Answer:
a) y₁ = ½ a, b) y₂ = 4 y₁, c) y₃ = 9 y₁
Explanation:
For this exercise we can use the accelerated motion relationships.
Let's set a reference system where the x axis is parallel to the plane and its positive side is going down the plane.
y = y₀ + v₀ t + ½ a t²
in that case where we throw the marble is the zero point, y₀ = 0, as part of rest its initial velocity is zero v₀ = 0 and a is the acceleration along the inclined plane
y = ½ a t²
a) in the first second t = 1
y₁ = ½ a
b) in the next second of movement
t = 2 s
y₂ = ½ a 2²
y₂ = 4 ½ a
y₂ = 4 y₁
c) for the next second
t = 3 s
y₃ = ½ a 3²
y₃ = 9 ½ a
y₃ = 9 y₁
Amy is in-line skating. Her mass is 50 kg. She is rolling forward (north) on a flat section of road at 10 m/s. The rolling friction acting on the wheels of her skates is 10 N backward (south). Air resistance creates a 15 N force also acting backward (south) on her. If these are the only horizontal forces acting on Amy. What is her horizontal acceleration as a result of these forces?
Answer:
The right solution is "0.50 m/s²". A further explanation is provided below.
Explanation:
The given values are:
Mass,
m = 50 kg
Speed,
v = 10 m/s
Rolling friction acting backward (south),
f = 10 N
Air resistance acting backward (south),
[tex]F_T[/tex] = 15 N
The total force acting will be:
⇒ [tex]F = -f-F[/tex]
On substituting the given values, we get
⇒ [tex]=-10-15[/tex]
⇒ [tex]=-25 \ N[/tex]
Now,
⇒ [tex]a = \frac{F}{m}[/tex]
On substituting the given values, we get
⇒ [tex]=\frac{-25}{50}[/tex]
⇒ [tex]=-0.50 \ m/s^2[/tex]
The horizontal acceleration will be "0.50 m/s²" because the (-)ve sign indicates it in south direction.
The sides of a right triangle that has any given vector for the hypotenuse are called _____
A. Scalar
B. Component
C. Resultant
D. Vector
Answer:
They are the resultant vector.
1. With the exception to water, matter (expands, contracts) when it gets
hotter. *
A)Expands
B)Contracts
An accelerometer has a damping ratio of 0.5 and a natural frequency of 18,000 Hz. It is used to sense the relative displacement of a beam to which it is attached. (a)If an impact to the beam imparts a vibration at 4500 Hz, calculate the dynamic error and phase shift in the accelerometer output. (b)Calculateits resonance frequency.(c)What isthe maximumpossiblemagnitude ratio that the system can achieve
Answer:
A) i) Dynamic error ≈ 3.1%
ii) phase shift ≈ -12°
B) 79971.89 rad/s
Explanation:
Given data :
Damping ratio = 0.5
natural frequency = 18,000 Hz
a) Calculate the dynamic error and phase shift in accelerometer output at an impart vibration of 4500 Hz
i) Dynamic error
This can be calculated using magnitude ratio formula attached below is the solution
dynamic error ≈ 3.1%
ii) phase shift
This phase shift can be calculated using frequency dependent phase shift formula
phase shift ≈ -12°
B) Determine resonance frequency
Wr = 2[tex]\pi[/tex] ( 18000 [tex]\sqrt{0.5}[/tex] ) = 79971.89 rad/s
C) The maximum magnitude ratio that the system can achieve
A storage tank has the shape of an inverted circular cone with height 12 m and base radius of 4 m. It is filled with water to a height of 10 m. Find the work required to empty the tank by pumping all of the water to the top of the tank. (The density of water is 1000 kg/m3. Assume g
Answer:
Work required to empty the tank by pumping all of the water to the top of the tank = 1674700 Kgm/s^2
Explanation:
Volume of Circular cone = V = (1/3)πr2h
where r is the radius in meters
and h is the height in meters
Substituting the given values in above equation, we get -
V = [tex]\frac{1}{3} * 3.14 * 4^2 * 10 = 167.47[/tex] cubic meters.
The force required will be equal to the mass of water in the cone
[tex]= 167.47 * 1000[/tex]
= 167470 Kg
Weight = Mass * g
= 167470 * 10
= 1674700 Kgm/s^2
Find the least frequency of incident light that will knock electrons out of the surface of a metal with a work function of 3eV.
Answer:
The frequency must be: [tex]725\,\,10^{12} \,Hz[/tex]
Explanation:
If the work function of the metal ([tex]\phi[/tex]) is 3 eV, then we can use the formula for the kinetic energy of an ejected electron:
[tex]KE= h*f-\phi[/tex]
considering for the minimum KE = 0, and using the Plank constant h in eV s as: 4.14 * 10 ^(-15) eV s, to solve for the frequency:
[tex]h*f=\phi\\4.14*10^{-15} * f = 3\\f=3*10^{15} /4.14\,\,\frac{1}{s} \\f=725\,10^{12} \,Hz[/tex]
How much energy is required to move 2 electrons through a potential difference of 1.0 x 10^ 2 volts?
The energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is -3.2 x 10⁻¹⁷ joules.
What is the potential difference?The potential difference, also known as voltage, is the difference in electrical potential energy per unit of charge between two points in an electrical circuit or an electric field.
In simpler terms, the potential difference is the amount of energy required to move a unit of electric charge from one point to another in an electric field or an electrical circuit. It is measured in volts (V) and can be calculated using the equation:
V = W/Q
Where V = is the potential difference,
W = is the work done in moving the charge Q from one point to another,
Q =is the amount of charge that is moved.
Potential difference is an essential concept in electrical engineering and physics, as it governs the flow of electric current in a circuit and determines the behavior of electrical devices such as resistors, capacitors, and batteries.
Here in this question,
The energy required to move an electron through a potential difference is given by the formula:
E = qV
Where
E = is the energy required,
q = is the charge of the electron,
V = is the potential difference.
The charge of one electron is -1.6 x 10⁻¹⁹coulombs.
Therefore, for two electrons, the total charge is:
q = 2 x (-1.6 x 10⁻¹⁹coulombs) = -3.2 x 10⁻¹⁹ coulombs
The potential difference is given as 1.0 x 10² volts.
Now, the energy required to move 2 electrons through a potential difference of 1.0 x 10² volts is:
E = (-3.2 x 10⁻¹⁹ coulombs) x (1.0 x 10² volts) = -3.2 x 10⁻¹⁷ joules.
Note: that the negative sign indicates that the electrons lose potential energy as they move through the potential difference.
Therefore, The required energy is -3.2 x 10⁻¹⁷ joules.
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Jason leaves his dorm and drives 3 kilometers north to a fruit shop.
Then he turns and drives 4 kilometers south to a restaurant. Then he
drives 5 kilometers east to an art classroom. The total distance that
Jason traveled is kilometers.
The total distance that Jason traveled is 12 kilometers.(3+4+5)
Explain two ways of magnetising an object
Answer:
There are two methods generally used to magnetize permanent magnets: static magnetization and pulse magnetization.
what is costant error
Answer:
In a scientific experiment, a constant error -- also known as a systematic error -- is a source of error that causes measurements to deviate consistently from their true value.
Explanation:
How do spectroscopes help with studying of distance stars?
It measures the distance of faraway stars.
It collects sound waves from distant stars.
It breaks apart lights that comes from stars faraway.
It gives a clear image of the shape of stars.
Answer:
You take the light from a star, planet or galaxy and pass it through a spectroscope, which is a bit like a prism letting you split the light into its component colours. "It lets you see the chemicals being absorbed or emitted by the light source. From this you can work out all sorts of things," says Watson
Cameron is standing on the edge of a 60 m high cliff. He horizontally throws a football
with an initial velocity of 10 m/s. How long does it take for the football to hit the
ground?
Answer:34.6 m/s
Explanation: It is asking how long meaning the answer is in time
help plz! what vibrates in following types of wave motion 1)light wave 2)sound waves 3)x-rays 4)water waves
Answer:
I believe it's 2) sound waves
Explanation:
With sound waves, the energy travels along in the same direction as the particles vibrate. This type of wave is known as a longitudinal wave, so named because the energy travels along the direction of vibration of the particles.
A car comes to a bridge during a storm and finds the bridge washed out. The driver must get to the other side, so he decides to try leaping it with his car. The side the car is on is 21.3 m above the river, whereas the opposite side is a mere 2.3 m above the river. The river itself is a raging torrent 54.0 m wide. 1. How fast should the car be traveling just as it leaves the cliff in order to just clear the river and land safely on the opposite side? 2. What is the speed of the car just before it lands safely on the other side?
Answer:
Explanation:
1 ) Let the initial horizontal velocity of car be v .
For vertical displacement
vertical displacement h = 21.3 - 2.3 = 19 m
Time taken to fall by 19 m be t
19 = 1/2 x 9.8 t² ( initial downward velocity is zero )
t = 1.97 s
This is also the time taken to cover horizontal displacement of 54 m which is width of river .
horizontal speed v = 54 / 1.97 m /s
v = 27.41 m /s
2 )
At the time of landing on other side , car will have both vertical and horizontal speed .
vertical speed
v = u + gt
= 0 + 9.8 x 1.97 = 19.31 m /s
horizontal speed will remain same as the initial speed = 27.41 m /s
Resultant speed = √ ( 27.41² + 19.31² )
= √ ( 751.3 + 372.87)
= 33.52 m /s
To understand the relationships between the parameters that characterize a wave. It is of fundamental importance in many areas of physics to be able to deal with waves. This problem will lead you to understand the relationship of variables related to wave propagation: frequency, wavelength, velocity of propagation, and related variables. Note that these are kinematic variables that relate to the wave's propagation and do not depend on its amplitude.
1. Traveling waves propagate with a fixed speed usually denoted as v (but sometimes c). The waves are called __________ if their waveform repeats every time interval T.
a. transverse
b. longitudinal
c. periodic
d. sinusoidal
2. If the velocity of the wave remains constant, then as the frequency of the wave is increased, the wavelength __________.
a. decreases
b. increases
c. stays constant
Answer:
1) c.
2) a.)
Explanation:
1)
At any wave, if its waveform repeats itself every time interval T, it is said that the wave is periodic, with a period T, which is the time needed to complete an entire cycle. The other options refer at the way in the waves propagates (longitudinal or transversal) and to the type of waveform (sinusoidal), so the right answer is c).2)
At any wave that propagates at a constant speed, there exists a fixed relationship between the velocity v, the frequency f and the wavelength λ, as follows:[tex]v = \lambda * f (1)[/tex]
So in order to v keep constant, if the frequency is increased, the wavelength will decrease in the same proportion, so a) is the right answer.What is the acceleration of a car that goes from 0 MS to 60 MS and six seconds
As you change the value of the battery voltage, how does this change the current through the circuit and the resistance of the resistor
Answer:
Current increases
Resistance remains same
Explanation:
As you change the value of the battery voltage
We know that the current is dependent on the supply of power , so the current increases but the resistance remains same.
A painter sits on a scaffold that is connected to a rope passing over a pulley. The other end of the rope rests in the hands of the painter who wants to lift the scaffold. She plans to pull downward on the loose end of the rope, thinking that the scaffold will then rise vertically with her along for the ride. The scaffold has a mass of 52 kg, and her mass is 63 kg. The painter pulls downward on the rope with a force of 600.0 N, while she and the scaffold are hanging from the other end above the ground.
Required:
a. What is the net acceleration on the system consisting of the painter and the scaffold?
b. What is the magnitude of the normal force exerted on the painter by the scaffold?
Solution :
a). From Newtons second law,
F = ma
The total tension force is 2T.
∴ 2T - (m + M)g = (m+ M)a
Then
[tex]$a=\frac{2T-(m+M)g}{m+M}$[/tex]
[tex]$a=\frac{2\times 600-(52+63)9.8}{52+63}$[/tex]
[tex]$=0.63 \ m/s^2$[/tex]
b). From the person,
F = ma
T - Mg + N = Ma
or N = Ma + Mg - T
= (63 x 9.8) + (52 x 9.8) - 600
= 617.4 + 509.6 - 600
= 527 N
Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.
Which describes the length of the power cord when Shanti gets off the train?
cannot be determined
less than 1.2 m
more than 1.2 m
equal to 1.2 m
Answer:
D. equal to 1.2
Explanation:
on edg
The length of the power cord will be equal to 1.2 m.
Describe about the length of power cord? The train is moving at a speed of 90 km /hr. Train was moving but the person in the train can be considered to be at rest. Shanti is the person travelling on the train. Her cord can be used only by her and the cord length of the phone will be 1.2 m.The length can be measured through the distance.The unit of length is meter.As we know the concept of motion and rest, there only the train in motion, shanti was at rest and shanti's power cord were also in the rest. Power cord length will be determined only at the time of manufacturing.If the power cord length to be change then the crimping process.So, the length will not change suddenly.
The length of the power cord when shanti gets off the train is equal to 1.2 m.
The Correct answer is Option D.
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Which statements accurately describe sound waves? Check all that apply.
Sound waves are transverse waves.
Sound waves require a medium to transfer energy.
Sound waves start with a vibration.
Energy from a vibration creates a mechanical wave.
Sound can be heard as soon as an object begins vibrating.
Answer:
the answers are
b) sound waves require a medium to transfer energy
c) sound waves start with a vibration
d) energy from a vibration creates a mechanical wave
Explanation:
i just did this and got 100%, edge 2021
Following statements are true for sound wave,
Sound waves require a medium to transfer energy.Sound waves start with a vibration.Energy from a vibration creates a mechanical wave.In sound waves,
Energy is transferred through vibration of air particles or particles of a solid through which the sound travels.As sound passes through air (or any fluid medium), the particles of air do not vibrate in a transverse mannerA transverse wave is a wave whose oscillations are perpendicular to the direction of the wave's advance.Learn more:
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A ball accelerates at 2 m/s what is the velocity after 5sec?
Answer:
10m
Explanation:
assuming the ball was at 0 m/s before, after 5 seconds the ball would be traveling at 10m
Candice is examining a cell under a microscope. She has identified a cell wall, a nucleus, and a chloroplast. What type of organism does this most likely belong to?
A. A plant B. An animal C. A fungus D. A bacterium
Answer:
A plant
Explanation:
because animals don't have cell walls, and fungus and bacteria dont have chloroplasts
A 0.25 kg beach ball rolling at a speed of 7 m/s collides with a heavy exercise ball at rest. The beach ball bounces straight back with a speed of 4 m/s. That is the change in momentum of the beach ball? What is the impulse exerted on the beach ball? What is the impulse exerted on the exercise ball?
The impulse exerted on the beach ball is 2.75 kgm/s.
The impulse exerted on the exercise ball is - 2.75 kgm/s.
What is impulse?
This is the force applied to an object that acts over a period of time.
The impulse exerted on the beach ball is the change in the momentum of the ball and it is calculated as follows;
J = ΔP
J = m(v - u)
J = 0.25(7 - (-4))
J = 0.25(7 + 4)
J = 2.75 kgm/s
The impulse exerted on the exercise ball is equal in magnitude but opposite in direction to the beach ball.
Thus, the impulse exerted on the exercise ball is - 2.75 kgm/s.
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In separate experiments, four different particles each start from far away with the same speed and impinge directly on a gold nucleus. The masses and charges of the particles are particle 1: mass m0, charge q0 particle 2: mass 2m0, charge 2q0 particle 3: mass 2m0, charge q0/2 particle 4: mass m0/2, charge 2q0 Rank the particles according to the distance of closest approach to the gold nucleus, from smallest to largest.
A. 1, 2, 3, 4
B. 4, 3, 2,1
C. 3, 1 and 2 tie, then 4
D. 4, 1 and 2 tie, then 1
Answer:
Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)
Kinetic energy at the beginning: KEi =[tex]mv^{2}[/tex]/2
Potential energy at the end: Uf =(Qq/r)*k
Kinetic energy at the beginning: KEf = 0
From energy conservation:
r = 2*Qk/[tex]v^{2}[/tex]*q/m
Ranking of the closest approach to the gold nucleous comes from q/m:
- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest
- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking
- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4
Explanation:
Potential energy at the beginning: Ui =Qq/R*k ≈ 0 (particles are far apart)
Kinetic energy at the beginning: KEi =[tex]mv^{2}[/tex]/2
Potential energy at the end: Uf =(Qq/r)*k
Kinetic energy at the beginning: KEf = 0
From energy conservation:
r = 2*Qk/[tex]v^{2}[/tex]*q/m
Ranking of the closest approach to the gold nucleous comes from q/m:
- particle 3 has the smallest q/m = 1/4 ⋅q0/m0 and, hence, comes the closest
- particle 4 has the largest q/m = 4⋅q0/m0 and, hence, comes the last in ranking
- particles 1 and 2 have equal q/m=q0/m0 and, hence, come tie in between particles 3 and 4
hii guys!! ~ i've comeback again!! could you please help me with an assignment?? i have a couple of i know questions!! thank you for helping me, hehe.
tenise can run the 100 meter dash in 9.77 seconds. what is her speed??
Answer:
Speed = 10.24 m/s.
Explanation:
Given the following data;
Distance = 100m
Time = 9.77
To find her speed;
Speed can be defined as distance covered per unit time. Speed is a scalar quantity and as such it has magnitude but no direction.
Mathematically, speed is given by the equation;
[tex]Speed = \frac{distance}{time}[/tex]
Substituting into the equation, we have;
[tex]Speed = \frac{100}{9.77}[/tex]
Speed = 10.24 meter per seconds.
g A large tank, at 400 kPa and 450 K, supplies air to a converging-diverging nozzle of throat area 4 cm2 and exit area 5 cm2. For what range of back pressures will the flow (a) be entirely subsonic; (b) have a shock wave inside the nozzle; (c) have oblique shocks outside the exit (i.e., over-expanded exit flow); and (d ) have supersonic expansion waves outside the exit (under-expanded exit flow)
Answer:
A) ≥ 325Kpa
B) ( 265 < Pe < 325 ) Kpa
C) (94 < Pe < 265 )Kpa
D) Pe < 94 Kpa
Explanation:
Given data :
A large Tank : Pressures are at 400kPa and 450 K
Throat area = 4cm^2 , exit area = 5cm^2
a) Determine the range of back pressures that the flow will be entirely subsonic
The range of flow of back pressures that will make the flow entirely subsonic
will be ≥ 325Kpa
attached below is the detailed solution
B) Have a shock wave
The range of back pressures for there to be shock wave inside the nozzle
= ( 265 < Pe < 325 ) Kpa
attached below is a detailed solution
C) Have oblique shocks outside the exit
= (94 < Pe < 265 )Kpa
D) Have supersonic expansion waves outside the exit
= Pe < 94 Kpa
A substance of mass 200g measures 50cm*4cm*8cm. Calculate it density giving your answer in SI unit
Answer:
0.2kg/0.0016m^3
Explanation:
200g is 0.2 kg
50*4*8 is equivalent to 1600 cubic centimeters which is also equivalent to 0.0016 cubic meters.
The SI unit for density is kg per cubic meter
So now we have 0.2kg/0.0016m^3
I hoped this helped, and if it did please give me brainliest
a. What is an electric field line?
b. What are equipotential sufaces are they necessarily physical surffaces?
c. Are they necessarily physical surfaces?
Answer:
Please see below as the answers are self-explanatory
Explanation:
a)
A electric field line is an imaginary line, which has the property that the electric field vector is tangent to it at any point. It starts from positive charges (since the electric field by convention it has the direction of the trajectory that would take a positive test charge, so it always goes away from positive charges) and ends in negative charges.b)
Since the potential difference between two points represents the work per unit charge needed for a charge to move between these points, a equipotential surface is the one over which it is not needed to do work to move a charge from any point on the surface to any other point, which means that all points are at the same potential.c)
Equipotential surfaces are not necessarily physical surfaces, they can be defined in vaccum for instance.As an example, any spherical surface concentric with a point charge, is an equipotential surface, and it can be a real surface or a fictitious one.What principle does a heat engine take advantage of in order to use heat to perform work?
A. Magnets can convert motion into electrical energy
B. The speed of atoms increases when heat is added
C. Fluids expand when heat is added to them
D. Energy can be created from nothing
You are explaining why astronauts feel weightless while orbiting in the space shuttle. Your friends respond that they thought gravity was just a lot weaker up there. Convince them and yourself that it isn't so by calculating the acceleration of gravity 608 km above the Earth's surface in units of g. (The mass of the Earth is 5.97 x 1024 kg, and the radius of the Earth is 6380 km.)
Answer:
The answer is "83.1%".
Explanation:
Given:
[tex]\text{Mass of the earth}\ (M_E) = 5.97 \times 10^{24}\ \ kg\\\\\text{Radius of the earth}\ (R_E) = 6380 \ km = 6.38 \times 10^{6}\ \ m\\\\\text{acceleration of gravity}\ (g_E) = 608 \ \ km= 608,000 \ \ m \\\\G= 6.67 \times 10^{-11} \ \ \frac{N \cdot n^2}{kg^2}[/tex]
Using formula:
[tex]\to g_E = G \frac{M_E}{(R_E +h)^2}[/tex]
[tex]\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6.38 \times 10^{6}+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km} =6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,380,000+ 608,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{(6,988,000)^2}\\\\\to g_{608 \ km}=6.67\times10^{-11}\frac{5.97 \times 10^{24}}{4.88 \times 10^{13}}\\\\[/tex]
[tex]\to g_{608 \ km}=6.67\times10^{-24}\frac{5.97 \times 10^{24}}{4.88}\\\\\to g_{608 \ km}=6.67\times \frac{5.97}{4.88}\\\\\to g_{608 \ km}=6.67\times 1.22336066\\\\\to g_{608 \ km}= 8.15 \ \frac{m}{s^2}[/tex]
Calculating the gravity on the Earth’s surface:
[tex]\to \frac{g_{608 \ km} }{ g_{\ earth \ surface}}[/tex] [tex]= \frac{8.15}{9.8} \times 100=83.1 \%[/tex]
