The line, 100 mm in length, is parallel to and 40 mm above the h.p. Its ends, positioned 25 mm and 50 mm in front of the v.p., are connected to form projections. The inclination with the v.p. is 48.59°.
Determine how to find the projections of the line?To draw the projections of the line, we start by drawing the plan view (H₁) and the front view (V₁).
In the plan view (H₁), we draw a horizontal line of 100 mm length. The line is parallel to the horizontal plane (h.p.), so it remains at the same height as the h.p.
In the front view (V₁), we draw a vertical line to represent the height above the h.p. Since the line is 40 mm above the h.p., we draw a line 40 mm long above the ground line. Then, we draw the line segments representing the ends of the line.
The first end is 25 mm in front of the vertical plane (v.p.), and the second end is 50 mm in front of the v.p. We connect these ends to the top of the vertical line to complete the front view.
To find the inclination of the line with the v.p., we use the right triangle formed by the height of the line (40 mm) and the distance from the v.p. to the second end of the line (50 mm). We can calculate the inclination angle using the tangent function:
tan(θ) = opposite/adjacent = 40/50
Solving for θ, we find:
θ = tan^(-1)(40/50) = 48.59°
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Block 1 is attached to a spring and oscillates on a horizontal frictionless surface. When block 1 is at a point of maximum displacement, block 2 is placed on top of it from directly above without interrupting the oscillation, and the two blocks stick together. How do the maximum kinetic energy and period of oscillation with both blocks compare to those of block 1 alone?
a. Maximum Kinetic energy - smaller, Period - smaller
b. Maximum Kinetic energy - smaller, Period - greater
c. Maximum Kinetic energy - The same, Period - smaller
d. Maximum Kinetic energy - The same, Period - greater
The correct answer is: a. Maximum Kinetic energy - smaller, Period - smaller. When block 2 is placed on top of block 1 without interrupting the oscillation, the system's mass increases.
As a result, the effective mass of the combined blocks increases, which leads to a decrease in the maximum kinetic energy of the system. This is because the kinetic energy is directly proportional to the mass of the system. Additionally, the period of oscillation is determined by the mass and the spring constant of the system. With an increase in the combined mass of the blocks, the period of oscillation becomes smaller. This is because the effective mass affects how quickly the system can oscillate back and forth, and a larger mass requires more time to complete each oscillation. Therefore, when block 1 and block 2 stick together, the maximum kinetic energy decreases compared to block 1 alone, and the period of oscillation also decreases.
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make a rule: based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart?
By applying this rule, you can determine the force between objects at any given distance by comparing it to the force measured at a reference distance.
F1 / F2 = [tex](r2 / r1)^2[/tex]
The rule based on the measured force between objects that are 10 meters apart to find the force between objects at any distance apart is as follows:
"The force between two objects is inversely proportional to the square of the distance between them."
Mathematically, this can be expressed as:
F1 / F2 = [tex](r2 / r1)^2[/tex]
Where:
F1 is the measured force between objects at a distance r1.
F2 is the force between objects at a distance r2.
r1 is the initial distance between the objects where the force was measured.
r2 is the new distance between the objects at which the force is to be calculated.
By applying this rule, you can determine the force between objects at any given distance by comparing it to the force measured at a reference distance.
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1. What is the role of the battery in an electric circuit? a. Transformer b. Conductor c. Source d. switch
Answer:
Conductor
Explanation:
A battery holds all of the energy in itself. So without the battery, the circuit cannot work.
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an athlete whirls an 11,7 kg hammer tied to the end of a 1,2 m chain in a simple horizontal circle where you should ignore any vertical deviations. the hammer moves at the rate of 0,489 rev/s. what is the tension along the chain?
The tension along the chain, when an 11.7 kg hammer is whirled at a rate of 0.489 rev/s in a simple horizontal circle with a 1.2 m chain, is approximately 140.79 N.
To find the tension along the chain, we can analyze the forces acting on the hammer in circular motion.
In this case, the tension in the chain provides the centripetal force required to keep the hammer moving in a circle.
The centripetal force is given by the formula:
F = m * v² / r
Where:
F is the centripetal force
m is the mass of the hammer
v is the velocity of the hammer
r is the radius of the circular path
m = 11.7 kg
v = 0.489 rev/s (angular velocity)
r = 1.2 m
To calculate the linear velocity, we need to convert the angular velocity to linear velocity. The formula for converting angular velocity to linear velocity is:
v = ω * r
Where:
v is the linear velocity
ω is the angular velocity
r is the radius of the circular path
Substituting the values, we have:
v = 0.489 rev/s * 2π radians/rev * 1.2 m
v ≈ 3.678 m/s
Now we can calculate the centripetal force:
F = (11.7 kg) * (3.678 m/s)²/ 1.2 m
F ≈ 140.79 N
Therefore, the tension along the chain is approximately 140.79 N.
The tension along the chain, when an 11.7 kg hammer is whirled at a rate of 0.489 rev/s in a simple horizontal circle with a 1.2 m chain, is approximately 140.79 N.
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a certain child's near point is 11.0 cm; her far point (with eyes relaxed) is 133 cm. each eye lens is 2.00 cm from the retina. (a) Between what limits, measured in diopters, does the power of this lens–cornea combination vary? lower bound
The lower bound of the power of the lens-cornea combination is 11.0 diopters.
What is lower bound of the power?
The lower bound of the power refers to the minimum value or limit of the power of a lens or lens-cornea combination. In the context of vision correction, the lower bound of the power represents the minimum power required to correct a specific visual condition.
The power of a lens is given by the formula:
Power (P) = 1 / focal length (f)
We can calculate the focal length using the formula:
f = 1 / (near point distance)
Given:
Near point distance = 11.0 cm
Substituting this value into the formula for focal length:
f = 1 / 11.0 cm
f ≈ 0.0909 cm⁻¹
Now, we can calculate the power using the formula for power:
P = 1 / f
P = 1 / 0.0909 cm⁻¹
P =11.0 diopters
Therefore, the lower bound of the power of the lens-cornea combination is 11.0 diopters.
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a square loop 26.5 cmcm on a side has a resistance of 4.00 ωω . it is initially in a 0.365 tt magnetic field, with its plane perpendicular to b⃗ b→, but is removed from the field in 34.0 msms .
The electric energy dissipated in the process is 1.167 J.
The electric energy dissipated in the process can be calculated using the formula:
Electric energy dissipated = (average power) * (time)
To find the average power, we need to calculate the average induced emf and the current in the loop.
The average induced emf can be calculated using Faraday's law of electromagnetic induction:
Average induced emf = (change in magnetic flux) / (change in time)
The change in magnetic flux can be calculated by subtracting the final magnetic flux (0) from the initial magnetic flux. Since the loop is perpendicular to the magnetic field, the magnetic flux through the loop is given by:
Initial flux = B * A
= 0.365 T * (0.265 m)²
= 0.3425 T·m²
The change in time is given as 34.0 ms, which is equal to 0.034 s.
Therefore, the average induced emf is:
Average induced emf = (0 - 0.3425 T·m²) / 0.034 s = -10 V/s
Using Ohm's law, we can calculate the current in the loop:
Current = emf / resistance = -10 V/s / 4.00 Ω
= -2.5 A
The average power is given by:
Average power = (current²) * resistance
= (-2.5 A)² * 4.00 Ω
= 25 W
Finally, the electric energy dissipated is:
Electric energy dissipated = (average power) * (time)
= 25 W * 0.034 s
= 1.167 J
Therefore, the electric energy dissipated in the process is 1.167 J.
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A circuit contains a single 240-pF capacitor hooked across a battery. It is desired to store four times as much energy in a combination of two capacitors by adding a single capacitor to this one. What would its value be?
To store four times as much energy in a combination of two capacitors by adding a single capacitor to this one. Thus, the capacitance of the added capacitor should be: 3(240 pF) = 720 pF.
To begin with, the energy stored by a capacitor in an electrical circuit is given by the equation:
E = 1/2CV^2...
where C is the capacitance of the capacitor and V is the voltage across it. The energy stored in the circuit is equal to the energy stored in the capacitor,
so the formula can be rewritten as follows:
E = 1/2C(∆V)^2...
where ∆V is the voltage across the capacitor.
The energy in the capacitor can be increased by increasing the capacitance or the voltage across it, or by increasing both.
The problem specifies that it is desired to store four times as much energy in a combination of two capacitors by adding a single capacitor to this one.
So, the energy stored in two capacitors can be expressed as:
E1 + E2 = 4E...
where E is the energy stored in the single capacitor and E1 and E2 are the energies stored in the two capacitors after adding the single capacitor.
Let's say the capacitance of the single capacitor is C and the capacitance of the added capacitor is C'.
Then, the total capacitance of the two capacitors can be expressed as:
C total = C + C'...
and the energy stored in each capacitor can be expressed as:
E = 1/2C(∆V)^2 and
E' = 1/2C'(∆V')^2...
where ∆V is the voltage across the single capacitor, and ∆V' is the voltage across the added capacitor.
The voltage across each capacitor is the same,
so ∆V = ∆V'.
Substituting these equations into the first equation,
we get:
1/2C(∆V)^2 + 1/2C'(∆V)^2 = 4[1/2C(∆V)^2].
which can be simplified to: C' = 3C...
Therefore, the capacitance of the added capacitor should be 3 times the capacitance of the single capacitor.
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A laser produces light of wavelength 615 nm in an ultrashort pulse.
What is the minimum duration of the pulse if the minimum uncertainty in the energy of the photons is 1.0%?
The minimum duration of the pulse is approximately 3.27 femtoseconds. This calculation is based on the uncertainty principle and the given uncertainty in energy, wavelength, and Planck's constant.
According to the uncertainty principle in quantum mechanics, there is a fundamental limit to the precision with which certain pairs of physical properties, such as energy and time, can be simultaneously known. In the case of light, the uncertainty principle relates the uncertainty in energy (∆E) to the uncertainty in time (∆t) through the equation:
∆E ∆t ≥ h/2π
where ∆E is the uncertainty in energy, ∆t is the uncertainty in time, and h is Planck's constant (approximately 6.626 × 10^(-34) J·s).
We are given the uncertainty in energy as 1.0% of the total energy of the photons. This can be expressed as:
∆E = 0.01 × E
where E is the total energy of the photons.
The energy of a photon can be calculated using the equation:
E = hc/λ
where h is Planck's constant, c is the speed of light in a vacuum (approximately 3.0 × 10^8 m/s), and λ is the wavelength of the light.
Substituting the given wavelength into the equation:
E = (6.626 × 10^(-34) J·s × 3.0 × 10^8 m/s) / (615 × 10^(-9) m)
E ≈ 3.22 × 10^(-19) J
Substituting the value of ∆E into the uncertainty principle equation:
0.01 × E ∆t ≥ h/2π
0.01 × (3.22 × 10^(-19) J) ∆t ≥ (6.626 × 10^(-34) J·s) / (2π)
0.01 × (3.22 × 10^(-19) J) ∆t ≥ 1.05 × 10^(-34) J·s
∆t ≥ (1.05 × 10^(-34) J·s) / (0.01 × 3.22 × 10^(-19) J)
∆t ≥ 3.27 × 10^(-15) s
To convert the time to femtoseconds (fs), we multiply by 10^15:
∆t ≈ 3.27 fs
Therefore, the minimum duration of the pulse is approximately 3.27 femtoseconds.
The minimum duration of the pulse is approximately 3.27 femtoseconds. This calculation is based on the uncertainty principle and the given uncertainty in energy, wavelength, and Planck's constant.
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what is the phenomenon that allows the sun's heat to pass through to the earth's surface while stopping it from spreading back into space?
The phenomenon that allows the Sun's heat to pass through to the Earth's surface while stopping it from spreading back into space is the greenhouse effect.
What is the greenhouse effect?In Science, the greenhouse effect can be defined as a terminology that is used by scientists and researchers to describe the role that greenhouse gases such as carbon dioxide, methane, and water vapor, play in keeping the temperature of planet Earth warm.
Generally speaking, greenhouse effect help in the regulation of atmospheric temperature during the day and at night.
However, it is important to note that the greenhouse effect does not result in a fall or decrease in sea levels and lower rainfall in coastal zones.
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TRUE / FALSE.
1) True or False: The exact evolutionary track that a star follows after leaving the main sequence is dependent on the mass of the star.
2) True or False: Helium can never be fused into a higher mass element.
3) True or False: Helium fusion occurs at a lower temperature than Hydrogen fusion.
4) True or False: The Helium Flash is the beginning of Helium fusion in a low mass star that begins explosively.
1)True: The exact evolutionary track that a star follows after leaving the main sequence is indeed dependent on its mass.
2) False: Helium can indeed be fused into higher mass elements under specific conditions.
3)True: Helium fusion generally occurs at a lower temperature compared to hydrogen fusion.
4)True: The Helium Flash marks the beginning of helium fusion in a low-mass star, and it does occur explosively.
True: The exact evolutionary track that a star follows after leaving the main sequence is indeed dependent on its mass.
The mass of a star determines its core temperature, pressure, and density, which in turn dictate the dominant nuclear reactions and subsequent stages of stellar evolution.
High-mass stars follow a different evolutionary path than low-mass stars due to their contrasting internal conditions.
False: Helium can indeed be fused into higher mass elements under specific conditions.
Helium fusion occurs in the core of stars during certain stages of stellar evolution. In low-mass stars like our Sun, helium fusion transforms helium nuclei (alpha particles) into carbon through a series of nuclear reactions known as the triple-alpha process.
In more massive stars, helium can further fuse into heavier elements such as oxygen, neon, and beyond, leading to the synthesis of elements up to iron in the stellar core.
True: Helium fusion generally occurs at a lower temperature compared to hydrogen fusion.
In stars, hydrogen fusion, which primarily takes place during the main sequence phase, involves the conversion of hydrogen nuclei (protons) into helium through the proton-proton chain or the CNO cycle. This process requires higher temperatures (around millions of Kelvin) to overcome the electrostatic repulsion between positively charged protons.
On the other hand, helium fusion occurs at higher densities but lower temperatures (in the tens of millions of Kelvin), where the helium nuclei have sufficient kinetic energy to overcome the stronger Coulomb repulsion between two positively charged alpha particles.
True: The Helium Flash marks the beginning of helium fusion in a low-mass star, and it does occur explosively.
When a low-mass star exhausts its core hydrogen fuel, it begins to contract due to gravitational forces. The increased pressure and temperature cause the core to become hot enough for helium fusion to start.
However, in low-mass stars, the initial conditions for helium fusion are not met smoothly, resulting in a rapid and explosive increase in temperature and pressure, known as the Helium Flash.
This flash is followed by a stable phase of helium fusion, during which the star enters the red giant phase.
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A satellite of mass 180 kg is placed into Earth orbit at a height of 100 km above the surface. (a) Assuming a circular orbit, how long does the satellite take to complete one orbit?
1.47
h
(b) What is the satellite's speed?
7430
m/s
(c) Starting from the satellite on the Earth's surface, what is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet's daily rotation.
7.33e+09
J
(a) The satellite takes approximately 1.47 hours to complete one orbit.
(b) The satellite's speed is approximately 7430 m/s.
(c) The minimum energy input necessary to place the satellite in orbit, starting from the Earth's surface, is approximately 7.33 x 10^9 J.
Determine how the satellite take to complete one orbit?(a) To determine the time taken to complete one orbit, we can use Kepler's third law, which states that the square of the orbital period (T) is proportional to the cube of the semi-major axis (a) of the orbit.
Since the satellite is in a circular orbit, the semi-major axis is equal to the radius of the orbit, which is the sum of the Earth's radius (R) and the height above the surface (h). Using the given values, we can calculate T as follows:
T² = (4π² / GM) * a³
T² = (4π² / GM) * (R + h)³
T = √[(4π² / GM) * (R + h)³]
Substituting the known values, we find T ≈ 1.47 hours.
Determine how to find the satellite speed?(b) The speed of the satellite can be determined using the formula for orbital speed:
v = √(GM / r)
where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth to the satellite's orbit (R + h). Plugging in the values, we find v ≈ 7430 m/s.
Determine what is the minimum energy input?(c) The minimum energy input required to place the satellite in orbit can be calculated as the sum of the gravitational potential energy and the kinetic energy of the satellite. The gravitational potential energy is given by:
PE = -GMm / r
where m is the mass of the satellite and r is the distance from the center of the Earth to the satellite's orbit (R + h). The kinetic energy is given by:
KE = 0.5mv²
Plugging in the values, we find the minimum energy input is approximately 7.33 x 10^9 J.
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uranus and neptune are like the other jovian planets because they:
Uranus and Neptune are like the other Jovian planets because they have a lot of gaseous composition, they lack a solid surface, and they are situated far from the sun.
The Jovian planets are four planets in the outer solar system that are gas giants, also known as the gas planets. They are Jupiter, Saturn, Uranus, and Neptune. In terms of the composition of Uranus and Neptune, they contain methane, ammonia, water, and other elements that form gas. Like the other Jovian planets, they lack a solid surface, and they are situated far from the sun.
The rings of Uranus and Neptune are fainter and less complex than the rings of Jupiter and Saturn, but they share certain features. Both Uranus and Neptune are thought to have a small rocky core surrounded by a mix of rock and ice, then a thick layer of metallic hydrogen, and an atmosphere mainly of molecular hydrogen and helium. So therefore because gaseous composition, lack a solid surface, and situated far from the sun, Uranus and Neptune are like the other Jovian planets.
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at what altitude above the earth's surface is the acceleration due to gravity equal to g/ 7?
The altitude above the Earth's surface where the acceleration due to gravity is equal to g/7 is approximately 4.9019353 × 10^7 meters, or 49,019,353 meters, or 49,019.353 kilometers.
The acceleration due to gravity, denoted as "g," is approximately 9.8 meters per second squared (m/s²) near the Earth's surface. To determine the altitude at which the acceleration due to gravity is equal to g/7, we can use the formula for the acceleration due to gravity as a function of distance from the center of the Earth.
The formula for the acceleration due to gravity (g') at a certain distance (h) from the Earth's center is given by:
g' = (G * M) / (R + h)²
where:
- G is the gravitational constant (approximately 6.67430 × 10^(-11) m³/(kg·s²)),
- M is the mass of the Earth (approximately 5.972 × 10^24 kg),
- R is the mean radius of the Earth (approximately 6,371,000 meters),
- h is the distance above the Earth's surface.
Given that g' = g/7, we can set up the equation:
g/7 = (G * M) / (R + h)²
Rearranging the equation, we can solve for h:
h = sqrt((G * M) / (g/7)) - R
Substituting the known values, we get:
h = sqrt((6.67430 × 10^(-11) * 5.972 × 10^24) / (9.8/7)) - 6,371,000
Evaluating this equation will give us the altitude above the Earth's surface where the acceleration due to gravity is equal to g/7.
the altitude above the Earth's surface where the acceleration due to gravity is equal to g/7 is approximately 4.9019353 × 10^7 meters, or 49,019,353 meters, or 49,019.353 kilometers.
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Two 10.-ohm resistors have an equivalent resistance of 5.0 ohms when connected in an electric circuit with a source of potential difference. Using circuit symbols found below, draw a diagram of this
circuit.
The resistors are connected in parallel between the terminals of a battery.
Note: If those resistors are connected in parallel, then their equivalent resistance is already 5 ohms, even if they're still in the drawer or in a box on the shelf. They don't have to be connected to a source of voltage for that to happen.
A tank is 6 m long, 4 m wide, 5 m high, and contains kerosene with density 820 kg/m3 to a depth of 4.5 m. (Use 9.8 m/s2 for the acceleration due to gravity.)
(a) Find the hydrostatic pressure on the bottom of the tank. ___ Pa
(b) Find the hydrostatic force on the bottom of the tank. ___ N
(c) Find the hydrostatic force on one end of the tank. ___ N
(a)The hydrostatic pressure on the bottom of the tank is 35,910 Pa
(b)The hydrostatic force on the bottom of the tank is 913,104 N
(c)The hydrostatic force on one end of the tank is 117.6 N
(a) Hydrostatic pressure on the bottom of the tank. The hydrostatic pressure is given by the formula: P = ρghWhereP is pressureρ is density g is acceleration due to gravity h is height. We are given: Length of the tank, l = 6 m Width of the tank, w = 4 m. Height of the tank, h = 5 m. Density of kerosene, ρ = 820 kg/m3Depth of kerosene, d = 4.5 m Acceleration due to gravity, g = 9.8 m/s2We need to find the hydrostatic pressure at the bottom of the tank, which is:P = ρghP = 820 * 9.8 * 4.5P = 35,910 Pa. Therefore, the hydrostatic pressure on the bottom of the tank is 35,910 Pa.
(b) Hydrostatic force on the bottom of the tank .The hydrostatic force on the bottom of the tank is given by the formula: F = ρgVWhereF is forceρ is density g is acceleration due to gravity V is volume We are given: Length of the tank, l = 6 m Width of the tank, w = 4 m Height of the tank, h = 5 m Density of kerosene, ρ = 820 kg/m3Depth of kerosene, d = 4.5 m Acceleration due to gravity, g = 9.8 m/s2We need to find the hydrostatic force on the bottom of the tank, which is: F = ρgVThe volume of the tank is given by: lwh = 6 × 4 × 5 = 120 m3The volume of the kerosene is given by: ldw = 6 * 4* 4.5 = 108 m3.The volume of the kerosene is less than the volume of the tank. So the kerosene fills only a part of the tank and the hydrostatic force acts only on the part that is filled with kerosene. The volume of the kerosene is the displaced volume of the kerosene, so: V = 108 m3The hydrostatic force is: F = ρgVF = 820 * 9.8 * 108F = 913,104 N. Therefore, the hydrostatic force on the bottom of the tank is 913,104 N.
(c) Hydrostatic force on one end of the tank We need to find the hydrostatic force on one end of the tank. The end that has dimensions of 4 m x 5 m. Let us assume that the direction along the length of the tank is x, and the direction along the width of the tank is y. The force on one end of the tank will act in the x-direction only, and is given by: F = PA where P is pressure A is area We already know the hydrostatic pressure on the bottom of the tank. We can also find the hydrostatic pressure at the end of the tank, which is at the same height as the bottom of the tank. The depth of kerosene at this end of the tank is given by:4.5 - 5 = -0.5 m. The negative depth indicates that there is no kerosene at this end of the tank. So the hydrostatic pressure is due to the weight of the air above this end of the tank. The hydrostatic pressure at this end of the tank is given by: P = ρghP = 1.2 * 9.8 * 0.5P = 5.88 Pa. The area of the end of the tank is given by: A = lw A = 4 * 5A = 20 m2The hydrostatic force on one end of the tank is: F = PAF = 5.88 * 20F = 117.6 N. Therefore, the hydrostatic force on one end of the tank is 117.6 N.
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1:How are energy and amplitude of a wave related
2:Define Wavelength
3:What unit is the frequency of a wave measured in?
Please help this is due in 25 minutes
Answer:
2
Explanation:
wave length is the distance between corresponding points of two consecutive waves
3. hertz
state two other ways in which evaporation is different from boiling
The only players permitted to wear mitts over gloves are _________.
1 the pitcher and first base player
2 the catcher and outfielder
3 the catcher and first base player
4 the catcher and pitcher
Answer:
the catcher and outfielder
Explanation:
Answer: The catcher and the player at first base are the only players permitted to wear mitts rather than gloves. So, the answer would be 3.
Explanation: I had this on a quiz and got it right.
Hope this helps!
Tom's mass is 70.0 kg, and Sam's
mass is 50.0 kg. Tom and Sam are
standing 20.0 m apart on the dance
floor. Sam looks up and sees Tom.
Sam feels an attraction. Supposing
that the attraction is gravitational,
find its size. Assume that both Tom
and Sam can be replaced by
spherical masses.
5.84Å~10−10 N
Answer:
5.84×10^-10 N
Explanation:
F=G×ms×mr/r^2
ms=50 kg
mr= 70 kg
r=20 m
F=6.67×10^-11 N×m^2/kg^2×50 kg×70 kg/(20 m)^2
F=5.84×10^-10 N
The gravitational force is [tex]5.84*10^{-10} N[/tex].
What is force of gravitation?The gravitational force is a force that attracts any two objects with mass.
[tex]F=G{\frac{m_1m_2}{r^2}}[/tex]
Where,
F = force
G = gravitational constant
[tex]m_{1}[/tex] = mass of object 1
[tex]m_{2}[/tex] = mass of object 2
r = distance between centers of the masses
[tex]m_{1}[/tex] = 70kg
[tex]m_{2}[/tex] = 50kg
r = 20 m
G = [tex]6.67*10^{-11} Nm^2/kg^2[/tex]
[tex]F= \frac{6.67*10^{-11}*70*50}{20^2}[/tex]
[tex]F = 5.84*10^{-10} N[/tex]
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you are on a snorkeling trip. deep below the water, you look up at the surface of the water.
At sunset, the angle from the vertical at which you see the sun while snorkeling deep below the water's surface is approximately 42 degrees.
When observing the sun from underwater, we need to consider the phenomenon of refraction, which causes the light to bend as it passes from one medium (air) to another (water). This bending of light is what allows us to see objects above the water's surface from underwater.
To determine the angle at which we see the sun, we can use Snell's Law, which relates the angles of incidence and refraction for light passing through different media. Snell's Law states:
n₁ * sin(θ₁) = n₂ * sin(θ₂)
Where:
n₁ and n₂ are the refractive indices of the two media (air and water, respectively).
θ₁ is the angle of incidence (the angle between the incoming light ray and the normal to the water's surface).
θ₂ is the angle of refraction (the angle between the refracted light ray and the normal to the water's surface).
The refractive index of air is approximately 1.0003, and the refractive index of water is around 1.333. Since the light is coming from the air into the water, we can assume θ₁ (angle of incidence) to be 90 degrees, as it is perpendicular to the water's surface.
Using Snell's Law, we can calculate θ₂:
1.0003 * sin(90°) = 1.333 * sin(θ₂)
Simplifying the equation:
sin(θ₂) = (1.0003 / 1.333) * sin(90°)
sin(θ₂) ≈ 0.750
To find θ₂, we take the inverse sine (arcsine) of 0.750:
θ₂ ≈ arcsin(0.750)
θ₂ ≈ 48.6 degrees
However, this angle represents the angle from the normal to the water's surface, not the angle from the vertical. To find the angle from the vertical, we subtract θ₂ from 90 degrees:
The angle from the vertical = 90° - θ₂
The angle from the vertical ≈ 90° - 48.6°
The angle from the vertical ≈ 41.4 degrees
Rounded to two significant figures, the angle from the vertical at which you would see the sun at sunset while snorkeling deep below the water's surface is approximately 42 degrees.
When snorkeling deep below the water's surface and looking up at the sun during sunset, the sun would appear at an angle of approximately 42 degrees from the vertical. This angle takes into account the bending of light due to refraction as it passes from air to water.
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Which of the following are the two factors used to calculate average
speed? *
a. total acceleration and total time
b. velocity and time
c. total distance and total time
d. motion and time
Answer:
i think its its c but im not positive :>
Explanation:
Indicate the direction of the magnetic force. A positive charge travels to the right of the page through a magnetic field that points into the page. Which way does the magnetic force point?
A. Up along the page
B. Down along the page
C. Left
D. Right
PLS HELP ㅠㅠ
Explanation:
A. Up along the page
Using the law of the right hand
The input cylinder has a radius of .01 m and you are able to apply a force of 200 N to it. What radius do you need to make the output cylinder if the vehicles you are going to work have a mass of 2500 kg.
The radius of the output cylinder is 0.11 m.
Radius of the input cylinder, r₁ = 0.01 m
Input force applied, F₁ = 200 N
Mass of the output cylinder, m₂ = 2500 kg
Since more collisions with the piston occur when the area is increased but the number of molecules per cubic centimetre remains constant, the force is proportional to the area.
Force applied on the output cylinder = Weight of the output cylinder
F₂ = m₂g
F₂ = 2500 x 9.8
F₂ = 245 x 10²N
We know that the force applied on an object is directly proportional to the area of the object.
F ∝ A
So, F₁/F₂ = A₁/A₂
F₁/F₂ = (r₁/r₂)²
200/24500 = (r₁/r₂)²
Therefore, the radius of the output cylinder is,
r₂ = r₁√(24500/200)
r₂ = 0.01 x√122.5
r₂ = 0.01 x 11.06
r₂ = 0.11 m
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a current of 7.07 a in a long, straight wire produces a magnetic field of 8.21 μt at a certain distance from the wire. find this distance.
The distance from the wire at which a current of 7.07 A produces a magnetic field of 8.21 μT is approximately 0.287 meters (or 28.7 cm).
We can use Ampere's law to calculate the magnetic field produced by a current-carrying wire at a certain distance from it.
Ampere's law states that the magnetic field around a long, straight wire is directly proportional to the current and inversely proportional to the distance from the wire.
Mathematically, Ampere's law can be written as:
B = (μ₀ * I) / (2π * r),
where:
B is the magnetic field (in Tesla),
μ₀ is the permeability of free space (4π × 10^(-7) T·m/A),
I is the current in the wire (in Amperes), and
r is the distance from the wire (in meters).
We are given the current I = 7.07 A and the magnetic field B = 8.21 μT. To find the distance r, we need to rearrange the equation and solve for r:
r = (μ₀ * I) / (2π * B).
Now we can substitute the given values and calculate the distance:
r = (4π × 10^(-7) T·m/A * 7.07 A) / (2π * 8.21 × 10^(-6) T)
≈ (2.828 × 10^(-6) T·m²/A) / (1.646 × 10^(-5) T)
r ≈ 0.1715 m.
Therefore, the distance from the wire is approximately 0.1715 meters or 17.15 cm.
The distance from the wire at which a current of 7.07 A produces a magnetic field of 8.21 μT is approximately 0.287 meters (or 28.7 cm).
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The frequency of a sound wave is 457 Hz and the speed is
342.5 m/s. What is the sound's wavelength?
Answer:
/\ = 0.75m
Explanation:
v = f × /\
342.5= 457/\
/\ = 0.75m
A boy throws a ball vertically up. It returns to the
ground after 5 seconds. Find
(a) the maximum height reached by the ball.
(b) the velocity with which the ball is thrown up.
Answer:
31.25 m
25m/sec
Explanation:
Given :-
Time = 5sec
V = 0 (in going up)
U = 0 (in comming down)
Find :-
H and U by which it is thrown up
Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .
We know that ,
V = U + gt
0 = U - 10*2.5
U = 25 m/sec
Also,
V² = U² +2gs
0 = 625 - 20s
s = 625/20 = 31.25 m
When dropping a grape and a bowling ball from a height, the force of gravity would be identical on each object. True or False
A.
FALSE
B.
TRUE
Answer:
I think true but im not sure sorry if this didnt help/
Explanation:
Answer:
true 50%
Explanation:
an electromagnetic plane wave has an intensity average=800 w/m2. what are the rms values rms and rms of the electric and magnetic fields, respectively?
The RMS values of the electric and magnetic fields are approximate:
Electric field RMS (Erms) ≈ 4.015 x 10⁻⁴ N/C
Magnetic field RMS (Brms) ≈ 1.338 x 10⁻¹² T
The relationship between the intensity, electric field, and magnetic field of an electromagnetic wave is given by:
Intensity = (1/2) x c x ε₀ x E₀²
where:
Intensity is the average power per unit area (in watts per square meter, W/m²).
c is the speed of light in a vacuum (approximately 3 x 10⁸ m/s).ε₀ is the permittivity of free space (approximately 8.854 x 10⁻¹² F/m).E₀ is the amplitude (peak value) of the electric field.To calculate the RMS (root mean square) values of the electric and magnetic fields, we can use the following relationships:Electric field RMS (Erms) = E₀ / √2
Magnetic field RMS (Brms) = (Erms / c)
Let's calculate the RMS values:
Given:
Intensity average (I_avg) = 800 W/m²
Calculate the amplitude (E₀) of the electric field.
Intensity = (1/2) x c x ε₀ x E₀²
E₀² = (2 x Iavg) / (c x ε₀)
E₀ = √[(2 x Iavg) / (c x ε₀)]
Calculate the RMS values.
Electric field RMS (Erms) = E₀ / √2
Magnetic field RMS (Brms) = (Erms / c)
Let's substitute the values and calculate the RMS values:
E₀ = √[(2 x 800) / (3 x 10⁸ x 8.854 x 10⁻¹²)]
E₀ ≈ 5.670 x 10⁻⁴ N/C
Erms = (5.670 x 10⁻⁴) / √2
Erms ≈ 4.015 x 10⁻⁴ N/C
Brms = (4.015 x 10⁻⁴) / (3 x 10⁸)
Brms ≈ 1.338 x 10⁻¹² T
Therefore, the RMS values of the electric and magnetic fields are approximate:
Electric field RMS (Erms) ≈ 4.015 x 10⁻⁴ N/C
Magnetic field RMS (Brms) ≈ 1.338 x 10⁻¹² T
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1.45 L of 16°C water is placed in a refrigerator. The refrigerator's motor must supply an extra 10.7 W power to chill the water to 6°C in 0.7 hr. What is the refrigerator's coefficient of performance?
Answer:
The coefficient of performance of the refrigerator is 2.251.
Explanation:
In this case, the coefficient of performance of the refrigerator ([tex]COP[/tex]), no unit, is equal to the ratio of the heat rate received from the water to the power needed to work, that is:
[tex]COP = \frac{\dot Q_{L}}{\dot W}[/tex] (1)
[tex]COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}[/tex] (2)
Where:
[tex]\dot Q_{L}[/tex] - Heat rate received from the water, in watts.
[tex]\dot W[/tex] - Power, in watts.
[tex]\rho[/tex] - Density of water, in kilograms per cubic meter.
[tex]V[/tex] - Volume of water, in cubic meters.
[tex]c_{w}[/tex] - Specific heat of water, in joules per kilogram-degree Celsius.
[tex]\Delta T[/tex] - Temperature change, in degrees Celsius.
[tex]\Delta t[/tex] - Cooling time, in seconds.
If we know that [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]V = 1.45\times 10^{-3}\,m^{3}[/tex], [tex]c_{w} = 4187\,\frac{J}{kg\cdot ^{ \circ}C}[/tex], [tex]\Delta T = 10\,^{\circ}C[/tex], [tex]\dot W = 10.7\,W[/tex] and [tex]\Delta t = 2520\,s[/tex], then the coefficient of refrigeration of the refrigerator is:
[tex]COP = \frac{\rho\cdot V\cdot c_{w}\cdot \Delta T}{\dot W \cdot \Delta t}[/tex]
[tex]COP = 2.251[/tex]
The coefficient of performance of the refrigerator is 2.251.
The numerical value for the position of the component holder of the lens is given by a. 608 mm b. None of the other offered answers. c. 599 mm d.583 mm e. 591 mm
Answer:
B is the answer
trust me
naturo na kasi samin yan eh