Answer:
3. It has the same net charge as 11 protons.
Explanation:
An electrically neutral object contains the same number of protons and electrons. Therefore, if 11 electrons are removed from it, there will be 11 more protons compared to the number of electrons in the object. Thus, the object It has the same net charge as 11 protons.
Atoms are electrically neutral, but removing electrons results in a charged object with a net charge equivalent to the number of protons present.
Atoms are electrically neutral, meaning that the overall electric charge is zero because the number of protons (positive charge) equals the number of electrons (negative charge). When an atom loses electrons, it becomes positively charged, and when it gains electrons, it becomes negatively charged. In this case, removing 11 electrons from a neutrally charged object will result in a net charge equivalent to having the same net charge as 11 protons.
The difference between the full wave rectifier with the center tap transformer, and the full wave bridge rectifier is_______________. Question 9 options: in full wave bridge rectifier two didoes conduct during each half of the cycle whereas one diode conducts during each half cycle in the center tapped full wave rectifier. transformer configuration is simpler in the full wave bridge rectifier. that the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer.
Answer:
That the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer.
Explanation:
See Image
Final answer:
In the full wave bridge rectifier, two diodes conduct during each half of the cycle, whereas in the center-tapped full wave rectifier, one diode conducts during each half cycle. The peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer. The transformer configuration is simpler in the full wave bridge rectifier.
Explanation:
The difference between the full wave rectifier with the center tap transformer and the full wave bridge rectifier is that in the full wave bridge rectifier, two diodes conduct during each half of the cycle, whereas in the center-tapped full wave rectifier, one diode conducts during each half cycle. Additionally, the peak inverse voltage on the diodes in the bridge full wave rectifier is half of the full wave rectifier with a center tap transformer. The transformer configuration is simpler in the full wave bridge rectifier.
A pitcher throws a 0.15 kg baseball so that it crosses home plate horizontally with a speed of 20 m/s. It is hit straight back at the pitcher with a final speed of 25 m/s. Assume the direction of the initial motion of the baseball to be positive.(a) What is the impulse delivered to the ball? (b) Find the average force exerted by the bat on the ball if the two are in contact for 2.0?
Answer:
a.-6.75 kgm/s
b.[tex]3375 N[/tex]
Explanation:
We are given that
Mass of baseball=0.15 kg
Initial speed=[tex]u=20m/s[/tex]
Final speed=[tex]v=-25m/s[/tex]
a.We know that
Impulse=Change in momentum=[tex]\Delta p=mv-mu=m(v-u)[/tex]
Momentum=[tex]mass\times velocity[/tex]
Using the formula
Impulse=[tex]0.15(-25-20)=-6.75 kgm/s[/tex]
b.Time=[tex]2\times 10^{-3} s[/tex]
Force=[tex]\frac{Impulse}{time}[/tex]
Using the formula
Average force exerted by the bat on the ball=[tex]\frac{-6.75}{2\times 10^{-3}}[/tex] N
Average force exerted by the bat on the ball=[tex]3375N[/tex]
A 47 gram golf ball is driven from the tee with an initial speed of 52 m/sec and rises to a height of 24.6 m. (a) Neglect air resistance and determine the kinetic energy of the ball at its highest point. (b) What is its speed when it is 8 m below its highest pint
Answer:
a) 52.2 J
b) 48.77 m/s
Explanation:
a)47 g = 0.047 jg
The kinetic (and total mechanical energy) of the ball at the ground is
[tex]E = mv^2/2 = 63.544 J[/tex]
The potential energy of the ball at its highest point 24.6m is. Let g = 9.8m/s2
[tex]E_h = mgh = 0.047*9.8*24.6 = 11.33 J[/tex]
Since the potential energy at the highest height is less than the total mechanical energy on ground, the difference must be kinetic energy
[tex]E_k = E - E_h = 63.544 - 11.33 = 52.2J[/tex]
b) 8m below 24.6m is 16.6m. The potential energy at this point is
[tex]E_{p8} = mgh = 0.047*9.8*16.6 = 7.64 J[/tex]
And so the kinetic energy at this point is
[tex]E_{k8} = E - E_{p8} = 63.544 - 7.64 = 55.9 J[/tex]
So the speed is
[tex]mv^2/2 = 55.9[/tex]
[tex]v^2 = 2*55.9/0.047 = 2378.64[/tex]
[tex]v = \sqrt{2378.64} = 48.77 m/s[/tex]
The kinetic energy of the golf ball at its highest point is zero and the speed of the ball when it is 8 m below its highest point can be determined using the principle of conservation of mechanical energy.
Explanation:To determine the kinetic energy of the golf ball at its highest point, we can use the principle of conservation of energy. Since the ball is at its highest point, its gravitational potential energy is maximum, and its kinetic energy is minimum. Therefore, the kinetic energy of the golf ball at its highest point is zero.
To find the speed of the ball when it is 8 m below its highest point, we can use the principle of conservation of mechanical energy. We can equate the initial mechanical energy of the ball to its final mechanical energy. The initial mechanical energy is the sum of the initial kinetic energy and the initial potential energy. The final mechanical energy is the sum of the final kinetic energy and the final potential energy. By solving this equation, we can find the speed of the ball when it is 8 m below its highest point.
Juanita lifts a round box and a square box to a shelf. The gravitational potential energy (GPE) for the round box increases by 50 J. The GPE for the square box increases by 100 J. On which box did Juanita do more work? Explain your reasoning.
Juanita did more work on the square box
Explanation:
According to the law of conservation of energy, the work done in lifting an object is equal to the increase in gravitational potential energy of the object. This is due to the fact that the work done on the object is converted into potential energy.
The potential energy of an object (GPE) is given by
[tex]GPE=mgh[/tex]
where
m is the mass of the object
g is the acceleration of gravity
h is the height of the object
In this problem, we have two objects:
- The roud box is lifted and its GPE increases by 50 J --> this means that the work done by Juanita on the box is 50 J
- Thr square box is lifted and its GPE increases by 100 J --> this means that the work done by Juanita on the box is 100 J
Therefore, Juanita did more work on the square box.
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The square box received more work by Juanita than the round box based on the increase in gravitational potential energy.
The square box received more work done by Juanita compared to the round box. When the gravitational potential energy increase for the square box is 100 J, which is greater than the 50 J increase in the round box, it indicates that Juanita did more work on the square box.
___ twisted pair wire is used in environments that have a noticeable amount of electromagnetic interference.
Answer:
shielded
Explanation:
shielded twisted pair wire is used in environments that have a noticeable amount of electromagnetic interference.
When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time of the impact about 10 times as great as for a stiff-legged landing. In this way the average force your body experiences is ________.a. less than 1/10 as great. b. more than 1/10 as great. c. about 1/10 as great.d. about 10 times as great.
Answer:
c. about 1/10 as great.
Explanation:
While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.
This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.
Mathematically:
[tex]F\propto\frac{d}{dt} (p)[/tex]
[tex]\Rightarrow F=\frac{d}{dt} (m.v)[/tex]
since mass is constant
[tex]F=m\frac{d}{dt}v[/tex]
when [tex]dt=10t[/tex]
then,
[tex]F'=m.\frac{v}{10\times t}[/tex]
[tex]F'=\frac{1}{10} \times \frac{m.v}{t}[/tex]
[tex]F'=\frac{F}{10}[/tex] the body will experience the tenth part of the maximum force.
where:
[tex]\frac{d}{dt} =[/tex] represents the rate of change in dependent quantity with respect to time
[tex]p=[/tex] momentum
[tex]m=[/tex] mass of the person jumping
[tex]v=[/tex] velocity of the body while hitting the ground.
Connor is a basketball player who makes, on average, 65% of her free throws. Asume each shot is independent. SHOW ALL WORK FOR FULL CREDIT. A) What is the probability Connor will miss three straight free throws before she makes one? (If you use your calculator to get your answer, make sure to show the key strokes in which you got your answer). Write your answer in Standard notation. B) During a season, Connor takes 150 free throws. What is the probability that he will make at least 100 out of 150 of these throws? C) What is the probability that he will make no more than 110 out of 150 free throws?
Answer:
DUNNO
Explanation:
BIT BAD FOR 65% WOULD GET -100%
Sabe-se que um alqueire paulista equivale a 24200 metros quadrados. Uma chácara retangular tem um alqueire e mede 100m de frente. Quanto ela mede de fundo?
Answer:
b = 242 m
Explanation:
A = 24200 m²
a = 100 m
b = ?
A seguinte fórmula é aplicada
A = a*b
⇒ b = A / a
⇒ b = (24200 m²) / (100 m)
⇒ b = 242 m
A conductor is placed in an external electrostatic field. Theexternal field is uniform before the conductor is placed within it.The conductor is completely isolated from any source of current orcharge.
PART A)
Which of the following describes the electricfield inside this conductor?
a. It is in thesame direction as the original external field.
b. It is in theopposite direction from that of the original externalfield.
c. It has adirection determined entirely by the charge on itssurface.
d. It is alwayszero.The charge density inside theconductor is:
a. 0
b. non-zero;but uniform
c. non-zero;non-uniform
d. infinitePART C)
Assume that at some point just outside thesurface of the conductor, the electric field has magnitudeE and is directed toward thesurface of the conductor. What is the charge density eta on the surface of the conductor at thatpoint?
Express your answer in terms ofE and epsilon_0.
The Electric Field Inside a Conductor is always zero, and the charge density inside a conductor is always zero. The charge density on the surface of the conductor can be expressed as eta = E / epsilon_0, where E is the magnitude of the electric field just outside the surface of the conductor and epsilon_0 is the permittivity of free space.
The electric field inside a conductor placed in an external electrostatic field is always zero (option d). This is because when an external electric field is applied to a conductor, the charges inside the conductor rearrange themselves in such a way that the net electric field inside the conductor becomes zero.
The charge density inside the conductor is also always zero (option a). In equilibrium, the charges in a conductor reside on its surface, creating an electric field within the conductor that is zero.
If the electric field just outside the surface of the conductor has a magnitude E and is directed toward the surface, the charge density (eta) on the surface of the conductor can be expressed as eta = E / (epsilon_0), where epsilon_0 is the permittivity of free space.
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In a population of Mendel's garden peas, the frequency of dominant yellow-flowered plants is 50%. The population is in Hardy-Weinberg equilibrium. What is the frequency of the homozygous recessive genotype in the population? 0.71 0.25 0.5 The frequency cannot be determined from the data provided.
The frequency of the homozygous recessive genotype in the population is approximately 0.71.
Explanation:The frequency of the homozygous recessive genotype in the population can be calculated using the Hardy-Weinberg equation. In this case, the frequency of the dominant yellow-flowered plants is given as 50% or 0.5. To calculate the frequency of the homozygous recessive genotype (qq), we need to find the value of q. The equation for the Hardy-Weinberg equilibrium is p² + 2pq + q² = 1.
Given that the frequency of the dominant allele (p) is 0.5, we can substitute the value of p into the equation and solve for q:
0.5² + 2(0.5)(q) + q² = 1
Simplifying the equation gives:
0.25 + q + q² = 1
Combining like terms:
q² + q - 0.75 = 0
Using the quadratic formula to solve for q, we find that q ≈ 0.71 or -1.21. Since the frequency of a trait cannot be negative, we can disregard the negative value. Therefore, the frequency of the homozygous recessive genotype in the population is approximately 0.71.
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A 52 kg pole vaulter running at 11 m/s vaults over the bar. Her speed when she is above the bar is 1.2 m/s. The acceleration of gravity is 9.8 m/s 2 . Find her altitude as she crosses the bar. Neglect air resistance, as well as any energy absorbed by the pole. Answer in units of m.
Answer:
6.1 m
Explanation:
m = Mass of person = 52 kg
h = Altitude
v = Velocity
Kinetic energy of the person on the ground
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}52\times 11^2\\ =3146\ J[/tex]
Kinetic energy of the person at the top
[tex]\dfrac{1}{2}52\times 1.2^2\\ =37.44\ J[/tex]
At the top the potential energy is given by
[tex]mgh=52\times 9.8h[/tex]
Balancing the energy of the system
[tex]3146=37.44+52\times 9.8h\\\Rightarrow h=\dfrac{3146-37.44}{52\times 9.8}\\\Rightarrow h=6.1\ m[/tex]
Her altitude is 6.1 m
If a bucket with water is hung to a spring balance and then a weight with the help of a thread is fully immersed inside the bucket such that it does not touch sides of the bucket, will the spring balance reading change or remain same?
Answer:
Explanation:
The weight fully immersed will displaced water of equal volume to itself. The weight of this water displaced = volume of water displaced × density of water × acceleration due to gravity. This increase in weight will lead the spring balance reading to change; increase because there is an increase in mass.
A particle moves along the x-axis according to x(t)=10t−2t²m. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?
Answer:
a) v(2) = 2m/s, v(3) = -2m/s
b) speed at t = 2s is 2m/s
speed at t = 3s is 2m/s
c) 0 m/s
Explanation:
We can take the derivative of x(t) to find the equation of velocity
v(t) = x'(t) = 10 - 4t
(a) v(2) = 10 - 4*2 = 10 - 8 = 2 m/s
v(3) = 10 - 4*3 = 10 - 12 = -2 m/s
(b) The speed would be the same as velocity without the direction
speed at t = 2s is 2m/s
speed at t = 3s is 2m/s
(c) The average velocity between t = 2s and t = 3s is distance it travels over period of time
[tex]v_a = \frac{s(3) - s(2)}{\Delta t} = \frac{10*3 - 2*3^2 - (10*2 - 2*2^2)}{3 - 2}[/tex]
[tex]v_a = \frac{12 - 12}{1} = 0/1 = 0 m/s[/tex]
Final answer:
The instantaneous velocity at t = 2 s is 2 m/s and at t = 3 s is -2 m/s. The instantaneous speed at both times is 2 m/s. The average velocity between t = 2 s and t = 3 s is 12 m/s.
Explanation:
(a) To find the instantaneous velocity, we need to find the derivative of the position function x(t) with respect to time. The derivative of x(t) = 10t - 2t² is v(t) = 10 - 4t. Substituting t = 2 and t = 3 into v(t), we get v(2) = 10 - 4(2) = 2 m/s and v(3) = 10 - 4(3) = -2 m/s.
(b) The instantaneous speed is the magnitude of the instantaneous velocity. Since speed is always positive, the speed at t = 2 s and t = 3 s is 2 m/s for both.
(c) The average velocity between t = 2 s and t = 3 s is given by the change in position divided by the change in time. The change in position is x(3) - x(2) = (10(3) - 2(3)²) - (10(2) - 2(2)²) = 12 m, and the change in time is 3 s - 2 s = 1 s. Therefore, the average velocity is 12 m/1 s = 12 m/s.
Based on the data thomson collected in his experiments using cathode rays, the concept of atomicc structure was modified. What were the four things validated by his cathode ray expirement?
Answer:
Based on the data thomson collected in his experiments using cathode rays, the concept of atomicc structure was modified. What were the four things validated by his cathode ray expirement?
Explanation:
Cathode Ray tubes' Experiment of J.J. Thomson, showed that atoms contain electrons or tiny negative charged subatomic particles.
Based on the Thomson's cathode ray tubes experiment, He validated four things which are as following
1. Cathode rays have mass.
2. Matter contains positive and negative charge.
3. Particles of cathode rays are fundamental to all matter.
4. An atom is divisible.
Water behind a dam has a certain amount of stored energy that can be released as the water falls over the top of the dam. It may be enough energy to turn a mill wheel or an electricity-generating turbine. Choose the term that best describes the type of energy stored in the water at the top of the dam.
Answer:
Gravitational potential energy
Explanation:
Gravitational potential energy is the type of energy an object has due to its position in a gravitational field. Water behind a dam possesses gravitational potential energy due to it being at a higher level than the water on the other side of the dam. When the water falls the gravitational potential energy is converted to kinetic energy, leading to the turning of the turbines to generate electricity.
An old manuscript reveals that a landowner in the time of King Arthur held 3.00 acres of plowed land plus a livestock area of 25.0 perches by 4.00 perches. What was the total area in (a) the old unit of roods and (b) the more modern unit of square meters? Here, 1 acre is an area of 40 perches by 4 perches, 1 rood is an area of 40 perches by 1 perch, and 1 perch is the length 16.5 ft.
Answer:
(a) Total area is 14.5 roods
(b) Total area is 14674.522 square meters
Explanation:
Area occupied by land = 3 acres
1 acre = 40 perches by 4 perches = 160 square perches
3 acres = 3×160 = 480 square perches
Area occupied by livestock = 25 perches by 4 perches = 100 square perches
Total area = 480 + 100 = 580 square perches
1 rood = 4 perches by 1 perch = 4 square perches
580 square perches = 580/4 = 14.5 roods
(b) Total area = 580 square perches
1 perch = 16.5ft = 16.5/3.2808 = 5.03 meters
580 square perches × (5.03 meters/1 perch)^2 = 580 ×25.3009 square meters = 14674.522 square meters
The total area owned by the landowner is 580 perches, which is equivalent to 14.5 roods. When converted to the modern unit of measurement, this totals to approximately 17516 square meters.
Explanation:To solve this problem, we first need to convert each area to the same unit. In this case, we can convert all areas to perches. According to the old manuscript, the landowner had 3 acres of plowed land. Given that 1 acre is equivalent to 160 perches (40 perches by 4 perches), the plowed land was 480 perches (3 x 160).
The livestock area is 25 perches by 4 perches, which totals to 100 perches. Therefore, the total area owned by the landowner is 580 perches (480 perches + 100 perches).
(a) To convert this to roods, we divide by 40 (since 1 rood is 40 perches by 1 perch) which equals to 14.5 roods
(b) To convert perches to square meters, we need to know that 1 perch is 16.5 ft. and 1 square foot is approximately 0.0929 square meters. So, 1 perch is 325.125 sq ft (16.5 ft * 16.5 ft). Therefore, 1 perch is about 30.2 square meters (325.125 ft * 0.0929), and 580 perches equate to approximately 17516 square meters.
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The water molecule has a dipole with the negative portion
A) localized on one of the hydrogens
B) localized between the hydrogen atoms
C) pointing toward the oxygen atom
D) pointing from the oxygen through the hydrogen atoms
In a water molecule, the dipole's negative portion is located on the oxygen atom and points towards the hydrogen atoms due to the difference in electronegativity between these elements.
Explanation:The water molecule, H2O, is a polar molecule, meaning it has a net dipole due to the presence of polar bonds, which result from a significant difference in electronegativities of the atoms involved. In a water molecule, the oxygen atom is more electronegative than hydrogen and therefore pulls the shared electrons toward itself. This creates a charge separation where the oxygen side of the molecule becomes partially negative, and the hydrogen side becomes partially positive. Looking at the provided options, the most accurate statement would be that the negative portion of the dipole is D) pointing from the oxygen through the hydrogen atoms
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A friend of yours who takes her astronomy class very seriously challenges you to a contest to find the thinnest crescent moon you can find just after new moon? What time of day is best for looking for this very thin crescent?
Answer:
after the sun sets or just as it is setting
Explanation:
a crescent moon is thin and reflects less sunlight during the daylight sky so it becomes difficult to spot, but can be spotted when the sun is setting or just sets.
The best time to find the thinnest crescent moon just after the new moon is in the early evening, just after sunset, when the moon starts to reflect sunlight towards the earth, showing the crescent shape. The appearance of the Moon's surface can vary significantly with its phase and this can be better viewed through binoculars.
To find the thinnest crescent moon just after new moon, the best time to look is usually in the early evening just after sunset. The Moon, moving eastward each day in its 30 days cycle around the Earth, moves roughly 12° in the sky each day. A day or two after the new phase, the thin crescent first appears, as we begin to see a small part of the Moon's illuminated hemisphere reflecting a little sunlight toward us.
Because the Moon is moving eastward away from the Sun, it rises later and later each day. Therefore, after a new moon, the thin crescent will be seen in the west just after sunset.
Keep in mind that the bright crescent increases in size on successive days as the Moon moves farther and farther around the sky away from the direction of the Sun.
Bear in mind that the brighter the Moon is in the night sky, the harder it is to see the faint flashes of meteors. Furthermore, as seen through a good pair of binoculars, the appearance of the Moon's surface changes dramatically with its phase, revealing more topographic details when sunlight streams in from the side, causing topographic features to cast sharp shadows.
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A space vehicle is traveling at 5320 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 98 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation ?
Answer:
5398.4km/h
Explanation:
IN THIS CASE THE MOMENTUM IS CONSERVED. THE VALUE OF MOMENTUM OF ONE COMBINED ROCKET WILL BE SAME AS OF TWO COMBINED.
Let mass of module be m
then
mass of motor = 4m (four times the mass of rocket module)
total mass = m + 4m = 5m
combined velocity = V = 5320kph
Let
absolute (relative to earth)motor velocity after disengagement = v
then
rocket module velocity (relative to earth) after disengagement = v+98 (relative velocity = 98)
momentum conservation equation
combined momentum = module momentum + motor momentum
(m+4m)V = m(v+98) + 4m*v
5mV = 98m+mv + 4mv
5V = 98+v + 4v (m cancels out)
5V - 98 = 5v
((5*5320)-98)/5 = v
v = 5300.4 km/h
velocity of rocket module relative to earth = v +98
= 5300.4 + 98
= 5398.4km/h
(a) If the maximum acceleration that is tolerable for passengers in a subway train is 1.34 m/s² and subway stations are located 806 m apart, what is the maximum speed a subway train can attain between stations? (b) What is the travel time between stations? (c) If a subway train stops for 20 s at each station, what is the maximum average speed of the train, from one start-up to the next? (d) Graph x, ν, and a versus t for the interval from one start-up to the next.
Answer:
The correct answer is 32.9 m/s
Explanation:
To solve this, we list out the known and the unknown variables as follows
Maximum allowable acceleration = 1.34 m/s²
Distance between sttions = 806 m
Therefore from the equation of motion
v = ut + 0.5·×at²
Where v = final velocity
u = initial velocity
S = distance covered
t = time
a = acceleration
Also v² = u² + 2·a·S
where u is the initial velocity, which we can take as u = 0, then
v² = 2·1.34·S = 2.68S m²/s² then
Also the train has to decelerate from maximum speed to stop at the next tran station wherev = 0, thus v² = u² -2·1.34·Z, so u² = 2.68Z
since u² = 2.68S from the previous calculation, then for v = 0
2.68S = 2.68Z thus S = Z which and to reach the next subway station S + Z must be = 806 m, then S = 806 m ÷ 2 = 403 m
and v² = 2.68S m²/s² = 1080.04 m²/s²
v = 32.9 m/s
The maximum speed a subway train can attain between stations is 32.9 m/s
The maximum speed a subway train can attain between stations is approximately 51.73 m/s. The travel time between stations is about 15.6 seconds. The maximum average speed of the train, from one start-up to the next, is approximately 27.62 m/s.
Explanation:To determine the maximum speed a subway train can attain between stations, we need to use the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement. Plugging in the values, we have v² = 0 + 2(1.34 m/s²)(806 m). Solving for v, we find that the maximum speed the subway train can attain between stations is approximately 51.73 m/s.
The travel time between stations can be calculated using the equation t = s/v, where t is the time, s is the displacement, and v is the velocity. Plugging in the values, we have t = 806 m / 51.73 m/s ≈ 15.6 seconds.
The maximum average speed of the train, from one start-up to the next, is given by the equation v_avg = (2s)/(t + t_stop), where v_avg is the average speed, s is the displacement, t is the travel time, and t_stop is the time the train stops at each station. Plugging in the values, we have v_avg = (2 * 806 m) / (15.6 s + 20 s) ≈ 27.62 m/s.
To graph x, ν, and a versus t for the interval from one start-up to the next, we would need specific values for x and a over time. However, we can consider a simplified case where a is constant, resulting in a linear graph of a versus t. The graph of v versus t would have a constant positive slope equal to the acceleration, and the graph of x versus t would be a curve with a positive concave upwards shape.
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A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19 s s later. You may ignore air resistance. Part A Part complete If the height of the building is 21.0 m m , what must the initial speed be of the first ball if both are to hit the ground at the same time? v v = 9.53 m/s m/s SubmitPrevious Answers Correct Part B Part complete Consider the same situation, but now let the initial speed v 0 v0 of the first ball be given and treat the height h h of the building as an unknown. What must the height of the building be for both balls to reach the ground at the same time for v 0 v0v_0 = 8.70 m/s m/s . h h = 6.51 m m SubmitPrevious Answers CorrectPart C Part complete If v 0 v0 is greater than some value v max vmax , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v max vmax . v max vmax = 11.7 m/s m/s SubmitPrevious Answers Correct Part D Part complete If v 0 v0 is less than some value v min vmin , a value of h h does not exist that allows both balls to hit the ground at the same time. Solve for v min vmin . v min vmin = 5.83 m/s m/s SubmitPrevious Answers Correct Provide Feedback Next
A) Initial velocity of ball 1: 9.53 m/s upward
B) Height of the building: 6.48 m
C) Maximum velocity: 11.7 m/s
D) Minimum velocity: 5.83 m/s
Explanation:
A)
The y-position of the 1st ball at time t is given by the equation for free fall motion:
[tex]y_1 = h + v_0 t - \frac{1}{2}gt^2[/tex] (1)
where
h = 21.0 m is the initial height of the ball, the height of the building
[tex]v_0[/tex] is the initial velocity of the ball, upward
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
The y-position of the 2nd ball instead, dropped from the roof 1.19 s later, is given by
[tex]y_2 = h-\frac{1}{2}g(t-1.19)^2[/tex]
where
h = 21.0 m is the initial height of the ball, the height of the building
t' = 1.19 s is the delay in time of the 2nd ball (we can verify that at t = 1.19 s, then [tex]y_2=h[/tex], so the ball is still on the roof
The 2nd ball reaches the ground when [tex]y_2=0[/tex], so:
[tex]0=h-\frac{1}{2}g(t-1.19)^2\\0=(21.0)-4.9(t^2-2.38t+1.42)\\4.9t^2-11.66t-14.04=0[/tex]
Which has two solutions:
t = -0.88 s (negative, we discard it)
t = 3.26 s (this is our solution)
The 1st ball reaches the ground at the same time, so we can substitute t = 3.26 s into eq.(1) and [tex]y_1=0[/tex], so we find the initial velocity:
[tex]0=h+v_0 t -\frac{1}{2}gt^2\\v_0 = \frac{1}{2}gt-\frac{h}{t}=\frac{1}{2}(9.8)(3.26)-\frac{21.0}{3.26}=9.53 m/s[/tex]
B)
In this case, the height of the building h is unknown, while the initial velocity of ball 1 is known:
[tex]v_0 = 8.70 m/s[/tex]
When the two balls reach the ground at the same time, there position is the same, so we can write:
[tex]y_1=y_2\\h+v_0 t - \frac{1}{2}gt^2 = h-\frac{1}{2}g(t-1.19)^2[/tex]
Solving the equation, we find:
[tex]v_0t=1.19gt-\frac{1}{2}g(1.19)^2\\t=\frac{0.5g(1.19)^2}{1.19g-v_0}=2.34 s[/tex]
This is the time at which both balls reache the ground; and substituting into the eq. of ball 2, we find the height of the building:
[tex]0=h-\frac{1}{2}g(t-1.19)^2\\h=0.5g(t-1.19)^2=0.5(9.8)(2.34-1.19)^2=6.48 m[/tex]
C)
If [tex]v_0[/tex] is greater than some value [tex]v_{max}[/tex], then there is no value of h such that the two balls hit the ground at the same time. This situation occurs when the demoninator of the formula found in part b:
[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}[/tex]
becomes negative: in that case, the time becomes negative, so no solution is possible.
The denominator becomes negative when
[tex]1.19g-v_0 < 0[/tex]
Therefore when
[tex]v_0>1.19g=(1.19)(9.8)=11.7 m/s[/tex]
So, if the initial velocity of ball 1 is greater than 11.7 m/s, the two balls cannot reach the ground at the same time.
D)
There is also another condition that must be true in order for the two balls to reach the ground at the same time: the time at which ball 1 reaches the ground must be larger than 1.19 s (because ball 2 starts its motion after 1.19 s). This means that the following condition must be true
[tex]t=\frac{0.5g(1.19)^2}{1.19g-v_0}>1.19[/tex]
Solving the equation for [tex]v_0[/tex], we find:
[tex]0.5g(1.19)^2>1.19(1.19g-v_0)\\6.94>13.88-1.19v_0\\1.19v_0>6.94[/tex]
Which gives
[tex]v_0>5.83 m/s[/tex]
Therefore, the minimum speed of ball 1 at the beginning must be 5.83 m/s.
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During a lunar mission, it is necessary to increase the speed of a spacecraft by 2.76 m/s when it is moving at 400 m/s relative to the Moon. The speed of the exhaust products from the rocket engine is 1100 m/s relative to the spacecraft. What fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?
Explanation:
Formula to calculate initial mass of given spacecraft is as follows.
[tex]v_{f} - v_{i} = v_{rel} \times ln(\frac{M_{i}}{M_{f}})[/tex]
The given data is as follows.
[tex]v_{f} - v_{i}[/tex] = 2.76 m/s
[tex]v_{rel}[/tex] = 1100 m/s
[tex]\frac{M_{f}}{M_{i}} = e^\frac{-dv}{v_{rel}}[/tex]
So, [tex]\frac{M_{i} - M_{f}}{M_{i}} = 1-e^-(\frac{2.76}{1100})[/tex]
= [tex]2.51 \times 10^{-3}[/tex]
Thus, we can conclude that [tex]2.51 \times 10^{-3}[/tex] fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase.
Stefan's Law says:a. the energy radiated by a blackbody is proportional to T³. c. the hotter a star's surface, the bluer it looks to us. E = mc². d. that if the Sun's temperature were doubled, it would give off 16X more energy. d. that doubling the star's temperature would also double its peak wavelength.
Answer:
d. that if the Sun's temperature were doubled, it would give off 16X more energy
Explanation:
Stefan's Law ;
According to this law ,the energy emitted by a body which have temperature T(in K) is proportional to the forth power of the temperature .This is apllicable for the black bodies(These bodies absorb all the incident radiation or we can say that these are perfect absorber bodies).
[tex]Energy\ \alpha\ T^4\\E \alpha\ T^4\\[/tex]
If we double the temperature then the energy will become 16 times of the initial energy.
That is why the option d is correct.
A light-year is _________. a. about 10 trillion kilometers the time it takes light to travel around the Sun b. about 300,000 kilometers per second the time it takes light to reach the nearest star
Answer:
about 10 trillion kilometers
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
t = Seconds in one year = [tex]365.25\times 24\times 60\times 60[/tex]
1 Light year
[tex]1\ ly=ct\\\Rightarrow 1\ ly=3\times 10^8\times 365.25\times 24\times 60\times 60=9.46728\times 10^{15}\ m\\ =9.46728\times 10^{15}\times 10^{-3}\\ =9.46728\times 10^{12}\ km\approx 10\ trillion\ km[/tex]
The answer is a. about 10 trillion kilometers
A distant star is moving toward the earth at a speed of 1/4 the speed of light. Compared to the light from a flashlight on the earth, the speed of the light from the star would be__________
Answer: THE SAME
Explanation: visible light is an electromagnetic wave, which has the properties of both magnet and electrical. The speed of light in air has Generally been estimated to be
299,792 kilometers per second. For for a distant start moving towards the Earth at a speed one-fourth (1/4th) of 299,792kilometers per second compared to a flashing light from the Earth,the speed of the light from the star is expected to be the same.
Consider a spherical Gaussian surface and three charges: q1 = 1.60 μC , q2 = -2.61 μC , and q3 = 3.67 μC . Find the electric flux through the Gaussian surface if it completely encloses (a) only charges q1 and q2, (b) only charges q2 and q3, and (c) all three charges.
Answer:
Explanation:
Guass Law: Also known as "Gauss's flux theorem" is the total of the electric flux "φ" out of a closed surface is equal to the charge "Q" enclosed divided by the permittivity εο. Solution is attached.The electric flux through a Gaussian surface can be calculated using Gauss's law. (a) Calculate electric flux for q1 and q2, (b) Calculate electric flux for q2 and q3, (c) Calculate electric flux for all three charges.
Explanation:The electric flux through a Gaussian surface can be calculated using Gauss's law. Gauss's law states that the total electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.
(a) To find the electric flux through the Gaussian surface enclosing only charges q1 and q2, we need to calculate the net charge enclosed by the surface, which is the sum of the two charges. Then, we divide this sum by the permittivity of free space to obtain the electric flux.
(b) Following the same procedure, we can find the electric flux through the Gaussian surface enclosing only charges q2 and q3.
(c) To find the electric flux through the Gaussian surface enclosing all three charges, we calculate the net charge enclosed by the surface, which is the sum of all three charges. Again, we divide this sum by the permittivity of free space to obtain the electric flux.
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Calculate the potential energy of a 300 gram volleyball that is 15 meters in the air. (g = 9.8 m/s2)
explain why if you can, thank you :)
Answer:
The book gravitational potential energy is 44.1 Joules fear
Explanation:
Given:
Mass of volleyball, m = 300g = 300 ÷ 1000 = 0.3 kg
height, h = 15 m
To Find:
Potential Energy = ?
Solution:
Gravitational Potential Energy :
Gravitational potential energy is energy an object possesses because of its position in a gravitational field.
Formula is given by
[tex]\textrm{Gravitational Potential Energy}= m\times g\times h[/tex]
Where,
m = mass
g = acceleration due to gravity = 9.8 m/s²
h = height
Substituting the values we get
[tex]\textrm{Gravitational Potential Energy}= 0.3\times 9.8\times 15\\\\\textrm{Gravitational Potential Energy}=44.1 Joules[/tex]
The book gravitational potential energy is 44.1 Joules.
The potential energy of a 300 gram volleyball that is 15 meters in the air is 44.1 Joules.
To calculate the gravitational potential energy (PEg), you would use the formula PEg = mgh,
where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above the ground.
Since the mass of the volleyball is given as 300 grams, you must first convert this mass into kilograms, resulting in 0.3 kg. Then, use the given values with g = 9.8 m/s² to find the potential energy:
PEg = (0.3 kg) × (9.8 m/s2) × (15 m)
PEg = 4.41 kg×m²/s²
PEg = 44.1 Joules
A capacitor consists of a set of two parallel plates of area A separated by a distance d. This capacitor is connected to a battery that maintains a constant potential difference V across the plates. If the separation between the plates is doubled, the electrical energy stored in the capacitor will be:_______
a- doubled
b- unchanged
c- quadrupled
d- quartered
e- halved
Answer:e
Explanation:
Given
Area of parallel plates is A
distance between plates is d
Potential difference between Plates is V
Capacitance is given by
[tex]C=\frac{\epsilon _0A}{d}[/tex]
If separation is doubled then capacitance become half
[tex]C'=\frac{\epsilon _0A}{2d}[/tex]
[tex]C'=\frac{C}{2}[/tex]
Electrical energy stored in the capacitor is given by
[tex]E=\frac{1}{2}CV^2[/tex]
When distance is doubled
[tex]E'=\frac{1}{2}\times \frac{C}{2}\times V^2[/tex]
[tex]E'=\frac{E}{2}[/tex]
Therefore Energy is halved
If the separation between the plates is doubled, the electrical energy stored in the capacitor will be: d. The electrical energy stored in the capacitor will be quartered.
To understand why the energy stored in the capacitor is quartered when the separation between the plates is doubled, let's consider the formula for the capacitance of a parallel plate capacitor and the energy stored in a capacitor.
The capacitance C of a parallel plate capacitor is given by:
[tex]\[ C = \frac{\varepsilon_0 A}{d} \][/tex]
The energy U stored in a capacitor is given by:
[tex]\[ U = \frac{1}{2} C V^2 \][/tex]
[tex]\[ C' = \frac{\varepsilon_0 A}{2d} = \frac{1}{2} \frac{\varepsilon_0 A}{d} = \frac{1}{2} C \][/tex]
[tex]\[ U' = \frac{1}{2} C' V^2 = \frac{1}{2} \left(\frac{1}{2} C\right) V^2 = \frac{1}{4} C V^2 = \frac{1}{4} U \][/tex]
So, when the separation between the plates is doubled, the capacitance is halved, and since the energy is directly proportional to the capacitance, the energy stored in the capacitor is quartered.
write a statement that warns people about the presence of peroxide in hair dyes
Answer:
Hydrogen peroxides are known to be an active ingredient in hair dyes. Hydrogen peroxide is a damaging chemical although it has been added in diluted amounts in hair dyes. Some hair dye companies have already started looking for alternatives for hydrogen peroxide to use in their products, concerning the damaging effects of the chemical.
Hydrogen peroxide can cause hair loss, dermatitis and scalp burns.
Concerning the harmful chemicals in dyes, many people consider it is best to use henna if you really want a colour change for your hair.
An object of mass 2.5 kg has a momentum < 3, 6, 7 > kg m/s. At this instant the object is acted on a by a force < 50, 50, 100 > N for 5 x 10-3 s . What is the momentum of the object at the end of this time interval?
Answer:
The momentum of the object at the end is [tex](3.25i+6.25j+7.5k)\ kg-m/s[/tex]
Explanation:
Given that,
Mass of object = 2.5 kg
Momentum [tex]p= 3i+6j+7k[/tex]
Force [tex]F=50i+50j+100k[/tex]
Time [tex]t=5\times10^{-3}\ s[/tex]
We need to calculate the momentum of the object at the end
Using formula of impulse
[tex]J=\Delta p[/tex]
[tex]J=m\Delta v[/tex]...(I)
[tex]J=F\times\Delta t[/tex]....(II)
From equation (I) and (II)
[tex]F\times\Delta t=m\Delta v[/tex]
[tex]F\times \Delta t=m(v_{f}-v_{i})[/tex]
[tex]mv_{f}=F\times \Delta t+mv_{i}[/tex]
Put the value into the formula
[tex]p_{f}=(50i+50j+100k)\times5\times10^{-3}+3i+6j+7k[/tex]
[tex]p_{f}=0.25i+0.25j+0.5k+3i+6j+7k[/tex]
[tex]p_{f}=(3.25i+6.25j+7.5k)\ kg m/s[/tex]
Hence, The momentum of the object at the end is [tex](3.25i+6.25j+7.5k)\ kg-m/s[/tex]
To determine the object's momentum after the force is applied, calculate the change in momentum from the applied force over the time interval and add it to the initial momentum. The final momentum of the object is <3.25, 6.25, 7.5> kg·m/s.
Explanation:The student asked: What is the momentum of the object at the end of this time interval? To calculate the final momentum, we must use the formula p = p_0 + F ∙ Δt, where p_0 is the initial momentum, F is the force applied, and Δt is the time interval.
With the given values:
Initial momentum: p_0 = <3, 6, 7> kg·m/sForce: F = <50, 50, 100> NTime: Δt = 5 × 10^-3 sCalculating the change in momentum due to the force:
Change in momentum: Δp = F ∙ Δt = <50, 50, 100> N ∙ 5 × 10^-3 s = <0.25, 0.25, 0.5> kg·m/sAdding the initial momentum to the change in momentum gives the final momentum:
Final momentum: p = <3, 6, 7> + <0.25, 0.25, 0.5> = <3.25, 6.25, 7.5> kg·m/sTherefore, the object's momentum at the end of the time interval is <3.25, 6.25, 7.5> kg·m/s.