Answer:
a) W = 139.4 kJ
b) W = 200J
Explanation:
a) given;
Force F = 410kN
distance d = 34cm = 0.34m
Workdone = force × distance = Fd
Substituting the values;
Workdone = 410kN × 0.34m = 139.4 kJ
b) given;
Force F = 4000N
distance d = 5cm = 0.05m
Workdone = force × distance = Fd
Substituting the values;
Workdone = 4000N × 0.05m = 200J
A 50-kilogram student is running and has 225 joules of kinetic energy.What is the students speed in meters per second
Answer:3m/s
Explanation:
mass(m)=50kg
Kinetic energy =225 joules
Kinetic energy=(m x (velocity)^2)/2
225=(50 x (velocity)^2)/2
Cross multiplying we get
225x2=50x(velocity)^2
450=50x(velocity)^2
Divide both sides by 50
450/50=(50x(velocity)^2)/50
9=(velocity)^2
Taking the square root of both sides we get
√9=velocity
3=velocity
Velocity=3m/s
A 65-kg skier grips a moving rope that is powered by an engine and is pulled at a constant speed to the top of a 230 hill. The skier is pulled a distance x = 320 m along the incline and it takes 2.0 minutes to reach the top of the hill. If the coefficient of kinetic friction between the snow and skis is µk = 0.10, what horsepower engine is required if 30 such skiers are on the rope at one time? 1 hp = 746 W
Answer:
The required power by the engine is 33.0 hp
Explanation:
Solution
Newton's second law says that, the net force Fnet on an object of mass m will accelerates the object
Where
Fnet = ma
a = acceleration
θ = angle of incline,
m = mass of the 30 skiers,
f = frictional force
N = normal force
mg sinθ, mg cos θ are components of weight skier
F = the force applied by engine
Now,
The skier mass is 65 kg
We calculate the mass of the 30 skier
m = 30 (65kg) = 1950 kg
Calculate the net force acting on the skiers along the x-axis
Fnet, x=ma
Now,
F-mg sin θ - f = 0
F= mg sin θ + f -----(1)
The kinetic frictional force is denoted by
f = μk N ------(2)
μk = The coefficient of the kinetic friction
We now, calculate the net force acting on the skiers along y axis
Fnet, y = ma
N- mg cos θ = 0
so,
N = mg cos θ
This value is substituted in equation (2)
f = μk mg cos θ
we substitute the value for equation (1)
F = mg sin θ + μk mg cos θ
mg = sinθ + μk cos θ)-----(3)
The next step is to calculate the work done by the engine in pulling the skiers, the incline top by applying the equation 3
W = Fx
= mg ( sinθ + μk cos θ)x
x = the displacement
we now substitute 1950 kg for m, 23° for θ, 0.10 for μk and 320m for x
so,
W = mg ( sinθ + μk cos θ)x
= (1950 kg) (9.81 m/s²) (sin 23° + (0.10) cos 23°) (320 m)
= 2.99 * 10 ^6 J
Then,
The time from minute to s is converted
t =(2.0min) ( 60sec/1.0min) = 120 sec
Now we calculate the power needed by the engine to pull the skiers at the incline top
Thus,
P = W/t
we substitute 2.955 * 10 ^6 J for W and 120 s for t
we have,
P = 2.955 * 10 ^ 6 J/ 120 s
= ( 2.4625 * 10 ^ 4 W) (1.0 hp/746 W)
= 33.0 hp
In conclusion the required power by the engine is 33.0 hp
To calculate the horsepower of the engine required for 30 skiers to be pulled up a hill, we need to consider the work done against friction and gravity. By plugging in the given values into the appropriate equations, we can find the required horsepower of the engine.
Explanation:To calculate the horsepower of the engine required for 30 skiers to be pulled up a hill, we need to consider the work done against friction and gravity. First, we find the net force of the skier parallel to the incline by subtracting the force due to friction from the component of the skier's weight parallel to the incline. Then, we calculate the work done against friction and gravity using the formula W = Fd. Finally, we convert the work done to horsepower using the conversion factor 1 hp = 746 W.
The work done against friction and gravity is:
W = F_net * d
The power required is:
P = W / t
And finally, we convert the power to horsepower:
Power (hp) = P / 746
By plugging the given values into these equations and multiplying the final power by 30 (for 30 skiers), we can find the required horsepower of the engine.
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A biker can ride at 12 m/s on a level road when there is no wind and at 7.5 m/s on a level road when there is a head wind of 5 m/s. Assume that fluid drag and resistance due to friction are the only forces acting on the biker. The friction force is given by the product of rolling resistance and bike velocity. The density of ambient air is 1.2 kg/m3 If the biker delivers 10.2500 W of power while riding the bike, determine the value of LaTeX: C_D A
Answer:
The answer is "[tex]1.94 \ m^2[/tex]".
Explanation:
Formula:
[tex]F= \frac{1}{2} \times C_D \times density \times area \times velocity^2\\\\Power (P) = F \times velocity \\\\P = \frac{C_D}{2} \times density \times area \times velocity^3\\[/tex]
Given value:
[tex]\ P = 10.25 \ W \\\\\ density = 1.2 \ kg/m^3 \\\\\ velocity= 2.5 \\\\[/tex]
[tex]10.25 = \frac{C_D A}{2} \times 1.2 \times 2.5^3\\\\C_D A= \frac{10.25 \times 2 }{1.2 \times 2.5^3}\\\\C_D A = 1.94 m^2\\[/tex]
Which method is used to calculate the slope on a distance time graph
The slope of the line is calculated by dividing the change in distance over the change in time.
A certain sound is recorded by a microphone. The same microphone then detects a second sound, which is identical to the first one except that the amplitude of the pressure fluctuations is larger. In addition to the larger amplitude, what distinguishes the second sound from the first one? View Available Hint(s) A certain sound is recorded by a microphone. The same microphone then detects a second sound, which is identical to the first one except that the amplitude of the pressure fluctuations is larger. In addition to the larger amplitude, what distinguishes the second sound from the first one? It is perceived as higher in pitch. It is perceived as louder. It has a higher frequency. It has a longer wavelength.
Answer:
It is perceived as louder.
Explanation:
Amplitude affects the loudness of sound.
Frequency and wavelength affect the pitch of the sound.
Two electrical motors, one with a power output of 250 watts and one with a power output of 500 watts are both connected to a standard home outlet. Which motor draws greater current?
Answer:
The 500 watts motor
Explanation:
Power is the workdone per unit time, it can be expressed as;
P = VI
P = I^2 × R
Where;
P = power
V = potential difference
I = current
R = resistance
Making current the subject of formula;
I = P/V
I^2 = P/R
Since they are both connected to the same standard home outlet. The potential difference across them is thesame( V is constant). Therefore the higher the power the higher the current.
I ∝ P
So the electrical motors with higher power 500 watts draws greater current.
(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2l. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.44 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.
Answer:
The diameter of the second pipe is [tex]D_2 = 1.095 D[/tex]
Explanation:
From the question we are told that
The length of the connected pipe is [tex]d = 2L[/tex]
The pressure drop for the first pipe is [tex]\Delta p __{1}} = 1.44* \Delta p__{2}}[/tex]
The diameter of the pipe is [tex]D[/tex]
The rate at which the fluid flows for laminar flow is mathematically represented as
[tex]\r m = \frac{\pi D^4 \Delta p}{128 \mu L}[/tex]
Where L is the length of the pipe
[tex]\mu[/tex] is the dynamic viscosity
[tex]\Delta p[/tex] is the difference in pressure
[tex]\r m[/tex] is the flow rate of the fluid
From the equation of continuity
[tex]\r m_ 1 = \r m_2[/tex]
Where [tex]\r m_1[/tex] is the flow rate in pipe one
[tex]\r m_2[/tex] is the flow rate in pipe two
So
[tex]\frac{\pi D^4 \Delta p_1}{128 \mu L} = \frac{\pi D^4_2 \Delta p}{128 \mu L}[/tex]
Where [tex]D_2[/tex] is the diameter of the second pipe
=> [tex]\frac{\pi D^4 (1.44 \Delta p_2)}{128 \mu L} = \frac{\pi D^4_2 \Delta p}{128 \mu L}[/tex]
=> [tex]1.44 D^4 = D_2 ^4[/tex]
[tex]D_2 =\sqrt[4]{ \frac{D ^4 }{1.44 } }[/tex]
[tex]D_2 = 1.095 D[/tex]
Every chemical element goes through natural exponential decay, which means that over time its atoms fall apart. The speed of each element's decay is described by its half-life, which is the amount of time it takes for the number of radioactive atoms of this element to be reduced by half.
The half-life of the isotope beryllium-11 is 14 seconds. A sample of beryllium-11 was first measured to have 800 atoms. After t seconds, there were only 50 atoms of this isotope remaining, Write an equation in terms of t that models the situation.
The equation modeling the exponential decay of the Beryllium-11 isotope is 50 = 800(1/2)^(t/14). This lets you find out the time that has passed since there were initially 800 atoms.
Explanation:The situation described is an example of exponential decay, modeled by the formula N = N0(1/2)^(t/h). In this case, N0 is the original number of atoms (800), N is the remaining number of atoms (50), t is the time that has passed, and h is the half-life of the isotope (14 seconds for Beryllium-11).
So the equation modelling this scenario is 50 = 800(1/2)^(t/14). This equation will let you solve for the time t (in seconds) that has passed since the sample initially contained 800 atoms of Beryllium-11.
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A steel drill making 180 rpm is used to drill a hole in a block of steel. The mass of the steel block and the drill is 180 g. If the entire mechanical work is used up in producing heat and the rate of rise in temperature of the block and the drill is 0.5 °C/sec. Find (a) the rate of working of the drill in Watts and (b) torque required to drive the drill. Given: specific heat capacity of steel = 420 J/(kgK)
Answer:
a) 37.8 W
b) 2 Nm
Explanation:
180 g = 0.18 kg
We can also convert 180 revolution per minute to standard angular velocity unit knowing that each revolution is 2π and 1 minute equals to 60 seconds
180 rpm = 180*2π/60 = 18.85 rad/s
We can use the heat specific equation to find the rate of heat exchange of the steel drill and block:
[tex]\dot{E} = mc\Delta \dot{t} = 0.18*420*0.5 = 37.8 J/s[/tex]
Since the entire mechanical work is used up in producing heat, we can conclude that the rate of work is also 37.8 J/s, or 37.8 W
The torque T required to drill can be calculated using the work equation
[tex]E = T\theta[/tex]
[tex]\dot{E} = T\dot{\theta} = T\omega[/tex]
[tex]T = \frac{\dot{E}}{\omega} = \frac{37.8}{18.85} = 2 Nm[/tex]
Calculating the rate of working of a drill involves using the specific heat capacity, mass of the block, and rate of temperature increase to find the power. The torque can be derived from this power and the angular velocity of the drill.
The question involves calculating the rate of working of the drill in Watts and the torque required to drive the drill based on the information that all mechanical work is converted to heat that raises the temperature of the steel block and drill at 0.5 0C/s. Given the mass of the steel block and drill is 180 g and the specific heat capacity of steel is 420 J/(kgK), we can use the formula power (P) = mcigtriangleup T/igtriangleup t, where m is the mass, c is the specific heat capacity, is the temperature change, and is the change in time. To find the torque, we can use the power-torque relationship P = au imes w, where P is power in watts, au is the torque in Newton-meters, and w is the angular velocity in radians per second. As the drill makes 180 revolutions per minute (rpm), we'll convert that to radians per second to use in the power-torque equation.
A 60kg60 kg board that is 6 m6 m long is placed at the edge of a platform, with 4 m4 m of its length extending over the edge. The board is held in place by blocks of masses M1M1 and M2M2 placed with their centers of mass on either end. If M2=30kgM2=30kg , what is the minimum value of M1M1 needed to keep the board from falling off the platform? 30kg
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The minimum mass of [tex]M_1 = 90\ kg[/tex] correct option is E
Explanation:
Free body diagram of the set up in the question is shown on the third uploaded image
The mass of board is [tex]M = 60kg[/tex]
The length of the board is [tex]L = 6 \ m[/tex]
The length extending over the edge is [tex]L_e = 4 \ m[/tex]
The second mass is [tex]M_2 = 30kg[/tex]
Now to obtain [tex]M_1[/tex] we take moment about the edge of the platform
[tex]M_1 g L_1 = Mg \frac{L}{2} + M_2 g L_2[/tex]
[tex]M_1 L_1 = M \frac{L}{2} + M_2 L_2[/tex]
Substituting value
[tex]M_1 (2) = (60)(1) + (30)(4)[/tex]
[tex]M_1 = 90 \ kg[/tex]
The minimum value of M1 needed to keep the board from falling off the platform is 90 kg.
From the information given, we are to find:
the mass (M1) of placed on the left side edge of the boardGiven that:
the mass of the board = 60 kgthe length of the board = 6 mIf the mass on the right side = 30 kg, and the length of the board L1 = 2mThen, the length L2 which extend over the edge = 4mConsider the center of gravity in the board that lies at the length of the board midpoint.
Then, the distance (D) of the gravity center from the platform end = 3 - 2
= 1 m
∴
Considering the moment about the platform end, the mass (M1) placed on the left side edge of the board can be computed as:
[tex]\mathbf{M_1gL_1 = MgD + M_2gL_2} \\ \\ \mathbf{M_1L_1 = MD + M_2gL_2} \\ \\ \mathbf{M_1(2) = 60 \ kg \times 1 + 30 \ kg \times (4)} \\ \\ \mathbf{ M_1 =\dfrac{60 \ kg + 120 kg }{2} } \\ \\ \mathbf{ M_1 =\dfrac{180 \ kg}{2} } \\ \\ \mathbf{ M_1 =90 \ kg }[/tex]
Therefore, we can conclude that the minimum value M1 needed to keep the board from falling off the platform is 90 kg.
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A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. The drawing shows a spherical reservoir that contains 7.41 x 105 kg of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter of the delivery pipes.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The gauge pressure that water has at the House A [tex]P_A = 257020.68 Pa[/tex]
The gauge pressure that water has at the House B [tex]P_B = 188454 \ Pa[/tex]
Explanation:
From the question we are told that
The mass of water when full is [tex]m_f = 7.41* 10^{5} kg[/tex]
Generally the volume of water in this tank is mathematically represented as
[tex]V = \frac{m }{\rho}[/tex]
Where [tex]\rho[/tex] is the density of water with a value of with a value of [tex]\rho = 1000 kg /m^3[/tex]
substituting values
[tex]V = \frac{7.41 *10^5}{10^3}[/tex]
[tex]V = 741 m^3[/tex]
This volume is the volume of a sphere since the tank is spherical so
[tex]V = \frac{4 \pi ^3}{3}[/tex]
making r the subject of the formula
[tex]r =\sqrt[3]{ \frac{741 *3 }{4\pi} }[/tex]
[tex]r = 5.6134 m[/tex]
Now we can use this parameter to obtain the diameter
So
[tex]d = 2 * r[/tex]
substituting values
[tex]d = 2 * 5.6134[/tex]
[tex]d = 11.23m[/tex]
The pressure the water has at faucet in House A is mathematically evaluated as
[tex]P_A = \rho g h_A[/tex]
This height is obtained as follows
[tex]h_A = d+ 15[/tex]
The value 15 is gotten from the diagram
so
[tex]h_A = 15 + 11.23[/tex]
[tex]h_A = 26.22 m[/tex]
Now substituting values
[tex]P_A = 26.23 * 9.8 * 1000[/tex]
[tex]P_A = 257020.68 Pa[/tex]
The pressure the water has at faucet in House B is mathematically evaluated as
[tex]P_B = \rho g h_B[/tex]
This height is obtained as follows
[tex]h_B = d+ 15[/tex]
The value 15 is gotten from the diagram
so
[tex]h_B = d + 15 -h[/tex]
substituting values
[tex]h_B =11.23 + 15 -7[/tex]
[tex]h_A = 19.23 m[/tex]
Now substituting values
[tex]P_B = 19.23 * 9.8 * 1000[/tex]
[tex]P_B = 188454 \ Pa[/tex]
1. Electromagnetic radiation is more common than you think. Radio and TV stations emit radio waves when they broadcast their programs; microwaves cook your food in a microwave oven; dentists use X rays to check your teeth. Even though they have different names and different applications, these types of radiation are really all the same thing: electromagnetic (EM) waves, that is, energy that travels in the form of oscillating electric and magnetic fields. Which of the following statements correctly describe the various applications listed above?A. All these technologies use radio waves, including low-frequency microwaves. B. All these technologies use radio waves, including high-frequency microwaves. C. All these technologies use a combination of infrared waves and high-frequency microwaves. D. Microwave ovens emit in the same frequency band as some wireless Internet devices. E. The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies. F. All these technologies emit waves with a wavelength in the range 0.10 to 10.0 m. G. All the technologies emit waves with a wavelength in the range 0.01 to 10.0 km.2. Despite their extensive applications in communication systems, radio waves a not the only form of EM waves present in our atmosphere. Another form of EM radiation plays an even more important role in our life (and the life of our planet) sunlight. The sun emits over a wide range of frequencies; however, the fraction of its radiation that reaches earth's surface is mostly in the visible spectrum. (Note that about 35% of the radiation coming from the sun is absorbed directly by the atmosphere before even reaching the earth's surface). The earth, then, absorbs this radiation and reemits it as infrared waves. Based on this information, which of the following statements is correct? A. The earth absorbs visible light and emits radiation with a shorter wavelength. B. The earth absorbs visible light and emits radiation with a longer wavelength. C. The earth absorbs visible light and emits radiation with a lower frequency. D. The earth absorbs visible light and emits radiation with a higher frequency.
For the first question
Answers:
b.) All these technologies use radio waves, including high-frequency microwaves
d.) Microwave ovens emit in the same frequency band as some wireless Internet devices.
e.) The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies listed above.
f.) All these technologies emit waves with a wavelength in the range of 0.10 to 10.0 m.
For second question
Answer:
B. The earth absorbs visible light and emits radiation with a longer wavelength
C. The earth absorbs visible light and emits radiation with a lower frequency.
Both options are correct
Explanation:
In the EM spectrum, wavelenght reduces with increasing energy and frequency increases with increasing energy. The order of increasing energy is radio wave, microwave, infrared rays, visible rays, ultraviolet rays, xrays, gamma rays.
If the earth absorbs visible rays of higher energy, and radiates infrared rays of lower energy, then this means that the radiated radiation will have a longer wavenght, and a lower frequency.
1) The statements that correctly describe the various applications given in the question are;
B. All these technologies use radio waves, including high-frequency microwaves
D. Microwave ovens emit in the same frequency band as some wireless Internet devices.
E. The radiation emitted by wireless Internet devices has the shortest wavelength of all the technologies.
F. All these technologies emit waves with a wavelength in the range 0.10 to 10.0 m.
2) The correct statements on the EM spectrum in the question are;
B. The earth absorbs visible light and emits radiation with a longer wavelength.
C. The earth absorbs visible light and emits radiation with a lower frequency.
1) We are told about different waves being used by different technologies such as;
- Radio and TV emitting radio waves
- microwave oven using microwaves to cook food
- Dentists checking teeth with x-rays
All these waves are electromagnetic waves.
Looking at the options, options B, D, E and F are true because;
The given technologies all make use of radio waves as well as high frequency microwaves. Also, from research it is known that the frequency band emitted by microwaves is the same as that of some internet devices since some of the waves emitted by internet devices are the shortest wavelengths.The wavelength of the waves emitted by the given technologies are usually within the range of 0.1m to 10m2) The correct options are Options B and C because;
In the EM radiation spectrum, when the energy increases, the wavelength usually reduces and the frequency usually increases. The order of radio waves according to increase in energy is;
radio waves < infrared radiation < visible rays < ultraviolet rays < xrays < gamma rays.
Now, if we say that the earth absorbs visible rays of higher energy, and radiation of infrared rays with lower energy, then from our definition, we can conclude that radiated waves will have a longer wavelength, and a lower frequency.
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The chart shows data for a moving object.
Which conclusion is best supported by the information in
the chart?
The object has negative displacement
The object has negative acceleration
The object does not have displacement
The object is not accelerating
Answer:the object is not accelerating
Explanation:
There object is not accelerating since there's no increase in velocity
Final answer:
Based on the descriptions of graphical data, we can conclude that the object has negative acceleration because its velocity is decreasing over time while it is moving in the positive direction, or its negative velocity is increasing over time. Also, since the object is moving in the negative direction, it experiences a negative displacement.
Explanation:
The question asks about the motion of an object based on different graphical representations. To determine the nature of this motion, we can analyze the provided diagrams. In this case, we have descriptions of how the velocity changes over time, which allows us to infer acceleration, and acceleration is defined as the change in velocity over time. According to the information, the object experiences a decrease in positive velocity, which indicates a negative acceleration, as acceleration is in the opposite direction to the velocity. Furthermore, it is mentioned that the object has a negative velocity which increases in magnitude, coherently resulting in negative acceleration as well.
Acceleration is a vector quantity, meaning it has both direction and magnitude. Since we are discussing a coordinate system where to the right is considered positive, any acceleration towards the left will be considered negative. The graphs that indicate an object having a constant negative velocity also imply that the object has negative displacement over the time period, because it's moving in the negative direction of the coordinate system.
Which plants are usually the first to live in soil?
The plants that are typically the first to live in soil are known as pioneer plants or pioneer species.
What are the first plants?Pioneer plants or pioneer species are the plants that are often the first to live in soil. These plants are distinguished by their propensity to colonize bare or disturbed soil, such as on newly formed land or in the wake of a natural disaster.
The process of primary succession, which is the gradual formation of plant and animal communities in a region devoid of soil or organic matter, depends critically on pioneer plants.
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g A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in a magnetic field so that the normal to the plane of the coil makes an angle of 30.0° with the direction of the field. When the magnetic field is increased from 250 µT to 700 µT in 0.300 s, an emf of magnitude 60.0 mV is induced in the coil. What is the total length of wire in the coil?
Answer:
L = 182.4 m
Explanation:
Given:-
- The number of turns of the coil, N = 50
- The shape of the coil = square
- The angle between the coil and magnetic field, θ = 30°
- The change in magnetic field, ΔB = ( 700 - 250 ) μT
- The time duration in which magnetic field changes, Δt = 0.3 s
- The induced emf, E = 60.0 mV
Solution:-
- The problem at hand is an application of Faraday's law. The law states that the induced emf ( E ) is proportional to the negative rate of change of magnetic flux ( ΔФ / Δt ) and number of turns of the coil ( N ).
- The Faraday's law is mathematically expressed as:
E = - N* ( ΔФ / Δt )
Where,
- The flux ( Ф ) through a current carrying with an cross-sectional area ( A ) at a normal angle ( θ ) to the direction of magnetic field ( B ) is given by the following relationship.
Ф = B*A*cos ( θ )
- We need the rate of change of magnetic flux ( ΔФ / Δt ) for the Faraday's law. I.e the induced emf ( E ) is proportional to rate of change in magnetic field ( ΔB / Δt ), rate of change of angle between the coil and magnetic field ( Δθ / Δt ) or rate of change of cross-sectional area of the coil under the influence of magnetic field.
- To determine the exact relationship. We will derive the multi-variable function of flux ( Ф ) with respect to time "t":
Ф ( B , A , θ ) = B*A*cos ( θ )
- The first derivative would be ( Use chain and product rules )
( ΔФ / Δt ) = ΔB / Δt*A*cos ( θ ) + B*ΔA/Δt*cos ( θ ) - B*A*sin ( θ )*Δθ/Δt
- For the given problem the only dependent parameter that is changing is magnetic field ( B ) with respect to time "t". Hence, ( ΔA/Δt = Δθ/Δt = 0 ):
ΔФ / Δt = (ΔB/Δt)*A*cos ( θ )
- Substitute the rate of change of magnetic flux ( ΔФ / Δt ) into the expression for Faraday's Law initially stated:
E = - N*(ΔB/Δt)*A*cos ( θ )
- Plug in the values and evaluate the Area of the square coil:
A = - E / ( N*(ΔB/Δt)*cos ( θ ) )
A = - 0.06 / ( 50*[ (250-700)*10^-6/0.3 ] *cos ( 30° ) )
A = - 0.06 / -0.07216
A = 0.8314 m^2
- The square coil has equal sides ( x ). The area of a square A is given by:
A = x^2
x = √0.8314
x = 0.912 m
- The perimeter length of a single coil in terms of side length "x" is given as:
P = 4x
Whereas for a coil of N turns the total length ( L ) would be:
L = N*P
L = 4Nx
L = 4 * 50 * 0.912
L = 182.4 m ... Answer
A person throws a pumpkin at a horizontal speed of 4.0 — off a cliff. The pumpkin travels 9.5 m horizontally
before it hits the ground. We can ignore air resistance.
Answer: Vertical displacement = -27.6m
And takes 2.375 s
A pumpkin thrown at a horizontal speed of 4.0 m/s and travels 9.5 m horizontally before hitting the ground will take 2.375 seconds to hit the ground.
How to find time?Given the following information:
Horizontal speed of the pumpkin = 4.0 m/s
Horizontal distance traveled before hitting the ground = 9.5 m
Ignoring air resistance
Use the following formula to calculate the time it takes the pumpkin to hit the ground:
time = horizontal distance / horizontal speed
time = 9.5 m / 4.0 m/s = 2.375 seconds
Therefore, it takes the pumpkin 2.375 seconds to hit the ground.
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The binding energies of K-shell and L-shell electrons in a certain metal are EK and EL, respectively, If a Kαx ray from this metal is incident on a crystal and gives a first-order Bragg reflection at an angle θ measured relative to parallel planes of atoms, what is the spacing between these parallel planes? State your answer in terms of the given variables, using h and c when needed.
Answer:
The separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
Explanation:
The relationship between energy and wavelength is expressed below:
E = hc/λ
λ = hc/EK - EL
Considering the condition of Bragg's law:
2dsinθ = mλ
For the first order Bragg's law of reflection:
2dsinθ = (1)λ
2dsinθ = hc/EK - EL
d = hc/2sinθ(EK - EL)
Where 'd' is the separation distance between the parallel planes of an atom, 'h' is the Planck's constant, 'c' is the velocity of light, θ is the angle of reflection, 'EK' is the energy of the K shell and 'EL' is the energy of the K shell.
Therefore, the separation distance between the parallel planes of an atom is hc/2sinθ(EK - EL)
(Solidification) You are performing a double slit experiment very similar to the one from DLby shining a laser on two narrow slits spaced 7.5 × 10-5meters apart. However, by placing a piece of crystal in one of the slits, you are able to make it so that the rays of light that travel through the two slits are πout of phasewith each other (that is to say,Δφ!=휋).If you observe that on a screen placed 4meters from the two slits that the distance between the bright spot closest to the center of the interference pattern and the center of the pattern is 1.5 cm, what is the wavelength of the laser? (Just as in DL, you may use the small angle approximation sinθ ≈ θ ≈ tanθ.)
Answer:
The wavelength of the laser, λ = 5.625 * 10⁻⁷ m
Explanation:
Separation of the narrow slits, d = 7.5 * 10⁻⁵ m
The distance between the screen and the two slits, d = 4m
The distance between the bright spot and the center of the pattern, Y = 1.5 cm
Y = 1.5 * 10⁻² m
To calculate the wavelength, λ, of the laser we will use the relationship:
[tex]Y = \frac{\lambda L}{2d} \\1.5 * 10^{-2} = \frac{\lambda * 4}{2*7.5*10^{-5} }\\\lambda = \frac{1.5 * 10^{-2} * 15 * 10^{-5} }{4}[/tex]
λ = 5.625 * 10⁻⁷ m
A young child hold a string attached to a balloon. What is the reaction force to the balloon pulling up on the earth?
Answer:
As the Ballon pulls up, the distance between the Ballon and the center of the earth increases, the gravitational pull reduces and the gravity potential energy increases.
In what direction must a force be applied so that the forces on the 1 kg object are balanced
Answer:
towards the object
Explanation:
Final answer:
To balance the forces on the 1 kg object, a force should be applied in the downward direction.
Explanation:
The forces on the 1 kg object can be balanced by applying a force in the opposite direction to the net force acting on the object. In this case, the net force is the sum of the weight of the object and the tension in the string. Since the weight acts downward and the tension in the string acts upward, the force should be applied in the downward direction to balance the forces on the 1 kg object.
A 5 kg spinning ring of inner radius 30cm and outer radius 40cm is dropped on a flat disk, where it interlocks and causes the disk to spin. Before the ring was dropped, it was spinning at 60 Hz. If the disk has a mass of 2kg and a radius of 1 meter, how much force needs to be applied to the rim of the disk (at a 90 degree angle) to bring it to a stop in 5 seconds
horizontal circular platform rotates counterclockwise about its axis at the rate of 0.919 rad/s. You, with a mass of 73.5 kg, walk clockwise around the platform along its edge at the speed of "1.05" m/s with respect to the platform. Your 20.5 kg poodle also walks clockwise around the platform, but along a circle at half the platform's radius and at half your linear speed with respect to the platform. Your 18.5 kg mutt, on the other hand, sits still on the platform at a position that is 3/4 of the platform's radius from the center. Model the platform as a uniform disk with mass 90.7 kg and radius 1.91 m. Calculate the total angular momentum of the system
Answer:
The total angular momentum is 292.59 kg.m/s
Explanation:
Given that :
Rotation of the horizontal circular platform [tex]\omega[/tex] = 0.919 rad/s
mass of the platform (m) = 90.7 kg
radius (R) = 1.91 m
mass of the poodle [tex]m_p[/tex] = 20.5 kg
Your mass [tex]m'[/tex] = 73.5 kg
speed v = 1.05 m/s with respect to the platform
[tex]V_p = \frac{1.05}{2} \ m /s \\ \\ = 0.525 \ m /s \\ \\r = \frac{R}{2}[/tex]
r = 0.955
Mass of the mutt [tex]m_m[/tex] = 18.5 kg
[tex]r' = \frac{3}{4} \ R[/tex]
Your angular momentum is calculated as:
Your angular velocity relative to the platform is [tex]\omega' = \frac{v}{R} = \frac{1.05}{1.91 } = 0.5497 \ rad/s[/tex]
[tex]Actual \ \omega_y = \omega - \omega ' = (0.919 - 0.5497) \ rad/s = 0.3693 \ rad/s[/tex]
[tex]I_y = m'R^2 = 73.5 *1.91^2= 268.14 \ kgm^2[/tex]
[tex]L_Y = I_y*\omega_y= 268.14*0.3683= 98.76 \ kg.m/s[/tex]
For poodle :
[tex]Relative \ \ \omega' = \frac{V_p}{R/2} = 0.550 \ rad/s[/tex]
Actual [tex]\omega_p = \omega - \omega' = 0.919 -0.550 = 0.369 \ rad/s[/tex]
[tex]I_p = m_p(\frac{R}{2} )^2 = 20.5(\frac{1.91}{2} )^2 = 18.70 \ kgm^2[/tex]
[tex]L_p = I_p *\omega_p = 18.70*0.369 = 6.9003 \ kgm/s[/tex]
[tex]I_M = m_m(\frac{3}{y4} R)^2 = 18.5 (\frac{3}{4} 1.91)^2 = 37.96 \ kgm^2[/tex]
[tex]L_M = I_M \omega = 37.96 * 0.919= 34.89 \ kg.m/s[/tex]
Disk [tex]I = \frac{mr^2}{2} = \frac{90.7*1.91^2}{2}= 165.44 \ kgm^2[/tex]
[tex]L_D = I \omega = 165.44*0.919 =152.04 \ kg.m/s[/tex]
Total angular momentum of system is:
L = [tex]L_D +L_Y+L_P+L_M[/tex]
= (152.04 + 98.76 + 6.9003 + 34.89) kg.m/s
= 292.59 kg.m/s
[03.02]
Which statement correctly describes the relationship between current, voltage, and resistance? If we (1 point)
Group of answer choices
decrease the resistance, and do not change the voltage, the current will decrease
decrease the voltage, and do not change the resistance, the current will also decrease
increase the resistance, and do not change the current, the voltage will remain the same
increase the current, and increase the resistance, the voltage will also decrease
Answer:
Decrease the voltage,and do not change the resistance,the current will also decrease
Explanation:
Decrease the voltage,and do not change the resistance, the current will also decrease, because voltage is directly proportional to current
Answer:
B.If we decrease the voltage, and do not change the resistance, then the current will also decrease.
Explanation:
i just take the (pre)test
hope it's help :)☺︎
A single-turn circular loop of wire of radius 45 mm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.10 s time interval, the magnitude of the field increases uniformly from 250 to 350 mT. (a) Determine the emf induced in the loop (in V). (Enter the magnitude.) V (b) If the magnetic field is directed out of the page, what is the direction of the current induced in the loop
Answer:
Magnitude of induced emf is 0.00635 V
Explanation:
Radius of circular loop r = 45 mm = 0.045 m
Area of circular loop [tex]A=\pi r^2[/tex]
[tex]A=3.14\times 0.045^2=0.00635m^2[/tex]
Magnetic field is increases from 250 mT to 350 mT
Therefore change in magnetic field [tex]dB=250-350=100mT[/tex]
Emf induced is given by
[tex]e=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}[/tex]
[tex]e=-0.00635\times \frac{100\times 10^{-3}}{0.10}=-0.00635V[/tex]
Magnitude of induced emf is equal to 0.00635 V
wyatt is designing a hollow cylindrical metal can with volume 1000 cm3.the materialused to make the circular top and bottom of the can costs twice as much as the material used tomake the side of the can. what dimensions should wyatt choose in order to minimize the cost ofthe can?show all your work, round off the numerical part of your final answer to four (4) decimal places,and express your final answer in the form of a complete sentence, using the correct units.
Answer:
Explanation:
Let r be the radius of circular top and h be the height of the cylinder
Given
π r² h = 1000
Let cost of making side of can be p per unit area , the cost of making top and bottom will be 2p per unit area.
total cost
C = 2 x π r² x 2p + 2πrh x p
= 2(1000 / h) x 2p + 2π x (1000 / π) x [tex]\frac{1}{\sqrt{h} }[/tex] x h x p
= 2(1000 / h) x 2p + 2π x (1000 / π) x √h x p
differenciating
dC / dh = 2(- 1000 / h²) x 2p + 2π x (1000 / π) x [tex]\frac{1}{2\sqrt{h} }[/tex] x p = 0 for minimum cost
- 4 / h² + 1 / √h = 0
h³ = 16
h = 2.519 cm .
π r² h = 1000
π r² x 2.519 = 1000
r = 11.24 cm
The cylinder will have height of 2.519 cm and radius of 11.24 cm.
Wyatt should design the can with a radius of approximately 3.4198 cm and a height of approximately 27.2837 cm to minimize cost. This considers the volume constraint of 1000 cm³ and the different material costs for the top/bottom and side. Detailed steps involve deriving these dimensions mathematically.
Let's denote the radius of the base of the cylinder by r and the height by h. The volume V of the cylinder is given by V = πr²h. Given that the volume is 1000 cm³, we have:
Step 1: V = πr²h = 1000 cm³
Therefore, h = 1000 / (πr²).
Step 2: Next, we need to express the cost of the material. The cost for the top and bottom parts is double the cost of the side. The surface area A of the can is:
Area of the top and bottom (each): [tex]A_t[/tex] = πr²
Area of the side: [tex]A_s[/tex] = 2πrh
Total area A:
A = 2πr² + 2πrh
Considering the cost factor, the effective cost area C is:
C = 2(2πr²) + 2πrh
C = 4πr² + 2πrh
Step 3: Substitute h from the volume equation:
C = 4πr² + 2πr(1000 / (πr²))
C = 4πr² + 2000 / r
Step 4: To minimize the cost, take the derivative of C with respect to r and set it to zero:
[tex]\frac{dC}{dr}[/tex] = 8πr - 2000 / r² = 0
8πr = 2000 / r²
8πr³ = 2000
r³ = 250 / π
r ≈ 3.4198 cm
Step 5: Calculate the height h:
h = 1000 / (πr²)
h ≈ 1000 / (π * (3.4198)²)
h ≈ 27.2837 cm
In summary, Wyatt should design the can with a radius of approximately 3.4198 cm and a height of approximately 27.2837 cm to minimize the cost.
Wyatt should choose a radius of approximately 3.4198 cm and a height of approximately 27.2837 cm for the metal can to minimize the cost, given the volume constraint of 1000 cm³.
A visitor to a lighthouse wishes to determine the height of the tower. She ties a spool of thread to a small rock to make a simple pendulum, which she hangs down the center of a spiral staircase of the tower. The period of oscillation is 6.40 s. What is the height of the tower
Answer:
l = 10.16 m
Explanation:
In this case, we have the period of oscillation of the pendulum is 6.4 s. It is required to find the height of the tower.
We know that the pendulum executes SHM. Let l is the height of the tower. The time period of simple pendulum is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
g is acceleration due to gravity
We need to rearrange the above equation such that,
[tex]l=\dfrac{T^2g}{4\pi ^2}\\\\l=\dfrac{(6.4)^2\times 9.8}{4\pi ^2}\\\\l=10.16\ m[/tex]
So, the height of the tower is 10.16 m.
Consider a rough asphalt roadway with 17 cm curbs and one 3.75 m lane in each direction and a 1.45 m shoulder on each side. The road has a longitudinal slope of 0.75% and is crowned in the center with a transverse slope of 2.25%. The local ordinance requires no ponding in the roadway (only on the shoulders) for a storm with an intensity of 90 mm/hr. A) Determine the maximum distance between curb inlets.
Answer:
Explanation:
Half width - 3.75 m
Rainfall intensity - 90 mm/hr
Longitudinal slope - 0.75%
Cross slope - 2.25%
Allowable limit of gutter flow, Q road = (Design constant x Intensity of rain x Area) / 360
= (0.91×90×(9×L1×10∧-4)/360 = 0.000204 L1
Using the standard design chart, allowable limit of gutter flow =0.018 m3/s
Therefore,0.018 = 0.000204 L1
L1 = 0.018 / 0.000204 = 88.24
Inlet spacing to be adapted is 88m
If a system is isolated the total energy within that system is constant. Consider the case of a hot cup of coffee: when it cools down, the energy goes into the surrounding air, causing the air to warm. If the coffee and the air are placed in an isolated chamber, the total energy of the system (coffee + air) is constant because the energy lost by the coffee is exactly equal to the energy gained by the air. The only way for the total energy of a system to change is when it is not isolated, meaning that external forces or sources of energy interact with the system. If we want to define a system where the total energy is conserved, which one of the choices below would be the best system to consider?A. The personB. The person and the EarthC. The Earth
Answer:b-The person and the earth
Explanation:
The best system will be the person and earth as a system
It is so because whatever the energy required by the person is derived from earth resource and thus by depleting the earth resource person is getting is required form of energy .
for example person want thermal energy , for this he need coal which is earth resource and thus all the energy is derived from earth and is being converted into some other form of energy.
A point source is fixed 1.0 m away from a large screen. Call the line normal to the screen surface and passing through the center of the point source the z axis. When a sheet of cardboard in which a square hole 0.020 m on a side has been cut is placed between the point source and the screen, 0.50 m from the point source with the hole centered on the z axis, a bright square shows up on the screen. If, instead, a second sheet of cardboard with a similar square hole is placed between the point source and screen, 0.25 m from the point source with the hole centered on the z axis, the bright square it casts on the screen is identical to the bright square from the first sheet. What is the length of the side of the hole in this sheet?
Answer:
The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]
Explanation:
From the question we are told that
The distance of the point source from the screen is [tex]d = 1.0 m[/tex]
The length of a side of the first square hole is [tex]L_1 = 0.020 \ m[/tex]
The distance of the cardboard from the point source is [tex]D_1 = 0.50\ m[/tex]
The distance of the second cardboard from the point source is [tex]D_2 = 0.25 \ m[/tex]
Let take the [tex]\alpha_{max }[/tex] as the angle at which the light is passing through the edges of the cardboards square hole
Since the bright square casted on the screen by both square holes on the individual cardboards are then it means that
[tex]\alpha_{max} __{1}} = \alpha_{max} __{2}}[/tex]
This implies that
[tex]tan (\alpha_{max} __{1}}) = tan (\alpha_{max} __{2}})[/tex]
Looking at this from the SOHCAHTOA concept
[tex]tan (\alpha_{max} __{1}}) = \frac{opposite}{Adjacent}[/tex]
Here opposite is the length of the side of the first cardboard square hole
and
Adjacent is the distance of the from the first cardboard square hole to the point source
And for
[tex]tan (\alpha_{max} __{2}}) = \frac{opposite}{Adjacent}[/tex]
Here opposite is the length of the side of the second cardboards square hole (let denote it with [tex]L_2[/tex])
and
Adjacent is the distance of the from the second cardboards square hole to the point source
So
[tex]tan (\alpha_{max} __{1}}) = \frac{0.020}{0.50}[/tex]
And
[tex]tan (\alpha_{max} __{2}}) = \frac{L_2}{0.25}[/tex]
Substituting this into the above equation
[tex]\frac{0.020}{0.50} = \frac{L_2}{0.25}[/tex]
Making [tex]L_2[/tex] the subject
[tex]L_2 = \frac{0.25 *0.020}{0.50}[/tex]
[tex]L_2 = 0.01m[/tex]
Since it is a square hole the sides are the same hence
The length of the side of the hole in the second cardboard sheet is [tex]L_2 = 0.01m[/tex]
In a Tesla coil, a long solenoid with length l and cross-sectional area A is closely wound with N 1 turns of wire. A coil with N 2 turns surrounds it at its center. You measure the mutual inductance. Then your friend gives you a new solenoid made out of twice as much wire, so that it has twice as many turns and is twice as long. How much larger is the mutual inductance M with the new solenoid instead of the old one
The mutual inductance of a Tesla coil system with two solenoids will double if the length and number of turns of the primary solenoid are both doubled, assuming all other factors remain constant.
When we are given a Tesla coil configuration with a solenoid of length l and cross-sectional area A, closely wound with N1 turns of wire, and another coil with N2 turns surrounding it at its center, we can calculate the mutual inductance based on the properties of the solenoids. If a new solenoid is introduced that has twice as many turns and is twice as long, the mutual inductance M of the system will be affected.
To understand how the mutual inductance changes, let's remember that for a closely wound solenoid, the mutual inductance can be calculated by a formula incorporating the number of turns, permeability of the core material, cross-sectional area, and the length of the solenoid. The mutual inductance is directly proportional to the product of the number of turns of each coil, the magnetic permeability of the core, and the area of the cross-section, and inversely proportional to the length of the solenoid. Therefore, if we double the length l and the number of turns N1 while keeping all other factors constant, the mutual inductance will also double.