Answer:
R1 = 4.8Ω
Explanation:
The loop circuit has an initial voltage of V = IR
I = 2 A , R1 = R
V = 2R1
with the current reduced to 1.5A with an additional 1.6Ω resistor
the total resistance of the circuit is 1.6 + R1
the voltage of the two scenarios has to be equal , since the same voltage flows through the circuit
therefore V = 2R1
from Ohms law V = IR
2R1= 1.5 (1.6 + R1)
2R1 = 2.4 + 1.5R1
collecting like terms
2R1 - 1.5R1 = 2.4
0.5R1 = 2.4
R1 = [tex]\frac{2.4}{0.5}[/tex]
R1 = 4.8Ω
Answer:
4.8 Ω
Explanation:
From Ohm's Law,
Using,
I = E/(R+r)................. Equation 1
E = I(R+r)................. Equation 2
Where I = current, E = emf, R = external resistance, r = internal resistance
Given: I = 2 A, R = R1, r = 0 Ω
Substitute into equation 2
E = 2(R1)
E = 2R1.
When an additional 1.6 Ω resistor is added in series,
E = 1.5(R1+1.6)
2R1 = 1.5R1+2.4
2R1-1.5R1 = 2.4
0.5R1 = 2.4
R1 = 2.4/0.5
R1 = 4.8 Ω
Which of the following characterizes a beta ray? Choose all that apply. is electromagnetic radiation is a product of natural radioactive decay is attracted to the positively charged plate in an electric field is attracted to the negatively charged plate in an electric field is composed of electrons
Explanation:
When a radioactive substance decays then the fast moving electrons emitted by it is known as beta ray. Basically, a number of beta particles are ejected by a beta ray.
Symbol of a beta particle is [tex]^{0}_{-1}e[/tex]. A beta ray is a natural decay of a radioactive element. As we know that opposite charges get attracted towards each other. So, a beta ray gets attracted towards a positively charged plate.
Therefore, we can conclude that following are the characterizes a beta ray:
a product of natural radioactive decay.is attracted to the positively charged plate in an electric field. is composed of electrons.Final answer:
Beta rays are negatively charged, attracted to the positively charged plate in an electric field, and are composed of electrons. They are a product of natural radioactive decay and are lighter and much less massive than alpha particles.
Explanation:
Beta rays are characterized by specific properties that distinguish them from other types of radiation produced during natural radioactive decay. According to Ernest Rutherford's research, which involved observing the behavior of radiation in magnetic and electric fields, beta particles are known to be negatively charged and relatively light. This means they are attracted to the positively charged plate in an electric field and are significantly deflected due to their lighter mass compared to alpha particles.
Beta rays are not electromagnetic radiation; this term is reserved for gamma rays, which are uncharged and therefore unaffected by electric fields. Beta rays are indeed a product of natural radioactive decay, specifically during a process known as beta decay, in which a nucleus emits an electron or a positron. Since they are composed of high-energy electrons, the identification that beta rays are composed of electrons is also correct.
An unstrained horizontal spring has a length of 0.43 m and a spring constant of 238 N/m. Two small charged objects are attached to this spring, one at each end. The charges on the objects have equal magnitudes. Because of these charges, the spring stretches by 0.013 m relative to its unstrained length. Determine the possible algebraic signs and the magnitude of the charges. (a) the possible algebraic signs
Answer:
The charges on the spring are 1.23E-5 C and they have the same sign
Explanation:
Given
Coulomb laws states that the force exerted by a charge q on another charge Q at a distance r is given by
F = kqQ/r²
Where k = 8.99 * 10^9 Nm²/C²
r = 0.43 + 0.013
r = 0.443m
The force on the spring is calculated as;.
F = kx where x is the stretch length of the spring and k is the spring constant
The force acting on the spring = 238 * 0.013
F = 3.094N
By comparison;
F = kqQ/r² becomes
3.094 = F = kqQ/r²
kqQ/r² = 3.094 (Considering that a = Q)
kq²/r² = 3.094
8.99 * 10^9 * q²/0.443 = 3.094 -- make q² the subject of formula
q² = 3.094 * 0.443/8.99*10^9
q² = 1.524629588431E−10
q = √1.524629588431E−10
q = 0.000012347589191545C
q = 1.23E-5 C
The charges on the spring are 1.23E-5 C and they have the same sign
A car is rounding a circular curve of radius r on a banked turn. As the drawing indicates, there are two forces acting on the car, its weight mg and the normal force FN exerted on it by the road. Which force, or force component, provides the centripetal force that keeps the car moving on the circular path?
1. The vertical component, FNcosθ of the normal force.
2. The horizontal component, FNsinθ of the normal force.
3. Both the normal force, FN, and the weight, mg, of the car.
4. The normal force, FN.
5. The weight, mg, of the car.
6. The horizontal component, FNsinθ of the norma.l
Final answer:
The horizontal component of the normal force, represented as FNsinθ, provides the centripetal force necessary for a car to round a circular curve on a banked turn without friction.
Explanation:
When a car is rounding a circular curve on a banked turn, the force that provides the centripetal force necessary to keep the car moving on the circular path is the horizontal component of the normal force exerted on it by the road. This can be represented as FNsinθ, where θ is the banking angle. The weight of the car, mg, acts vertically downwards and does not contribute to the centripetal force in this ideal frictionless scenario. For ideal banking, where the angle is perfect for the speed and radius of the turn, the net external force equals the horizontal centripetal force required for circular motion. Therefore, the horizontal component of the normal force is the only force component that acts towards the center of the curvature, providing the centripetal acceleration.
How does sugar affect the attention of small children?
The average distance of the planet mercury from the sun is 0.39 times the average distance of the earth from the sun. How long is a year on mercury in units of earth years?
Answer:
[tex]T_1=0.24y[/tex]
Explanation:
Using Kepler's third law, we can relate the orbital periods of the planets and their average distances from the Sun, as follows:
[tex](\frac{T_1}{T_2})^2=(\frac{D_1}{D_2})^3[/tex]
Where [tex]T_1[/tex] and [tex]T_2[/tex] are the orbital periods of Mercury and Earth respectively. We have [tex]D_1=0.39D_2[/tex] and [tex]T_2=1y[/tex]. Replacing this and solving for
[tex]T_1^2=T_2^2(\frac{D_1}{D_2})^3\\T_1^2=(1y)^2(\frac{0.39D_2}{D_2})^3\\T_1^2=1y^2(0.39)^3\\T_1^2=0.059319y^2\\T_1=0.24y[/tex]
Final answer:
A year on Mercury is approximately 88 Earth days long, which means it is about 0.241 Earth years, due to Mercury's average distance from the Sun being 0.39 times that of Earth's.
Explanation:
The student's question relates to the orbital period of Mercury compared to Earth's, given its average distance from the Sun. Mercury’s orbit around the Sun takes approximately 88 Earth days, which constitutes a Mercury year. This is calculated using Kepler's third law of planetary motion, which relates the orbital period of a planet to its average distance from the Sun (orbital semi-major axis).
Because Mercury is 0.39 times as far from the Sun as Earth is, its orbital period is significantly shorter than Earth's. Earth's average distance from the Sun is approximately 1 astronomical unit (AU), making it the basis for measuring distances in our solar system. Therefore, a year on Mercury, in Earth years, is 88/365, or about 0.241 Earth years.
Two people, one of mass 78 kg and the other of mass 59 kg, sit in a rowboat of mass 88 kg. With the boat initially at rest, the two people, who have been sitting at opposite ends of the boat 2.9 m apart from each other, now exchange seats.
How far will the boat move?
Answer:
The boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.
Explanation:
The boat only moves because the centre of mass changes a bit if the two people on opposite ends of the boat exchange seats.
The boat moves a distance of the change in centre of mass
Noting that the weight of the boat acts at the centre of the boat at 1.45m from both ends.
For convention, we call the original position of the 59 kg person as x=0
This means,
59 kg person is at x = 0 m
88 kg of the boat acts at x = 1.45 m from the end of the 59 kg person.
78 kg person is at x = 2.90 m
Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)
For the initial setup,
X = [(59×0) + (88×1.45) + (78×2.90)]/(59+88+78)
X = (353.8/225)
X = 1.572 m
(Don't forget that this is 1.572 m from the end we designated x=0 m)
When the people exchange positions,
59 kg person is now at the other end of the boat with x = 2.90 m
88 kg of the boat still acts at the centre of the boat at x = 1.45 m
And 78 kg person is now at the end of the boat with x = 0 m
Centre of mass = X = (Σ mᵢxᵢ)/(Σ mᵢ)
X = [(59×2.90) + (88×1.45) + (78×0)]/(59+88+78)
X = (298.7/225)
X = 1.328 m
(This is 1.328 m from the end we designated x=0 m from the start)
So, the centre of mass moves a distance of (1.572 - 1.328) towards the end of the boat we designated x=0 m from the start.
Hence, the boat moves 0.244 m towards the end where the 59 kg person was at the start of the calculations.
Hope this Helps!!!
SPEAR is a storage ring at the Stanford Linear Accelerator which has a circulating beam of electrons that are moving at nearly the speed of light (2.998 108 m/s). If a similar ring is about 92.0 m in diameter and has a 0.40 A beam, how many electrons are in the beam
Answer:
[tex]2.4\times 10^{12}[/tex]
Explanation:
We are given that
Speed of light,v=[tex]2.998\times 10^8 m/s[/tex]
Diameter of ring,d=92 m
Radius,r=[tex]\frac{d}{2}=\frac{92}{2}=46 m[/tex]
Current, I=0.40 A
We have to find the number of electrons in the beam.
We know that
Current,I=[tex]\frac{q}{t}[/tex]
Where q= ne
[tex]e=1.6\times 10^{-19} C[/tex]
Using the formula
[tex]0.40=\frac{1.6\times 10^{-19}n}{\frac{2\pi r}{v}}[/tex]
[tex]0.40=\frac{1.6\times 10^{-19}n\times v}{2\pi r}[/tex]
[tex]0.40=\frac{1.6\times 10^{-19}n\times 2.998\times 10^8}{2\pi\times 46}[/tex]
[tex]n=\frac{0.40\times 2\pi\times 46}{1.6\times 10^{-19}\times 2.998\times 10^8}=2.4\times 10^{12}[/tex]
a 4kg block is attatched to a vertical sspring constant 800n/m. the spring stretches 5cm down. how much elastic potential energy is stored in the system
The Potential energy stored in the system is 1 J
Explanation:
Given-
Mass, m = 4 kg
Spring constant, k = 800 N/m
Distance, x = 5cm = 0.05m
Potential energy, U = ?
We know,
Change in potential energy is equal to the work done.
So,
[tex]U = \frac{1}{2} k (x)^2\\\\[/tex]
By plugging in the values we get,
[tex]U = \frac{1}{2} * 800 * (0.05)^2\\ \\U = 400 * 0.0025\\\\U = 1J\\[/tex]
Therefore, Potential energy stored in the system is 1 J
Two trains start from towns 224 mi apart and travel towards each other on parallel tracks. They pass each other 1.6 hr later. If one train travels 10 mph faster than the other, find the speed of each train.
Answer: 65mph, 75mph
Explanation:
Let us assume x to be the speed of the slower train, in mph (miles per hour).
Then the speed of the other train is (x+10) mph, according to the question.
We then would have an equation like this
1.6x + 1.6(x+10) = 224.
This is because, the first addend in the left side is the distance covered by the slower train.
The second addend in the left side is the distance covered by the faster train.
The sum is 224 miles, because they together covered all the distance to the moment when they meet each other.
1.6x + 1.6x + 16 = 224
3.2x + 16 = 224
3.2x = 224 - 16
3 2x = 208
x = 208/3.2
x = 65
Thus the speed of the slower train is 65mph, and that of the other train is 65 + 10 = 75mph
Final answer:
By setting up and solving a linear equation, the speed of the slower train is determined to be 65 mph, while the faster train's speed is 75 mph as one travels 10 mph faster than the other.
Explanation:
To solve the problem of the two trains travelling on parallel tracks towards each other, we must first define the variables for their speeds. Let's say one train travels at x mph and the other at (x + 10) mph. Since they are moving towards each other, you can add their speeds together to find how fast the distance between them is closing. The total closing speed is then x + (x + 10) mph, which simplifies to 2x + 10 mph.
The total distance to be covered by the two trains until they pass each other is 224 miles. We know they pass each other after 1.6 hours, so using the formula Distance = Speed x Time, we can set up the equation: 224 miles = (2x + 10 mph) x 1.6 hours. Solving for x gives us the speed of the slower train.
Performing the algebraic steps:
224 = (2x + 10) x 1.6
224 = 3.2x + 16
208 = 3.2x
x = 65 mph.
Therefore, the speed of the slower train is 65 mph and the speed of the faster train is 75 mph.
A small metal ball is given a negative charge, then brought near (i.e., within about 1/10 the length of the rod) to end A of the rod ) What happens to end A of the rod when the ball approaches it closely this first time?
a.It is strongly repelled.
b.It is strongly attracted.
c.It is weakly attracted.
d.It is weakly repelled.
e.It is neither attracted nor repelled.
Answer:
Option B, it is strongly attracted
Explanation:
A Test Charge Determines Charge on Insulating and Conducting Balls, and the points made regarding conductors, it can be ascertained that in conductors, the electrons are free to move about. This means that when a charge is brought near to a conductor, the opposite charges all navigate to the sharpest point closest the charge and a strong attraction is created.
This shows that the rod will be strongly attracted. The density of the charges on the rod is mostly concentrated at the sharpest point.
End A of the rod will be strongly attracted to the negatively charged metal ball because of the process of charge induction, where opposite charges attract.
Explanation:When the negatively charged metal ball is brought near to end A of the rod, end A of the rod will be strongly attracted to the negatively charged ball. This is because of the principle of charge induction. When a charged body is brought near to another body, it will cause the charges in that body to redistribute. Opposite charges attract, so the near side of the rod (end A) will have a positive charge induced on it, and this positive charge will be attracted to the negative charge on the ball. So, the correct answer is option b. It is strongly attracted.
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Which of the following statements about diffusion is true?
"1. It requires integral proteins in the cell membrane. 2. It is very rapid over long distances. 3. It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration. 4. It is an active process in which molecules move from a region of lower concentration to one of higher concentration. 5. It requires an expenditure of energy by the cell."
Answer:
3. It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration.
Explanation:
Diffusion can be defined as the movement of solute or gaseous molecules from the region in which they have higher concentration to the regions in which they have lower concentration until they become evenly distributed across the two regions.
Diffusion is a passive process, that is, it requires no energy.
Option 3 is the correct option.
Answer:
It is a passive process in which molecules move from a region of higher concentration to a region of lower concentration
Explanation:
Diffusion is the movement of a substance from an area of high concentration to an area of low concentration. It is an important process for living things.
A single substance tends to move from an area of high concentration to an area of low concentration until the concentration is equal across a space
To identify a diatomic gas (X2), a researcher carried out the following experiment: She weighed an empty 5.1-L bulb, then filled it with the gas at 1.00 atm and 20.0 ∘C and weighed it again. The difference in mass was 5.9 g . Identify the gas.
Answer: Nitrogen gas
Explanation:
Using ideal Gas's law
PV = nRT
where
Pressure of gas, P= 1atm
Volume of gas, V= 5.1L
no of moles of gas, n=
Ideal gas constant, R= 0.0821
Temperature of gas, T= 20°C = 20+273 = 293K
also, n= (mass/molar mass)
mass of the gas m = 5.9g
Molar mass of the gas = ?
So, PV = (mRT/M)
We're looking for molar mass M, then
M = mRT/PV
M = (5.9 * 0.0821 * 293)/(1 * 5.1)
M = 141.93/5.1
M = 27.8g/mol ~ 28g/mol
Since the gas is diatomic, then we say,
Atomic mass of gas = 1/2 * molar mass
Atomic mass = 1/2 * 28
Atomic mass = 14
Therefore, the gas is nitrogen.
A 0.40-kg block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the spring stretches for 2.0 cm relative to its unstrained length. When the block is released, it moves with an acceleration of 8.0 m/s2. What is the spring constant of the spring
Answer:
160N/m
Explanation:
According to Hooke's law which states that the extension of an elastic material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically,
F = ke where
F is the applied force
k is the spring constant
e is the extension
From the formula k = F/e
Since the body accelerates when the block is released, F = ma according to Newton's second law of motion.
The spring constant k = ma/e where
m is the mass of the block = 0.4kg
a is the acceleration = 8.0m/s²
e is the extension of the spring = 2.0cm = 0.02m
K = 0.4×8/0.02
K = 3.2/0.02
K = 160N/m
The spring constant of the spring is therefore 160N/m
Final answer:
The spring constant of the spring is calculated using Hooke's Law and Newton's second law of motion. By multiplying the mass of the block by its acceleration, we found the force, and then divided the force by the displacement to get the spring constant, which is 160 N/m.
Explanation:
To determine the spring constant of the spring, we need to apply Hooke's Law, which states that the force exerted by an ideal spring is directly proportional to its displacement from the equilibrium position (F = -kx), where 'F' is the force, 'k' is the spring constant, and 'x' is the displacement. Since the block is on a frictionless surface and we know the acceleration (a = 8.0 m/s2) and the mass (m = 0.40 kg), we can first find the force using Newton's second law (F = ma), and then use that force to calculate the spring constant 'k'.
The force exerted by the spring can be calculated as:
F = m * a
= 0.40 kg * 8.0 m/s²
= 3.2 N
The displacement (x) from the equilibrium position is given as 2.0 cm, which is 0.020 m in SI units. Using Hooke's Law, the spring constant can be calculated:
k = F / x
= 3.2 N / 0.020 m
= 160 N/m
Water molecules attracting other water molecules is called
Answer:
Water molecules attracting other water molecules is called cohesive attraction.
Explanation:
They are basically two forces in liquids that determine their wetting characteristics, they are cohesive and adhesive forces.
Cohesion is the attraction between molecules of same liquid example water and water, while adhesion is attraction between molecules of different liquids example alcohol and water.
Therefore, Water molecules attracting other water molecules is called cohesive attraction.
Why do the pvc plastic wells used in this weeks elisa need to be sticky on the inner walls and how may your results change if a different non-adherent plastic was used instead?
Answer:
The stickiness in the inner walls allows them to be easily coated with the desired antigens, this translates in the use of a smaller amount of antigen. If the walls weren't sticky there's a possibility the antigen won't stick to them and therefore the result of the ELISA can be a false negative.
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Consider the following possible stages in the evolution of a star like our sun: black dwarf, giant, main-sequence, planetary nebula, supernova, white dwarf. Rank the stages in the order they occur. Leave out any stages that will not occur in the evolution of a star of similar mass to the sun
Answer:
Main sequence
Giant
Planetary nebula
White dwarf
Black dwarf
Explanation:
Main Sequence:
Stars are called main sequence stars when their core temperature reaches up to 10 million kelvin and their start the nuclear fusion reactions of hydrogen into helium in the core of the star. For example sun is known as to be in the stage of main sequence as the nuclear fusion reactions are happening in its core.
Giant:
Next step is the Giant phase. When the stars run out of their fuel that is hydrogen for the nuclear fusion reactions then they convert into Giant stars. Giant stars have the larger radius and luminosity then the main sequence stars.
Planetary nebula:
Planetary nebula consists of glowing gases and plasma, it ejects from the red giant stars that run out of their fuel.
White dwarf:
When the stars run out of their fuel then they shed the outer layer planetary nebula, the remaining core part that left behind is called as white dwarf. It's the most dense part as the most of the mass is concentrated in this part.
Black dwarf:
When the white dwarf cool down completely that it no longer emitt heat and light then it is called as black dwarf.
These were the possible stages that includes in the evolution of stars
Toy car a is moving at a speed of 1 m/s towards toy car b that is at rest. Toy car a has a mass of 3 kg and car b has a mass of 2 kg. They couple together after collison. At what speed are they moving after the collison?
Answer:
0.6m/s
Explanation:
Totally inelastic collision
m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)
Where v (v₁ and v₂) is the initial velocity of the objects
v(final) is the final velocity of the objects stuck together
Toy car a , m₁ = 3kg, v₁ = 1m/s
Toy car b , m₂ = 2kg, v₂ = 0m/s
m₁v₁ + m₂v₂ = ( m₁ + m₂)v(final)
3(1) + 2(0) = (3 + 2) v(final)
3 = 5 v(final)
v(final) = 0.6m/s
Explanation:
Below is an attachment containing the solution.
A high-speed dart is shot from ground level with a speed of 150 m/s at an angle 30° above the horizontal. What is the vertical component of its velocity after 4.0 s if air resistance is neglected?
Explanation:
The dart is project with 150 m/s from a point at an angle of 30⁰
The vertical component of velocity = 150 sin 30 = 75 m/s
Thus initial vertical velocity is = 75 m/s
The velocity after 4 s can be calculate by
v = u - g t
here u is the initial velocity and t is the time , g is the acceleration due to gravity .
Thus v = 75 - 10 x 4 = 35 m/s
The velocity after 4.0 seconds will be "35 m/s".
According to the question,
Speed,
150 m/sAngle,
30°Vertical component of velocity,
[tex]150 \ Sin 30^{\circ} = 75 \ m/s[/tex]After 4 seconds, the velocity will be:
→ [tex]v = u-gt[/tex]
By substituting the values, we get
[tex]= 75-10\times 4[/tex]
[tex]= 75-40[/tex]
[tex]= 35 \ m/s[/tex]
Thus the answer above is right.
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Astronomers estimate that new stars form in our galaxy at the rate of about
Explanation:
The quantity at which the speed of formation of the star depends is called the "star formation rate". Astronomers estimate that in our Galaxy the star formation rate is about 3 solar masses per year According to "solar mass per year" the mass of interstellar gas and dust related to about 3 times the mass of the Sun goes into stars each year).However, the whole of the mass doesn't necessarily go in the formation of 1 star.Stars form in the Milky Way at a rate of about 1 solar mass per year, which means it would take a few hundred billion years for all gas to be turned into stars, far exceeding the universe's age of 14 billion years. Pulsars and supernova events influence this star-forming process and the interstellar medium.
Explanation:Astronomers have estimated that new stars form in our galaxy, the Milky Way, at a rate of about 1 solar mass per year. If we consider the amount of interstellar gas available to form new stars, and no new gas was added, we can calculate how long it would take for all the gas to be converted into stars. Given that the Milky Way contains about a few hundred billion solar masses of gas, it would take a few hundred billion years for all the interstellar gas to be used up at the current rate of star formation, which is significantly longer than the age of the universe (14 billion years).
Stars, including pulsars, form and die within the galaxy at varying rates. For example, one new pulsar is born approximately every 25 to 100 years, aligned with the rate of supernovae occurrences. Supernova explosions contribute to the recycling of interstellar material, affecting star formation and the interstellar medium.
In class we calculated the range of a projectile launched on flat ground. Consider instead, a projectile is launched down-slope of angle with an initial velocity of magnitude directed at an angle above the horizontal. How far will this projectile travel horizontally before it lands
Answer:
With an initial speed of 10m/s at an angle 30° below the horizontal, and a height of 8m, the projectile travels 7.49m horizontally before it lands.
Explanation:
Since the horizontal motion is independent from the vertical motion, we can consider them separated. The horizonal motion has a constant speed, because there is no external forces in the horizontal axis. On the other hand, the vertical motion actually is affected by the gravitational force, so the projectile will be accelerated down with a magnitude [tex]g[/tex].
If we have the initial velocity [tex]v_o[/tex] and its angle [tex]\theta[/tex], we can obtain the vertical component of the velocity [tex]v_{oy}[/tex] using trigonometry:
[tex]v_{oy}=v_osin\theta[/tex]
Therefore, if we know the height at which the projectile was launched, we can obtain the final velocity using the formula:
[tex]v_{fy}^{2} =v_{oy}^{2}+2gy\\\\ v_{fy}=\sqrt{v_{oy}^{2}+2gy }[/tex]
Next, we compute the time the projectile lasts to reach the ground using the definition of acceleration:
[tex]g=\frac{v_{fy}-v_{oy}}{\Delta t} \\\\\Delta t= \frac{v_{fy}-v_{oy}}{g}=\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]
Finally, from the equation of horizontal motion with constant speed, we have that:
[tex]x=v_{ox}\Delta t= v_{ox}\frac{\sqrt{v_{oy}^{2}+2gy} -v_{oy}}{g}[/tex]
For example, if the projectile is launched at an angle 30° below the horizontal with an initial speed of 10m/s and a height 8m, we compute:
[tex]v_{ox}=10\frac{m}{s} cos30=8.66\frac{m}{s}\\v_{oy}=10\frac{m}{s} sin30=5\frac{m}{s}\\\\x=8.66\frac{m}{s} \frac{\sqrt{(5\frac{m}{s}) ^{2}+2(9.8\frac{m}{s^{2}})8m}-5\frac{m}{s} }{9.8\frac{m}{s^{2} } } =7.49m[/tex]
In words, the projectile travels 7.49m horizontally before it lands.
The only force acting on a 3.0 kg canister that is moving in an xy plane has a magnitude of 5.0 N. The canister initially has a velocity of 3.6 m/s in the positive x direction, and some time later has a velocity of 7.0 m/s in the positive y direction. How much work is done on the canister by the 5.0 N force during this time
Answer:
The work done on the canister by the 5.0 N force during this time is
54.06 Joules.
Explanation:
Let the initial kinetic energy of the canister be
KE₁ = [tex]\frac{1}{2} mv_1^{2}[/tex] = [tex]\frac{1}{2} *3*3.6^{2}[/tex] = 19.44 J in the x direction
Let the the final kinetic energy of the canister be
KE₂ = [tex]\frac{1}{2} mv_2^{2}[/tex] = [tex]\frac{1}{2} *3*7.0^{2}[/tex] = 73.5 J in the y direction
Therefore from the Newton's first law of motion, the effect of the force is the change of momentum and the difference in energy between the initial and the final
= 73.5 J - 19.44 J = 54.06 J
Explanation:
Below is an attachment containing the solution.
You are observing a binary star system and obtain a series of spectra of the light from the two stars. In this spectrum, most of the absorption lines shift back and forth as expected from the Doppler Effect. A few lines, however, do not shift at all, but remain at the same wavelength. How could we explain the behavior of the non-shifting lines?
Answer: non-shifting lines indicate non moving star.
Explanation: when a star is moving toward the detector, the wavelength will decrease - there will be a blue shift.
When it's moving away from the earth or detector, the wavelength will increase - there will be a red shift.
Identifiable patterns of absorption lines that appear shorter or longer wavelengths than normal indicate that the star is moving
In visualisation, the bottom spectrum shows the normal position of absorption line for a star that is not moving toward or away from the earth.
Non-shifting lines in the binary star system's spectrum are attributed to absorption by interstellar clouds, which unlike the stars, do not move relative to us. The narrowness of these lines indicates the low pressure of the absorbing gas.
Explanation:When observing a binary star system and noting the Doppler Effect in the spectral lines, if certain lines do not shift, it indicates they originate from something that is not moving with respect to us. Most of the absorption lines shift due to the motion of the binary stars, which causes Doppler shifts as the stars move toward or away from us. This movement results in the spectral lines being blue-shifted when the star is approaching us, and red-shifted when it's receding. However, the lines that remain constant are likely due to the absorption by interstellar clouds located between Earth and the stars. The non-shifting lines are also much narrower, suggesting that the absorbing gas is at a very low pressure. This non-movement of specific lines helped in discovering the presence of interstellar materials, as their spectral lines do not participate in the Doppler shifts associated with the stars' orbits.
Block A, with a mass of 10 kg, rests on a 35 incline. The coefficient of static friction is 0.40. An attached string is parallel to the incline and passes over a massless, frictionless pulley at the top. What is the largest mass MB, attached to the dangling end, for which A remains at rest?
The maximum mass of block B, for which block A remains at rest on a 35-degree incline given a static friction coefficient of 0.40 and a 10kg mass of block A, is approximately 1.98 kg.
Explanation:To solve this question, we will need to use the principles of static friction, inclined planes, and gravitational force. Static friction is what keeps block A from sliding down the incline. It has to overcome the downward force due to gravity on block A which is a component of the weight of block A acting downwards the inclined plane.
To determine the maximum mass of block B, we can equate the static frictional force to the net force acting downward on the inclined plane. The static frictional force is [tex]\mu N[/tex] where μ is the coefficient of static friction and N is the normal force. N = [tex]mAgcos\theta[/tex] where mA and g are the mass and acceleration due to gravity respectively and θ is the angle of the inclined plane. So, static frictional force = [tex]\mu mAgcos\theta[/tex]. The downward force is the sum of components of the weight of block A acting downwards and the weight of block B. So, [tex]mAgcos\theta (\mu)[/tex] = mAg*[tex]sin\theta[/tex] + mBg.
From this equation, you can solve for the mass of Block B: MB = [tex]mA\mu cos\theta[/tex] - mA*[tex]sin\theta[/tex].
Plugging in the given numbers, we get:
MB = 10(0.4*cos(35 degrees) - sin(35 degrees)).
This gives us approximately 1.98 kg as the maximum mass of block B before block A begins to slide.
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To find the largest mass for block MB, we first need to calculate the gravitational force and the static friction on block A, using the given values for mass, gravitational acceleration, coefficient of static friction and incline angle. The tension in the string, created by block MB, has to balance these forces. By setting the net forces equal, we can calculate the value of MB.
Explanation:The problem revolves around understanding the concept of static friction and forces acting on an inclined plane. Firstly, let's calculate the force due to gravity acting on block A. This is simply F_gravity = mg sin θ where m=10 kg is the mass of the block, g=9.8 m/s² is the gravitational acceleration, and θ is 35°. Next, we need to calculate the frictional force that prevents the block from sliding. Since block A is at rest, static friction is at work here, and we can use the formula F_friction = μN, where μ=0.40 (the coefficient of static friction) and N is the normal force acting on the block, which equals mg cos θ.
To keep block A at rest, the tension T in the string due to the dangling mass MB must balance both gravity and friction. Therefore: T=F_gravity + F_friction. The weight of the dangling mass brings about the tension in the string, and hence, T=MBg. From these, we can calculate the largest mass of MB for which A remains at rest.
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Two boxes of masses m=35kg and m2=45kg, are hung vertically from opposite ends of a rope passing over a rigid horizontal metal rod. They system starts moving from rest. Assuming that friction between the rod and the rope is negligible, determine the magnitude of
(a) the acceleration of the boxes
(b) the tension in the rope
(c) the magnitude of each box's displacement after 0.5s
Answer:
a) 5.51m/s² b) 192.94N c) 1.38m each
Explanation:
Given two boxes of masses m1 = 35kg and m2 = 45kg hung vertically from opposite ends of a rope passing over a rigid horizontal metal rod, we will analyze the forces acting on each body.
According Newton's second law, Force = mass ×acceleration
The forces acting on body of mass m1 are the tension (T) and the frictional force (Ff) which opposes the tension.
Taking the sum of horizontal forces acting on mass m1, we will have;
T +(-Ff) = m1a
T - Ff = m1a... (1)
For the mass m2, the forces acting on the body are in the vertical direction and this forces are the weight (W) acting downwards and the tension(T) acting upwards. The sum of the forces in the body is given as ;
W + (-T) = m2a
W-T = m2a ...(2)
Since W = mg, equation 2 will become;
m2g - T = m2a...(2)
Solving equation 1 and 2 simultaneously to get the tension and the acceleration, we have;
T - Ff = m1a ... 1
m2g - T = m2a ... 2
Since friction is negligible, Ff = 0
Adding the two equation will give;
m2g-Ff = m2a+m1a
Since Ff =0
m2g = (m2+m1)a
a = m2g/m1+m2
a = 45(9.8)/45+35
a =441/80
a = 5.51m/s²
b) Substituting a = 5.51 into equation 1 to get the tension T in the rope will give;
T = m1a
T = 35×5.51
T = 192.94N
c) since velocity = displacement/time
Displacement = velocity × time
To get the velocity, since acceleration = velocity/time,
Velocity = acceleration ×time
Velocity = 5.51× 0.5
Velocity = 2.76m/s
Displacement of each box will be the same since they are moving with the same acceleration.
Displacement = 2.76m/s × 0.5s
Displacement of each boxes = 1.38m
The acceleration of the boxes is 0.98 m/s^2, the tension in the rope is 376.3 N, and the displacement of each box after 0.5 seconds is 0.12 m.
Explanation:This question deals with the physics of motion, particularly involving concepts of mass and friction. Given that the two boxes with different masses m1=35kg and m2=45kg are hung vertically from opposite ends of a rope over a rigid horizontal metal rod, and stating negligible friction, we can compute:
The acceleration (a) of the boxes using the equation: a = (m2 - m1)*g / (m1 + m2) where g is the acceleration due to gravity. Gravity is approximately 9.8 m/s^2. Substituting the given values, a = (45 - 35)*9.8 / (35 + 45) = 0.98 m/s^2. The tension (T) in the rope using the formula T = m1 * (g + a) or T = m2 * (g - a). Both formulas lead to the same result. Substituting the given values for the lighter 35kg box, T = 35*(9.8 + 0.98) = 376.3 N. The magnitude of each box's displacement after 0.5 seconds can be calculated using the equation s=0.5*at^2. Substituting the values, s = 0.5 * 0.98 * (0.5)^2 = 0.12 m.Learn more about Motion here:
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A super ball is dropped from a height of 100 feet. Each time it bounces, it rebounds half the distance it falls. How many feet will the ball have traveled when it hits the ground for the fourth time
The total distance travelled by the ball after the fourth impact is 275 feet.
Explanation:
Given-
Height, h = 100 feet
Rebounds half the distance
Distance in feet for the fourth time, x = ?
For the first time, the distance travelled by the ball is, x = 100 feet
For the second time, it will bounce up to 50 feet and fall upto 50 feet( half of 100 feet)
So, the distance travelled after the second impact, x = 100 + 50 + 50 = 200 feet
For the third time, it will bounce up to 25 feet and fall upto 25 feet( half of 50 feet)
So, the distance travelled after the third impact, x = 200 + 25 + 25 = 250 feet
For the fourth time, it will bounce up to 12.5 feet and fall upto 12.5 feet( half of 25 feet)
So, the distance travelled after the fourth impact, x = 250 + 12.5 + 12.5 = 275 feet
Therefore, total distance travelled by the ball after the fourth impact is 275 feet.
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A sound wave traveling at 343 m/s takes 10.0 s to go from a speaker to a detector. How far apart are the two devices?
A.) 3.43 x 10^3
B.) 3.43 x 10^2
C.) 3.43 x 10^-1
D.) 3.43 x 10^4
Answer:
[tex]\boxed {3.43 x 10^{3}}[/tex]
Explanation:
We know that speed is defined as distance moved per unit time hence expressed as [tex]v=\frac {d}{t}[/tex] where v is speed in m/s, d is distance in m and t is time in seconds. Making d the subject of the above formula then
[tex]d=vt[/tex]
Substituting 343 m/s for d and 10 s for t then
[tex]d= 343\times10= 3430= 3.43 x 10^{3}[/tex]
Therefore, the distance between speaker and deter is [tex]\boxed {3.43 x 10^{3}}[/tex]
Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mass of 5.40 10-2 kg and a charge of 5.10 10-5 C. The particle has a speed of 2.00 m/s on surface A. A nonconservative outside force is applied to the particle, and it moves to surface B, arriving there with a speed of 3 m/s. How much work is done by the outside force in moving the particle from A to B
Answer:
0.247 J = 247 mJ
Explanation:
From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.
So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)
= 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)
= 0.135 J + 0.11220 J
= 0.2472 J
≅ 0.247 J = 247 mJ
The ball was kicked in the air and it iss about to hit the . if horizonta componenet of its final velocity is 10m/s and vertical component of its final velocity is -10m/s .what is the magnotide of th final vellocity of the ball?
Answer:
The magnitude of the final velocity is 14.14 m/s
Explanation:
The horizontal component of the final velocity and vertical component of the final velocity, forms a perpendicular vector. To determine the magnitude of the final velocity, we sum the two vectors.
To add two perpendicular vector, Pythagoras principle is used.
[tex]V^2 = V_x^2 +V_y^2\\\\V = \sqrt{V_x^2 +V_y^2}\\\\V = \sqrt{(10)^2 +(-10)^2} =14.14 m/s[/tex]
The magnitude of the final velocity is 14.14 m/s
One kind of baseball pitching machine works by rotating light and stiff rigid rod about a horizontal axis until the ball is moving toward the target. Suppose a 144 gg baseball is held 81 cm from the axis of rotation and released at the major league pitching speed of 81 mph.
a. What is the ball's centripetal acceleration just before it is released?
b. What is the magnitude of the net force that is acting on the ball just before it is released?
Answer:
(a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]
(b). The magnitude of the net force is 232.9 N.
Explanation:
Given that,
Mass of baseball = 144 g
Speed = 81 mph = 36.2 m/s
Distance = 81 cm
(a). We need top calculate the ball's centripetal acceleration just before it is released
Using formula of centripetal acceleration
[tex]a=\dfrac{v^2}{r}[/tex]
Where, v = speed
r = radius
Put the value into the formula
[tex]a=\dfrac{(36.2)^2}{81\times10^{-2}}[/tex]
[tex]a=1617.82\ m/s^2[/tex]
[tex]a=16.17\times10^{2}\ m/s^2[/tex]
(b). We need to calculate the magnitude of the net force that is acting on the ball just before it is released
Using formula of force
[tex]F=\dfrac{mv^2}{r}[/tex]
Put the value into the formula
[tex]F=\dfrac{144\times10^{-3}\times(36.2)^2}{81\times10^{-2}}[/tex]
[tex]F=232.9\ N[/tex]
Hence, (a). The ball's centripetal acceleration is [tex]16.17\times10^{2}\ m/s^2[/tex]
(b). The magnitude of the net force is 232.9 N.
Enrico is having trouble telling the difference between the sound of a tuba and the sound of a piccolo. Even though a piccolo produces much briefer, faster sound waves than does a tuba, he has trouble picking out the differences in the _____ of these sounds. Please choose the correct answer from the following choices, and then select the submit answer button.
Answer: Pitch
Explanation:
Pitch of sound is defined as the factor that monitors the sound quality through produced vibrations rate.It helps in determination of sounds tone in terms highness or lowness.
According to the question,Enrico is finding difficulty in judging the difference between pitch sound of tuba and piccolo as per their tone in terms of high or low.
Enrico is having trouble telling the difference between the sound of a tuba and the sound of a piccolo. Even though a piccolo produces much briefer, faster sound waves than does a tuba, he has trouble picking out the differences in the pitch of these sounds.
Pitch: Pitch is the perceptual attribute of sound that allows us to distinguish between different frequencies. The sound of a tuba and a piccolo are different primarily because they produce sound waves at different frequencies. A tuba produces lower frequency sound waves (lower pitch), while a piccolo produces higher frequency sound waves (higher pitch). If Enrico is having trouble telling the difference between the sound of a tuba and a piccolo, it suggests he is having trouble distinguishing between their pitches.
Loudness: Loudness refers to the perceived volume or intensity of a sound. While the tuba and piccolo can be played at different volumes, the primary distinguishing factor between them is not loudness but pitch.
Hue: Hue is a term used in the context of color, not sound. It refers to the distinct characteristic of color that allows us to differentiate between colors such as red, blue, and green.
Amplitude: Amplitude refers to the height of the sound wave and is related to the loudness or volume of the sound. While amplitude can affect how loud a sound is, it does not directly differentiate between the characteristic sounds of a tuba and a piccolo.
Therefore, the appropriate term for distinguishing between the sounds of different instruments, such as a tuba and a piccolo, is pitch.
The complete question is:
Enrico is having trouble telling the difference between the sound of a tuba and the sound of a piccolo. Even though a piccolo produces much briefer, faster sound waves than does a tuba, he has trouble picking out the differences in the of these sounds.
O pitch
O loudness
O hue
O amplitude