelerated 0.2 m/s?
The relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.
What is relative velocity?The relative velocity of an object is the velocity of the object observed with respect to rest frame of another object.
Distance traveled by the objectThe distance traveled by each object is determined from the prouduct of velocity and time of motion.
d = vt
where;
v is velocityt is timeThus, if the ball moved the same distance when it was on the flat bed truck, then the relative velocity of the ball with respect to the truck is same as that of the truck with respect to the ball at equal time.
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A turntable is switched from 11.3 rad/s to 1.9 rad/s, and the platter goes through an angle of 17.95 radian in reaching the new angular speed. What is the angular acceleration of the platter?
Hi there!
We can use the angular equivalent of the following kinematic equation:
[tex]\omega_f^2 = \omega_i^2 + 2\alpha \theta[/tex]
ωf = final angular velocity (1.9 rad/s)
ωi = initial angular velocity (11.3 rad/s)
α = angular acceleration (? rad/s²)
θ = angular displacement (17.95 rad)
We can rearrange the equation to solve for angular acceleration.
[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta\\\\\alpha = \frac{\omega_f^2 - \omega_i^2}{2\theta}[/tex]
Plug in the given values and solve.
[tex]\alpha = \frac{1.9^2 - 11.3^2}{2(17.95)} = \boxed{3.456 \frac{rad}{s}}[/tex]
A body is piloted at a point. A force of 10 N is applied at a distance of 30 cm from the pivot. Find the amount of force about the pivot .
Question :-
A Body is Pivoted at a Point. A Force of 10 N is Applied at a Distance of 30 cm from the Pivot. Find the Amount of Force about the Pivot.Answer :-
Amount of Force is 3 Nm .Explanation :-
As per the provided information in the given question, The Force is given as 10 Newton . The Distance is given as 30 cm [ 0.3 m ] . And, we have been asked to calculate the Amount of Force .
For calculating the Force , we will use the Formula :-
[tex] \bigstar \: \: \: \boxed{ \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} \: } [/tex]
Therefore , by Substituting the given values in the above Formula :-
[tex] \dag \: \: \: \sf {Moment \: of \: Force \: = \: Force \: \times \: Distance} [/tex]
[tex] \longmapsto \: \: \: \sf {Moment \: of \: Force \: = \: 10 \: \times \: 0.3} [/tex]
[tex] \longmapsto \: \: \: \textbf {\textsf {Moment \: of \: Force \: = \: 3}} [/tex]
Hence :-
Amount of Force = 0.3 Nm .[tex] \underline {\rule {185pt}{4pt}} [/tex]
1. A listener stands 20.0 m from a speaker that pumps out music with a power output of 100.0 W.
a. What is surface area being vibrated by the time the sound reaches the listener (you may
have to look up the equation for the area of a sphere).
b. What is the intensity of the sound wave as it reaches the person listening?
c. What is the relative intensity of the sound as heard by the listener?
2. A police car siren has a frequency of 512 Hz. They are chasing a suspect at 65.0 m/s.
a. What is the speed of sound if the air temperature is 15.0o?
b. If the suspect is running away from the police car at 10.0 m/s, what is the frequency of
the sound he hears?
(1.a) The surface area being vibrated by the time the sound reaches the listener is 5,026.55 m².
(1.b) The intensity of the sound wave as it reaches the person listening is 0.02 W/m².
(1.c) The relative intensity of the sound as heard by the listener is 103 dB.
(2.a) The speed of sound if the air temperature is 15⁰C is 340.3 m/s.
(2.b) The frequency of the sound heard by the suspect is 614.3 Hz.
Surface area being vibratedThe surface area being vibrated by the time the sound reaches the listener is calculated as follows;
A = 4πr²
A = 4π x (20)²
A = 5,026.55 m²
Intensity of the soundThe intensity of the sound is calculated as follows;
I = P/A
I = (100) / (5,026.55)
I = 0.02 W/m²
Relative intensity of the sound[tex]B = 10log(\frac{I}{I_0} )\\\\B = 10 \times log(\frac{0.02}{10^{-12}} )\\\\B = 103 \ dB[/tex]
Speed of sound at the given temperature[tex]v= 331.3\sqrt{1 + \frac{T}{273} } \\\\v = 331.3\sqrt{1 + \frac{15}{273} } \\\\v = 340.3 \ m/s[/tex]
Frequency of the soundThe frequency of the sound heard is determined by applying Doppler effect.
[tex]f_o = f_s(\frac{v \pm v_0}{v \pm v_s} )[/tex]
where;
-v₀ is velocity of the observer moving away from the source-vs is the velocity of the source moving towards the observerfs is the source frequencyfo is the observed frequencyv is speed of sound[tex]f_0 = f_s(\frac{v-v_0}{v- v_s} )[/tex]
[tex]f_0 = 512(\frac{340.3 - 10}{340.3 - 65} )\\\\f_0 = 614.3 \ Hz[/tex]
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When you jump, you exert a pushing force against the ground. Gravity pulls you back down. Why can a person jump higher on the moon than on Earth?
Answer:
This is because the force of gravity is much less on the moon than on the earth, therefore the person wont be pulled down much and will jump higher
Answer: This is because the force of gravity is much less on the moon than on the earth, therefore the person won't be pulled down much and will jump higher.
Friction between our feet and the surface we walk on is desirable. True False
Answer:
True
Explanation:
A class of 10 students taking an exam has a power output per student of about 200W. Assume the initial temperature of the room is 20 C and that its dimensions are 6.0 m by 15.0 m by 3.0 m. What is the temperature of the room at the end of 1.0 h if all the energy remains in the air in the room and none is added by an outside source? The specific heat is 837 J/kg C, and its density is about 1.3 x 10^-3 g/cm^3
For a class of 10 students taking an exam has a power output per student of about 200W, the temperature is mathematically given as
dT=25C
What is the temperature of the room at the end of 1.0 h if all the energy remains in the air in the room and none is added by an outside source?Generally, the equation for the temperature is mathematically given as
dT=Q/mc
Where
Q=Pt
Q=2000*3600
Q=7200000J
And m
m=pv
m=1.3*270
m=351kg
Therefore
dT=7200000J/351*837
dT=25C
In conclusion, The temperature of the room at the end of 1.0 h
dT=25C
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A heat engine exhausts 6000 J of heat while performing 2000 J of useful work.
Determine the efficiency of the engine
Answer:
Explanation:
The efficiency can be calculated by:
[tex]\eta = \frac{Q_{E}-Q_{W}}{Q_{E}}=\frac{6000-2000}{6000}=\frac{2}{3}[/tex]
or we can say it becomes approximately 66.7%
A proton in the nucleus of an atom has an electrical charge of:
neutral
-
+
zero
Answer:
proton is positively charged changechar
Explanation:
A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figure below. The string goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is k = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the string does not stretch and does not slip on the pulley, and that the mass of the string is negligible. Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown.
The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
Angular Speed of the pulley
The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;
K.E = P.E
[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]
[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]
Substitute the given parameters and solve for the angular speed;
[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]
Linear speed of the blockThe linear speed of the block after travelling 0.7 m;
v = ωR₂
v = 35.39 x 0.03
v = 1.1 m/s
Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
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17. How long will it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s? (A) 0.6s (B) 1.2 s (C) 2.4 s (D) 3.6 s (E) 4.8 s
Answer:
(B)
Explanation:
Time = change of velocity ÷ acceleration
= (6-0) ÷ 5
= 1.2
1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
To calculate the time it takes for an object to change its velocity from 0 to 6 m/s, we can use the formula:
time = change in velocity / acceleration
Given that the change in velocity (Δv) is 6 m/s and the acceleration (a) is 5 m/s², we can plug these values into the formula:
time = 6 m/s / 5 m/s²
time = 1.2 seconds
Therefore, 1.2 s it take for an object accelerating at a constant rate of 5 m/s to change its velocity from 0 to 6 m/s.
Hence, the correct option is D.
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What are some INDIRECT applications of centripetal force in real life?
Answer:
The circular turning of roads
Explanation:
please mark brainliest
Answer:
Explanation:
The circular turning of roads
Driving on Curves