Answer:
m = 4
Explanation:
The expression that explains the constructive interference of a diffraction pattern is
a sin θ = m λ
where a is the width of the slit and λ the wavelength
sin θ = m λ / a
The maximum value is for when the sine is 1, let's substitute
1 = m λ/a
m = a /λ
let's reduce the magnitudes to the SI system
a = 2.1 um = 2.1 10⁻⁶
lam = 475 nm = 475 10⁻⁹ m
let's calculate
m = 2.1 10⁻⁶ / 475 10⁻⁹
m = 4.42
with m must be an integer the highest value is
m = 4
If you could see stars during the day, this is what the sky would look like at noon on a given day. The Sun is near the stars of the constellation Gemini. Near which constellation would you expect the Sun to be located at sunset?
Answer:
The sun will be located near the Gemini constellation at sunset
An electric heater is constructed by applying a potential different of 120V across a nichrome wire that has a total resistant of 8 ohm .the current by the wire is
Answer:
15amps
Explanation:
V=IR
I=V/R
I = 120/8
I = 15 amps
Which of the following biotic organisms makes its own energy from inorganic substances?
producers
consumers
decomposers
minerals
Answer:
producers make its own energy frominorganic substances.
Two 60.o-g arrows are fired in quick succession with an initial speed of 82.0 m/s. The first arrow makes an initial angle of 24.0° above the horizontal, and the second arrow is fired straight upward. Assume an isolated system and choose the reference configuration at the initial position of the arrows.
(a) what is the maximum height of each of the arrows?
(b) What is the total mechanical energy of the arrow-Earth system for each of the arrows at their maximum height?
Answer:
a) The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters, b) Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Explanation:
a) The first arrow is launch in a parabolic way, that is, horizontal speed remains constant and vertical speed changes due to the effects of gravity. On the other hand, the second is launched vertically, which means that velocity is totally influenced by gravity. Let choose the ground as the reference height for each arrow. Each arrow can be modelled as particles and by means of the Principle of Energy Conservation:
First arrow
[tex]U_{g,1} + K_{x,1} + K_{y,1} = U_{g,2} + K_{x,2} + K_{y,2}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,2}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{x,1}[/tex], [tex]K_{x,2}[/tex] - Initial and final horizontal translational kinetic energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,2}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
Now, the system is expanded and simplified:
[tex]m \cdot g \cdot (y_{2} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 2}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{2}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,2}^{2})[/tex]
[tex]y_{2}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,2}^{2}}{g}[/tex]
Where:
[tex]y_{1}[/tex]. [tex]y_{2}[/tex] - Initial and final height of the arrow, measured in meters.
[tex]v_{y,1}[/tex], [tex]v_{y,2}[/tex] - Initial and final vertical speed of the arrow, measured in meters.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
The initial vertical speed of the arrow is:
[tex]v_{y,1} = v_{1}\cdot \sin \theta[/tex]
Where:
[tex]v_{1}[/tex] - Magnitude of the initial velocity, measured in meters per second.
[tex]\theta[/tex] - Initial angle, measured in sexagesimal degrees.
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the initial vertical speed is:
[tex]v_{y,1} = \left(82\,\frac{m}{s} \right)\cdot \sin 24^{\circ}[/tex]
[tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} \approx 33.352\,\frac{m}{s}[/tex] and [tex]v_{y,2} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{2} - y_{1} = \frac{1}{2}\cdot \frac{\left(33.352\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{2} - y_{1} = 56.712\,m[/tex]
Second arrow
[tex]U_{g,1} + K_{y,1} = U_{g,3} + K_{y,3}[/tex]
Where:
[tex]U_{g,1}[/tex], [tex]U_{g,3}[/tex] - Initial and final gravitational potential energy, measured in joules.
[tex]K_{y,1}[/tex], [tex]K_{y,3}[/tex] - Initial and final vertical translational kinetic energy, measured in joules.
[tex]m \cdot g \cdot (y_{3} - y_{1}) + \frac{1}{2}\cdot m \cdot (v_{y, 3}^{2} -v_{y, 1}^{2}) = 0[/tex]
[tex]g \cdot (y_{3}-y_{1}) = \frac{1}{2}\cdot (v_{y,1}^{2}-v_{y,3}^{2})[/tex]
[tex]y_{3}-y_{1} = \frac{1}{2}\cdot \frac{v_{y,1}^{2}-v_{y,3}^{2}}{g}[/tex]
If [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{y,1} = 82\,\frac{m}{s}[/tex] and [tex]v_{y,3} = 0\,\frac{m}{s}[/tex], the maximum height of the first arrow is:
[tex]y_{3} - y_{1} = \frac{1}{2}\cdot \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]y_{3} - y_{1} = 342.816\,m[/tex]
The first arrow reaches a maximum height of 56.712 meters, whereas second arrow reaches a maximum height of 342.816 meters.
b) The total energy of each system is determined hereafter:
First arrow
The total mechanical energy at maximum height is equal to the sum of the potential gravitational energy and horizontal translational kinetic energy. That is to say:
[tex]E = U + K_{x}[/tex]
The expression is now expanded:
[tex]E = m\cdot g \cdot y_{max} + \frac{1}{2}\cdot m \cdot v_{x}^{2}[/tex]
Where [tex]v_{x}[/tex] is the horizontal speed of the arrow, measured in meters per second.
[tex]v_{x} = v_{1}\cdot \cos \theta[/tex]
If [tex]v_{1} = 82\,\frac{m}{s}[/tex] and [tex]\theta = 24^{\circ}[/tex], the horizontal speed is:
[tex]v_{x} = \left(82\,\frac{m}{s} \right)\cdot \cos 24^{\circ}[/tex]
[tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex]
If [tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]y_{max} = 56.712\,m[/tex] and [tex]v_{x} \approx 74.911\,\frac{m}{s}[/tex], the total mechanical energy is:
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (56.712\,m)+\frac{1}{2}\cdot (0.06\,kg)\cdot \left(74.911\,\frac{m}{s} \right)^{2}[/tex]
[tex]E = 201.720\,J[/tex]
Second arrow:
The total mechanical energy is equal to the potential gravitational energy. That is:
[tex]E = m\cdot g \cdot y_{max}[/tex]
[tex]m = 0.06\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]y_{max} = 342.816\,m[/tex]
[tex]E = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (342.816\,m)[/tex]
[tex]E = 201.720\,J[/tex]
Both arrows have a total mechanical energy at their maximum height of 201.720 joules.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Complete question:
A force F is applied to the block as shown (check attached image). With an applied force of 1.5 N, the block moves with a constant velocity.
Approximately what applied force is needed to keep the box moving with a constant velocity that is twice as fast as before? Explain
Answer:
The applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the initially applied force.
Explanation:
Given;
magnitude of applied force, F = 1.5 N
Apply Newton's second law of motion;
F = ma
[tex]F = m(\frac{v}{t} )\\\\F = \frac{m}{t} v\\\\Let \ \frac{m}{t} \ be \ constant = k\\F = kv\\\\k = \frac{F}{v} \\\\\frac{F_1}{v_1} = \frac{F_2}{v_2}[/tex]
The applied force needed to keep the box moving with a constant velocity that is twice as fast as before;
[tex]\frac{F_1}{v_1} = \frac{F_2}{v_2} \\\\(v_2 = 2v_1, \ and \ F_1 = 1.5N)\\\\\frac{1.5}{v_1} = \frac{F_2}{2v_1} \\\\1.5 = \frac{F_2}{2}\\\\F_2 = 2*1.5\\\\F_2 = 3 N[/tex]
Therefore, the applied force that is needed to keep the box moving with a constant velocity that is twice as fast as before, is 3 N
Force is directly proportional to velocity, to keep the box moving at the double of initial constant velocity, we must also double the value of the applied force.
An ac circuit consist of a pure resistance of 10ohms is connected across an ae supply
230V 50Hz Calculate the:
(i)Current flowing in the circuit.
(ii)Power dissipated
Plz check attachment for answer.
Hope it's helpful
How would the magnetic field lines appear for a bar magnet cut at the midpoint, with the two pieces placed end to end with a space in between such that the cut edges are closest to each other? What would the general shape of the field lines look like? What would the field lines look like in between the two pieces?
Answer:
Explanation:
check this out and rate me
what tools use cut wood
Answer:
hand saws
power saws
Circular Saw
Explanation:
that is all that i know
You illuminate a slit with a width of 77.7 μm with a light of wavelength 721 nm and observe the resulting diffraction pattern on a screen that is situated 2.83 m from the slit. What is the width, in centimeters, of the pattern's central maximum
Answer:
The width is [tex]Z = 0.0424 \ m[/tex]
Explanation:
From the question we are told that
The width of the slit is [tex]d = 77.7 \mu m = 77.7 *10^{-6} \ m[/tex]
The wavelength of the light is [tex]\lambda = 721 \ nm[/tex]
The position of the screen is [tex]D = 2.83 \ m[/tex]
Generally angle at which the first minimum of the interference pattern the light occurs is mathematically represented as
[tex]\theta = sin ^{-1}[\frac{m \lambda}{d} ][/tex]
Where m which is the order of the interference is 1
substituting values
[tex]\theta = sin ^{-1}[\frac{1 *721*10^{-9}}{ 77.7*10^{-6}} ][/tex]
[tex]\theta = 0.5317 ^o[/tex]
Now the width of first minimum of the interference pattern is mathematically evaluated as
[tex]Y = D sin \theta[/tex]
substituting values
[tex]Y = 2.283 * sin (0.5317)[/tex]
[tex]Y = 0.02 12 \ m[/tex]
Now the width of the pattern's central maximum is mathematically evaluated as
[tex]Z = 2 * Y[/tex]
substituting values
[tex]Z = 2 * 0.0212[/tex]
[tex]Z = 0.0424 \ m[/tex]
A student is conducting an experiment that involves adding hydrochloric acid to various minerals to detect if they have carbonates in them. The student holds a mineral up and adds hydrochloric acid to it. The acid runs down the side and onto the student’s hand causing irritation and a minor burn. If they had done a risk assessment first, how would this situation be different? A. It would be the same, there is no way to predict the random chance of acid dripping off the mineral in a risk assessment. B. The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. C. The student would be safer because he would have been wearing goggles, but his hand still would not have been protected. D. The student would not have picked up the mineral because he would know that some of the minerals have dangerous chemicals in them.
By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
What is experiment ?An experiment would be a technique used to confirm or deny a hypothesis, as well as assess the likelihood or effectiveness of something that has never been tried before.
What is hydrochloric acid?Hydrochloric acid is a kind of compound in which hydrogen and chlorine element is present.
Maintain a safe distance between your hands and your body, mouth, eyes, as well as a face when utilizing lab supplies and chemicals.
By the experiment "By the experiment "The student would have no injuries because he would know hydrochloric acid is dangerous and would be wearing gloves when using it. "
To know more about experiment and hydrochloric acid
https://brainly.com/question/13770820
#SPJ3
describe the relation among density, temperature, and volume when the pressure is constant, and explain the blackbody radiation curve
Answer:
in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body,
Explanation:
Let's start for ya dream gas
PV = nRT
Since it indicates that the pressure is constant, we see that the volume is directly proportional to the temperature.
The density of is defined by
ρ = m / V
As we saw that volume increases with temperature, this is also true for solid materials, using linear expansion. Therefore in all cases with increasing temperature the density should decrease.
Black body radiation is a construction that maintains a constant temperature and a hole is opened, this hole is called a black body, since all the radiation that falls on it is absorbed or emitted.
This type of construction has a characteristic curve where the maximum of the curve is dependent on the tempera, but independent of the material with which it is built, to explain the behavior of this curve Planck proposed that the diaconate in the cavity was not continuous but discrete whose energy is given by the relationship
E = h f
what is thermodynamic?
Answer:
Thermodynamics is a branch of physics which deals with the energy and work of a system. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiment.
Answer:
thermodynamics is the branch of physics which deals with the study of heat and other forms energy and their mutual relationship(relation ship between them)
Explanation:
i hope this will help you :)
What is the relationship between the magnitudes of the collision forces of two vehicles, if one of them travels at a higher speed?
Explanation:
The collision forces are equal and opposite. Therefore, the magnitudes are equal.
As an ice skater begins a spin, his angular speed is 3.14 rad/s. After pulling in his arms, his angular speed increases to 5.94 rad/s. Find the ratio of teh skater's final momentum of inertia to his initial momentum of inertia.
Answer:
I₂/I₁ = 0.53
Explanation:
During the motion the angular momentum of the skater remains conserved. Therefore:
Angular Momentum of Skater Before Pulling Arms = Angular Momentum of Skater After Pulling Arms
L₁ = L₂
but, the formula for angular momentum is:
L = Iω
Therefore,
I₁ω₁ = I₂ω₂
I₂/I₁ = ω₁/ω₂
where,
I₁ = Initial Moment of Inertia
I₂ = Final Moment of Inertia
ω₁ = Initial Angular Velocity = 3.14 rad/s
ω₂ = Final Angular velocity = 5.94 rad/s
Therefore,
I₂/I₁ = (3.14 rad/s)/(5.94 rad/s)
I₂/I₁ = 0.53
The buoyant force on an object placed in a liquid is (a) always equal to the volume of the liquid displaced. (b) always equal to the weight of the object. (c) always equal to the weight of the liquid displaced. (d) always less than the volume of the liquid displaced.
Answer:
(c) always equal to the weight of the liquid displaced.
Explanation:
Archimedes principle (also called physical law of buoyancy) states that when an object is completely or partially immersed in a fluid (liquid, e.t.c), it experiences an upthrust (or buoyant force) whose magnitude is equal to the weight of the fluid displaced by that object.
Therefore, from this principle the best option is C - always equal to the weight of the liquid displaced.
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 12.9 rad/s in 2.98 s.
(a) Find the magnitude of the angular acceleration of the wheel.
(b) Find the angle in radians through which it rotates in this time interval.
Explanation:
(a) Find the magnitude of the angular acceleration of the wheel.
angular acceleration = angular speed /timeangular acceleration = 12.9/2.98 = 4.329rad/s²(b) Find the angle in radians through which it rotates in this time interval.
angular speed = 2x3.14xf12.9rad = 2 x3.14rad = 6.28/12.9rad = 0.487Now we convert rad to angle
1 rad = 57.296°0.487 = unknown angleunknown angle =57.296 x 0.487 = 27.9°The angle in radians = 27.9°
If the velocity of a pitched ball has a magnitude of 47.0 m/s and the batted ball's velocity is 55.0 m/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.
Answer:
14.79 kgm/s
Explanation:
Data provided in the question
Let us assume the mass of baseball = m = 0.145 kg
The Initial velocity of pitched ball = [tex]v_i[/tex] = 47 m/s
Final velocity of batted ball in the opposite direction = [tex]v_f[/tex]= -55m/s
Based on the above information, the change in momentum is
[tex]\Delta P = m(v_f -v_i)[/tex]
[tex]= 0.145 kg(-55m/s - 47m/s)[/tex]
= 14.79 kgm/s
Hence, the magnitude of the change in momentum of the ball is 14.79 kg m/s
When moving to a new apartment, you rent a truck and create a ramp with a 244 cm long piece of plywood. The top of the moving ramp lies on the edge of the truck bed at a height of 115 cm. You load your textbooks into a wooden box at the bottom of the ramp (the coefficient of kinetic friction between the box and ramp is = 0.2). Then you and a few friends give the box a quick push and it starts to slide up the ramp. A) What angle is made by the ramp and the ground?B) Unfortunately, after letting go, the box only tables 80cm up the ramp before it starts coming back down! What speed was the box initially traveling with just after you stopped pushing it?
Answer:
A) θ = 28.1º , B) v = 2.47 m / s
Explanation:
A) The angle of the ramp can be found using trigonometry
sin θ = y / L
Φ = sin⁻¹ y / L
θ = sin⁻¹ (115/244)
θ = 28.1º
B) For this pate we can use the relationship between work and kinetic energy
W =ΔK
where the work is
W = -fr x
the negative sign is due to the fact that the friction force closes against the movement
Lavariacion of energy cineta is
ΔEm = ½ m v² - mgh
-fr x = ½ m v² - m gh
the friction force has the equation
fr = very N
at the highest part there is no speed and we take the origin from the lowest part of the ramp
To find the friction force we use Newton's second law. Where one axis is parallel to the ramp and the other is perpendicular
Axis y . perpendicular
N- Wy = 0
cos tea = Wy / W
Wy = W cos treaa
N = mg cos tea
we substitute
- (very mg cos tea) x = ½ m v²2 - mgh
v2 = m (gh- very g cos tea x)
let's calculate
v = Ra (9.8 0.80 - 0.2 9.8 0.0 cos 28.1)
v = RA (7.84 -1.729)
v = 2.47 m / s
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
Answer:
Explanation:
Find the average value of position x, momentump, and square of the mometum p2 for the ground and first excited states of the particle-in-a-box with mass m and box length L.
A particle of charge = 50 µC moves in a region where the only force on it is an electric force. As the particle moves 25 cm, its kinetic energy increases by 1.5 mJ. Determine the electric potential difference acting on the partice
Answer:
nvbnncbmkghbbbvvvvvvbvbhgggghhhhb
An archer shoots an arrow toward a 300-g target that is sliding in her direction at a speed of 2.10 m/s on a smooth, slippery surface. The 22.5-g arrow is shot with a speed of 37.5 m/s and passes through the target, which is stopped by the impact. What is the speed of the arrow after passing through the target
Answer:
The speed of the arrow after passing through the target is 30.1 meters per second.
Explanation:
The situation can be modelled by means of the Principle of Linear Momentum, let suppose that the arrow and the target are moving on the same axis, where the velocity of the first one is parallel to the velocity of the second one. The Linear Momentum model is presented below:
[tex]m_{a}\cdot v_{a,o} + m_{t}\cdot v_{t,o} = m_{a}\cdot v_{a,f} + m_{t}\cdot v_{t,f}[/tex]
Where:
[tex]m_{a}[/tex], [tex]m_{t}[/tex] - Masses of arrow and target, measured in kilograms.
[tex]v_{a,o}[/tex], [tex]v_{a,f}[/tex] - Initial and final speeds of the arrow, measured in meters per second.
[tex]v_{t,o}[/tex], [tex]v_{t,f}[/tex] - Initial and final speeds of the target, measured in meters per second.
The final speed of the arrow is now cleared:
[tex]m_{a} \cdot v_{a,f} = m_{a} \cdot v_{a,o} + m_{t}\cdot (v_{t,o}-v_{t,f})[/tex]
[tex]v_{a,f} = v_{a,o} + \frac{m_{t}}{m_{a}} \cdot (v_{t,o}-v_{t,f})[/tex]
If [tex]v_{a,o} = 2.1\,\frac{m}{s}[/tex], [tex]m_{t} = 0.3\,kg[/tex], [tex]m_{a} = 0.0225\,kg[/tex], [tex]v_{t,o} = 2.10\,\frac{m}{s}[/tex] and [tex]v_{t,f} = 0\,\frac{m}{s}[/tex], the speed of the arrow after passing through the target is:
[tex]v_{a,f} = 2.1\,\frac{m}{s} + \frac{0.3\,kg}{0.0225\,kg}\cdot (2.10\,\frac{m}{s} - 0\,\frac{m}{s} )[/tex]
[tex]v_{a,f} = 30.1\,\frac{m}{s}[/tex]
The speed of the arrow after passing through the target is 30.1 meters per second.
To understand the standard formula for a sinusoidal traveling wave.
One formula for a wave with a y displacement (e.g., of a string) traveling in the x direction is
y(x,t)=Asin(kxâÏt).
All the questions in this problem refer to this formula and to the wave it describes.
1) What is the phase Ï(x,t) of the wave?
Express the phase in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï
Ï(x,t)=
2) What is the wavelength λ of the wave?
Express the wavelength in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï.
λ=
3) What is the period T of this wave?
Express the period in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï.
T=
4) What is the speed of propagation v of this wave?
Express the speed of propagation in terms of one or more given variables ( A, k, x, t, and Ï) and any needed constants like Ï.
v=
Answer:
1) Φ=zero
2) λ = 2π / k
3) T = 2π / w
4) v = w / k
Explanation:
The equation of a traveling wave is
y = A sin (ka - wt + Ф)
Let's answer using this equation the different questions
1) we see that the equation given in the problem the phase is zero
2) wavelength
k = 2π /λ
λ = 2π / k
3) The perido
angular velocity is related to frequency
w = 2π f
frequency and period are related
f = 1 / T
w = 2 π / T
T = 2π / w
4) the wave speed is
v = λ f
λ = 2π / k
f = w / 2π
v = 2π /k w /2π
v = w / k
When you are told that the wind has a "Small Coriolis force" associated with it, what is that "small force" exactly
Answer:
Coriolis force is a type of force of inertia that acts on objects that is in motion within a frame of reference that rotates with respect to an inertial frame. Due to the rotation of the earth, circulating air is deflected result of the Coriolis force, instead of the air circulating between the earth poles and the equator in a straight manner. Because of the effect of the Coriolis force, air movement deflects toward the right in the Northern Hemisphere and toward the left in the Southern Hemisphere, eventually taking a curved path of travel.
Now moving horizontally, the skier crosses a patch of soft snow, where the coefficient of friction is μk = 0.160. If the patch is of width 62.0 m and the average force of air resistance on the skier is 160 N , how fast is she going after crossing the patch?
Answer:
14.1 m/s
Explanation:
From the question,
μk = a/g...................... Equation 1
Where μk = coefficient of kinetic friction, a= acceleration of the skier, g = acceleration due to gravity.
make a the subject of the equation
a = μk(g).................. Equation 2
Given: μk = 0.160, g = 9.8 m/s²
Substitute into equation 2
a = 0.16(9.8)
a = 1.568 m/s²
Using,
F = ma
Where F = force, m = mass.
Make m the subject of the equation
m = F/a................... Equation 3
m = 160/1.568
m = 102.04 kg.
Note: The work done against air resistance by the skier+ work done against friction is equal to the kinetic energy after cross the patch.
Assuming the initial velocity of the skier to be zero
Fd+mgμ = 1/2mv²........................Equation 4
Where v = speed of the skier after crossing the patch, d = distance/width of the patch.
v = √2(Fd+mgμ)/m)................ Equation 5
Given: F = 160 N, m = 102.04 kg, d = 62 m, g = 9.8 m/s, μk = 0.16
Substitute these values into equation 5
v = √[2[(160×62)+(102.04×9.8×0.16)]/102.04]
v = √197.57
v = 14.1 m/s
v = 9.86 m/s
A uniform 2.0-kg rod that is 0.92 m long is suspended at rest from the ceiling by two springs, one at each end. Both springs hang straight down from the ceiling. The springs have identical lengths when they are unstretched. Their spring constants are 29 N/m and 66 N/m. Find the angle that the rod makes with the horizontal.
Answer:
11.7°
Explanation:
See attached file
When the charges in the rod are in equilibrium, what is the magnitude of the electric field within the rod?
Answer: If we have equilibrium, the magnitude must be zero.
Explanation:
If the charges are in equilibrium, this means that the total charge is equal to zero.
And as the charges must be homogeneously distributed in the rod, we can conclude that the electric field within the rod must be zero, so the magnitude of the electric field must be zero
Bromine, a liquid at room temperature, has a boiling point
Yes it does ! The so-called "boiling point" is the temperature at which Bromine liquid can change state and become Bromine vapor, if enough additional thermal energy is provided. The boiling point is higher than room temperature.
The cost of energy delivered to residences by electrical transmission varies from $0.070/kWh to $0.258/kWh throughout the United States; $0.110/kWh is the average value.
Required:
At this average price, calculate the cost of:
a. leaving a 40-W porch light on for two weeks while you are on vacation?
b. making a piece of dark toast in 3.00 min with a 970-W toaster
c. drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer.
Answer:
Cost = $ 1.48
Cost = $ 0.005
Cost = $ 0.38
Explanation:
given data
electrical transmission varies = $0.070/kWh to $0.258/kWh
average value = $0.110/kWh
solution
when leaving a 40-W porch light on for two weeks while you are on vacation so cost will be
first we get here energy consumed that is express as
E = Pt .................1
here E is Energy Consumed and Power Delivered is P and t is time
so power is here 0.04 KW and t = 2 week = 336 hour
so
put value in 1 we get
E = 0.04 × 336
E = 13.44 KWh
so cost will be as
Cost = E × Unit Price .............2
put here value and we get
Cost = 13.44 × 0.11
Cost = $ 1.48
and
when you making a piece of dark toast in 3.00 min with a 970-W toaster
so energy consumed will be by equation 1 we get
E = Pt
power is = 0.97 KW and time = 3 min = 0.05 hour
put value in equation 1 for energy consume
E = 0.97 × 0.05 h
E = 0.0485 KWh
and we get cost by w\put value in equation 2 that will be
cost = E × Unit Price
cost = 0.0485 × 0.11
Cost = $ 0.005
and
when drying a load of clothes in 40.0 min in a 5.20 x 10^3-W dryer
from equation 1 we get energy consume
E = Pt
Power Delivered = 5.203 KW and time = 40 min = 0.67 hour
E = 5.203 × 0.67
E = 3.47 KWh
and
cost will by put value in equation 2
Cost = E × Unit Price
Cost = 3.47 × 0.11
Cost = $ 0.38
Stress is a factor that contributes to heart disease risk.true or false
A passenger jet flies from one airport to another 1,233 miles away in 2.4 h. Find its average speed. = ____ m/s
Speed = (distance) / (time)
Speed = (1,233 mile) / (2.4 hour)
Speed = 513.75 mile/hour
Speed = (513.75 mi/hr) x (1609.344 meter/mi) x (1 hr / 3600 sec)
Speed = (513.75 x 1609.344 / 3600) (mile-meter-hour/hour-mile-second)
Speed = 229.7 meter/second