Answer:
The mass density of the string is (0.3/L)kg/m
Explanation:
Mass density of the string = Mass/Length
Mass = 0.3kg
The length of the string is unknown so it is assumed to be L meter(s)
Therefore, mass deny of the string = 0.3kg/Lm = (0.3/L)kg/m
Bees often work together for the greater good of the hive. Sometimes they can get excess charge on them. Four pairs of bees, which can be represented as identical, conducting spheres, bump into each other in the hive. Rank the bees in the order of the size of the magnitude of charge transfer. (Enter your answer in decreasing order of the size of the magnitude of charge transfer. Use only ">" or "=" symbols. Do not include any parentheses around the letters or symbols.)
Answer:
B > C > A > D
Explanation:
As we see four pairs of bees; conducting spheres; it seems as below
A down 4
B up 8
C down 6
D down 3
so
B > C > A > D
After the process of survey, excavation, recording, mapping, and post-excavation analysis, all these efforts are largely wasted unless?
Answer: The results are Documented and published.
Explanation: The process of surveying, recording,mapping,and post-excavation analysis are activities that has to do with Real estates like Lands. All these activities must be documented and published according to the enabling laws guiding it,in all the States of the Federation their are Archeological guidelines which clearly states how all this activities are to be conducted.
Planet X has a diameter that is 3 times the diameter of Earth and a mass that is 30 times the mass of Earth. In SI units, what is the gravitational acceleration on planet X?
Final answer:
The gravitational acceleration on planet X is approximately 34.4 m/s².
Explanation:
The gravitational acceleration on planet X can be determined using the equation:
g = (G * M) / r^2
Where:
g is the gravitational acceleration
G is the gravitational constant (approximately 6.674 × 10^-11 N m^2/kg^2)
M is the mass of the planet
r is the radius of the planet
Since the diameter of planet X is 3 times the diameter of Earth, its radius would be 1.5 times that of Earth. The mass of planet X is 30 times the mass of Earth.
Let's assume the radius of Earth (rE) is 6371 km and the mass of Earth (ME) is 5.972 x 10^24 kg.
Using these values, the radius of planet X (rX) would be 1.5 * rE = 9556.5 km, and the mass of planet X (MX) would be 30 * ME = 1.7916 x 10^26 kg.
Now, we can plug these values into the equation to calculate the gravitational acceleration on planet X:
gX = (G * MX) / rX^2
gX = (6.674 x 10^-11 N m^2/kg^2 * 1.7916 x 10^26 kg) / (9556.5 km)^2
Converting km to meters and solving for gX gives us approximately 34.4 m/s².
One liter (1000cm3) of oil is spilled onto a smooth lake. If the oil spreads out uniformly until it makes an oil slick just one molecule thick, with adjacent molecules just touching, estimate the diameter of the oil slick. Assume the oil molecules have a diameter of 2 × 10-10 m.
Answer:
he diameter of the oil slick is 2523 m
Explanation:
given information?
V = 1 L = 1000 cm³ = 0.001 m³
h = 2 x 10⁻¹⁰ m
first we have to find the radius using the following equation
V = πr²h
r = √V/(πh)
= √(0.001)/(π x 2 x 10⁻¹⁰ )
= 1261.56 m
now, we can calculate the diameter of the oil slick
d = 2r
= 2 (1261.56)
= 2523 m
To estimate the diameter of oil slick, we can calculate the volume of one oil molecule and then divide the total volume of the spilled oil by the volume of one molecule. The diameter of the oil slick would be the diameter of one molecule multiplied by the square root of the number of oil molecules.
To estimate the diameter of the oil slick, we can first calculate the volume of one oil molecule. The volume of a sphere is given by the formula V = (4/3)πr^3, where r is the radius. Given the diameter of the oil molecule is 2 × 10-10 m, the radius would be half of that, which is 10-10 m. Plugging in the values, we get V = (4/3)π(10-10)^3 = 4.19 × 10-29 m3.
The total volume of the spilled oil is given as 1000 cm3, which is equal to 1000 × 10-6 m3. To find the number of oil molecules in the spilled oil, we can divide the total volume of oil by the volume of one molecule: 1000 × 10-6 m3 / 4.19 × 10-29 m3 = 2.39 × 1022.
Since the oil slick is one molecule thick, the diameter of the slick would be the diameter of one oil molecule multiplied by the square root of the number of oil molecules: 2 × 10-10 m × √ (2.39 × 1022) ≈ 4.89 × 106 m.
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A particle is being accelerated through space by a 10-N force. Suddenly the particle encounters a head-on second force of 10 N in the opposite direction. The particle with both forces acting on it_____________.
The particle with both forces acting on it will move at constant velocity
Explanation:
We can solve this problem by applying Newton's second law of motion, which states that the net force acting on a body is equal to the product between its mass and its acceleration:
[tex]F=ma[/tex]
where
F is the net force
m is the mass
a is the acceleration
For the particle in this problem, initially it has a forward force of
[tex]F_1 = 10 N[/tex]
Later, it encounters a second additional force in the opposite direction, therefore
[tex]F_2 = -10 N[/tex]
This means that the net force on the particle now is
[tex]F=F_1+F_2 = +10 +(-10) = 0[/tex]
As a consequence, the acceleration of the particle is zero:
[tex]a=0[/tex]
And this means that the particle moves with constant velocity.
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The particle will not accelerate due to the offsetting forces, it will maintain a constant velocity or remain at rest, pursuant to Newton's Third Law of motion.
Explanation:In the realm of Physics, the scenario outlined is a classic representation of Newton's Third Law of motion, which states that 'every action has an equal and opposite reaction'. Here, a particle is initially accelerated by a 10-N force. Then, it encounters a second force of the same magnitude but in the opposite direction. Essentially, these two forces cancel each other out because they are equal in magnitude, but opposite in direction. Therefore, the particle with both forces acting on it will not accelerate and will maintain a constant velocity, assuming it had a nonzero velocity to begin with. If it was initially at rest, it will remain at rest.
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Mr. F has taken in a lot of fluid and put out very little. During this period his heart rate has increased a little and his blood pressure has decreased a little. What do you infer from this? Select the best answer.A.His body is reducing his high blood pressure. This is a good thing.B. There isn't any problem, the clients blood pressure and heart rate are both within normal rangesC. His body is increasing blood osmolarityD. His body is raising blood pressure by both increasing blood volume with all this fluid intake and increasing heart rate. Something may be wrong with his blood volume.E. His increased fluid intake must be a response to blood osmolarity because his blood pressure is normal.
Final answer:
The best inference is that Mr. F's body is raising blood pressure by increasing blood volume with all the fluid intake and increasing heart rate. Something may be wrong with his blood volume.
Explanation:
The best inference from the given scenario is option D: His body is raising blood pressure by both increasing blood volume with all this fluid intake and increasing heart rate. Something may be wrong with his blood volume.
When Mr. F takes in a lot of fluid but puts out very little, it suggests that his body is retaining fluid. The increased fluid intake is causing an increase in blood volume, which in turn raises blood pressure. The slight increase in heart rate is also a compensatory mechanism to maintain adequate blood flow.
A orbiting satellite stays over a certain spot on the equator of (rotating) Earth. What is the altitude of the orbit (called a "synchronous orbit")?
The altitude of a geosynchronous orbit is [tex]3.59\cdot 10^7 m[/tex]
Explanation:
A geostationary (or geosynchronous) orbit is the orbit of a satellite that stays over the same spot on the equator of the rotating Earth.
This means that the period of a geostationary satellite is equal to the period of rotation of the Earth, which is 24 hours:
[tex]T=24 h \cdot 3600 s/h = 86400 s[/tex]
We can find the altitude of the orbit in the following way. First, we notice that the orbital speed of the satellite is given by
[tex]v=\frac{2\pi r}{T}[/tex]
where r is the radius of the orbit.
Then we also notice that the gravitational force between the satellite and the Earth is equal to the centripetal force, so we can write:
[tex]\frac{GMm}{r^2}=m\frac{v^2}{r}[/tex]
where
G is the gravitational constant
M is the mass of the Earth
m is the mass of the satellite
Re-arranging the equation,
[tex]\frac{GM}{r}=v^2[/tex]
And substituting the expression for the velocity,
[tex]\frac{GM}{r}=(\frac{2\pi r}{T})^2=\frac{4\pi^2 r^2}{T^2}[/tex]
Solving for r,
[tex]r=\sqrt[3]{\frac{GMT^2}{4\pi^2}}[/tex]
And substituting:
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}\\M=5.98\cdot 10^{24} kg\\T=86400 s[/tex]
we find:
[tex]r=\sqrt[3]{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})(86400)^2}{4\pi^2}}=4.225\cdot 10^7 m[/tex]
And since the radius of the Earth is
[tex]R=6.37\cdot 10^6 m[/tex]
The altitude of the satellite is
[tex]h=r-R=4.225\cdot 10^7 - 6.37\cdot 10^6 = 3.59\cdot 10^7 m[/tex]
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A satellite in a synchronous orbit stays over a certain spot on the equator as the Earth rotates beneath it. The altitude of a synchronous orbit can be calculated using the formula: altitude = (radius of the Earth) + (height of the geostationary orbit), which gives an altitude of approximately 42,157 kilometers.
Explanation:
In order for a satellite to stay over a certain spot on the equator of Earth, it needs to be in a synchronous orbit. A synchronous orbit is an orbit in which the satellite's orbital period matches the rotation period of the Earth. This means that the satellite stays above the same spot on the equator as the Earth rotates beneath it.
The altitude of a synchronous orbit can be determined using the formula:
altitude = (radius of the Earth) + (height of the geostationary orbit)
The radius of the Earth is approximately 6,371 kilometers, and the height of the geostationary orbit is approximately 35,786 kilometers. So the altitude of a synchronous orbit is:
altitude = 6,371 km + 35,786 km = 42,157 km
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A 20 g bullet is fired horizontally with a velocity of 250 m/s from a 1.5 kg rifle. If the rifle was held loosely, what would be the recoil velocity?
Answer:
3.33 m/s
Explanation:
given,
mass of bullet, m = 20 g = 0.02 Kg
speed of bullet, v = 250 m/s
mass of rifle, M = 1.5 kg
speed of recoil, u = ?
initial speed of the bullet and the rifle is zero.
using conservation of momentum
(M + m) V = m v + M u
(M + m) x 0 = 0.02 x 250 + 1.5 x u
1.5 u = -5
u = -3.33 m/s
negative sign represent that the recoil velocity is opposite to bullet velocity.
Hence, the recoil velocity is equal to 3.33 m/s
Which of the following expressions uses the correct conversion factor to convert 35.7 km into the equivalent distance in miles?A) 35.7 km × 1 km/1.609 milesB) 35.7 km × 1 mile/1.609 milesC) 35.7 km × 1 km/1.609 kmD) 35.7km × 1 mile/1.609 km
Answer: D) 35.7km × 1 mile/1.609 km
Explanation:
Given that;
We need to convert 35.7km to miles
And we know that 1.609km makes 1 mile = 1.609km/1mile
To convert it to mile we need to obtain the number of miles per kilometre
= 1/(1.609km/mile)
= 1mile/1.609km
Then we can now multiply the conversion rate by the number.
= 35.7km × 1mile/1.609km
= 22.19 miles
Note that the sign must cancel out to give miles.
Write the definition of a function print Grade, which takes one parameter containing a string value and returns nothing. The function prints "Grade: " followed by the string parameter.
Answer:
def print_Grade(grade):
print("Grade: ",grade)
Explanation:
The above code segment is a function that takes grade as a string parameter and returns nothing.
The line 1 starts with def keyword.
The keyword is used to start a function and the function will be uniquely identified with it.
The name of the declared function is print_Grade
Variable grade is then declared along with the function.
Line 2 prints "Grade: " without the strings along with the value of grade variable.
The code segment is written in Python
To define a function named printGrade that takes a string parameter and prints 'Grade: ' followed by the string, you'd create a function in a programming language like Python without a return statement, which outputs the concatenated string.
Explanation:In programming, when you are asked to write the definition of a function like printGrade, which takes a string parameter and prints it with some additional text, you are essentially writing a small part of a program. For example, in Python, a simple function that meets the requirements might look like this:
def printGrade(grade):This function definition includes the name printGrade, a single parameter grade, and a body that executes the print statement, which is returning nothing. When you call printGrade with a string argument, it will output the text "Grade: " concatenated with the string provided as an argument to the function.
You set your stationary bike on a high 80-N friction-like resistive force and cycle for 30 min at a speed of 8.0 m/s . Your body is 10% efficient at converting chemical energy in your body into mechanical work.
A.) What is your internal chemical energy change?
B.) How long must you bike to convert 3.8×105 J of chemical potential while staying at this speed? (This amount of energy equals the energy released by the body after eating three slices of bread.)
A) The change in internal chemical energy is [tex]1.15\cdot 10^7 J[/tex]
B) The time needed is 1 minute
Explanation:
First of all, we start by calculating the power output of you and the bike, given by:
[tex]P=Fv[/tex]
where
F = 80 N is the force that must be applied in order to overcome friction and travel at constant speed
v = 8.0 m/s is the velocity
Substituting,
[tex]P=(80)(8.0)=640 W[/tex]
The energy output is related to the power by the equation
[tex]P=\frac{E}{t}[/tex]
where:
P = 640 W is the power output
E is the energy output
[tex]t = 30 min \cdot 60 = 1800 s[/tex] is the time elapsed
Solving for E,
[tex]E=Pt=(640)(1800)=1.15\cdot 10^6 J[/tex]
Since the body is 10% efficient at converting chemical energy into mechanical work (which is the output energy), this means that the change in internal chemical energy is given by
[tex]\Delta E = \frac{E}{0.10}=\frac{1.15\cdot 10^6}{0.10}=1.15\cdot 10^7 J[/tex]
B)
From the previous part, we found that in a time of
t = 30 min
the amount of internal chemical energy converted is
[tex]E=1.15\cdot 10^7 J[/tex]
Here we want to find the time t' needed to convert an amount of chemical energy of
[tex]E'=3.8\cdot 10^5 J[/tex]
So we can setup the following proportion:
[tex]\frac{t}{E}=\frac{t'}{E'}[/tex]
And solving for t',
[tex]t'=\frac{E't}{E}=\frac{(3.8\cdot 10^5)(30)}{1.15\cdot 10^7}=1 min[/tex]
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A tennis pro charges $15 per hour for tennis lessons for children and $30 per hour for tennis lessons for adults. The tennis pro is practicing A) first-degree price discrimination. B) second-degree price discrimination. C) third-degree price discrimination. D) fourth-degree price discrimination. E) fifth-degree price discrimination.
Answer:
Third-degree price discrimination
Explanation:
Third degree price discrimination according to Investopedia.com occurs when companies price products and services differently based on the unique demographics of subsets of its consumer base, such as students, military personnel, or seniors.
Suppose a 1 Gbps point-to-point link is being set up between the Earth and a new lunar colony. The distance from the moon to Earth is approximately 385,000 km, and data travels over the link at the speed of light (3 x 10⁸ meters per sec.).
Calculate the minimum RTT (two-way propagation time) for the link.(In other words, how long would it take to send a single bit to the moon and immediately bounce it right back to Earth?)
Answer:
The time required to send the data from Earth to Moon will be 1.28s while for a two way communication, to send it back to the earth, it will take double time i.e. RTT = 2.56s
Explanation:
Distance between Earth and Moon = 385,000 km = 3.85 x 10⁸m
Speed of data travel = speed of light ≈ 3 x 10⁸m/s
As, v=d/t
t=d/v
[tex]t=\frac{3.85*10^{8} }{3*10^{8}}[/tex]
t=1.28s
RTT = Double of single way time taken = 2x1.28
RTT=2.56s
What type of material makes up the functional aspect of the transducer that creates the high-frequency sound?
a. Tungsten
b. Iron oxide
c. Silver/chromium alloy
d. Crystals
The type of material makes up the functional aspect of the transducer that creates the high-frequency sound is Crystals which is a ceramic.
Explanation:
High frequency of sound can be produced in the transducer with the help of materials like ceramic materials.These are non-metallic compounds. The ceramic materials are those that are brittle in nature. Crystals, glasses are some of the examples of ceramic materials.
Ceramic materials like crystals are used in transducers that makes them functionally fit to produce high frequency sounds. The materials that are given in the options tungsten, iron oxide and silver/chromium alloy are not fit for these type of materials manufacturing. These belongs to metals and metal oxides and alloy types of metals.
A sailboat runs before the wind with a constant speed of 2.8 m/s in a direction 52° north of west. How far (a) west and (b) north has the sailboat traveled in 35 min?
Explanation:
Let east be positive x axis and north be positive y axis
Velocity of boat = 2.8 m/s in a direction 52° north of west.
Velocity, v = -2.8 cos 52 i + 2.8 sin 52 j = -1.72 i + 2.21 j m/s
Time taken = 35 min = 35 x 60 = 2100 s
Displacement = Velocity x Time
Displacement = (-1.72 i + 2.21 j) x 2100
Displacement = -3612 i + 4633.50 j m
Displacement along west = 3612 m
Displacement along north = 4633.50 m
Final answer:
To find the westward and northward distances traveled by the sailboat, we decompose the boat's velocity into westward and northward components and then multiply each component by the travel time of 2100 seconds (35 minutes). Calculations involving trigonometry such as cosine for the westward component and sine for the northward component will yield the respective distances.
Explanation:
The question asks for the distance traveled westward and northward by a sailboat that runs before the wind with a constant speed of 2.8 m/s, heading 52° north of west. To solve this, we can decompose the sailboat's velocity into its westward and northward components using trigonometry. The total time traveled is 35 minutes, which is equivalent to 2100 seconds (35 min x 60 s/min).
Westward Component (a)
The westward component of the velocity can be found using the cosine function:
Vwest = V * cos(θ)
Where V is the speed of the boat, and θ is the angle north of west. Plugging in the given values:
Vwest = 2.8 m/s * cos(52°)
The distance traveled westward is the westward component of the velocity multiplied by the time:
Distancewest = Vwest * time
Northward Component (b)
Similarly, the northward component of the velocity is:
Vnorth = V * sin(θ)
The distance traveled northward is:
Distancenorth = Vnorth * time
By calculating these components, we can determine how far the sailboat has traveled in both the westward and northward directions.
When a test charge is brought near a charged object, we know from Coulomb's law that it will experience a net force (either attractive or repulsive, depending on the nature of the object's charge). A test charge may also experience an electric force when brought near a neutral object. Any attraction of a neutral insulator or neutral conductor to a test charge must occur through induced polarization. In an insulator, the electrons are bound to their molecules. Though they cannot move freely throughout the insulator, they can shift slightly, creating a rather weak net attraction to a test charge that is brought close to the insulator's surface. In a conductor, free electrons will accumulate on the surface of the conductor nearest the positive test charge. This will create a strong attractive force if the test charge is placed very close to the conductor's surface.
1. What is the nature of the force between balls A and B?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive.
2. What is the nature of the force between balls A and C?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive .
Answer:
1. (a)strongly attractive
2 (c) weakly attractive.
Explanation:
Consider three plastic balls (A, B, and C), each carrying a uniformly distributed charge equal to either +Q, -Q or zero, and an uncharged copper ball (D). A positive test charge (T) experiences the forces shown in the figure when brought very near to the individual balls. The test charge T is strongly attracted to A, strongly repelled from B, weakly attracted to C, and strongly attracted to D.
should be the concluding part to this question as presented above
1. What is the nature of the force between balls A and B?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive.
since there is an attractive force between ball A and the test charge, the charge on ball A must be negative
A = -Q
since ball B is repulsive to the test charged , then B must be positively charged
B = +Q
since A is negative and B is positive , then they will experienced a strong attraction
option A
2. What is the nature of the force between balls A and C?
a) strongly attractive.
b) strongly repulsive.
c) weakly attractive.
d) neither attractive nor repulsive .
Since C is weakly attracted to the test charge, we can say that Ball C will be weakly attracted to A, because it possess some weka charges
option C
Answer:its c trust me
Explanation:
Two equal-magnitude forces are applied to a door at the doorknob. The first force is applied perpendicular to the door, and the second force is applied at 60° to the plane of the door. Which force exerts the greater torque about the door hinge?
A. the first force (applied perpendicular to the door)
B. Both forces exert equal non-zero torques
C. Both forces exert zero torque
D. the second force
Answer:
option (A)
Explanation:
Torque is defined as the
Torque = Force x distance x SinФ
Where, Ф is the angle between force vector and the displacement
In case I:
torque = F x r x Sin 90 = F x r
In case II:
Torque = F x r x Sin 60 = 0.866 F r
So, the torque is more in first case.
Thus, option (a) is correct.
Thomas Newcomen was the first to produce a working steam engine. Why is the work of James Watt more widely known than the work of Newcomen?
Answer:
The steam engine of James watt is more efficient than Newcomen ans more suitable for the industrial revolution.
Explanation:
James Watt is more widely know for working steam engine because Watt has created better engine which is suitable for the industrial revolution. The steam engine of James watt is more efficient than Newcomen. Watt developed the condensing arrangement by using piston which lessen the initial pressure leading to effectively worked than Newcomen's
The steam engine made by James Watt has a separate condenser from the original design which increases the efficiency of the engine.
Steam EngineA machine that converts hot steam and heat energy into work is called the steam engine.
Thomas Newcomen invented the first useful steam engine in 1712. The Newcomen's engine was used to pump water out of mines.
James Watt came up with a revolution in the steam engine in 1765. He invented a steam engine with a separate condenser. A steam engine with a separate condenser results in improved efficiency and the size of the engine is reduced as compared to the previous design. The engine uses less coal.
Hence we can conclude that the steam engine invented by James Newcomen is more efficient as compared to the engine invented by Thomas Newcomen.
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A mountain is 1100 feet above sea level. A trench is 420 feet below sea level. What is the difference in elevation between the mountain top and the bottom of the trench?
Answer:1520feet
Explanation:
The difference in elevation between the mountain top and the bottom of the trench will be addition of the distance between the sea and mountain top and the distance between the sea level and bottom trench i.e 1100 + 420
= 1520feet
A rock is thrown from a 50.0-m-high cliff with an initial velocity of 7.0 m/s at an angle of 53.0 degrees above the horizontal. How far from the base of the cliff will the rock hit?
Answer:
4.8063m
Explanation:
Horzontal range is given by the formula;
R=(u²sin2θ)/g
u=7m/s, θ=53°, g=9.8m/s
[tex]R=\frac{7^{2}sin2*53 }{9.8}[/tex]
[tex]R=\frac{7^{2}sin106 }{9.8}[/tex]
[tex]R=\frac{49*sin2*53 }{9.8}[/tex]
[tex]R=\frac{47.102 }{9.8}[/tex]
R=4.8063m
A rock thrown at an angle of 53 degrees with an initial speed of 7.0 m/s from a 50-m-height cliff will hit the ground approximately 8.3 meters away from the base of the cliff.
Explanation:The problem regards the range of a projectile which is given by the formula R = (v²/g) * sin(2*Theta), where v is the initial velocity, g is the acceleration due to gravity, and Theta is the launch angle. Given the initial velocity v = 7.0 m/s, launch angle Theta = 53 degrees, and g = 9.8 m/s², it's a matter of substituting these values into the range formula: R = ((7.0 m/s)² / 9.8 m/s²) * sin(2 * 53 degrees).
After performing the calculations, you obtain that the rock will hit ground approximately 8.3 meters away from the base of the cliff.
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From the edge of a cliff, a 0.49 kg projectile is launched with an initial kinetic energy of 1670 J. The projectile's maximum upward displacement from the launch point is 160 m. What are the (a) horizontal and (b) vertical components of its launch velocity?
Answer:
(a) Horizontal component of the launch velocity = 60.7m/s
(b) Vertical component of the launch velocity = 56.0m/s
Explanation:
Initial Kinetic Energy = (IKE)= 1670J, mass (m) = 0.49kg, hmax = 160m
IKE = 1/2mu^2
1670×2/0.48 = u^2
u^2 = 6816.3
u = √6816.3 = 82.6m/s
hmax = u^2sin^2A/2g
160×2×9.8/6816.3 = sin^2A
Sin^2A = 0.46
SinA = √0.46 = 0.6782
A = inverse (sin 0.6782)
A = 42.7° (angle of inclination to the horizontal)
(a) Horizontal component (Ux) = ucosA = 82.6×cos42.7° = 82.6×0.7349 = 60.7m/s
(b) Vertical component (Uy) = usinA = 82.6×sin42.7° = 82.6×0.6782 = 56.0m/s
To find the horizontal and vertical components of the projectile's launch velocity, use the principle of conservation of energy.
Explanation:To find the horizontal and vertical components of the projectile's launch velocity, we'll use the principle of conservation of energy. The projectile's initial kinetic energy is equal to its potential energy at its maximum displacement.
The projectile's initial kinetic energy is given as 1670 J. At the maximum displacement of 160 m, the potential energy is given as m*g*h, where m is the mass of the projectile (0.49 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum displacement (160 m).
Using these values, we can solve for the vertical component of the launch velocity using the equation for potential energy: m*g*h = 0.49 kg * 9.8 m/s^2 * 160 m. The horizontal component of the launch velocity remains unchanged throughout the projectile's motion.
Children in a tree house lift a small dog in a basket 4.90m up to their house. If it takes 201J of work to do this, what is the combined mass of the dog and basket?
Answer:
4.18 kg.
Explanation:
Work done: Work is said to be done when ever a force move a body through a given distance. The S.I unit of work done is Joules (J)
Mathematically, work done is expressed as
W' = F×d.................... Equation 1
Where W' = work done, F = force , d = distance.
making F the subject of the equation,
F = W'/d............................. Equation 2
Note: The force need to lift the small dog n a basket = combined weight of the dog ans the basket.
Therefore,
W = F
Where W = combined weight of the dog and the basket.
Also
W = Mg
M = W/g............................. Equation 3
Where M = combined mass of the dog and the basket, g = acceleration due to gravity.
Given: W' = 201 J, d = 4.90 m.
Substitute into equation 2
F = 201/4.9
F = 41.02 N.
Since F = W = 41.02 N and g = 9.81 m/s²
Substitute these values into equation 3
M = 41.02/9.81
M = 4.18 kg.
Thus the combined mass of the dog and the basket = 4.18 kg.
Each day, Ted can wax 8 cars or wash 10 cars, and Tom can wax 3 cars or wash 5 cars. What is each man's opportunity cost of washing a car?
Answer:
1/2
Explanation:
This will lead us to simultaneous equation since there are two person involved doing the same job each day.
For Ted;
Since Ted can wax 8 cars or wash 10 cars in a day. Mathematically, we have
8x + 10w = 1... (1)
Where x is for waxing cars, w is for washing cars
Similarly for Tom,
Tom can wax 3 cars or wash 5 cars in a day as well, this gives us;
3x + 5w = 1... (2)
Equating 1 and 2, we have;
8x + 10w = 1... (1)
3x + 5w = 1... (2)
Using elimination method, we will multiply eqn 1 by 3 and eqn 2 by 8 to have;
24x + 30w = 3
24x + 40w = 8
Subtracting eqn 4 from 5 we have;
30w - 40w = 3 - 8
-10w = -5
w = 5/10
w = 1/2
Therefore each man's opportunity cost of washing a car is 1/2
The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the electric field at twice the distance from the point charge?
A) At twice the distance, the strength of the field is E/2.B) At twice the distance, the strength of the field is 2E.C) At twice the distance, the strength of the field is 4E.D) At twice the distance, the strength of the field remains equal to E.E) At twice the distance, the strength of the field is E/4.
Answer:
E
Explanation:
Using Coulomb's law equation
Force of the charge = k qQ /d²
and E = F/ q
substitute for F
E = ( K Qq/ d² ) / q
q cancel q
E = KQ / d²
so twice the distance of the from the point charge will lead to the E ( electric field ) decrease by a 4 = E/4. E is inversely proportional to d²
Here's a basketball problem: A 87.2 kg basketball player is running in the positive direction at 7.0 m/s. She is met head-on by a 102.0 kg player traveling at 5.2 m/s toward her. If the 102.0 kg player is knocked backwards at 2.9 m/s, what is the resulting velocity of the 87.2 kg player?
The problem is about the conservation of linear momentum in a collision between two basketball players. The combined momentum before the collision equals the combined momentum after the collision. By setting these two equal and solving for the unknown, we find the velocity of the 87.2 kg player after the collision.
Explanation:In this problem, we are dealing with the conservation of linear momentum. The combined momentum of the two basketball players before the collision should equal the combined momentum after the collision given that no external forces are acting on the system.
Before the collision, the momentum is calculated as the mass of the 1st player times her velocity (87.2 kg x 7.0 m/s in the positive direction) plus the mass of the 2nd player times his velocity (102.0 kg x 5.2 m/s in the negative direction, because he is moving in the opposite direction).
After the collision, the momentum equals the mass of the 2nd player times his velocity (102.0 kg x -2.9 m/s) plus the mass of the 1st player times her final velocity (87.2 kg x v, where v is the velocity we're trying to find).
Setting these two expressions for momentum equal to one another and solving for v, we find the velocity of the 87.2 kg player after the collision.
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The resulting velocity of the 87.2 kg basketball player after the collision is calculated to be approximately 4.3 m/s in the positive direction.
To solve this problem, we use the principle of conservation of momentum. The total momentum before and after the collision must be the same.
The initial velocities and the masses of the players are:
Player 1: mass (m₁) = 87.2 kg, running at velocity (u₁) = 7.0 m/s in the positive direction
Player 2: mass (m₂) = 102.0 kg, running at velocity (u₂) = -5.2 m/s (toward the first player)
Therefore, the initial momentum of the system will be:
m₁ u₁ + m₂ u₂ = 87.2 kg × 7.0 m/s + 102.0 kg × (-5.2 m/s) = 80 kgm/s -(i)
After the collision:
Player 2 ends up moving with velocity (v₂) = -2.9 m/s (knocked backward)
We have to find the velocity of the Player (1) given as v₁.
Therefore, the final momentum of the system will be:
m₁ v₁ + m₂ v₂ = 87.2 kg × v₁ + 102.0 kg × (-2.9 m/s) = (87.2 v₁ - 295.8) kgm/s -(ii)
According to the principle of conservation of momentum, the initial momentum of the system before collision and the final momentum of the system after collision should be the same. So, equations (i) and (ii) must be equal:
80 kgm/s = (87.2 v₁ - 295.8) kgm/s
375.8 kgm/s = 87.2 v₁ kgm/s
or, v₁ = 375.8 kgm/s ÷ 87.2 kg = 4.3 m/s
The difference in arrival times of P and S waves at a seismic observatory can be used to calculate the distance to a quake but not the ____ from the station.
Answer:
Direction
Explanation:
The arrival of seismic waves namely P and S wave are essential as it helps in the determination of the distance from the recording seismic station. In order to cover all the possibilities that are related to an earthquake, seismologists draw a circle around the station but this does not help in obtaining the proper information. By the use of two seismic stations, it can draw two circles that intersect at two points, which again does not help in determining the exact epicenter location. So, at least three seismic stations are needed, by which the triangulation method can be implemented, and as the three circles drawn from the three seismic stations intersect at one common point, it represents the exact location of the earthquake epicenter, and the exact direction also is obtained from this.
Pacman the arcade game went out of order at fat max's arcade. Max is dragging it 5 meters across the ground and out of the store with a rope at a 45 degree angle. What is the force be exerts if he performs 3,500 joules of work
The magnitude of the force is 990 N
Explanation:
The work done by a force on an object is given by:
[tex]W=Fd cos \theta[/tex]
where
F is the magnitude of the force
d is the displacement
[tex]\theta[/tex] is the angle between the direction of the force and of the displacement
In this problem, we have the following:
W = 3500 J (work done by Max)
d = 5 m (displacement)
[tex]\theta=45^{\circ}[/tex] (direction at which the force is applied)
Solving for F, we find the magnitude of the force:
[tex]F=\frac{W}{dcos \theta}=\frac{3500}{(5)(cos 45^{\circ})}=990 N[/tex]
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To find the force Max exerts while moving the Pacman arcade game, use the formula Work = Force x Distance x cos(θ). Solve for force, giving Force = Work / (Distance x cos(θ)). Substituting the given values into the formula will provide the answer.
Explanation:The work done can be used in the equation that relates work, force, and displacement to identify the force exerted by Max. The relationship between these concepts is expressed as: Work = Force x Distance x cos(θ), where 'θ' is the angle at which the force is applied.
In this scenario, the work done (W) is 3,500 joules, the distance (d) is 5 meters, the angle (θ) is 45 degrees, and the force (F) is what we're trying to find.
Rearranging the formula to solve for Force, we get: Force = Work / (Distance x cos(θ)).
Substituting the given values and doing the math, Max's exerted force will be calculated. The cos(45) is approximately 0.7071, and substituting this in our formula might aid us in accurately finding the force exerted by Max while moving the Pacman arcade game.
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Releasing the clutch pedal reapplies the clamping force and reconnects the engine and transmission by firmly clamping the clutch disc between the ______ and the flywheel, allowing them to rotate as a unit.
Answer:
Preassure plate
Explanation:
The flywheel, clutch disc, pressure plate, throw-out bearing clutch fork, and the pilot bearing are the components of a clutch. It works when the flywheel bolts onto the crankshaft and therefore the pressure plate bolts back into the flywheel.
I hope you find this information useful and interesting! Good luck!
A driver driving along a highway at a steady 41 mph (60 ft/sec) sees an accident ahead and slams on the brakes. What constant deceleration is required to stop the car in 200 ft? To find out, carry out the following steps.
Answer:
constant deceleration required is 9 m/s²
Explanation:
Data provided in the question:
Initial Speed of the driver = 41 mph = 60 ft/s
Stopping distance = 200 ft
Now,
Since the car stops after 200 ft therefore final speed, u = 0 ft/s
from the Newton's equation of motion
we have
v² - u² = 2as
where,
v is the final speed
u is the initial speed
a is the acceleration
s is the distance
thus,
0² - 60² = 2a(200)
or
-3600 = 400a
or
a = - 9 m/s²
here, negative sign means deceleration
Hence,
The constant deceleration required is 9 m/s²
If a liquid twice as dense as mercury were used in a barometer, approximately how high would the column of that liquid be under normal sea-level conditions?
A fluid twice as dense as mercury would rise to approximately half the height of mercury in a barometer under normal sea-level conditions. This is due to the relation between hydrostatic pressure, density, and height of the fluid column.
Explanation:Under normal sea-level conditions, atmospheric pressure supports a column of mercury about 760 mm high. This occurs due to the hydrostatic pressure, which is essentially the pressure exerted by a fluid due to gravity. In the case of a liquid twice as dense as mercury, the height of the liquid column would be half that of mercury under the same conditions. This is because the hydrostatic pressure is directly proportional to the density and height of the fluid column, which implies that for a given pressure, if we increase the density, the height decreases correspondingly. So, a liquid twice as dense as mercury would rise to approximately 380 mm in the barometer.
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