Answer:
Velocity=1.1m/s
Amplitude=0.35m
Explanation:
Given:
time 't' = 2.9s
wavelength 'λ'= 5.5m
distance 'd'=0.7m
The time period 't' is the time b/w two successive waves. Therefore, the time it takes from the boat to travel from its highest point to its lowest is a half period.
So, T = 2 x 2.9 => 5.8 s
As we know that frequency is the reciprocal of time period, we have
f= 1/T = 1/5.8 =>0.2 Hz
In order to find how fast are the waves traveling, the velocity is given by
Velocity = f λ
V= 0.2 x 5.5 =>1.1m/s
The distance between the boat's highest point to its lowest point is double the amplitude.
Therefore , we can write
Amplitude 'A'= d/2 =>0.7/2 =>0.35m
A glass tube 1.50 meters long and open at one end is weighted to keep it vertical and is then lowered to the bottom of a lake. When it returns to the surface it is determined that at the bottom of the lake the water rose to within 0.133 meters of the closed end. The lake is 100 meters deep, the air temperature at the surface is 27 "C, atmospheric pressure is 1.01x10s N/m2, and the density of water is 998 kg/m3. a) What is the total pressure at the bottom of the lake
Complete Question
The complete question is shown on the first uploaded image
Answer:
The total pressure is [tex]P_T = 10.79*10^{5} N/m^2[/tex]
The temperature at the bottom is [tex]T_b = 284.2 \ K[/tex]
Explanation:
From the question we are told that
The length of the glass tube is [tex]L = 1.50 \ m[/tex]
The length of water rise at the bottom of the lake [tex]d = 1.33 \ m[/tex]
The depth of the lake is [tex]h = 100 \ m[/tex]
The air temperature is [tex]T_a = 27 ^oC = 27 +273 = 300 \ K[/tex]
The atmospheric pressure is [tex]P_a = 1.01 *10^{5} N/m[/tex]
The density of water is [tex]\rho = 998 \ kg/m^3[/tex]
The total pressure at the bottom of the lake is mathematically represented as
[tex]P_T = P_a + \rho g h[/tex]
substituting values
[tex]P_T = 1.01*10^{5} + 998 * 9.8 * 100[/tex]
[tex]P_T = 10.79*10^{5} N/m^2[/tex]
According to ideal gas law
At the surface the glass tube not covered by water at surface
[tex]P_a V_a = nRT_a[/tex]
Where is the volume of
[tex]P_a *A * L = nRT_a[/tex]
At the bottom of the lake
[tex]P_T V_b = nRT_b[/tex]
Where [tex]V_b[/tex] is the volume of the glass tube not covered by water at bottom
and [tex]T_b[/tex] i the temperature at the bottom
So the ratio between the temperature at the surface to the temperature at the bottom is mathematically represented as
[tex]\frac{T_b}{T_a} = \frac{d * P_T}{P_a * h}[/tex]
substituting values
[tex]\frac{T_b}{27} = \frac{0.133 * 10.79 *10^5}{1.01 *10^{5} * 1.5}[/tex]
=> [tex]T_b = 284.2 \ K[/tex]
A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 10 nC is located at a distance r = 10 cm from the center of the sphere. If the net flux through the surface of the sphere is 800 N.m2/C, find q1.
Answer:
q₁ = -2.92 nC
Explanation:
Given;
first point charge, q₁ = ?
second point charge, q₂ = 10 nC
net flux through the surface of the sphere, Φ = 800 N.m²/C
According to Gauss’s law, the flux through any closed surface (Gaussian surface), is equal to the net charge enclosed divided by the permittivity of free space.
[tex]\phi = \frac{q_{enc.}}{\epsilon_o}[/tex]
where;
Φ is net flux
[tex]q_{enc.}[/tex] net charge enclosed
ε₀ is permittivity of free space.
[tex]q_{enc.}[/tex] = Φε₀
= 800 x 8.85 x 10⁻¹²
= 7.08 x 10⁻⁹ C
[tex]q_{enc.}[/tex] = 7.08 nC
q₁ + q₂ = [tex]q_{enc.}[/tex]
q₁ = [tex]q_{enc.}[/tex] - q₂
q₁ = 7.08nC - 10 nC
q₁ = -2.92 nC
Three balls with the same radius 21 cm are in water. Ball 1 floats, with
half of it exposed above the water level. Ball 2, with a density 893
kg/m3 is held below the surface by a cord anchored to the bottom of the
container, so that it is fully submerged. Ball 3, of density 1320 kg/m3, is
suspended from a rope so that it is fully submerged. Assume the
density of water is 1000 kg/m3 in this problem.
A. Which is true for Ball 1?
B. What is the tension on the rope holding the second ball, in newtons?
C. What is the tension on the rope holding the third ball in N?
Ball 1 floats with half exposed above water level. Tension on rope holding Ball 2 is calculated using weight and buoyant force. Tension on rope holding Ball 3 is equal to buoyant force.
Explanation:A. Ball 1 is floating with half of it exposed above the water level.
This means that the buoyant force on the ball is equal to the weight of the ball.
Since the buoyant force is greater than the weight of the ball, the ball floats.
B. The tension on the rope holding Ball 2 can be found using the equation:
Tension = Weight - Buoyant force.
The weight of the ball is calculated by multiplying its volume by its density and acceleration due to gravity.
The buoyant force can be found by multiplying the volume of the ball submerged in water by the density of water and acceleration due to gravity.
C. The tension on the rope holding Ball 3 is the same as the buoyant force acting on it.
The buoyant force can be found by multiplying the volume of the ball submerged in water by the density of water and acceleration due to gravity.
Ball 1's density is 500 kg/m³. The tension on the rope holding Ball 2 is 41.15 N. The tension on the rope holding Ball 3 is 121.24 N.
Let's solve this problem step-by-step to better understand floating and submerged objects in water.
A.) Which is true for Ball 1?
Ball 1 floats with half of it exposed above the water level. This means the density of Ball 1 must be half the density of water. Since the density of water is 1000 kg/m³, the density of Ball 1 is:
500 kg/m³B.) What is the tension on the rope holding the second ball, in newtons?
Ball 2 has a density of 893 kg/m³ and is held below the surface of water. The buoyant force is equal to the weight of the volume of water displaced by Ball 2.
Calculate the volume of Ball 2: (Volume of a sphere = 4/3 π r³)
r = 21 cm r = 0.21 mVolume = (4/3) π (0.21)³ = 0.0388 m³Calculate the buoyant force:
Buoyant Force = Density of Water x Volume of Water x gBuoyant Force = 1000 kg/m³ x 0.0388 m³ x 9.8 m/s² Bouyant force = 380.24 NCalculate the weight of Ball 2:
Weight = Density of Ball 2 x Volume x gWeight = 893 kg/m³ x 0.0388 m³ x 9.8 m/s² = 339.09 NCalculate the tension in the rope:
Tension = Weight - Buoyant ForceTension = 339.09 N - 380.24 N = -41.15 N (negative indicates an upward force)Tension on the rope for Ball 2: 41.15 NC.) What is the tension on the rope holding the third ball in N?
Ball 3 has a density of 1320 kg/m³ and is also fully submerged, suspended by a rope.
Calculate the weight of Ball 3:
Weight = Density of Ball 3 x Volume x gWeight = 1320 kg/m³ x 0.0388 m³ x 9.8 m/s² = 501.48 NCalculate the tension in the rope:
Tension = Weight - Buoyant ForceTension = 501.48 N - 380.24 N = 121.24 NTension on the rope for Ball 3: 121.24 NThis example shows how to calculate buoyant forces and tensions for submerged and floating objects.
how many grams of nitrogen gas are needed to produce 34 g of ammonia
Answer:
28 grams
Explanation:
The equation for the reaction is
3H(2) + N(2) -> 2NH(3)
Then we have.
The molar mass, M of ammonia is 17 g/mol.
34 grams of ammonia, NH3 then would be
34 g / 17 g/mol
= 2 moles
2 moles of ammonia will be obtained from
(2 * 1) / 2
= 1 mole of nitrogen
The molar masses of nitrogen is 28 g/mol
2 moles of nitrogen corresponds to 1 * 28 = 28 grams.
A snail can crawl 160cm at an avg speed of 4cm/min. If it crawled at an avg speed of 5cm/min instead, how much sooner would it take to reach the destination?
Answer:
8 minutes sooner
Explanation:
Average speed of snail= 4cm/min
Distance to be covered = 160cm
Time taken for the journey = distance/speed
Time taken for the journey = 160/4
Time taken for the journey = 40 min
If it crawed an average speed of 5cm/min
Distance = 160 cm
Time for the journey = distance/speed
Time for the journey = 160/5
Time for the journey = 32 min
Its going to take the snail 40 min - 32 min to Kno how sooner it will taje it if the average speed is 5cm/min
40 min - 32 min = 8 min
g Two cars, car 1 and car 2 are traveling in opposite directions, car 1 with a magnitude of velocity v1=13.0 m/s and car 2 v2= 7.22 m/s. If car 1’s exhaust system is loud enough to be heard by car 2 and the frequency fe produced from the exhaust is 2.10 kHz. What frequencies would be heard by car 2 when the cars are approaching, passing, and retreating from one another?
Answer:
When they are approaching each other
[tex]f_a = 2228.7 \ Hz[/tex]
When they are passing each other
[tex]f_a = 2100Hz[/tex]
When they are retreating from each other
[tex]f_a = 1980.7 Hz[/tex]
Explanation:
From the question we are told that
The velocity of car one is [tex]v_1 = 13.0 m/s[/tex]
The velocity of car two is [tex]v_2 = 7.22 m/s[/tex]
The frequency of sound from car one is [tex]f_e = 2.10 kHz[/tex]
Generally the speed of sound at normal temperature is [tex]v = 343 m/s[/tex]
Now as the cars move relative to each other doppler effect is created and this can be represented mathematically as
[tex]f_a = f_o [\frac{v \pm v_o}{v \pm v_s} ][/tex]
Where [tex]v_s[/tex] is the velocity of the source of sound
[tex]v_o[/tex] is the velocity of the observer of the sound
[tex]f_o[/tex] is the actual frequence
[tex]f_a[/tex] is the apparent frequency
Considering the case when they are approaching each other
[tex]f_a = f_o [\frac{v + v_o}{v - v_s} ][/tex]
[tex]v_o = v_2[/tex]
[tex]v_s = v_1[/tex]
[tex]f_o = f_e[/tex]
Substituting value
[tex]f_a = 2100 [\frac{343 + 7.22}{ 343 - 13} ][/tex]
[tex]f_a = 2228.7 \ Hz[/tex]
Considering the case when they are passing each other
At that instant
[tex]v_o = v_s = 0m/s[/tex]
[tex]f_o = f_e[/tex]
[tex]f_a = f_o [\frac{v }{v } ][/tex]
[tex]f_a = f_o[/tex]
Substituting value
[tex]f_a = 2100Hz[/tex]
Considering the case when they are retreating from each other
[tex]f_a = f_o [\frac{v - v_o}{v + v_s} ][/tex]
[tex]v_o = v_2[/tex]
[tex]v_s = v_1[/tex]
[tex]f_o = f_e[/tex]
Substituting value
[tex]f_a = 2100 [\frac{343 - 7.22}{343 + 13} ][/tex]
[tex]f_a = 1980.7 Hz[/tex]
After landing on an unexplored Klingon planet, Spock tests for the direction of the magnetic field by firing a beam of electrons in various directions and by observing the following: Electrons moving upward feel a magnetic force in the NW direction; Electrons moving horizontally North are pushed down; Electrons moving horizontally South-East are pushed upward. He naturally concludes that the magnetic field at this landing site is in which direction?
Answer:
Magnetic field is in south west direction .
Explanation:
Let us represent various direction by i , j, k . i representing east , j representing north and k representing vertically upward direction .
magnetic field is represented vectorially as follows
B = B₀ ( - i - j )
In the first case velocity of electron
v = v k
Force = q ( v x B )
= -e [ vk x B₀ ( - i - j ) ]
= evB₀ ( j -i )
Direction of force is north -west .
In the second case velocity of electron
v = vj
Force = -e [ vj x B₀ ( - i - j ) ]
= - evB₀ k
force is downward
In the third case, velocity of electron
v = v( -j +i )
Force = -e [ v( -j +i ) x B₀ ( - i - j ) ]
= 2 evB₀ k
Force is upward.
An object is moving in the absence of a net force. Which of the following best describes the object’s motion? A. The object will slow down at a constant rate until coming to rest B. The object will stop moving and remain at rest until acted on by a net force C. The object will continue to move at a constant speed but in a circular path D. The object will continue to move with a constant velocity
Answer:
D. The object will continue to move with a constant velocity
Explanation:
According to Newton's first law also known as law of inertia, states that an object at rest will remain at rest or, if in motion, will remain in motion at constant velocity unless acted on by a net external force.
Therefore, An object moving in the absence of a net force will continue to move at a constant velocity
In the absence of a net force, an object will continue to move with a constant velocity.
Explanation:The correct answer is D. The object will continue to move with a constant velocity. In the absence of a net force, an object will continue to move at a constant velocity. This means that the object will continue to move in a straight line at the same speed without slowing down or changing direction.
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A piece of glass has a temperature of 72.0 degrees Celsius. The specific heat capacity of the glass is 840 J/kg/deg C. A liquid that has a temperature of 40.0 degrees Celsius is poured over the glass, completely covering it, and the temperature at equilibrium is 57.0 degrees Celsius. The mass of the glass and the liquid is the same. Determine the specific heat capacity of the liquid
Answer:
741 J/kg°C
Explanation:
Given that
Initial temperature of glass, T(g) = 72° C
Specific heat capacity of glass, c(g) = 840 J/kg°C
Temperature of liquid, T(l)= 40° C
Final temperature, T(2) = 57° C
Specific heat capacity of the liquid, c(l) = ?
Using the relation
Heat gained by the liquid = Heat lost by the glass
m(l).C(l).ΔT(l) = m(g).C(g).ΔT(g)
Since their mass are the same, then
C(l)ΔT(l) = C(g)ΔT(g)
C(l) = C(g)ΔT(g) / ΔT(l)
C(l) = 840 * (72 - 57) / (57 - 40)
C(l) = 12600 / 17
C(l) = 741 J/kg°C
Kyle lays a mirror flat on the floor and aims a laser at the mirror. The laser beam reflects off the mirror and strikes an adjacent wall. The plane of the incident and reflected beams is perpendicular to the wall. The beam from the laser strikes the mirror at a distance a = 16.3 cm a=16.3 cm from the wall. The reflected beam strikes the wall at a height b = 32.5 cm b=32.5 cm above the surface of the mirror. Find the angle of incidence θ i θi at which the laser beam strikes the mirror.
Answer:
26.64°
Explanation:
Given:
a = 16.3 cm
b = 32.5 cm
Angle when laser beam reflects off the mirror and strike the wall =
θ [tex] = tan^-^1(\frac{b}{a}) [/tex]
[tex] = tan^-^1(\frac{32.5}{16.3}) [/tex]
= 63. 36°
For angle of reflection, we have:
θr = 90° - 63.36°
θr = 26.64°
Since angle of incidence, θi is equal to angle of reflection θr, the angle of incidence θi at which the laser beam strikes the mirror is =
θi = θr = 26.64°
To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 1 The pupil of the eye is the circular mm in dim light. Find the angular resolution of the eye for 550 nm opening through which light enters. wavelength light at those extremes. In which light can you see more sharply, dim or bright
Correct question is;
To regulate the intensity of light reaching our retinas, our pupils1 change diameter anywhere from 2 mm in bright light to 8 mm in dim light. Find the angular resolution of the eye for 550 nm wavelength light at those extremes. In which light can you see more sharply, dim or bright?
Answer:
We'll see more sharply in dim light
Explanation:
If we consider diffraction through a circular aperture, then angular resolution is given by;
θ = 1.22λ/D
where:
θ is the angular resolution (radians) λ is the wavelength of light
D is the diameter of the lens' aperture.
Thus,
at diameter = 2mm = 2 x 10^(-3) m = 2 x 10^(6) nm
θ = (1.22 * 550)/(2 x 10^(6))
θ = 335.5 x 10^(-6) radians
Now, we need to convert this to arc seconds.
Thus;
1 arc second = 4.85 x 10^(-6) radians
So,θ = 335.5 x 10^(-6) radians = [335.5 x 10^(-6)]/[4.85 x 10^(-6)]
= 69.18 arc seconds
at diameter = 8mm = 8 x 10^(-3) m = 8 x 10^(6) nm
θ = (1.22 * 550)/(8 x 10^(6))
θ = 83.875 x 10^(-6) radians
Now, we need to convert this to arc seconds.
Thus;
1 arc second = 4.85 x 10^(-6) radians
So,θ = 83.875 x 10^(-6) radians = [83.875 x 10^(-6)]/[4.85 x 10^(-6)]
= 17.3 arc seconds
From the values of angular resolution gotten, we see that sharpness of image increases with increasing angular resolution. Thus, objects are sharper in dim light.
A 969-kg satellite orbits the Earth at a constant altitude of 99-km. (a) How much energy must be added to the system to move the satellite into a circular orbit with altitude 195 km? 469 Incorrect: Your answer is incorrect. How is the total energy of an object in circular orbit related to the potential energy? MJ (b) What is the change in the system's kinetic energy? MJ (c) What is the change in the system's potential energy?
Answer:
1.3*10^14 J
Explanation:
The energy of the satellite that orbits the earth is given by the second Newton law:
[tex]F=ma_c\\\\-G\frac{mM_s}{r^2}=m\frac{v^2}{r}\\\\v^2=\frac{GM}{r}\\\\E_T=K+U=G\frac{mM_s}{2r}-G\frac{mM}{r}=-G\frac{mM}{2r}[/tex]
where you have taken into account the centripetal acceleration of the satellite.
m: mass of the satellite
M_s: mass of the sun = 1.98*10^30 kg
G: Cavendish's constant = 6.67*10^-11 m^3/kg s^2
r: distance to the center of the Earth = Earth radius + distance satellite-Earth surface
To find the needed energy, you first compute the energy for a constant altitude of 99km:
r = 6.371*10^6m + 99*10^3m = 6.47*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.47*10^6)}\\\\E_T=-9.88*10^{15} \ J[/tex]
Next, you calculate the energy for an altitude of 195km:
r = 6.371*10^6m + 195*10^{3}m = 6.56*10^6 m
[tex]E_T=-(6.67*10^-11 m^3 /kg.s^2)\frac{(969kg)(1.98*10^{30}kg)}{2(6.56*10^6)}\\\\E_T=-9.75*10^{15} \ J[/tex]
Finally, the energy required to put the satellite in the new orbit is:
-9.75*10^15 J - (-9.88*10^15 J) = 1.3*10^14 J
Which statements are true about the flow of blood in the body? Check all that apply.
Answer:
i need some explanation
Explanation:
The correct statements are D, C, and E. Blood flows from the heart to the lungs to pick up oxygen and then to the rest of the body to deliver oxygen, sugar, and nutrients while collecting carbon dioxide.
Understanding the flow of blood in the body is essential. Here are the correct statements regarding blood circulation:
D.) Blood flows from the heart to the lungs to pick up oxygen.C.) Blood picks up carbon dioxide from the cells of the body.E.) Blood delivers sugar and nutrients to cells in the body.The heart pumps oxygen-poor blood to the lungs through the pulmonary circuit where it releases carbon dioxide and picks up oxygen.
The oxygen-rich blood is then pumped through the systemic circuit to the rest of the body, delivering oxygen, sugar, and nutrients to the cells and collecting carbon dioxide to be expelled during the next circulation.
Therefore, the correct statements are D, C, and E.
Complete Question
Which statements are true about the flow of blood in the body? Check all that apply.
A. Blood picks up oxygen from the cells of the body.
B. Blood delivers carbon dioxide to cells in the body.
C. Blood picks up carbon dioxide from the cells of the body.
D. Blood flows from the heart to the lungs to pick up oxygen.
E. Blood delivers sugar and nutrients to cells in the body.
F. Blood flows from the lungs to the heart to pick up oxygen.
Vectors A and B lie in the x-y plane. Vector A has a magnitude of 17.6 and is at an angle of 120.5° counter-clockwise from the x-axis. Vector B has a magnitude of 21.7 and is 240.3° from the x-axis. Resolve A and B into components, and express in unit vector form below.
The unit vector forms of the given vectors is required.
The required vectors are [tex]A=-8.93\hat{i}+15.16\hat{j}[/tex] and [tex]B=-10.75\hat{i}-18.84\hat{j}[/tex]
VectorsMagnitude of vector A = [tex]|A|=17.6[/tex]
Angle vector A makes with positive x axis counter clockwise = [tex]\theta_1=120.5^{\circ}[/tex]
Magnitude of vector B = [tex]|B|=21.7[/tex]
Angle vector B makes with positive x axis counter clockwise = [tex]\theta_2=240.3^{\circ}[/tex]
The vectors need to be resolved in order to write in the unit vector forms.
The vectors are
[tex]A=|A|(\cos\theta_1\hat{i}+\sin\theta_1\hat{j})\\\Rightarrow A=17.6(\cos120.5\hat{i}+\sin120.5\hat{j})\\\Rightarrow A=-8.93\hat{i}+15.16\hat{j}[/tex]
[tex]B=|B|(\cos\theta_2\hat{i}+\sin\theta_2\hat{j})\\\Rightarrow B=21.7(\cos240.3\hat{i}+\sin240.3\hat{j})\\\Rightarrow B=-10.75\hat{i}-18.84\hat{j}[/tex]
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Final answer:
To resolve the vectors into components, apply trigonometric functions -- cosine for x components and sine for y components -- to their magnitudes and angles, then express them in unit vector form with i and j.
Explanation:
When resolving vectors A and B into their components in the x-y plane, the general method involves using trigonometry, specifically, the cosine and sine functions for the x and y components, respectively. Given that vector A has a magnitude of 17.6 and an angle of 120.5° from the x-axis, its components can be calculated as follows:
Ax = A * cos(θ) = 17.6 * cos(120.5°)
Ay = A * sin(θ) = 17.6 * sin(120.5°)
Similarly, vector B with a magnitude of 21.7 and an angle of 240.3° from the x-axis has components:
Bx = B * cos(θ) = 21.7 * cos(240.3°)
By = B * sin(θ) = 21.7 * sin(240.3°)
The resulting components should then be written in unit vector form by attaching the unit vectors i (for the x-axis) and j (for the y-axis) to the respective components.
Katie rubs a balloon against her hair. Electrons from her hair travel to the balloon, giving the balloon a negative charge and her hair a positive charge.
When the negatively charged balloon is brought near the strands of Katie's hair, they move to get closer to the balloon, without the balloon actually touching them. This shows that
A. electric attraction is a force that can only act on contact.
B. Katie's hair would move in this way with or without the balloon.
C. electric attraction is a force that can act at a distance.
D. particles in the air must be pulling Katie's hair toward the balloon.
Answer:
c. electric attraction is a force that can act at a distance.
Explanation:
stuisland
Final answer:
A negatively charged balloon attracting positively charged hair strands without contact illustrates that c. electric attraction can act at a distance.
Explanation:
When Katie rubs the balloon against her hair, electrons move from her hair to the balloon, resulting in the balloon having a negative charge and her hair having a positive charge. If the negatively charged balloon is brought near Katie's hair and the hair strands move toward it without direct contact, this demonstrates electric attraction as a force that can act at a distance.
Therefore, electric attraction is a force that can act at a distance. This example, similar to the effect observed when someone touches a Van de Graaff generator, shows charge separation and induction. It validates the scientific concept that electric forces can operate between charged objects even when they are not in physical contact.
An imaginary cubical surface of side L has its edges parallel to the x-, y- and z-axes, one corner at the point x = 0, y = 0, z = 0 and the opposite corner at the point x=L, y=L, z=L. The cube is in a region of uniform electric field E⃗ =E1i^+E2j^, where E1 and E2 are positive constants. Calculate the electric flux through the cube face in the plane x = 0 and the cube face in the plane x=L. For each face the normal points out of the cube.
Find the given attachment for solution
The electric flux trough x = 0 plane is = - EL² and the electric flux trough x = l plane is = EL².
What is electric flux?Although an electric field cannot flow by itself, electric flux in electromagnetism is a measure of the electric field passing through a specific surface.
An electric field surrounds an electric charge, such as a solitary electron in space. Field lines have no physical significance and are merely a graphic representation of field strength and direction.
The number of "lines" per unit area, also known as the electric flux density, is inversely proportional to the electric field strength. The total number of electric field lines passing through a surface determines the amount of electric flux.
Hence, electric flux trough x = 0 plane is = - EL² and the electric flux trough x = l plane is = EL².
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The simple pendulum above consists of a bob hanging from a light string. You wish to experimentally determine the frequency of the swinging pendulum. (a) By checking the line next to each appropriate item on the list below, select the equipment that you would need to do the experiment. ____ Meter Stick ____ Protractor ____ Additional string ____ Stopwatch ____ Photogate ____ Additional masses (b) Describe the experimental procedure that you would use. In your description, state the measurements you would make, how you would use the equipment to make them, and how you would determine the frequency from those measurements. (c) You next wish to discover which parameters of a pendulum affect its frequency. State one parameter that could be varied, describe how you would conduct the experiment, and indicate how you would analyze the data to show whether there is a dependence. (d) After swinging for a long time, the pendulum eventually comes to rest. Assume that the room is perfectly thermally insulated. How will the temperature of the room change while the pendulum comes to rest
A) the item required for the experiment is a stopwatch.
B) You are looking for time savings. To get this, divide the time by 10. The result is the period. The formula is given as
F = [tex]\frac{1}{T}[/tex] Where F is the frequency i.e. the number of cycles per second and T is the number of seconds per cycle.
C) One parameter of the pendulum that can be altered in order to affect the frequency of the pendulum is its length. A pendulum with a longer string will have a lower frequency.
The one with a shorter length will have a higher frequency
D) In an environment that is thermally insulated perfectly, it means that any heat generated within the room is trapped within it. As the pendulum comes to rest, the room will experience a slight increase in temperature due to the conversion of mechanical energy to thermal energy.
Uses of the pendulum experimentsUnderstanding the physics of pendulums helps one to get a better grasp of gravity, inertia, and centripetal force.
Pendulums are used for the construction or engineering of clocks, metronomes, sismometers, amusement park rides.
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The simple pendulum above consists of a bob hanging from a light string the experiments is :
A. Stopwatch
B. You are looking for time savings. To get this, divide the time by 10.
C. The one parameter of the pendulum that can be modified in arrange to influence the recurrence of the pendulum is its length.
D. In an environment that's thermally protects superbly
"Simple Pendulum"Answer A:
The item that is required for the experiment is a stopwatch.Answer B:
The experimental procedure that you would use is :
You are looking for time savings. To get this, divide the time by 10. The result is the period. Formula is given as :F = 1/10Where :
F is the frequency and T is the number of seconds per cycle.Answer C:
The one parameter of the pendulum that can be modified in arrange to influence the recurrence of the pendulum is its length. A pendulum with a longer string will have a lower frequency. The one with a shorter length will have a better frequency.Answer D:
In an environment that's thermally protects superbly, it implies that any warm created inside the room is caught inside it. As the pendulum comes to rest, the room will encounter a slight increment in temperature due to the change of mechanical vitality to warm energy.Uses of the pendulum experiments :
Understanding the material science of pendulums helps one to induce distant better a much better, a higher, a stronger ,an improved an improved get a handle on of gravity, inactivity, and centripetal force. Pendulums are utilized for the development or building of clocks, metronomes, sismometers, entertainment stop rides.Learn more about pendulum experiments :
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A small, positively charged ball is moved close to a large, positively charged ball. Which describes how the small ball likely responds when it is released?
It will move toward the large ball because like charges repel.
It will move toward the large ball because like charges attract.
It will move away from the large ball because like charges repel.
It will move away from the large ball because like charges attract.
Answer:
C
Explanation:
i took the test
A small, positively charged ball is moved close to a large, positively charged ball. "It will move away from the large ball because like charges repel." The correct option is A.
Charge Interaction: The behavior of charged objects is governed by the fundamental principle that opposite charges attract each other, and like charges repel each other.
Coulomb's Law: Coulomb's Law describes the electrostatic force between two charged objects. It states that the force of attraction or repulsion between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Repulsion: Because both the small and large balls have a positive charge, they will exert a repulsive force on each other when they are in close proximity.
Movement Away: When the small ball is released near the large ball, it will experience this repulsive force, causing it to move away from the large ball.
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A mass is hung from a spring and set in motion so that it oscillates continually up and down. The velocity v of the weight at time t is given by the equation v=−2 cos(3πt) with v measured in feet per second and t measured in seconds. Determine the maximum velocity of the mass and the amount of time it takes for the mass to move from its lowest position to its highest position.
Answer:
the maximum velocity of the mass v (max) = 2 ft/s
the amount of time it takes for the mass to move from its lowest position to its highest position∆t = 1/3 seconds = 0.33 seconds
Explanation:
Given the velocity equation;
v=−2 cos(3πt)
The maximum velocity would be at cos(3πt) = 1 or cos(3πt) = -1
v (max) = -2 × -1 = 2 ft/s
The time taken for the mass to move from lowest position to highest position
At Lowest position, vertical velocity equals zero.
At highest position, vertical velocity equals zero.
The time taken for the mass to move from one v = 0 to the next v = 0
Cos(π/2) = 0 and
Cos(3π/2) = 0
For the first;
Cos(3πt) = cos(π/2)
3πt1 = π/2
t1 = π/2(3π)
t1 = 1/6 second
For the second;
Cos(3πt) = cos(3π/2)
3πt2 = 3π/2
t2 = 3π/2(3π)
t2 = 1/2 second
∆t = t2 - t1 = 1/2 - 1/6 = 3/6 - 1/6 = 2/6 = 1/3 seconds
∆t = 1/3 seconds
A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV. The beam is then entered into a region between two parallel metal plates with potential difference 120 V and a separation 8 mm, perpendicular to the direction of the field. What magnitude of magnetic field is needed so that the alpha particles emerge undeflected from between the plates?
Answer:
The magnetic field required required for the beam not to be deflected is [tex]B = 0.0036T[/tex]
Explanation:
From the question we are told that
The charge on the particle is [tex]q = +2e[/tex]
The mass of the particle is [tex]m = 6.64 *10^{-27} kg[/tex]
The potential difference is [tex]V_a = 1.8 kV = 1.8 *10^{3} V[/tex]
The potential difference between the two parallel plate is [tex]V_b = 120 V[/tex]
The separation between the plate is [tex]d = 8 mm = \frac{8}{1000} = 8*10^{-3}m[/tex]
The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV
[tex]KE_b = PE_b[/tex]
Generelly
[tex]KE_b = \frac{1}{2} m v^2[/tex]
And [tex]PE_b = q V_a[/tex]
Equating this two formulas
[tex]\frac{1}{2} mv^2 = q V_a[/tex]
making v the subject
[tex]v = \sqrt{\frac{q V_a}{2 m} }[/tex]
Substituting value
[tex]v = \sqrt{\frac{ 2* 1.602 *10^{-19} * 1.8 *10^{3}}{2 * 6.64 *10^{-27}} }[/tex]
[tex]v = 41.65*10^4 m/s[/tex]
Generally the electric field between the plates is mathematically represented as
[tex]E = \frac{V_b}{d}[/tex]
Substituting value
[tex]E = \frac{120}{8*10^{-3}}[/tex]
[tex]E = 15 *10^3 NC^{-1}[/tex]
the magnetic field is mathematically evaluate
[tex]B = \frac{E}{v}[/tex]
[tex]B = \frac{15 *10^{3}}{41.65 *10^4}[/tex]
[tex]B = 0.0036T[/tex]
Problem (2) A 16 kg cylinder, initially at rest, is held by a cord connected to a grooved drum whose mass is 20 kg. The drum has an outer radius �% = 250 mm and an inner radius of �& = 160 mm. If the drum experiences a constant frictional moment of 3 N∙m at O, how far has the cylinder dropped when it has a downward velocity of 2 m/s? Neglect the mass of the cord and treat the drum as a thin disk. Use the Principle of Work and Energy.
Answer:
Explanation:
The solution to the problem is given in the pictures attached below; the three pictures explains the problem fully and I hope it helps you. Thank you
A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. True False
Answer:
The question above is repeated twice.
Removing the repetition, we have: A plane flies toward a stationary siren at 1/4 the speed of sound. Then the plane stands still on the ground and the siren is driven toward it at 1/4 the speed of sound. In both cases, a person sitting in the plane will hear the same frequency of sound from the siren. True or False?
The correct answer to the question is "False"
Explanation:
The question above, illustrates a phenomenon referred to as "Doppler effect"
The Doppler effect only changes the frequency of the sound which explains how wavelength changes when a wave source is moving toward or away from an object. The Doppler effect occurs when a source of waves and/or observer move relative to each other.
When a sound source is moving toward the observer (a person sitting in the plane) in the case above, the observer will hear a higher pitch as the source approaches. That is, the plane stands still on the ground and the siren is driven toward it.This is due to a decrease in the amplitude of the sound wave.
However, If the observer moves toward the stationary source, the observed frequency is higher than the source frequency. In this case, A plane flies toward a stationary siren.
λ = v/f = vT,
where T is the period,
The relationship between frequency, speed, and wavelength is:
f = v/λ
v represents the speed of sound through the medium.
Doppler effect depends on things moving, as the observer moves, the frequency becomes higher as the distance decreases. If the observer moves and the distance becomes larger, it means that the sound frequency becomes lower.
Final answer:
The Doppler effect explains why a person sitting in a plane moving toward or away from a stationary siren at 1/4 the speed of sound perceives the same frequency of sound. This phenomenon is true due to the relative motion between the source of sound and the observer.
Explanation:
True
The phenomenon described in the question is related to the Doppler effect in physics. When a source of sound and an observer are in motion relative to each other, the frequency of the sound waves changes due to this motion. In this case, when the plane is moving toward or away from the siren at 1/4 the speed of sound, the observer perceives the same frequency of sound from the siren.
Consider an incident normal shock wave that reflects from the end wall of a shock tube. The air in the driven section of the shock tube (ahead of the incident wave) is at p, = 0.01 atm and TI = 300 K. The pressure ratio across the incident shock is 1050. With the use of Eq. (7.23), calculate a. The reflected shock wave velocity relative to the tube b. The pressure and temperature behind the reflected shock
Answer:
Find the given attachments for complete solution
Blood flow rates in the umbilical cord can be found by measuring the Doppler shift of the ultrasound signal reflected by the red blood cells. If the source emits a frequency f, what is the measured reflected frequency fR? Assume that all of the red blood cells move directly toward the source. Let c be the speed of sound in blood and v be the speed of the red blood cells.
Answer:
fR = f(c + v)/c
Explanation:
The speed of a wave is its frequency x wavelenght. Therefore,
Frequency is speed of wave over the wavelength.
Since the source (ultrasound machine) is stationary, and the receiver red blood cell is moving towards it. The wavelenght of the wave sent out towards the observer is c/f
The speed of the reflected sound wave is (c + v), so that the reflected frequency fR is given by
fR = f(c + v)/c
The measured reflected frequency ( fR) in Doppler-shifted ultrasound, when blood cells move towards the source, is calculated using the formula fR = f * (c + v) / c, where f is the original frequency, c is the speed of sound in blood, and v is the velocity of blood cells.
The question relates to the principle of the Doppler effect and its application in calculating the velocity of blood flow using Doppler-shifted ultrasound. The Doppler effect occurs when a wave source and an observer are in relative motion, resulting in a change in the observed frequency. Specifically, when the blood cells move towards the ultrasound source, the frequency of the reflected ultrasound increases. This change in frequency can be measured for diagnostic purposes.
To find the measured reflected frequency fR when blood cells are moving towards the source, you can use the formula:
fR = f * (c + v) / c
Where,
f = original frequency of the ultrasound
c = speed of sound in blood (or human tissue)
v = velocity of blood cells relative to the ultrasound source
A sound is traveling through the air with a temperature of 35•C. The sound wave has a wavelength of 0.75 meters. What is the frequency of the sound
Answer:
f = 687.85 Hz
Explanation:
Given that,
The wavelength of sound wave, [tex]\lambda=0.75\ m[/tex]
We need to find the frequency of the sound wave. The relation between wavelength and frequency is given by :
[tex]v=f\lambda[/tex]
v is speed of sound at T = 35°C = 308.15 K
[tex]v=331+0.6T\\\\v=331+0.6\times 308.15 \\\\v=515.89\ m/s[/tex]
So,
[tex]f=\dfrac{v}{\lambda}\\\\f=\dfrac{515.89}{0.75}\\\\f=687.85\ Hz[/tex]
So, the frequency of the sound is 687.85 Hz.
Final answer:
To calculate the frequency of a sound wave, use the formula frequency = speed of sound / wavelength. For this specific scenario, the frequency of the sound wave is 3400 Hz.
Explanation:
The frequency of the sound wave can be calculated using the formula:
frequency = speed of sound / wavelength
Given that the speed of sound in air is 340 m/s and the wavelength is 0.10 m, we can plug in the values to find the frequency:
frequency = 340 m/s / 0.10 m = 3400 Hz
Therefore, the frequency of the sound wave is 3400 Hz.
What is the distance from the center of the Moon to the point between Earth and the Moon where the gravitational pulls of Earth and Moon are equal? The mass of Earth is 5.97 × 1024 kg, the mass of the Moon is 7.35 × 1022 kg, the center-to-center distance between Earth and the Moon is 3.84 × 108 m, and G = 6.67 × 10-11 N ∙ m2/kg2. Group of answer choices 4.69 × 107 m 4.69 × 106 m 3.83 × 106 m 3.45 × 108 m
Answer:
3.83*10^7 m
Explanation:
Assume that the distance at which gravitational force due to both Earth and moon is zero and it l is given by force force balance
F(moon) = F(earth)
F(moon) = GM(moon) / r²
F(earth) = GM(earth) / (d - r)²
If F(moon) = F(earth), then
GM(moon) / r² = GM(earth) / (d - r)²
7.35*10^22 / r² = 5.97*10^24 / (3.84*10^8 - r)²
now, we take the square root of both sides, we have
2.71*10^11 / r² = 24.4*10^11 / (3.84*10^8 - r) =>
2.71 / r² = 24.4 / (3.84*10^8 - r)
if we cross multiply, we have
24.4r = 1.04064*10^9 - 2.71r
24.4r + 2.71r = 1.04064*10^9
27.11r = 1.04064*10^9
r = 1.04064*10^9 / 27.11
r = 3.83*10^7 m
Acceleration is measured in m/s/s (meters per second squared).
True or False
Answer:
true
Explanation:
Answer:true
Explanation:
Acceleration is measured in meters per second squared
Tish scarf is 17.75 inches long and knits 2 3/8 per minute. Emma scarf is 4 inches knits 3 3/4 inches per minute. After how many minutes will Emma’s scarf be longer than tish scarf
Answer:
t = 10 minutes
Emma’s scarf will be longer than tish scarf after 10 minutes
Explanation:
Given;
The length of tish scarf after t minutes can be written as;
L1 = 17.75 + 2 3/8 ×t
L1 = 17.75 + 2.375t
The length of emma scarf after t minutes can be written as;
L2 = 4 + 3 3/4 × t
L2 = 4 + 3.75t
For emma scarf to be longer than tish;
L2 >/= L1
At L2 = L1
4+3.75t = 17.75 + 2.375t
Solving for t
3.75t - 2.375t = 17.75 - 4
1.375t = 13.75
t = 13.75/1.375
t = 10 minutes
Emma’s scarf will be longer than tish scarf after 10 minutes
A 88.6-kg wrecking ball hangs from a uniform heavy-duty chain having a mass of 26.9kg . (Use 9.80m/s2 for the gravitational acceleration at the earth's surface.)
Part A
Find the maximum tension in the chain.
Tmax = N
Part B
Find the minimum tension in the chain.
Tmin = N
Part C
What is the tension at a point three-fourths of the way up from the bottom of the cha
Answer:
Tension maximum =1131.9 N
Tension minimum =868.28 N
Tension at 3/4= 1065.995 N
Explanation:
a)
Given Mass of wrecking ball M1=88.6 Kg
Mass of the chain M2=26.9 Kg
Maximum Tension Tension max=(M1+M2) × (9.8 m/s²)
=(88.6+26.9) × (9.8 m/s²)
=115.5 × 9.8 m/s²
Tension maximum =1131.9 N
b)
Minimum Tension Tension minimum=Mass of the wrecking ball only × 9.8 m/s²
=88.6 × 9.8 m/s²
Tension minimum =868.28 N
c)
Tension at 3/4 from the bottom of the chain =In this part you have to use 75% of the chain so you have to take 3/4 of 26.9
= (3/4 × 26.9)+88.9) × 9.8 m/s²
= (20.175+88.6) × 9.8 m/s²
=(108.775) × 9.8 m/s²
=1065.995 N
Final answer:
The maximum tension in the chain is 1131.9 N, occurring at the top, while the minimum tension is 263.62 N at the bottom. The tension at a point three-fourths the way up from the bottom is 935.465 N.
Explanation:
To find the maximum and minimum tension in the chain, we need to consider the system's configuration, and the force due to gravity. The maximum tension occurs at the top of the chain, where it supports the entire weight of the wrecking ball and the chain. The minimum tension occurs at the bottom of the chain, where it only needs to support the chain's weight. To find the tension at a point three-fourths of the way up from the bottom, we need to consider the weight of the portion of the chain below that point and the wrecking ball's weight.
Maximum tension (Tmax) is the sum of the weight of the wrecking ball and the entire chain:
Tmax = (mass of ball + mass of chain) × gravitational acceleration
Tmax = (88.6 kg + 26.9 kg) × 9.80 m/s²
Tmax = 115.5 kg × 9.80 m/s²
Tmax = 1131.9 N
Minimum tension (Tmin) is just the weight of the chain:
Tmin = mass of chain × gravitational acceleration
Tmin = 26.9 kg × 9.80 m/s²
Tmin = 263.62 N
Tension at three-fourths the way up:
We calculate the weight of the top one-fourth of the chain plus the wrecking ball:
Tension at three-fourths the way up = (mass of one-fourth of the chain + mass of ball) × gravitational acceleration
Tension at three-fourths = ((26.9 kg / 4) + 88.6 kg) × 9.80 m/s²
Tension at three-fourths = (6.725 kg + 88.6 kg) × 9.80 m/s²
Tension at three-fourths = 935.465 N
Calculate How much energy is transferred as useful energy
=A 98% efficient kettle that has a total input of 2000J
Answer:1960j
Explanation:
total input energy=2000j
98% of total input energy is useful
98% of 2000
98/100 x 2000
(98 x 2000) ➗ 100
196000 ➗ 100=1960
1960j is useful
The amount of useful transferred energy is 1960 J.
What is energy?A body's capacity for work is measured in terms of energy. It cannot be produced or eliminated. There are numerous types of energy, including thermal, electrical, fusion, electrical, and nuclear. Energy has the ability to change its forms.
What is efficiency?Efficiency is essentially a measurement of the amount of labour or energy that can be saved throughout a process. In other words, it's similar to comparing the energy input and output in any particular system. For instance, we observe that many processes result in the loss of effort or energy like vibration or waste heat.
Given parameters:
Total input energy; I =2000 Joule.
efficiency of the kettle; η = 98%.
We have to find useful output energy of the kettle: O = ?
We know that: output energy = efficiency × input energy
= 98% × 2000 J.
= 98/100 x 2000 J.
= (98 x 2000) ÷ 100 J.
= 196000 ÷ 100 J
= 1960 J.
Hence, the useful transferred energy is 1960 J.
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