Answers
Answer:
D(t)= 11x2.09t cos( 20pi,t)
Explanation:
Pls attached file
Related Questions
An electron, moving toward the west, enters a uniform magnetic field. Because of this field the electron curves upward. The direction of the magnetic field is
Answers
Answer:
The magnetic field's direction is towards the north
Explanation:
The force on a positive charge in a uniform magnetic field is represented by the right hand rule. To determine the direction of the force, place your right hand with your palm up, and your thumb at 90° to the other fingers. If the fingers represent the magnetic field, and the thumb the direction of the positive charge, then the palm will push up in the direction of the force. But a negative charge like an electron pushes in exactly the opposite direction. So if you follow this rule, you will find that the magnetic field points towards the north.
The direction of the magnetic field is towards the North. This can be
determined using the right hand rule by Fleming.
The right hand rule states that to determine the direction of the magnetic
force, the right thumb should be pointed in in the direction of the velocity,
index finger in the direction of the magnetic field and middle finger in the
direction of magnetic force.
When this is applied, we will discover that the index finger will point towards
the north region.
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A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0 m away. (a)How wide on the screen is the central bright fringe
Answers
Answer:
0.0127m
Explanation:
Using
Ym= (1)(633x10^-9m)(2m) / (0.1x10^-3m) = 0.0127m
A centrifuge's angular velocity is initially at 159.0 radians/second to test the stability
of a high speed drill component. It then increases its angular velocity to 999.0
radians/second. If this is achieved in 4,100.0 radians what is the angular acceleration
of the centrifuge?
Answers
Answer:
118.6 rad/s²
Explanation:
Δθ = 4100.0 rad
ω₀ = 159.0 rad/s
ω = 999.0 rad/s
Find: α
ω² = ω₀² + 2αΔθ
(999.0 rad/s)² = (159.0 rad/s)² + 2α (4100.0 rad)
α = 118.6 rad/s²
What would be the correct value of m in the interference equation if the growth process were started again and the diamond layer were grown to three times the thickness t
Answers
Answer:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
Explanation:
To calculate the correct value of m if the thickness t is grown 3 times again we can deduce that:
2*(3u)*t = m* lambda
Making m the subject of the formula will give the formula:
m= 6*u*t/ Lambda
Given: Lambda= 633*10^9 while u and t are unknown
Therefore the value of m can be calculated given the formula below:
m= 6*u*t/ 633*10^9
A spaceship is moving past Earth at 0.99c. The spaceship fires two lasers. Laser A is in the same direction it is traveling, and Laser B is in the opposite direction. How fast will the light from each laser be traveling according to an observer on Earth?
Answers
Answer:
Vx' = (Vx - u) / (1 - Vx *u / c^2) velocity transformation formula
In both cases we wish to measure the velocity in the frame of the earth which is moving at speed u = -.99 c relative to the spaceship
VA' = (c + .99c) / (1 - (-.99 c * c) / c^2) = 1.99c / 1.99 = c
VB' = (-c + .99c) / (1 - (-c * -.99c) / c^2) = .01 c / .01 = c
In both cases an observer on earth will observe the light traveling at speed c.
An apple falls from a tree and hits your head with a force of 9J. The apple weighs 0.22kg. How far did the apple fall?
Answers
Answer:
The apple fell at a distance of 4.17 m.
Explanation:
Work is defined as the force that is applied on a body to move it from one point to another. When a force is applied, an energy transfer occurs. Then it can be said that work is energy in motion.
When a net force is applied to the body or a system and this produces displacement, then that force is said to perform mechanical work.
In the International System of Units, work is measured in Joule. Joule is equivalent to Newton per meter.
The work is equal to the product of the force by the distance and by the cosine of the angle that exists between the direction of the force and the direction that travels the point or the object that moves.
Work=Force*distance* cosine(angle)
On the other hand, Newton's second law says that the acceleration of a body is proportional to the resultant of forces on it acting and inversely proportional to its mass. This is represented by:
F=m*a
where F is Force [N], m is Mass [kg] and a Acceleration [m / s²]
In this case, the acceleration corresponds to the acceleration of gravity, whose value is 9.81 m / s². So you have:
Work= 9 JF=m*a=0.22 kg*9.81 m/s²= 2.1582 Ndistance= ?angle=0 → cosine(angle)= 1Replacing:
9 J= 2.1582 N* distante* 1
Solving:
[tex]distance=\frac{9J}{2.1582 N*1}[/tex]
distance= 4.17 m
The apple fell at a distance of 4.17 m.
Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diameter and 6 meters long is carried on the back of a truck and is used to fuel tractors. The axis of the tank is horizontal. The opening on the tractor tank is 5 meters above the top of the tank in the truck. Find the work done in pumping the entire contents of the fuel tank into the tractor.
Answers
Answer:
work done in pumping the entire fuel is 1399761 J
Explanation:
weight per volume of the gasoline = 6600 N/m^3
diameter of the tank = 3 m
length of the tank = 6 m
The height of the tractor tank above the top of the tank = 5 m
The total volume of the fuel is gotten below
we know that the tank is cylindrical.
we assume that the fuel completely fills the tank.
therefore, the volume of a cylinder =
where r = radius = diameter ÷ 2 = 3/2 = 1.5 m
volume of the cylinder = 3.142 x x 6 = 42.417 m^3
we then proceed to find the total weight of the fuel in Newton
total weight = (weight per volume) x volume
total weight = 6600 x 42.417 = 279952.2 N
therefore,
the work done to pump the fuel through to the 5 m height = (total weight of the fuel) x (height through which the fuel is pumped)
work done in pumping = 279952.2 x 5 = 1399761 J
A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answers
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant voltage by a battery as they are pulled apart to twice their original separation, which is small compared to the dimensions of the plates. The amount of charge on the plates is now equal to:__________.
a) 4 Q.
b) 2 Q.
c) Q.
d) Q/2.
e) Q/4.
Answers
Answer:
D. Q/2
Explanation:
See attached file
Calculate the wavelengths of the first five members of the Lyman series of spectral lines, providing the result in units Angstrom with precision one digit after the decimal point.
Answers
Answer:
Explanation:
The formula for hydrogen atomic spectrum is as follows
energy of photon due to transition from higher orbit n₂ to n₁
[tex]E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV[/tex]
For layman series n₁ = 1 and n₂ = 2 , 3 , 4 , ... etc
energy of first line
[tex]E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})[/tex]
10.2 eV
wavelength of photon = 12375 / 10.2 = 1213.2 A
energy of 2 nd line
[tex]E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})[/tex]
= 12.08 eV
wavelength of photon = 12375 / 12.08 = 1024.4 A
energy of third line
[tex]E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})[/tex]
12.75 e V
wavelength of photon = 12375 / 12.75 = 970.6 A
energy of fourth line
[tex]E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})[/tex]
= 13.056 eV
wavelength of photon = 12375 / 13.05 = 948.3 A
energy of fifth line
[tex]E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})[/tex]
13.22 eV
wavelength of photon = 12375 / 13.22 = 936.1 A
The temperature coefficient of resistivity for copper is 0.0068 (C°)-1. If a copper wire has a resistance of 104 Ω at 20°C, what is its resistance 80°C?
Answers
Answer:
R₈₀ = 146.43 Ω
Explanation:
The resistance of a resistor depends upon many factors. One of the main factors of the change in resistance of a resistor is the change in temperature. The formula for the resistance at a temperature other than 20°C is given as follows:
R₈₀ = R₀(1 + αΔT)
where,
R₈₀ = Resistance of wire at 80°C = ?
R₀ = Resistance of wire at 20° C = 104 Ω
α = Temperature coefficient of resistance for copper = 0.0068 °C⁻¹
ΔT = T₂ - T₁ = 80°C - 20°C = 60°C
Therefore,
R₈₀ = (104 Ω)[1 + (0.0068°C⁻¹)(60°C)]
R₈₀ = 146.43 Ω
A commercial aircraft is flying westbound east of the Sierra Nevada Mountains in California. The pilot observes billow clouds near the same altitude as the aircraft to the south, and immediately turns on the "fasten seat belt" sign. Explain why the aircraft experiences an abrupt loss of 500 meters of altitude a short time later.
Answers
Answer:
Billow clouds provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents.
Explanation:
Billow clouds are created in regions that are not stable in a meteorological sense. They are frequently present in places with air flows, and have marked vertical shear and weak thermal separation and inversion (colder air stays on top of warmer air). Billow clouds are formed when two air currents of varying speeds meet in the atmosphere. They create a stunning sight that looks like rolling ocean waves. Billow clouds have a very short life span of minutes but they provide a visible signal to aviation interests of potentially dangerous turbulent sky since they indicate instability in air currents, which although may not affect us on the ground but is a concern to aircraft pilots. The turbulence due to the Billow wave is the only logical explanation for the loss of 500 m in altitude of the plane.
The cart now moves toward the right with an acceleration toward the right of 2.50 m/s2. What does spring scale Fz read? Show your calculations, and explain.
Answers
Complete Question
The complete question is shown on the first uploaded image
Answer:
The spring scale [tex]F_2[/tex] reads [tex]F_2 = 2.4225 \ N[/tex]
Explanation:
From the question we are told that
The first force is [tex]F_1 = 10.5 \ N[/tex]
The acceleration by which the cart moves to the right is [tex]a = 2.50 \ m/s^2[/tex]
The mass of the cart is m = 3.231 kg
Generally the net force on the cart is
[tex]F_{net} = F_1 - F_2[/tex]
This net force is mathematically represented as
[tex]F_{net} = m * a[/tex]
So
[tex]m* a = 10 - F_2[/tex]
[tex]F_2 = 10.5 - 2.5 (3.231)[/tex]
[tex]F_2 = 2.4225 \ N[/tex]
Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.
Answers
Answer:
a
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
b
[tex]L = 1.183 *10^{-7} \ H[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 6 \ cm = \frac{6}{100} = 0.06 \ m[/tex]
The current it carries is [tex]I = 1.50 \ A[/tex]
The magnetic flux of the coil is mathematically represented as
[tex]\phi = B * A[/tex]
Where B is the magnetic field which is mathematically represented as
[tex]B = \frac{\mu_o * I}{2 * r}[/tex]
Where [tex]\mu_o[/tex] is the magnetic field with a constant value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
substituting value
[tex]B = \frac{4\pi * 10^{-7} * 1.50 }{2 * 0.06}[/tex]
[tex]B = 1.571 *10^{-5} \ T[/tex]
The area A is mathematically evaluated as
[tex]A = \pi r ^2[/tex]
substituting values
[tex]A = 3.142 * (0.06)^2[/tex]
[tex]A = 0.0113 m^2[/tex]
the magnetic flux is mathematically evaluated as
[tex]\phi = 1.571 *10^{-5} * 0.0113[/tex]
[tex]\phi = 1.78 *10^{-7} \ Weber[/tex]
The self-inductance is evaluated as
[tex]L = \frac{\phi }{I}[/tex]
substituting values
[tex]L = \frac{1.78 *10^{-7} }{1.50 }[/tex]
[tex]L = 1.183 *10^{-7} \ H[/tex]
Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3, 2, 1, 4 D. 3, 1, 2, 4
Answers
Given that,
Rank in order from largest to smallest the magnitude of the electric field at block dot.
Electric field :
Electric field is proportional to the charge divided by square of distance.
In mathematically,
[tex]E\propto\dfrac{q}{r^2}[/tex]
Where, q = charge
r = distance
If the charge is greater then electric field will be greater.
If the distance is greater then electric field will be smaller.
We need to find the electric field at black dot
According to figure,
(I). The electric field at black dot due to positive charge point q to left. the distance is r.
The electric field will be
[tex]E=\dfrac{kq}{r^2}[/tex]
The electric field will be largest.
(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{2r^2}[/tex]
The electric field will be smallest.
(III). The electric field at black dot due to positive charge point 2q to left. The distance is r.
Then, the electric field will be
[tex]E=\dfrac{k2q}{(r)^2}[/tex]
The electric field will be very largest.
(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.
Then, the electric field will be
[tex]E=\dfrac{kq}{(2r)^2}[/tex]
[tex]E=\dfrac{kq}{4r^2}[/tex]
The electric field will be very smallest.
So, The electric field from largest to smallest will be
[tex]E_{3}>E_{1}>E_{2}>E_{4}[/tex]
Hence, The ranking will be 3, 1, 2, 4.
(D) is correct option.
A particle with mass m = 700 g is found to be moving with velocity v vector (-3.50i cap + 2.90j cap) m/s. From the definition of the scalar product, v^2 = v vector. v vector.
a. What is the particle's kinetic energy at this time? J If the particle's velocity changes to v vector = (6.00i cap - 5.00j cap) m/s,
b. What is the net work done on the particle? J
Answers
Answer:
Explanation:
v₁² = v₁ . v₁
= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )
= 12.25 + 8.41
= 20.66 m /s
a ) kinetic energy = 1/2 m v₁²
= 1/2 x .7 x 20.66
= 7.23 J
b )
changed velocity v₂ = v₂.v₂
= (6i - 5 j ) . (6i - 5 j )
= 36 + 25
= 61 m /s
kinetic energy = 1/2 m v₂²
= 1/2 x .7 x 61
= 21.35 J
Work done = change in energy
= 21.35 - 7.23
= 14.12 J .
Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb. Assume that light is completely absorbed.
Answers
..........................................................
The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².
What is the Radiation pressure?Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.
Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.
E.g: Black-body radiation
Given the values are,
Wattage of bulb = W = 25 W
distance = d = 6.5 cm = 0.065 m
To know the Radiation Pressure,
It can be given by
P = I/c
Where, c = 299792458 m/s is the speed of light,
I is the intensity of radiation and given by
I = W/4πd²
Where W is the Wattage of bulb and d is the distance
I = 25/4π*0.065²
I = 470.872 w/m²
so, the radiation pressure becomes
P = 470.872/299792458
P = 1.5707x10⁻⁶ N/m²
Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²
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Three identical resistors are connected in series to a battery. If the current of 12 A flows from the battery, how much current flows through any one of the resistors
Answers
Answer:
Current that flows through any one of the resistors has a value of 12 amperes.
Explanation:
When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:
[tex]V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})[/tex]
[tex]i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}[/tex]
Where:
[tex]V_{batt}[/tex] - Voltage of the battery, measured in volts.
[tex]R_{1}[/tex], [tex]R_{2}[/tex], [tex]R_{3}[/tex] - Electric resistance of the first, second and third resistors, measured in ohms.
[tex]i[/tex] - Current, measured in amperes.
If [tex]R_{1} = R_{2} = R_{3} = R[/tex], then:
[tex]i = \frac{V_{batt}}{3\cdot R}[/tex]
Current that flows through any one of the resistors has a value of 12 amperes.
The current flows via any of the resistors should have a value of 12 amperes.
Ohm law:At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.
Here the circuit should be
vbatt = i.(R1 + R2+ R3)
i = Vbatt/R1 + R2 + R3
here
Vbatt means the voltage of the battery
R1,R2, and R3 means the resistance of the first, second and third resistors
I means the current
So, in the case when
R1 = R2 = R3 = R
So,
i = Vbatt/3.R
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A hot air balloon competition requires a balloonist to drop a ribbon onto a target on the ground. Initially the hot air balloon is 50 meters above the ground and 100 meters from the target. The wind is blowing the balloon at v= 15 meters/sec on a course to travel directly over the target. The ribbon is heavy enough that any effects of the air slowing the vertical velocity of the ribbon are negligible. How long should the balloonist wait to drop the ribbon so that it will hit the target?
Answers
Answer:
The wait time is [tex]t_w = 3.4723 \ s[/tex]
Explanation:
From the question we are told that
The distance of the hot air balloon above the ground is [tex]z = 50 \ m[/tex]
The distance of the hot air balloon from the target is [tex]k = 100 \ m[/tex]
The speed of the wind is [tex]v = 15 \ m/s[/tex]
Generally the time it will take the balloon to hit the ground is
[tex]t = \sqrt{ \frac{2 * z }{g} }[/tex]
where g is acceleration due to gravity with value [tex]g = 9.8 m/s^2[/tex]
substituting values
[tex]t = \sqrt{ \frac{2 * 50 }{9.8} }[/tex]
[tex]t = 3.194 \ s[/tex]
Now at the velocity the distance it will travel before it hit the ground is mathematically represented as
[tex]d = v * t[/tex]
substituting values
[tex]d = 15 * 3.194[/tex]
[tex]d = 47.916 \ m[/tex]
Now in order for the balloon to hit the target on the ground it will need to travel b distance on air before the balloonist drops it and this b distance can be evaluated as
[tex]b = k - d[/tex]
substituting values
[tex]b =100 -47.916[/tex]
[tex]b = 52.084 \ m[/tex]
Hence the time which the balloonist need to wait before dropping the balloon is mathematically evaluated as
[tex]t_w = \frac{b}{v}[/tex]
substituting values
[tex]t_w = \frac{52.084}{15}[/tex]
[tex]t_w = 3.4723 \ s[/tex]
A beach ball filled with air is pushed about 1 m below the surface of a swimming pool and released from rest. Which of the following statements are valid, assuming the size of the ball remains the same?
a) The buoyant force on the ball decreases as the ball approaches the surface of the pool.
b) As the ball rises in the pool, the buoyant force on it increases.
c) The buoyant force on the ball equals its weight and remains constant as the ball rises.
d) The buoyant force on the ball while it is submerged is approximately equal to the weight of the volume of water that could fill the ball.
e) When the ball is released, the buoyant force exceeds the gravitational force, and the ball accelerates upward.
Answers
Answer:
e is correct
Explanation:
When a ball is pushed below the surface of a pool, it is submerged when the buoyant force is approximately equal to the water's weight of the volume that could fill the ball. When the ball is released, the buoyant force becomes greater than the gravitational force so that the ball accelerates upward.
What is buoyant force?The buoyant force can be described as the upward force exerted on an object wholly or partially immersed in a fluid and is also called Upthrust. A body submerged partially or fully in a fluid due to the buoyant force appears to lose its weight.
The following factors affect buoyant force the density of the fluid, the volume of the fluid displaced, and the local acceleration due to gravity.
When an object immerses in water, the object experiences a force from the downward direction opposite to the gravitational pull, which causes a decrease in its weight. The difference in this pressure gives the upward force on the object, as buoyancy.
Therefore, options (d), (e) are correct.
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A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2
Answers
Answer:
650W/m²Explanation:
Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²
Given parameters
Power rating = 6.50Watts
Cross sectional area = 100cm²
Before we calculate the intensity, we need to convert the area to m² first.
100cm² = 10cm * 10cm
SInce 100cm = 1m
10cm = (10/100)m
10cm = 0.1m
100cm² = 0.1m * 0.1m = 0.01m²
Area (in m²) = 0.01m²
Required
Intensity of the sunlight I
I = P/A
I = 6.5/0.01
I = 650W/m²
Hence, the intensity of the sunlight in W/m² is 650W/m²
A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g glass ball charged to 4.7 nC is shot straight up at 4.8 m/s from the floor level. How high does the ball go if the ceiling voltage is +3.0x10^6V?
Answers
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
[tex]F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N[/tex]
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
[tex]W = F_{net} *h\\\\W = 15.328 *10^{-3} * h[/tex]
W = K.E
[tex]15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m[/tex]
Therefore, the ball traveled 0.827 m
The height at which the ball goes for the given parameters is; 0.827 m
What is the height of the ball?We are given;
distance between the metal plates; d = 3.1 m
mass of glass; m = 1.1g = 0.0011 kg
charge on the glass; q = 4.7 nC = 4.7 × 10⁻⁹ C
speed of the glass ball; v = 4.8 m/s
voltage of the ceiling; V = +3.0 × 10⁶ V
The repulsive force experienced by the ball is gotten from the formula;
F = qV/d
|F| = (4.7 × 10⁻⁹ × 3 × 10⁶)/3.1
|F| = 4.548 × 10⁻³ N
F = -4.548 × 10⁻³ N (negative because it is repulsive force)
The net horizontal force experienced by the ball is;
F_net = F - mg
F_net = (-4.548 × 10⁻³) - (0.0011 × 9.8)
F_net = -15.328 × 10⁻³ N
To get the height of the ball, we will use the formula;
F_net * h = ¹/₂mv²
h = (¹/₂ * 0.0011 * 4.8²)/(15.328 × 10⁻³)
We took the absolute value of F_net, hence it is not negative
h = 0.827 m
Read more about height of ball at; https://brainly.com/question/12446886
6. How would the measurements for potential difference and current change if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor? Explain your answer.
Answers
Answer:
Explanation:
Resistance is defined as the opposition to the flow of an electric current in a circuit. This means that a higher amount of resistance tends to reduce the amount of current flowing through the resistance. The lower the current, the greater the possibility for the resistor to allow current to pass through it. if a 200 Ω resistor was used in Circuit 1 instead of the 100 Ω resistor, then the current in the circuit will tends to increase since we are replacing the load with a lesser resistor and a smaller resistance tends to allow more current to flow through it
For the potential difference, a decrease in the resistance value will onl decrease the potential difference flowing in the circuit according to ohm's law. According to the law the pd in a circuit is directly proportional to the current which means an increase in the resistance value will cause an increase in the corresponding pd and vice versa.
Suppose you wish to make a solenoid whose self-inductance is 1.8 mH. The inductor is to have a cross-sectional area of 1.6 x 10-3 m2 and a length of 0.066 m. How many turns of wire are needed
Answers
Answer:
The number of turns of the wire needed is 243 turns
Explanation:
Given;
self inductance of the solenoid, L = 1.8 mH
cross sectional area of the inductor, A = 1.6 x 10⁻³ m²
length of the inductor, l = 0.066 m
The self inductance of long solenoid is given by;
L = μ₀n²Al
where;
μ₀ is permeability of free space = 4π x 10⁻⁷ H/m
n is number of turns per length
A is the area of the solenoid
l is length of the solenoid
[tex]n = \sqrt{\frac{L}{\mu_o Al} } \\\\n = \sqrt{\frac{1.8*10^{-3}}{(4\pi*10^{-7}) (1.6*10^{-3})(0.066)} } \\\\n = \sqrt{13562583.184} \\\\n = 3682.74 \ turns/m[/tex]
The number of turns is given by;
N = nL
N = (3682.74)(0.066)
N = 243 turns
Therefore, the number of turns of the wire needed is 243 turns
In a single-slit diffraction experiment, the width of the slit through which light passes is reduced. What happens to the width of the central bright fringe
Answers
Answer:
It becomes wider
Explanation:
Because The bigger the object the wave interacts with, the more spread there is in the interference pattern. Decreasing the size of the opening increases the spread in the pattern.
What do Equations 1 and 2 predict will happen to the single-slit diffraction pattern (intensity, fringe width, and fringe spacing) as the slit width is increased.
Equation 1:
Sinθ = mλ/ω
Equaiton 2:
I= Io [Sinθ (πωλ/πωλ/Rλ)
Answers
Answer:
the firtz agrees with the expression for the shape of the curve of diracion of a slit
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
where a is the width of the slit, t is the angle from the center of the slit, l is the wavelength and m is an integer that corresponds to the maximum diffraction.
the previous equation qualitatively describes the curve of the diffraction phenomenon the equation takes the form
I = I₀ [(sin ππ a y / R λ) / π a y / Rλ]²
I = I₀ ’[sin π a y /Rλ]²
I₀ ’= I₀ / (π a y /Rλ)²
By reviewing the two expressions given
equation 1
w sin θ = m λ
where w =a w is the slit width
we see that the firtz agrees with the expression for the shape of the curve of diracion of a slit
Equation 2
the squares are missing
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 850 m . After this point, its acceleration is that of gravity, downward.
Answers
Answer:
v = 73.75 m/s
Explanation:
It is given that,
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.
Let us assume we need to find the velocity of the rocket when it runs out of fuel.
Let v is the final speed. Using the third equation of kinematics as :
[tex]v^2-u^2=2as[/tex]
u = 0
[tex]v=\sqrt{2as} \\\\v=\sqrt{2\times 3.2\times 850}\\\\v=73.75\ m/s[/tex]
So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s
A Huge water tank is 2m above the ground if the water level on it is 4.9m high and a small opening is there at the bottom then the speed of efflux of non viscous water through the opening will be
Answers
Answer:
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Explanation:
Let assume the existence of a line of current between the water tank and the ground and, hence, the absence of heat and work interactions throughout the system. If water is approximately at rest at water tank and at atmospheric pressure ([tex]P_{atm}[/tex]), then speed of efflux of the non-viscous water is modelled after the Bernoulli's Principle:
[tex]P_{1} + \rho\cdot \frac{v_{1}^{2}}{2} + \rho\cdot g \cdot z_{1} = P_{2} + \rho\cdot \frac{v_{2}^{2}}{2} + \rho\cdot g \cdot z_{2}[/tex]
Where:
[tex]P_{1}[/tex], [tex]P_{2}[/tex] - Water total pressures inside the tank and at ground level, measured in pascals.
[tex]\rho[/tex] - Water density, measured in kilograms per cubic meter.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Water speeds inside the tank and at the ground level, measured in meters per second.
[tex]z_{1}[/tex], [tex]z_{2}[/tex] - Heights of the tank and ground level, measured in meters.
Given that [tex]P_{1} = P_{2} = P_{atm}[/tex], [tex]\rho = 1000\,\frac{kg}{m^{3}}[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], [tex]v_{1} = 0\,\frac{m}{s}[/tex], [tex]z_{1} = 6.9\,m[/tex] and [tex]z_{2} = 4.9\,m[/tex], the expression is reduced to this:
[tex]\left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m) = \frac{v_{2}^{2}}{2} + \left(9.807\,\frac{m}{s^{2}} \right)\cdot (4.9\,m)[/tex]
And final speed is now calculated after clearing it:
[tex]v_{2} = \sqrt{2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (6.9\,m-4.9\,m)}[/tex]
[tex]v_{2} \approx 6.263\,\frac{m}{s}[/tex]
The speed of efflux of non-viscous water through the opening will be approximately 6.263 meters per second.
Suppose your 50.0 mm-focal length camera lens is 51.0 mm away from the film in the camera. (a) How far away is an object that is in focus
Answers
Answer:
2.55m
Explanation:
Using 1/do+1/di= 1/f
di= (1/f-1/do)^-1
( 1/0.0500-1/0.0510)^-1
= 2.55m
why was the observation of the double-slit interference pattern more convincing evidence for the wave theory of light than the observation of diffraction
Answers
Answer:
The double slit experiment showed for the first time that light can be interfered, producing bands of light and dark fringes on a screen. This phenomenon was a unique and typical characteristic of waves.
Explanation:
Th double slit experiment by Thomas Young proved, and sealed for the first time the wave nature of light; showing that light just as any other wave can produce interference which was a unique, typical phenomenon of waves. The Interference of light was shown by allowing light to pass through narrow slits and superimpose on a wall or screen, at a distance away from the slit, producing a clear pattern of light and dark fringes. This was the first experiment to proof that darkness can be produced by the addition of light on light. Interference is accompanied by a spatial redistribution of the optical intensity without violation of power conservation. The phenomenon of interference proved the intuitive ideas of Huygens regarding the wave nature of light against Newton's particle nature of light theory.