Therefore, the amount of cosmic dust that covered Brooklyn on that day was 0.00002118 tons (to 1 significant figure).
Given that 60 tons of cosmic dust fell to Earth on a particular day and we need to find out how many tons of that dust covered Brooklyn on that day. The Earth’s surface is 196.9 million square miles and Brooklyn covers an area of 69.5 square miles.
To solve the problem, we need to find the fraction of the earth's surface area that Brooklyn covers and multiply it by the total dust that fell to earth on that day.
So, the fraction of the earth's surface area that Brooklyn covers is:
69.5/196.9 million = 3.5303 x 10^-7 (rounded to 4 significant figures)
Now, we multiply this fraction by the total amount of dust that fell to earth on that day:
3.5303 x 10^-7 x 60 tons = 0.00002118 tons (rounded to 1 significant figure)
In the problem, we are given the total amount of cosmic dust that fell to earth on a particular day, which is 60 tons. We are also given the surface area of the earth and the surface area of Brooklyn. We need to find out how much of that dust covered Brooklyn on that day.In order to solve this problem, we need to find the fraction of the earth's surface area that Brooklyn covers. To do this, we divide the surface area of Brooklyn by the surface area of the earth. This gives us a fraction which we can then use to find out how much of the dust covered Brooklyn on that day.Once we have the fraction, we can multiply it by the total amount of dust that fell to earth on that day. This gives us the amount of dust that covered Brooklyn on that day.
The final answer is 0.00002118 tons, which is rounded to 1 significant figure.
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a circuit has three resistors connected in series. resistor r2 has a resistance of 200 ohms and a voltage drop of 30 volts. what is the current in resistor r3?
The current through resistor R3 is 0.33 A. A circuit with three resistors connected in series is shown below: Circuit diagram of three resistors connected in series As per the given information, R2 has a resistance of 200 ohms and a voltage drop of 30 volts.
Therefore, the voltage drop across R1 is V1 = V - V2 - V3V = voltage supplied to the circuit = voltage drop across R1 + voltage drop across R2 + voltage drop across R3R1 = Resistance of resistor R1.R2 = Resistance of resistor R2 = 200 Ω.V3 = Voltage drop across resistor R3.I3 = Current through resistor R3.To calculate the current in resistor R3, let's follow the steps given below.Step 1: Find the voltage drop across R1.Using Ohm's Law, the voltage drop across R2 is V2 = IR2Substitute the values of V2 and R2 to get the value of current I.I = V2/R2I = 30/200I = 0.15 A
Using Kirchhoff's voltage law, the voltage drop across R1 isV1 = V - V2 - V3V = V1 + V2 + V3Substitute the values of V, V2, and V3 to get the value of V1.V1 = V - V2 - V3V1 = 100 - 30 - V1V1 = 70 VStep 2: Find the current through R3.Using Ohm's Law, the voltage drop across R3 is V3 = I3R3.Substitute the values of V3 and R3 to get the value of current I3.I3 = V3/R3I3 = (V - V1 - V2)/R3I3 = (100 - 70 - 30)/R3I3 = 0.33 A
Therefore, the current through resistor R3 is 0.33 A.
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A wildlife researcher is tracking a flock of geese. The geese fly 5.0 km due west, then turn toward the north by 50 ∘ and fly another 4.5 km .PART A .How far west are they of their initial position? dw=? PART B What is the magnitude of their displacement? d=?
Answer:
PART A:
The geese initially fly 5.0 km due west. This movement is entirely in the west direction, so the west component (dw) is equal to 5.0 km.
Therefore, the geese are 5.0 km west of their initial position.
PART B:
After flying 5.0 km due west, the geese turn toward the north by 50° and fly another 4.5 km.
To determine the displacement (d), we need to find the resultant of their west and north components.
West component: The initial movement of 5.0 km due west does not change after turning north. So the west component of displacement remains the same at 5.0 km.
North component: The geese fly 4.5 km in the north direction. Since they turn by 50° from west, we can use trigonometry to find the north component (dn).
dn = 4.5 km * sin(50°)
dn ≈ 3.454 km
The displacement (d) is the magnitude of the resultant of the west and north components. We can use the Pythagorean theorem to find the magnitude:
d = sqrt(dw^2 + dn^2)
d = sqrt((5.0 km)^2 + (3.454 km)^2)
d ≈ 6.076 km
Therefore, the geese are approximately 6.076 km away from their initial position, and their displacement is approximately 6.076 km.
The flock of geese is initially 5.0 km west of their starting position. Their displacement, considering both the westward and northward movements, is approximately 5.2 km.
In the given scenario, the geese first fly 5.0 km due west. This indicates a purely westward displacement. Therefore, the distance west of their initial position is 5.0 km.
Afterward, the geese turn toward the north by an angle of 50 degrees and continue flying for another 4.5 km. This northward displacement can be broken down into its vertical and horizontal components. The vertical component can be found by multiplying the distance flown (4.5 km) by the sine of the angle (50 degrees). The horizontal component can be found by multiplying the distance flown (4.5 km) by the cosine of the angle (50 degrees).
Calculating the vertical component: 4.5 km × sin(50°) ≈ 3.42 km
Calculating the horizontal component: 4.5 km × cos(50°) ≈ 2.90 km
To find the magnitude of the total displacement, we can use the Pythagorean theorem. The total displacement is the square root of the sum of the squares of the horizontal and vertical components.
Calculating the magnitude of displacement: √(5.0 km² + 2.90 km² + 3.42 km²) ≈ √39.97 km² ≈ 6.32 km
Therefore, the geese are approximately 5.0 km west of their initial position, and their displacement is approximately 6.32 km.
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A toy boat made of solid wood has a mass of 4.78 kgkg and density of 0.5 g/cm3g/cm3 . It is placed in water. Some steel weights are placed on top of it so it floats with 10 %% of its volume above the surface. Find the mass of steel placed on top of the boat
Therefore, the mass of steel placed on top of the boat is 0.000956 kg or 0.956g
To solve this problem, we need to determine the volume of the toy boat and the volume of water it displaces when floating.
Since the boat floats with 10% of its volume above the water surface, we can calculate the total volume of the boat using the given density.
Density is defined as mass divided by volume.
Rearranging the equation, we can find the volume of the boat:
Volume of the boat = Mass of the boat / Density of the boat
Volume of the boat = 4.78 kg / (0.5 g/cm³ * 1000 cm³/g) [Converting the density from g/cm³ to kg/cm³]
Volume of the boat = 4.78 kg / 0.5 kg/cm³
Volume of the boat = 9.56 cm³
Now, since 10% of the boat's volume is above the water surface, we can calculate the volume of water it displaces:
Volume of water displaced = 10% * Volume of the boat
Volume of water displaced = 10% * 9.56 cm³
Volume of water displaced = 0.1 * 9.56 cm³
Volume of water displaced = 0.956 cm³
To find the mass of the steel placed on top of the boat, we need to consider that the density of water is 1 g/cm³.
Mass of steel = Density of water * Volume of water displaced
Mass of steel = 1 g/cm³ * 0.956 cm³
Mass of steel = 0.956 g
Since we want the mass in kilograms, we convert grams to kilograms:
Mass of steel = 0.956 g / 1000 g/kg
Mass of steel = 0.000956 kg
Therefore, the mass of steel placed on top of the boat is 0.000956 kg or .0956g (rounded to three decimal places).
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c) what is the angle of incidence from glass when the reflected light in glass is linearly polarized?
The angle of incidence from the glass when the reflected light in the glass is linearly polarized is called Brewster's angle.
Brewster's angle is the angle of incidence at which light is polarized when it is reflected from a transparent surface, such as glass. The reflected light at this angle is entirely polarized and has no parallel component.What is polarization of light?When light waves propagate through space, the electric and magnetic fields at each point in the wave can oscillate in different directions.
The polarization of light is the orientation of the electric field vector that produces the electromagnetic wave as it propagates. The reflected light from a transparent surface, such as glass, is entirely polarized when the angle of incidence is Brewster's angle. When the angle of incidence is greater than Brewster's angle, both polarizations are reflected, and the reflected light is no longer linearly polarized.
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A tank contains 100 kg of salt and 2000 L of water. Pure water enters a tank at the rate 12 L/min. The solution is mixed and drains from the tank at the rate 13 L/min.
Let y be the number of kg of salt in the tank after t minutes.
The differential equation for this situation would be:
dy
y(0) =
A tank contains 60 kg of salt and 1000 L of water. A solution of a concentration 0.03 kg of salt per liter enters a tank at the rate 9 L/min. The solution is mixed and drains from the tank at the same rate.
Let y be the number of kg of salt in the tank after t minutes.
Write the differential equation for this situation
dy =
y(0) = 60
y' + ty^1/3 = tan(t), y(3) = – 5
a) Rewrite the differential equation, if necessary, to obtain the form y' = f(t, y)
F(t, x) = _______
b) Compute the partial derivative of f with respect to y. Determine where in the ty-plane both f(t, y) and its derivative are continuous.
c) Find the largest open rectangle in the ty-plane on which the solution of the initial value problem above is certain to exist for the initial condition. (Enter oo for infinity)
t interval is
y interval is
The interval of y is (-5,∞), as the solution exists until the concentration of salt reaches zero. The solution to the differential equation is given by,dy/dt = rate in - rate out. There is initially 100 kg of salt and 2000 L of water in the tank.
This means that the initial concentration of salt in the tank is:
100 kg / (1000 L + 2000 L)
= 0.03333 kg/L
As pure water is added to the tank at the rate of 12 L/min, and the mixed solution is drained out of the tank at the rate of 13 L/min. The volume of the solution in the tank after t minutes is given by,
(2000 + 12t) - 13t = 2000 - t.
The concentration of salt in the tank after t minutes is given by,
y = (100 - t)(0.03333)
The differential equation for this situation would be,
dy/dt = (0.03)(9) - (y/2000)(13)dy/dt + (13/2000)y
= 0.27
Rewrite the differential equation, if necessary, to obtain the form
y' = f(t, y):
dy/dt + (13/2000)y
= 0.27 - y(0)
= 60
The given differential equation is,
y' + ty^(1/3) = tan(t),
y(3) = -5a)
Rewriting the differential equation, we gety' = -ty^(1/3) + tan(t) - y...[1]b) The partial derivative of f with respect to y is,
df/dy = (1/3)ty^(-2/3) - 1
f and its derivative are continuous everywhere except for y = 0.
c) The largest open rectangle in the ty-plane on which the solution of the initial value problem above is certain to exist for the initial condition is obtained as follows:
The interval of t is (3,∞), as the initial condition is given at t = 3.
The interval of y is (-5,∞), as the solution exists until the concentration of salt reaches zero.
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A 20 kg object has 500 J of potential energy. How far off the ground is this object?
The object is approximately 2.55 meters off the ground.
To determine the height of the objectUtilizing the gravitational potential energy formula
Mass times gravitational acceleration times height equals potential energy
In this instance, the object has a mass of 20 kg and a potential energy of 500 J. On Earth, the gravitational acceleration is roughly 9.8 m/s².
Rearranging the formula, we can solve for the height:
height = Potential Energy / (mass * gravitational acceleration)
Substituting the values into the equation:
height = 500 J / (20 kg * 9.8 m/s²)
height ≈ 2.55 meters
Therefore, the object is approximately 2.55 meters off the ground.
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Question 4 (1 point) Diffraction gratings provide much brighter interference patterns since more light passes through them compared with double slits. O True O False Question 5 (1 point) A When the th
False statement regarding Diffraction and True statement regarding reflected light
4) False. **Diffraction gratings** do not provide much brighter interference patterns compared to double slits. In fact, more light passes through double slits than through diffraction gratings. Diffraction gratings consist of multiple closely spaced slits that diffract and spread out the light, resulting in individual interference maxima and minima that are less intense. On the other hand, double slits allow more light to pass through, resulting in brighter interference patterns.
5) True. When the thickness of a film in air is such that reflected light undergoes destructive interference, the statement is true. Destructive interference occurs when the path length difference between the two reflected rays is an odd multiple of half the wavelength of light. This leads to the cancellation of certain wavelengths of light, resulting in reduced or no reflected light. Therefore, if the film thickness satisfies the condition for destructive interference, the reflected light will be significantly diminished.
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Answer the following question about expanding Universe.
1. Describe the mechanism of a Type 1a supernova, explain how these have been used to
construct a "Hubble diagram" that extends to large redshifts, and describe what we learn
from it. [5 marks]
2. Does our understanding of the expanding Universe imply that some objects are receding
from us faster than the speed of light? Explain your answer. [3 points]
3. Describe the likely long-term fate of the Milky Way galaxy. [2 marks]
Answer:
2. no. the long - term fate of the Milky way galaxy is subject to the ongoing scientific study, but a number of predictions have been. the most significant of all the predictions is the collision of the Milky way and the Andromeda galaxy in about 4 to 5 billion year, leading to formation of a new large galaxy called Milkomeda. with time the Milky way will undergo stellar evolution, where stars will use up their nuclear energy and change into different stages. the galaxy will also record an increase in the quantity of black holes as well as black holes at it's center. interactions with small galaxies may lead to mergers and growth through a process called Galactic cannibalism. there are uncertainties about the influence of dark matter and it's energy on the-long- term fate of the Milky way.
our understanding will evolve with regards to new scientific discoveries.
3. the universe is expanding , but then this expansion does not have a finite speed or rather in other words it doesn't have any speed. the speed per- unit- distance of this expansion is equivalent to a frequency or an inverse of time. which implies that objects in the universe move at or below the speed of light but not exceeding the speed of light as the speed of light is considered the ultimate speed limit for bodies or objects moving in the universe.
Explanation:
2. a car traveling at 27 m/s runs out of gas while traveling up a slope. if the car coasts 85 m up the slope before starting to roll back down, what is the angle of incline?
The angle of inclination is 25.92°. The gravitational force is equal and opposite to the force component parallel to the inclined surface on a block placed on an inclined surface.
We can find the angle of inclination by using this information. This can be illustrated with the following formula;
mg sin θ = f
Here, m = Mass of car = 1,000 kg
g = acceleration due to gravity = 9.8 ms⁻²
sin θ = Opposite / Hypotenuse
= Height / Length
f = Force component parallel to slope
= Weight × sin θ
= mg sin θ
We'll need to utilize the following data:
Initial velocity, u = 27 m/s
Displacement, S = 85 m
Acceleration, a = -9.8 m/s²
By using the kinematic equation of motion, the time it takes to travel up the slope can be calculated;
v² = u² + 2as,
Where, v = Final velocity = 0 m/s
u = Initial velocity = 27 m/s
a = Acceleration = -9.8 m/s²
s = Displacement = 85 m.
After calculating for t, we can use this time value to calculate the angle of inclination by utilizing the aforementioned formula.
mg sin θ = f
Here, m = 1,000 kg,
g = 9.8 ms⁻²,
f = mgsinθsinθ = f / mg.
Now, solve for f, f = ma
Therefore, f = 1,000 kg × 9.8 m/s² × sin θ
The angle of inclination can now be calculated:
sinθ = f / mg
= 1,000 kg × 9.8 m/s² × sin θ / 1,000 kg × 9.8 m/s²
= sin θ= (u² - v²) / 2as
= (27 m/s)² / 2(-9.8 m/s²)(85 m)
= 25.92°
Therefore, the angle of inclination is 25.92°.
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Force (MxLxT) (MXL T-²) XLXT 7. Force is equal to mass x acceleration and is typically expressed in units of Newtons (kg m s). Acceleration is the rate of change of velocity. If gravitational acceleration is equal to 9.8 ms, then what is the gravitation force experienced by the mass of air in the box from Figure 1 (pg. 5) (see question 5)? (Note: again, this question does not involve a conversion but rather use of an equation) kg m s²- Newtons (N) 9.8m5² (kg) mg-2) Pressure (L¹ x M x T²) 8. Pressure is equal to force divided by the area over which the force is applied and is typically expressed in units of N m2 (Pa). If the box in Figure 1 (pg. 5) rests on the Earth's surface, what is the pressure exerted by the gravitational force over the bottom area of the box equal to 4 m²? Nm² = Pa 9. Pressure on weather maps is usually expressed in units of bars, where one bar (100,000 Pa) approximates the average sea-level pressure (101,325 Pa or nearly 100,000 Pa). Using this and other aids (see appendix), convert the following: 1 mb= Pa 101, 325
The mass of the air is 1000 kg and the gravitational acceleration is 9.8 m/s² so the gravitational force is 9800 N.
How to explain the informationThe gravitational force experienced by the mass of air in the box is equal to the mass of the air times the gravitational acceleration.
The pressure exerted by the gravitational force over the bottom area of the box is equal to the force divided by the area. The force is 9800 N and the area is 4 m², so the pressure is 2450 Pa.
1 mb = 101,325 Pa. This can be found by converting 1 mb to Pa using the conversion factor 1 mb = 101,325 Pa.
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What are advantages of charge-coupled devices (CCDs) over photographic film? Choose all that apply. a. The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly reduces exposure times. b. CCD chips can be designed to detect wavelengths of infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light. c. CCD chips can be designed to block out infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light. d.The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly increases exposure times. e. CCD chips can be designed to detect wavelengths of X-ray and microwave light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light.
Charge-coupled devices (CCDs) are electronic devices used for detecting electromagnetic radiation (light) and converting it into an electrical signal. The advantages of CCDs over photographic film include the following:
(a) The large CCD cameras that astronomers use can collect almost all the photons of incoming light that strike the chip, which greatly reduces exposure times. This option a is correct.
(b) CCD chips can be designed to detect wavelengths of infrared and ultraviolet light, allowing astronomers to gather data on some astronomical objects that are more visible at these wavelengths than in visible light is also correct. Thus option b is correct. Additionally, option d is incorrect because large CCD cameras can't increase exposure times but instead, reduce them. Option e is incorrect because CCD chips are not designed to detect wavelengths of X-ray and microwave light but instead for infrared and ultraviolet light as stated in option b. Thus, the options that are correct include options a and b.
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1.
(a) What are the period and amplitude of the function f(x) = sin(x)?
(b) What are the period and amplitude of the function g(x) = 5 sin(3x)?
(c)What are he period and amplitude of the function h(x) = 2 sin(x)?
2. A point starts at the point (3,0) on a circle centered at the origin and travels counter clockwise at a constant angular speed of 2 radians per second. Let t represent the number of seconds since the point started moving.
(a) Write an expression in terms of t to represent the number of radians the point has swept out since the point started moving.
(b)Write a formula that expresses the x-coordinate of the point in terms of the number of seconds t since the point started moving.
(c) Write a formula that expresses the y-coordinate of the point in terms of the number of seconds t since the point started moving.
(a) Period of the function f(x) = sin(x):`2π`. The amplitude of the function f(x) = sin(x) is 1.
(b) Period of the function g(x) = 5 sin(3x): `(2π)/3`. The amplitude of the function g(x) = 5.
(c) Period of the function h(x) = 2 sin(x): `2π`. The amplitude of the function h(x) = 2. 2.
(a) Expression in terms of t to represent the number of radians the point has swept out since the point started moving is `2tπ`.
(b) The formula that expresses the x-coordinate of the point in terms of the number of seconds t since the point started moving is `x = r cos(2tπ/T)` where r is the radius of the circle, and T is the period of rotation which is `2π/2π=1`second.
Substituting the given values: `x = 3 cos(2tπ)`.
(c) The formula that expresses the y-coordinate of the point in terms of the number of seconds t since the point started moving is `y = r sin(2tπ/T)` where r is the radius of the circle, and T is the period of rotation which is `2π/2π=1`second.
Substituting the given values: `y = 3 sin(2tπ)`.
The period and amplitude of the functions f(x), g(x), and h(x) are given as:(a) Period of f(x) = sin(x): 2π, amplitude = 1 Period of g(x) = 5sin(3x): `(2π)/3`, amplitude = 5Period of h(x) = 2sin(x): 2π, amplitude = 2 (b) The x-coordinate of the point in terms of t is x = 3 cos(2tπ). (c) The y-coordinate of the point in terms of t is y = 3 sin(2tπ).
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you are driving a vehicle at 55 mph on dry pavement, about how much total stopping distance do you need to bring it to a stop?
When driving at 55 mph on dry pavement, you'll need approximately 420 feet of total stopping distance to bring your vehicle to a stop.
The total stopping distance that you need to bring a vehicle to a stop when driving at 55 mph on dry pavement depends on many variables, including the following:
Reaction time, vehicle weight, braking system efficiency, tires, road conditions, and so on.
According to a generally accepted formula, the total stopping distance can be calculated as follows:
TSD = (DR + BD) + (RD × DD)Where:TSD = Total Stopping DistanceDR = Distance Traveled during Driver Reaction TimeBD = Braking Distance (Distance required to stop a vehicle, from the point where brakes are applied)RD = Rolling Distance (Distance traveled by the vehicle while its brakes are in operation)
DD = Distance Lost Due to Perception Time or Delay Time
For a vehicle traveling at 55 mph on dry pavement, the driver's reaction time is around 1.5 seconds (according to some sources).
DR is estimated to be approximately 203 feet (60 m), which is equivalent to 1.5 seconds of time at 55 mph. If the brakes on the vehicle are well-maintained, the BD for a typical car is around 216 feet (66 m).
Rolling distance is determined by the friction between the tires and the road surface and is determined by the tire tread and type, as well as the road surface texture.
It ranges from 155 to 175 feet (47 to 53 m) on dry pavement in most cases. If we use the upper limit of 175 feet for rolling distance, the total stopping distance can be calculated as follows:TSD = (DR + BD) + (RD × DD)TSD = (203 + 216) + (175 × 1.5)TSD = 419.5 feet
Therefore, when driving at 55 mph on dry pavement, you'll need approximately 420 feet of total stopping distance to bring your vehicle to a stop.
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(a) A 19.0 kg child is riding a playground merry-go-round that is rotating at 40.0 rev/min. What centripetal force (in N) must she exert to stay on if she is 2.50 m from its center? (Enter a number.)
A 19.0 kg child is rotating at 40.0 rev/min in a merry-go-round. The centripetal force that the child must exert to stay on the merry-go-round is 832.8 N.
The formula to determine the centripetal force,
Fc = mv^2/r
where,
m is the mass,
v is the velocity,
r is the radius of the rotation
Here,
mass of the child = 19 kg
distance of the child from the center of the merry-go-round = 2.5 m
The velocity can be determined using the given frequency of 40.0 rev/min, which is the same as 40.0/60 = 0.67 revolutions per second.
The circumference of the circular path is given by 2πr.
Therefore, the velocity v is given by,
v = 2πr * f
where,
f is the frequency
v = 2π * 2.5 * 0.67 = 10.5 m/s
Substituting the given values in the formula to determine the centripetal force, we get,
Fc = mv^2/r
Fc = 19 * 10.5^2/2.5
Fc = 832.8 N
Therefore, the centripetal force that the child must exert to stay on the playground merry-go-round is 832.8 N.
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a rock hits the ground at a speed of 15 m/s and leaves a hold 50 cm deep. after it hits the ground, what is the magnitude of the rock's (assumed) uniform acceleration?
The magnitude of the rock's (assumed) uniform acceleration is v² - 225.
Initial speed, u = 15 m/s
Displacement, s = 50 cm = 0.5 m
Magnitude of acceleration, a = ?
We know, v² - u² = 2as
Let's substitute the given values into the above formula. v² - u² = 2as (v is the final velocity)
Final velocity, v = ?u = 15 m/s (Initial velocity)
s = 0.5 m (Displacement)
a = ?
v² - u² = 2as (v² - u²)/2s = a(v+u)/2(a = (v² - u²)/2s)
(a = (v² - u²)/2s)(a = (v² - (15 m/s)²)/2(0.5 m))(a = (v² - 225)/1)(a = v² - 225)
Therefore, the magnitude of the rock's uniform acceleration is v² - 225, given that a rock hits the ground at a speed of 15 m/s and leaves a hold 50 cm deep after it hits the ground.
The magnitude of the rock's uniform acceleration is v² - 225.
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Penetration capabilities in...
- Radio Waves
- Microwaves
- Infrared
- Visible light
- Ultraviolet
- X-rays
- Gamma rays
Ultra = X-ray
Explanation:
how are things going to paint the roof for beginners painting and decorating the whole house so I'm going away with a few sports mates for a couple nights for beginners but will have a look painting at a time and see what is going to be Strong enough
1. A boy lifts a 5.1-kg block vertically 6.0 m at constant speed. The work done (in Joules) by the boy is Round of your answer to 1 decimal place. Do not include the unit.
2. A 54.2 kg diver jumps from a height of 2.2 m with an initial speed of 2.5 m/s. What is his speed (in m/s) entering the water?
(1)A boy lifts a 5.1-kg block vertically 6.0 m at constant speed the work done by the boy is approximately 299.9 J.(2) the speed of the diver entering the water is approximately 2.5 m/s.
1): The work done by the boy can be calculated using the formula:
Work = Force × Distance
Since the block is lifted vertically at a constant speed, the force applied by the boy must be equal to the weight of the block.
Weight = mass × acceleration due to gravity
Weight = 5.1 kg × 9.8 m/s^2 (acceleration due to gravity)
Weight ≈ 49.98 N
Therefore, the work done by the boy is:
Work = Force × Distance
Work = 49.98 N × 6.0 m
Work ≈ 299.9 Joules
Rounded to 1 decimal place, the work done by the boy is approximately 299.9 J.
(2) To find the speed of the diver entering the water, we can use the principle of conservation of energy. The initial potential energy of the diver at the top of the dive can be converted into kinetic energy just before entering the water.
The potential energy at the top of the dive is given by:
Potential energy = mass × gravity × height
Potential energy = 54.2 kg × 9.8 m/s^2 × 2.2 m
Potential energy ≈ 1198.36 J
The initial kinetic energy just before entering the water can be calculated as:
Kinetic energy = 0.5 × mass × velocity^2
Kinetic energy = 0.5 × 54.2 kg × (2.5 m/s)^2
Kinetic energy ≈ 169.75 J
According to the conservation of energy, the potential energy at the top should be equal to the kinetic energy just before entering the water. Therefore:
Potential energy = Kinetic energy
1198.36 J = 169.75 J
To find the speed (velocity) of the diver, we can rearrange the equation:
Velocity^2 = (2 × Kinetic energy) / mass
Velocity^2 = (2 × 169.75 J) / 54.2 kg
Velocity^2 ≈ 6.254
Taking the square root of both sides, we find:
Velocity ≈ 2.5 m/s
Therefore, the speed of the diver entering the water is approximately 2.5 m/s.
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When a single charge q is placed on one corner of a square, the electric field at the center of the square is F/q. If three other equal charges are placed on the other corners, the electric field at the center of the square due to these four equal charges is
a) F/(2q)
b) F/(4q)
c) 4F/q
d) F/q
e) zero
Please explain how you came to the correct answer.
The answer to the question is option (c) 4F/q.
The electric field due to a single charge q at the center of a square is given by
E = F/qWhere F = 1/4πε₀ * q / r²,
where r is the distance between the charge and the center of the square. The electric field of the single charge q has a magnitude of F/q when it is at the center of the square.
If three other charges, each of the same magnitude, are placed on the other corners of the square, the resultant electric field at the center of the square is the vector sum of the electric fields due to the four charges.Let the distance between the charges and the center of the square be a.
The force due to each charge is given by
F' = 1/4πε₀ * q / a²
The electric field due to each charge at the center of the square is given by
E' = F'/q = 1/4πε₀ * q / a²q.
The electric field at the center of the square due to these four charges is the vector sum of the electric fields due to the four charges. Since the charges are placed at the corners of a square and are equidistant from the center, the angle between any two fields is 90°.
Hence, the resultant electric field is given by
E = 2E' sin 45° + 2E' sin 135°= 2E' /√2 + 2E' /-√2= 4E' /√2= 4 (1/4πε₀ * q / a²q) /√2= 4F /q.
Hence, the option (c) is the correct answer.
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Consider a cell that is running under standard conditions: nissduni21saqduucu1saqducussd.
a) Is this cell a voltaic or an electrolytic cell, and how can you differentiate between the two?
b) Does current flow spontaneously in this cell under standard conditions?
c) What is the maximum potential of this cell?
d) If the cell is connected to a voltmeter, what would you observe in terms of initial voltage and changes over time?
e) What is the initial free energy of this cell at the point of construction?
f) Does the free energy of the cell change over time as the cell runs, and if so, how does it change?
The given information is incomplete and does not specify whether the cell is voltaic or electrolytic. Differentiating between the two requires understanding their fundamental characteristics.
a) To differentiate between a voltaic and an electrolytic cell, additional information such as the direction of electron flow, the presence of an external power source, and the nature of the electrode reactions is required.
b) The spontaneity of current flow in the cell depends on the overall cell potential, which is not given in the provided information.
c) The maximum potential of the cell cannot be determined without knowledge of the specific redox reactions occurring and the concentrations of species involved.
d) The behavior of the voltmeter connected to the cell would depend on the cell potential and how it changes over time, which is not fundamental provided.
e) The initial free energy of the cell at the point of construction cannot be calculated without more information about the chemical reactions and standard free energy changes.
f) Without further details, it is impossible to determine how the free energy of the cell changes over time as the cell operates. The specific reactions and concentrations involved would be necessary to make such an assessment.
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A particle rotates in a circle with centripetal acceleration a = 6.6 m/s². Part A What is a if the radius is doubled without changing the particle's speed? Express your answer with the appropriate un
A particle rotates in a circle with centripetal acceleration a = 6.6 m/s².if the radius is doubled without changing the particle's speed, the new centripetal acceleration will be 3.3 m/s².
The centripetal acceleration (a) of a particle moving in a circle is given by the equation:
a = v^2 / r
where v is the velocity of the particle and r is the radius of the circle.
If the radius is doubled without changing the particle's speed, it means that the velocity remains constant.
Let's denote the original radius as r₁ and the new radius as r₂ (which is twice the original radius).
Given:
Centripetal acceleration with original radius: a₁ = 6.6 m/s²
Velocity: v (constant)
For the original radius:
a₁ = v^2 / r₁
For the new radius:
a₂ = v^2 / r₂
Since the velocity remains constant, we can equate the two expressions for acceleration:
a₁ = a₂
v^2 / r₁ = v^2 / r₂
To solve for a₂, we can substitute r₂ = 2r₁:
a₂ = v^2 / (2r₁)
Thus, if the radius is doubled without changing the particle's speed, the new centripetal acceleration (a₂) will be half of the original acceleration (a₁):
a₂ = 6.6 m/s² / 2 = 3.3 m/s²
Therefore, if the radius is doubled without changing the particle's speed, the new centripetal acceleration will be 3.3 m/s².
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A jumbo jet of mass 4 x 10² kg travelling at a speed 5000 m/s lands on the airport. It takes 2 minutes t come to rest. Calculate the average force applied by th ground on the aeroplane. 12 Ans: -1.67x107
Answer:
I got 16680N
Explanation:
We can use this formula; F = ma
F = ma
F = (4 × 10²) × ((0-5000)/120)
F = (4 × 10²) × (-41.7)
F = -16680N
Therefore, Fₐᵥ = 16680N.
DRAW
the figure and show complete solution on each problem. please
answer all
IV. Find the momentum of a 60-g bullet whose kinetic energy is 270 J. V. A 60-ton car moving at 1.2 mile per hour is instantaneously coupled to a stationary 40-ton car. What is the speed of the couple
The momentum of the car is 5.7 Kgm/s
The final velocity is 0.3 m/s
What is the momentum?
Momentum is a fundamental concept in physics that describes the motion of an object. It is defined as the product of an object's mass and its velocity.
p = m * v
Where;
KE = 1/2mv^2
v = √2KE/M
v = √2 * 270/0.06
v = 95 m/s
Then;
p = mv
p = 0.06 * 95 m/s
= 5.7 Kgm/s
V = 1.2 m/hr or 0.5 m/s
mass = 60 ton and 40 ton or 54431.1 Kg and 36287.4 Kg
Momentum before collision = Momentum after collision
(54431.1 * 0.5 m/s) + 0 = (54431.1 + 36287.4 )v
v = 27215.55/90718.5
v = 0.3 m/s
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A star is approximately a blackbody. Use Stefan-Boltzmann's law to calculate the power output of a star that has a radius of 695 million meters and a surface temperature of 5778 K. (Please note that l
A star is approximately a blackbody. Using Stefan-Boltzmann's law, the power output of the star is approximately [tex]1.25 x 10^2^7[/tex]watts.
The power output of a star can be calculated using Stefan-Boltzmann's law, which relates the power emitted by a black body to its temperature and surface area. To calculate the power output, we'll use the given values of the star's radius and surface temperature.
First, let's convert the radius of the star to meters. The given radius is 695 million meters, which is equivalent to 6.95 x [tex]10^8[/tex] meters.
Next, we'll calculate the surface area of the star using the formula for the surface area of a sphere. The surface area (A) is given by A = 4π[tex]r^2[/tex], where r is the radius of the star.
A = 4π(6.95 x [tex]10^8)^2[/tex]
Calculating the value of A:
A ≈ 4 * 3.14159 * (6.95 x [tex]10^8)^2[/tex]
A ≈ 4 * 3.14159 * 4.82 x [tex]10^1^7[/tex]
A ≈ 7.28 x [tex]10^1^8[/tex] square meters
Now, let's convert the surface temperature of the star to Kelvin. The given surface temperature is 5778 K.
We can now calculate the power output (P) using Stefan-Boltzmann's law. The formula is P = σA[tex]T^4[/tex], where σ is the Stefan-Boltzmann constant (approximately 5.67 x [tex]10^-^8[/tex] [tex]W/m^2K^4[/tex]).
P = (5.67 x [tex]10^-^8[/tex]) * (7.28 x [tex]10^1^8)[/tex] * ([tex]5778^4[/tex])
Calculating the value of P:
P ≈ (5.67 x[tex]10^-^8[/tex]) * ([tex]7.28 x 10^1^8[/tex]) * ([tex]3.394 x 10^1^6[/tex])
P ≈ [tex]1.25 x 10^2^7[/tex] watts
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The half-life of radium-226 is 1590 years. (a) A sample of radium-226 has a mass of 100mg. Find a formula for the mass of the sample that remains after t years. (b) Find the mass after 1000 years correct to the nearest milligram. (c) When will the mass be reduced to 30mg ?
The formula for the mass remaining after t years for a sample of radium-226 with an initial mass of 100mg is given by [tex]$M(t) = 100 \times 0.5^{t/1590}$[/tex]. After 1000 years, the mass is approximately 87mg. The mass will be reduced to 30mg after approximately 2167 years.
(a) The decay of radium-226 follows an exponential decay model, where the amount of radium remaining decreases by half every 1590 years. The formula for the mass remaining after t years can be derived using the half-life concept. Let M(t) represent the mass remaining after t years, then the equation can be written as [tex]$M(t) = 100 \times 0.5^{t/1590}$[/tex]. Here, 100 represents the initial mass of the sample, and 0.5 is the decay constant derived from the half-life.
(b) To find the mass after 1000 years, we substitute t = 1000 into the formula: [tex]$M(1000) = 100 \times 0.5^{1000/1590}$[/tex]. Evaluating this expression gives us approximately 87mg.
(c) To determine when the mass will be reduced to 30mg, we need to solve the equation [tex]$M(t) = 30$[/tex] for t. Substituting M(t) and rearranging the equation gives us [tex]$100 \times 0.5^{t/1590} = 30$[/tex]. Solving this equation, we find t ≈ 2167 years. Therefore, it will take approximately 2167 years for the mass of the radium-226 sample to be reduced to 30mg.
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car mass 1000kg is travelling along a straight horizontal road at a speed of 20m/s when it brakes sharply then skids. Friction brings the car to
,rest. If the friction force between the tires and road is 9000N. Calculate the distance travelled by car befor it comes to rest
A car mass 1000kg is traveling along a straight horizontal road at a speed of 20m/s when it brakes sharply then skids. Friction brings the car to,rest. If the friction force between the tires and road is 9000N. The distance traveled by the car before it comes to rest (while skidding due to braking) is approximately 22.22 meters.
To calculate the distance traveled by the car before it comes to rest, we can use the equations of motion.
First, we need to find the acceleration of the car when it brakes sharply. The friction force acting on the car is equal to the product of the mass of the car and its acceleration:
Friction force = mass × acceleration
9000 N = 1000 kg × acceleration
acceleration = 9000 N / 1000 kg
acceleration = 9 m/s^2
Next, we can use the equation of motion that relates initial velocity, final velocity, acceleration, and distance:
v^2 = u^2 + 2as
Where:
v = final velocity (0 m/s, as the car comes to rest)u = initial velocity (20 m/s)a = acceleration (-9 m/s^2, negative because it acts in the opposite direction to the car's motion)s = distance traveledPlugging in the values, we get:
0^2 = 20^2 + 2(-9)s
0 = 400 - 18s
18s = 400
s = 400 / 18
s ≈ 22.22 m
Therefore, the distance traveled by the car before it comes to rest (while skidding due to braking) is approximately 22.22 meters.
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I n it in 20 minutes
JL.61 A producer of refrigerator compressors wants to implement
a just-in-time production line to support demand from a neighboring
appliance manufacturer. Demand from the applian
The producer of refrigerator compressors needs to determine the number of kanbans required for implementing a just-in-time production line to meet the demand for 150 compressors per day from a neighboring appliance manufacturer.
To calculate the number of kanbans required, we need to consider the production lead time, safety stock factor, and optimal production quantity. The production lead time is 5 days, which means that it takes 5 days to produce a batch of compressors once the production process starts.
The safety stock factor is 17%, indicating that the producer wants to maintain an additional 17% of the daily demand as safety stock to mitigate any unforeseen fluctuations. The optimal production quantity is 95 units, which is the batch size that minimizes setup costs.
To determine the number of kanbans, we first need to calculate the total demand during the production lead time. Since the demand is 150 compressors per day and the lead time is 5 days, the total demand during the lead time is 150 compressors/day * 5 days = 750 compressors. Adding the safety stock, the total demand becomes 750 compressors + (17% * 150 compressors) = 750 compressors + 25.5 compressors = 775.5 compressors.
Next, we divide the total demand by the optimal production quantity to get the number of kanbans required. The number of kanbans is calculated as 775.5 compressors / 95 compressors per kanban = 8.16 kanbans. Since we cannot have a fraction of a kanban, we round up to the nearest whole number. Therefore, the producer of compressors requires 9 kanbans to meet the demand of 150 compressors per day from the neighboring appliance manufacturer.
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The complete question is:
A producer of refrigerator compressors wants to implement a just-in-time production line to support demand from a neighboring appliance manufacturer. Demand from the appliance manufacturer is for 150 compressors a day. The production lead time is 5 days and the producer wants to have a 17% safety stock factor. This producer has also cut setup costs such that the optimal production quantity is 95 units. How many kanbans does this producer of compressors require?
Cosmos: The Electric Boy Assignment
1. What was Humphry Davy's experiment and how did it go wrong? What could he have done differently?
2. What did Humphry Davy notice about a wire with electricity running through it as he brought it near a compass?
3. What was Humphry Davy's next project for Michael Faraday and why did he give him that particular project?
4. What did Michael Faraday create as a result of his efforts? How did it work?
5. What did Michael Faraday notice when he moved a magnet in and out of a wire?
6. What were some of the materials Michael Faraday used to see if light would be affected by magnets? What ended up working in the end? What did it mean?
7. What did Michael Faraday notice when he sprinkled iron filings around current carrying wires? What did he think ultimately meant?
8. Why did Michael Faraday's contemporaries in science not believe his hypothesis about field forces? What did he need in order to convince them?
9. How do the effects of Michael Faraday's invention shape society even today?
Humphry Davy's most famous experiment involved the use of a voltaic pile (an early battery) to conduct electrolysis on various substances.
1. Humphry Davy's most famous experiment involved the use of a voltaic pile (an early battery) to conduct electrolysis on various substances. One of his experiments involved passing an electric current through a mixture of potassium and water. The experiment went wrong when he increased the power of the battery, causing a violent explosion due to the release of hydrogen gas. The explosion was caused by the high reactivity of potassium with water. To prevent this mishap, Davy could have used a smaller amount of potassium or diluted the solution to reduce the reactivity.
2. When Humphry Davy brought a wire with electricity running through it near a compass, he noticed that the needle of the compass was deflected from its usual north-south orientation. This observation indicated that an electric current produces a magnetic field around the wire, leading to the deflection of the compass needle.
3. Humphry Davy assigned Michael Faraday the task of finding a way to liquefy chlorine gas. He gave Faraday this project because he recognized Faraday's experimental skills and believed that his ingenuity and dedication would lead to a successful outcome.
4. As a result of his efforts, Michael Faraday succeeded in liquefyingchlorine gas and several other gases. He developed a method using high pressure and low temperatures to condense the gases into liquid form. This breakthrough allowed for further investigation and study of these substances.
5. Michael Faraday noticed that when he moved a magnet in and out of a wire, it induced an electric current in the wire. This phenomenon is known as electromagnetic induction, and it demonstrated the relationship between magnetism and electricity.
6. Michael Faraday experimented with various materials to test their response to magnetic fields. He tried substances such as glass, copper, and sulfur, but they did not show any significant effects. However, when he used a coil of wire, he observed that a current was induced in the wire when exposed to a changing magnetic field. This discovery led to the development of the concept of electromagnetic induction and its practical applications.
7. When Michael Faraday sprinkled iron filings around current-carrying wires, he observed that the filings arranged themselves in a pattern, forming circles around the wires. He realized that these patterns represented the lines of magnetic force around the wires. This finding suggested the existence of magnetic fields and provided evidence for Faraday's theory of field forces.
8. Michael Faraday's contemporaries in science initially did not believe his hypothesis about field forces because it went against the prevalent understanding of action at a distance. They adhered to the idea that forces acted only through direct contact between objects. To convince them, Faraday needed to provide experimental evidence and develop a coherent theoretical framework to explain the observed phenomena. He achieved this through his extensive experiments and the formulation of field theory, which established the concept of field forces acting at a distance.
9. Michael Faraday's inventions and discoveries in electromagnetism and electrochemistry have shaped society to this day. They laid the foundation for the development of modern electrical technology and power generation. Faraday's work led to the invention of electric motors, generators, and transformers, which are essential components of our electrical infrastructure. His principles of electromagnetic induction and field theory also underpin technologies such as wireless communication, electric lighting, and the functioning of modern electronics. Additionally, Faraday's emphasis on experimental investigation and his dedication to sharing scientific knowledge contributed to the advancement of scientific methodology and the popularization of science education.
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5. The East Campus Provost decides to order a new rope for the flagpole. To find out what length of rope is needed, the provost observes that the pole casts a shadow 14.6 meters long. The angle the su
The length of the rope required for the flagpole is approximately 16.8 meters.
Let 'l' be the length of the rope required for the flagpole, 'h' be the height of the flagpole, and 'θ' be the angle between the flagpole and the ground. It is given that the shadow cast by the flagpole is 14.6 meters long. Hence, using trigonometry, we get:tan θ = h/ltan θ = (14.6/l)l = 14.6/tan θHere, θ = 55° (approx). Hence,l = 14.6/tan 55°= 16.8 meters (approx).Therefore, the length of the rope required for the flagpole is approximately 16.8 meters. The angle that the sun makes with the ground is 35 degrees since it is given that the pole casts a shadow 14.6 meters long.
Distance is measured in length. A quantity with the dimension distance is length in the International System of Units of Measurement. The majority of measurement systems have a base unit for length from which all other units are derived. The meter is the base unit for length in the International System of Units (SI).
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use kepler's law, which states that the square of the time, t, required for a planet to orbit the sun varies directly with the cube of the mean distance, a, that the planet is from the sun. using the earth's distance of 1 astronomical unit (a.u.), determine the time, in earth years, for a planet to orbit the sun if its mean distance is 9.58 a.u. (round your answer to two decimal places.) t
The time for the planet to orbit the Sun is 1 Earth year. use kepler's law, which states that the square of the time, t, required for a planet to orbit the sun varies directly with the cube of the mean distance.
Using Kepler's law, which states that the square of the time required for a planet to orbit the Sun varies directly with the cube of the mean distance, we can determine the time for a planet with a mean distance of 9.58 astronomical units (a.u.) to orbit the Sun.
According to Kepler's law, the relationship period between the time (t) and the mean distance (a) is expressed as:
[tex]t^2 = k * a^3[/tex]
where k is a constant.
Given that the mean distance of the planet is 9.58 a.u. and the Earth's distance is 1 a.u., we can set up the following equation:
(1)² = k × (9.58)³
Simplifying the equation, we have:
1 = k × (9.58)³
To solve for the constant k, we divide both sides by (9.58)³:
k = 1 / (9.58)³
Now, we can substitute the mean distance of the planet (9.58 a.u.) into the equation to calculate the time (t):
t² = (1 / (9.58)³) × (9.58)³
Simplifying further, we get:
t² = 1
Taking the square root of both sides, we find:
t = 1
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a 500 g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle the tension in the string is 15 N. What is the speed of the ball at that point?
A 500 g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle the tension in the string is 15 N. the speed of the ball at the bottom of the vertical circle is approximately 9.49 m/s.
To determine the speed of the ball at the bottom of the vertical circle, we can make use of the tension in the string and the gravitational force acting on the ball. At the bottom of the circle, the tension in the string provides the centripetal force required to keep the ball moving in a circular path.
The centripetal force Is given by the formula:
F_c = m * (v^2 / r)
Where F_c is the centripetal force, m is the mass of the ball, v is the velocity of the ball, and r is the radius of the circle (1.5 m).
In this case, the tension in the string (F_c) is given as 15 N. Therefore, we can set up the equation:
15 N = (0.5 kg) * (v^2 / 1.5 m)
Simplifying the equati the speed of the ball at the bottom of the vertical circle is approximately 9.49 m/s.on, we find:
V^2 = (15 N * 1.5 m) / 0.5 kg
V^2 = 45 N*m / 0.5 kg
V^2 = 90 m^2/s^2
Taking the square root of both sides of the equation, we obtain:
V = √(90 m^2/s^2) ≈ 9.49 m/s
Hence, At this point, the tension in the string provides the necessary centripetal force to keep the ball moving in its circular path.
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