Answer:
6.02 dB increase
Explanation:
Let us take the initial power from the speaker P' = P Watt
then, the final power P = 4P Watt
for a given unit area, initial intensity (power per unit area) will be
I' = P Watt/m^2
and the final quadrupled sound will produce a sound intensity of
I = 4P Watt/m^2
Increase in loudness is gotten from the relation
ΔL = [tex]10log_{10} \frac{I}{I'}[/tex]
where
I = final sound intensity
I' = initial sound intensity
imputing values of the intensity into the equation, we have
==> [tex]10log_{10} \frac{4P}{P}[/tex] = [tex]10log_{10} 4[/tex] = 6.02 dB increase
70 kg man walks out on a 10 kg beam that rests on, but is not attached to, two supports. When the beam just starts to tip, what is the force exerted on the beam by the right support
Answer:
The force is [tex]F = 784 \ N[/tex]
Explanation:
From the question we are told that
The mass of the man is [tex]m = 70 \ kg[/tex]
The mass of the beam is [tex]m_b = 10 \ kg[/tex]
Now from the question we can deduce that when this beam start to tip that both the force exerted by the weight of the man and that of the beam is been supported by the right support so
The force exerted on the right support is mathematically evaluated as
[tex]F = (m + m_b) * g[/tex]
substituting values
[tex]F = (70 + 10 ) * 9.8[/tex]
[tex]F = 784 \ N[/tex]
The force exerted on the beam by the right support is 784 Newton.
Given the data in the question;
Mass of the man; [tex]m_m = 70kg[/tex]Mass of the beam; [tex]m_b = 10kg[/tex]Force exerted on the beam by the right support; [tex]F = W = \ ?[/tex]
When the beam just starts to tip, the right support holds up the combined mass of the man and the beam.
Hence;
[tex]M_{net} = m_m + m_b\\\\M_{net} = 70kg + 10kg\\\\M_{net} = 80kg[/tex]
Now, To determine the force exerted on the beam by the right support, we use the general formula for weight or equation of force of gravity which is expressed as:
[tex]F = W = m * g[/tex]
Where m is mass and g represents the acceleration due to gravity( [tex]9.8m/s^2[/tex] )
We substitute our values into the equation
[tex]F = 80kg * 9.8m/s^2\\\\F = 784kg.m/s^2\\\\F = 784N[/tex]
Therefore, the force exerted on the beam by the right support is 784 Newton.
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A woman is listening to her radio, which is 174 m from the radio station transmitter. (a) How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz
Explanation:
It is given that,
The distance between the radio and the radio station is 174 m
We need to find how many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz.
f = 1540 kHz
Wavelength,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1540\times 10^3}\\\\\lambda=194.8\ m[/tex]
Let there are n wavelengths of the radio waves. So,
[tex]n=\dfrac{d}{\lambda}\\\\n=\dfrac{174}{194.8}\\\\n=0.89\ \text{wavelengths}[/tex]
There are 0.89 wavelengths.
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It strikes a target 1.7 seconds later. What is the vertical distance from where the projectile was launched to where it hit the target.
Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± [tex]\frac{1}{2} at^2[/tex] ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± [tex]\frac{1}{2} at^2[/tex] ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
From the question;
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt - [tex]\frac{1}{2} at^2[/tex]
h = 26.54(1.7) - [tex]\frac{1}{2} (10)(1.7)^2[/tex]
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m
A wheel 2.40 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.40 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.
A. What is the tangential speed?
B. Total acceleration
C. Angular position of point P.
Answer:
Explanation:
Radius of wheel = 1.2 m
A )
To know angular speed after t sec , we use the formula
ω = ω₀ + α t , where ω₀ is initial velocity , α is angular acceleration
ω = 0 + 4.4 x 2
= 8.8 rad / s
v= ωR , v is tangential speed , ω is angular speed , R is radius of wheel .
= 8.8 x 1.2 = 10.56 m /s
B )
radial acceleration
Ar = v² / R
= 10.56² / 1.2
= 92.93 m /s²
Tangential acceleration
At = angular acceleration x radius
= 4.4 x 1.2 = 5.28 m /s²
Total acceleration
= √ ( At² + Ar² )
=√ (5.28² +92.93²)
= 93 m /s²
C )
θ = ωt + 1/2 α t² where θ is angular position after time t .
= 0 + .5 x 4.4 x 2²
= 8.8 rad
= 180x 8.8/ 3.14 = 504.45 degree
initial position = 57.3°
final position = 504 .45 + 57.3
= 561.75 °
= 561.75 - 360
= 201.75 ° .
Position of radius vector of point P will be at angle of 201.75 from horizontal axis .
At what rate must a cylindrical spaceship rotate if occupants are to experience simulated gravity of 0.50 gg? Assume the spaceship's diameter is 35 mm , and give your answer as the time needed for one revolut
Answer:
Time needed for one revolution is 0.38 s
Explanation:
The formula for the frequency of rotation of a spaceship, to create the desired artificial gravity, is as follows:
f = (1/2π)√(a/r)
where,
f = frequency of rotation = ?
a = artificial gravity required = 0.5 g
g = acceleration due to gravity on surface of Earth = 9.8 m/s²
r = radius of ship = 35 mm/2 = 17.5 mm = 17.5 x 10⁻³ m
Therefore,
f = (1/2π)√[(0.5)(9.8 m/s₂)/(17.5 x 10⁻³ m)]
f = 2.66 Hz
Now, for the time required for one revolution, is given as:
Time Period = T = 1/f
T = 1/2.66 Hz
T = 0.38 s
The time required for one revolution to simulate the desired gravity is 0.38 s.
The frequency can be calculate by the formula
[tex]\bold {f = (\dfrac {1}{2\pi})\sqrt{ar}}[/tex]
where,
f - frequency of rotation = ?
a- artificial gravity required = 0.5 g
g - gravitational acceleration on surface of Earth = 9.8 m/s²
r - radius of ship = 35 mm/2 = 17.5 mm = 17.5 x 10⁻³ m
Put the value in the equation,
[tex]\bold {f = \dfrac {1}{2\pi}\squrt {(0.5)(9.8\ m/s^2)}{(17.5 x 10^{-3} m)}}\\\\\bold {f = 2.66\ Hz}[/tex]
the time required for one revolution can be calculated as
[tex]\bold {T =\dfrac 1f}\\\\\bold {T = \dfrac 1{2.66}\ Hz}\\\\\bold {T = 0.38\ s}[/tex]
Therefore, the time required for one revolution to simulate the desired gravity is 0.38 s.
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Each charge is equidistant from the origin. In which direction is the net electric field at the point P on the y-axis?
Answer:
"Upwards and towards the left" is the right answer.
Explanation:
The magnitude of the field will be:
⇒ [tex]E=\frac{kq}{r^2}[/tex]
And direction -> for negative charges, to positive charges, except charges.
Charging across the y-axis. It would be up to the aggregate field. Because the x-axis needs to charge. Total production is to the west.Thus the net field is upwards as well as to the left.
g The force of kinetic friction for a particular pair of interacting objects is always _____ the force of static friction. less than greater than equal to None of the above
Answer:
less than
Explanation:
The force of kinetic friction for a particular pair of interacting objects is always less than the force of static friction.
The force of static friction between two surfaces is always higher than the force of kinetic friction.
Un bloque de 10 kg se encuentra sobre un plano rugoso inclinado 37º respecto a la horizontal, sobre él actúa una fuerza constante, horizontal, de módulo 50 N. Si el bloque desciende sobre el plano 5 m, lentamente, determine la cantidad de trabajo que realiza la fuerza de rozamiento (considere g = 10 m/s2).
Answer:
El trabajo realizado por la fuerza de rozamiento sobre el bloque tras recorrer este último una distancia de 5 metros sobre el plano es de 500.566 joules.
Explanation:
El fenómeno alrededor del bloque puede ser modelado por el Principio de Conservación de la Energía y el Teorema del Trabajo y la Energía. Al descender lentamente, significa que la aceleración neta experimentada por el bloque es aproximadamente cero. El diagrama de cuerpo libre del bloque se presenta a continuación como archivo adjunto. Las ecuaciones de equilbrio del sistema son:
[tex]\Sigma F_{x'} = P\cdot \cos \theta + m\cdot g \cdot \sin \theta - f = 0[/tex]
[tex]\Sigma F_{y'} = N + P\cdot \sin \theta -m\cdot g\cdot \cos \theta = 0[/tex]
Donde:
[tex]P[/tex] - Fuerza externa aplicada a la caja, medida en newtons.
[tex]m[/tex] - Masa del bloque, medida en kilogramos.
[tex]g[/tex] - Aceleración gravitacional, medidas en metros sobre segundo al cuadrado.
[tex]f[/tex] - Fuerza de rozamiento, medida en newtons.
[tex]N[/tex] - Fuerza normal del plano sobre la caja, medida en newtons.
[tex]\theta[/tex] - Ángulo de inclinación del plano, medido en grados sexagesimales.
Dado que todas las fuerzas son constantes, se puede emplear la definición de trabajo como el producto de la fuerza paralela a la dirección del movimiento y la magnitud de distancia recorrida en el movimiento, entonces la primera ecuación de equilibrio queda así al multiplicar cada lado por la distancia recorrida:
[tex]P\cdot \Delta s \cdot \cos \theta + m\cdot g \cdot \Delta s \cdot \sin \theta - W_{f} = 0[/tex]
Ahora, la cantidad de trabajo realizado por la fuerza de rozamiento es:
[tex]W_{f} = (P\cdot \cos \theta+m\cdot g\cdot \sin \theta)\cdot \Delta s[/tex]
Si [tex]P = 50\,N[/tex], [tex]m = 10\,kg[/tex], [tex]g = 10\,\frac{m}{s^{2}}[/tex], [tex]\theta = 37^{\circ}[/tex] and [tex]\Delta s = 5\,m[/tex], entonces el trabajo realizado por la fuerza de rozamiento es:
[tex]W_{f} = \left[(50\,N)\cdot \cos 37^{\circ}+(10\,kg)\cdot \left(10\,\frac{m}{s^{2}} \right)\cdot \sin 37^{\circ}\right]\cdot (5\,m)[/tex]
[tex]W_{f} = 500.566\,J[/tex]
El trabajo realizado por la fuerza de rozamiento sobre el bloque tras recorrer este último una distancia de 5 metros sobre el plano es de 500.566 joules.
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge at point P, which is on the x-axis, 9.83 mm from the origin
Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
[tex]\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}][/tex] (1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
[tex]\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°[/tex]
The distance r is:
[tex]r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m[/tex]
You replace the values of all parameters in the equation (1):
[tex]\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}[/tex]
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
How did the horizontal velocity vector component change during the flight of the cannonball in the simulation
Answer:
The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight
A piston absorbs 42 J of heat from its surroundings while being compressed from 0.0007 m3 to 0.0002 m3 at a constant pressure of 1.0 × 105 Pa. What are the correct values for heat and work for the piston?
Answer:
D
Explanation:
W = P∆V
Use the above equation and substitute, thanks
Two bullets are fired simultaneously parallel to a horizontal plane. The bullets have different masses and different initial velocities. Which one will strike the plane first?
a) The fastest one.b) The lightest one.c) The heaviest one.d) The slowest one.e) They strike the plane at the same time.
Answer:
Therefore, the answer is E. They strike the plane at the same time.
Explanation:
Here, it is seen that the time depends only on acceleration due to gravity (which is a constant) and vertical displacement, and not on velocity of the bullets or mass of the bullets.
Hence, the bullets that are fired simultaneously parallel to the horizontal plane will strike the plane at the same time.
using equation of motion for displacement
s= ut + ¹/₂gt²
here, g is the acceleration due to gravity along y- direction
U along y is 0
s = (0)t + ¹/₂gt²
s=¹/₂gt²
make t the subject of formula = [tex]\sqrt{\frac{2s}{g} }[/tex]
Consider two coils, with the first coil having twice as many loops as the second. Given the flux Φ though each loop of the first coil due to current in the second coil, what can be said about the flux through each loop of the second coil due to an equal current in the first coil?
Answer:
[tex]$ \phi_{21} = \frac{\phi_{12}}{2} $[/tex]
Which means that the flux through each loop of the second coil is half as much as flux through each loop of first coil.
Explanation:
The flux through each loop of the first coil due to current in the second coil is,
[tex]\phi_{12} = \phi[/tex]
The number of loops in the first coil is
no. of loops = 2N
Total flux passing through the first coil is
[tex]\phi_{12} = 2N\phi[/tex]
The flux through each loop of the second coil due to current in the first coil is,
[tex]\phi_{21} = \phi[/tex]
The number of loops in the second coil is
no. of loops = N
Total flux passing through the second coil is
[tex]\phi_{21} = N\phi[/tex]
Comparing both
[tex]\phi_{12} = \phi_{21} \\\\ 2N\phi = N\phi\\\\\phi_{21} = \frac{\phi_{12}}{2}[/tex]
Which means that the flux through each loop of the second coil is half as much as flux through each loop of first coil.
A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential of the surface of this sphere and find it to be 14 kV relative to infinity. Find the excess charge on this sphere.
Answer:
The excess charge is [tex]Q = 3.5 *10^{-7} \ C[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 45 \ cm = 0.45 \ m[/tex]
The potential of the surface is [tex]V = 14 \ kV = 14 *10^{3} \ V[/tex]
The radius of the sphere is
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.45}{2}[/tex]
[tex]r = 0.225 \ m[/tex]
The potential on the surface is mathematically represented as
[tex]V = \frac{k * Q }{r }[/tex]
Where k is coulomb's constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
given from the question that there is no other charge the Q is the excess charge
Thus
[tex]Q = \frac{V* r}{ k}[/tex]
substituting values
[tex]Q = \frac{14 *10^{3} 0.225}{ 9*10^9}[/tex]
[tex]Q = 3.5 *10^{-7} \ C[/tex]
An internal explosion breaks an object, initially at rest, into two pieces, one of which has 1.8 times the mass of the other.
Requried:
a. If 7230 J were released in the explosion, how much kinetic energy did the heavier piece acquire?
b. How much kinetic energy did the lighter piece acquire?
Answer:
a) The heavier piece has a translational kinetic energy of 4647.857 joules, b) The lighter piece has a translational kinetic energy of 2582.143 joules.
Explanation:
a) The object breaking can be described by means of the Principle of Energy Conservation, knowing that heavier piece has 1.8 times the mass of the lighter ([tex]m_{h} = 1.8\cdot m_{l}[/tex]), both are modelled as particle due to the absence of rotation and that energy liberated by explosion is transform into kinetic energy, the equation that describes the phenomenon is:
[tex]E_{ex} = K_{h} + K_{l}[/tex]
Where:
[tex]E_{ex}[/tex] - Energy liberated by the explosion, measured in joules.
[tex]K_{h}[/tex], [tex]K_{l}[/tex] - Translational kinetic energies of the heavier and lighter piece, respectively.
This expression is expanded by using the definition of translational kinetic energy and supposing the both parts are liberated at the same initial speed ([tex]v_{o}[/tex]). Then:
[tex]E_{ex} = \frac{1}{2}\cdot (m_{h} + m_{l})\cdot v_{o}^{2}[/tex]
As can be seen, the energy liberated by expression is directly proportional to the mass of the system. Hence, the kinetic energy can be estimated by simple rule of three:
[tex]K_{h} = \frac{m_{h}}{m_{h}+m_{l}}\times E_{ex}[/tex]
If [tex]m_{h} = 1.8\cdot m_{l}[/tex] and [tex]E_{ex} = 7230\,J[/tex], then:
[tex]K_{h} =\frac{1.8\cdot m_{l}}{2.8\cdot m_{l}}\times E_{ex}[/tex]
[tex]K_{h} = \frac{9}{14}\cdot (7230\,J)[/tex]
[tex]K_{h} = 4647.857\,J[/tex]
The heavier piece has a translational kinetic energy of 4647.857 joules.
b) The translational kinetic energy of the lighter piece is calculated by using the equation derived from the Principle of Energy Conservation:
[tex]K_{l} = E_{ex} - K_{h}[/tex]
Given that [tex]E_{ex} = 7230\,J[/tex] and [tex]K_{h} = 4647.857\,J[/tex], the translational kinetic energy of the lighter piece is:
[tex]K_{l} = 7230\,J - 4647.857\,J[/tex]
[tex]K_{l} = 2582.143\,J[/tex]
The lighter piece has a translational kinetic energy of 2582.143 joules.
A handheld glass rod can be charged by rubbing it with silk or a plastic bag while holding it in your hands. Would you conclude from this that glass is a conductor or an insulator? Why?
Answer:
a conductor is an object or type of material that allows the flow of charge (electrical current) in one or more directions.
Explanation:
. Materials made of metal are common electrical conductors.
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.1 kg⋅m2 and for arms and legs in is 0.90 kg⋅m2 . If she starts out spinning at 4.0 rev/s , what is her angular speed (in rev/s) when her arms and one leg open outward?
Answer:
The angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s
Explanation:
Given;
moment of inertia of a skater with arms out, [tex]I_{arms \ out}[/tex] = 3.1 kg.m²
moment of inertia of a skater with arms in, [tex]I_{arms \ in}[/tex] = 0.9 kg.m²
inward angular speed, [tex]\omega _{in}[/tex] = 4 rev/s
The angular momentum of the skater when her arms are out and one leg extended is equal to her angular momentum when her arms and legs are in.
[tex]L_{out} = L_{in}[/tex]
[tex]I_{out} \omega_{out} = I_{in} \omega_{in}\\\\\omega_{out} = \frac{ I_{in} \omega_{in} }{I_{out} } \\\\\omega_{out} = \frac{0.9*4}{3.1} \\\\\omega_{out} = 1.161 \ rev/s[/tex]
Therefore, the angular speed (in rev/s) when her arms and one leg open outward is 1.161 rev/s
A piece of iron rests on top of a piece of wood floating in a bathtub. If the iron is removed from the wood, and kept out of the water, what happens to the water level in the tub?
a. It goes up.
b. It does not change.
c. It goes down.
d. It is impossible to determine from the information given.
Answer:
It goes down.
The water level remain the same.
Explanation:
This can be explained using Archimedes principle which states that a body fully or partially submerged in a fluid is acted by an upward bouyant force which is equal to the weight of the fluid the body displaced.
The wood will only sink if the weight of the wood is greater than the weight of the fluid the wood displaced, but the weight of the wood is equal to the weight of fluid displaced, therefore the wood will float.
Therefore, the weight of the wood is the same as the weight of the fluid displaced, so the wood will be at the same level as the water.
If the iron is removed, the level of the water goes down because iron weight is bigger than the water displaced and it tends to increase the water level but since it is removed, the water level will decrease.
If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water knocks two rocks together, you'll barely hear the sound.
Match the words.
The air-water interface is an example of boundary. The( )portion of the initial wave energy is way smaller than the( )portion. This makes the( ) wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can( ) .
1. reflect more efficiently
2. transmitted
3. travel directly to your ears
4. boundary
5. reflected
6. discontinuity
Answer:
The air-water interface is an example of boundary. The transmitted portion of the initial wave energy is way smaller than the reflected portion. This makes the boundary wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can travel directly to your ear.
Explanation:
The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.
Answer: The air-water interface is an example of boundary. The (transmitted) portion of the initial wave is way smaller than the (reflected) portion. This makes the (transmitted) wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can (travel directly to your ears.)
Explanation:
The part of the sound wave that is transmitted across the boundary between air and water is much smaller than the part of the wave that is reflected. This is what makes it hard to hear your friend knocking two rocks together above the surface.
When you and the rocks are underwater, the sound that comes from knocking the rocks together can travel directly to your ears rather than having to be transmitted across mediums.
Which one of the following is the shortest length?
A)
100 meters
C)
104 millimeters
E)
10 nanometers
B)
10² centimeters
D)
105 micrometers
Answer:
Option E (10 nanometers) is the shortest lengthExplanation:
From,
1cm = [tex]10^{-2}m[/tex]
1mm = [tex]10^{-3}m[/tex]
1nanometer = [tex]10^{-9[/tex]
1micrometer = [tex]10^{-6[/tex]
Therefore,
A) [tex]10^0[/tex] meters = 1meter
B) [tex]10^2[/tex] cm = [tex]10^2 * 10^{-2} = 1meter[/tex]
C) [tex]10^4[/tex] mm = [tex]10^4 * 10^{-3} = 10meter[/tex]
D) [tex]10^5[/tex] micrometer = [tex]10^5 * 10^{-6} = 0.1meter[/tex]
E) [tex]10[/tex] nanometer = [tex]10 * 10^-9 = 1*10^{-8}[/tex]
Therefore 10nanometers is the shortest length
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The bases of developing convective cumulus clouds will be relatively higher at a location with a relatively ______ difference between the surface temperature and surface dew point temperature.
Answer:
large
Explanation:
Cumulus clouds is a term in metrology that defines the type of clouds which are characterized by its low altitude, puffy appearance, and fair-weather nature. They are generally considered as low-level clouds, with less than than 2,000m in altitude except they are the more vertical cumulus congestus form.
Thus, it can be noted that, the difference between the surface dew point temperature and the surface temperature is related to relative humidity. Hence, in a situation when there is a LARGE difference between the surface temperature and the surface dewpoint temperature, then the relative humidity is very low (e.g., 10%).
Therefore, the bases of developing convective cumulus clouds will be relatively higher at a location with a relatively LARGE difference between the surface temperature and surface dew point temperature.
A transformer has 480 primary turns and 7.8 secondary turns. (a) If Vp is 120 V (rms), what is Vs with an open circuit? If the secondary now has a resistive load of 17 Ω, what is the current in the (b) primary and (c) secondary?
Answer:
a) 1.95 V
b) 1.87 mA
c) 0.115 A
Explanation:
Given that
Number of primary turns, N(p) = 480
Number of secondary turns, N(s) = 7.8
Velocity of primary turns, V(p) = 120 V
Velocity of secondary turns, V(s) = ?
Current in the primary, I(p) = ?
Current in the secondary, I(s) ?
To solve this question, we would be using the formula
V(s)/V(p) = N(s)/N(s), now substituting the values, we have
V(s) / 120 = 7.8 / 480
V(s) = (7.8 * 120) / 480
V(s) = 936 / 480
V(s) = 1.95 V
To find the current in the primary, remember ohms law?
I = V/R
I(s) = V(s) / R(s)
I(s) = 1.95 / 17
I(s) = 0.115 A
Now, remember the relationship between current and voltage
I(p)/I(s) = V(s)/V(p)
I(p) / 0.115 = 1.95 / 120
I(p) = (1.95 * 0.115) / 120
I(p) = 0.22425 / 120
I(p) = 0.00187 A
I(p) = 1.87 mA
A positive charge moves in the direction of an electric field. Which of the following statements are true?
a. The potential energy associated with the charge decreases.
b. The electric field does positive work on the charge.
c. The electric field does negative work on the charge.
d. The potential energy associated with the charge increases.
e. The electric field does not do any work on the charge.
f. The amount of work done on the charge cannot be determined without additional information.
Answer:
The potential enwrgy associated with charge decreases.
The ele ric field does negative work on the charge.
Explanation:
Answer:
The potential energy associated with the charge decreases
The electric field does positive work on the charge.
Wind gusts create ripples on the ocean that have a wavelength of 3.03 cm and propagate at 3.37 m/s. What is their frequency (in Hz)?
Answer:
Their frequency is 111.22 Hz
Explanation:
Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration and is expressed in units of length (m).
Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
The propagation speed of a wave is the quantity that measures the speed at which the wave's disturbance propagates throughout its displacement. The speed at which the wave propagates depends on both the type of wave and the medium through which it propagates. Relate wavelength (λ) and frequency (f) inversely proportional using the following equation:
v = f * λ.
Then the frequency can be calculated as: f=v÷λ
In this case:
λ=3.03 cm=0.0303 m (1m=100 cm)v= 3.37 m/sReplacing:
[tex]f=\frac{3.37 \frac{m}{s} }{0.0303 m}[/tex]
Solving:
f=111.22 Hz
Their frequency is 111.22 Hz
A small object with mass 3.80 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.70 m centered at the origin. It starts at the point with position vector 2.70 m. Then it undergoes an angular displacement of 8.70 rad.
(a) What is its new position vector?
in meters
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity?
in m/s
(d) In what direction is it moving?
_____° from the +x direction.
(e) What is its acceleration?
in m/s2
(f) What total force is exerted on the object?
in N
Answer:
Explanation:
angular velocity
ω = 1.65 rad /s
radius R = 2.70 m
angular displacement = 8.70 rad
a )
New position vector in vector form
= R cos8.7 i + R sin8.7 j
= 2.7 cos8.7 i + 2.7 sin8.7 j
= 2.7 x .748 i + 2.7 x .663 j
= 2.01 i + 1.79 j
b )
8.7 radian = 180/π x 8.7 degree
= 498.72 degree
= 498.72 - 360
= 138.72 degree
It will be in second quadrant .
angle made with positive x - axis
= 138.72 degree .
c )
velocity
v = ω R
= 1.65 x 2.7
= 4.455 m /s
d )
It is moving in a direction making 138.72° with positive x direction .
e )
acceleration will be centripetal acceleration
= v²/ R
= 4.455² / 2.7
= 7.35 m /s²
f ) force = mass x acceleration
= 3.8 x 7.35
= 27.93 N .
Good day can I get some help please?
Answer:
432 J
Explanation:
When moving linearly:
Kinetic Energy = (1/2)mV^2
So here you have:
KE=(1/2)(6)(12^2)=(1/2)(6)(144)=432
The unit for energy is Joules (J), so your answer would be 432 J.
A student in the front of a school bus tosses a ball to another student in the back of the bus while the bus is moving forward at constant velocity. The speed of the ball as seen by a stationary observer in the street:_________
a. is less than that observed inside the bus.
b. is the same as that observed inside the bus
c. may be either greater or smaller than that observed inside the bus.
d. may be either greater, smaller or equal than that observed inside the bus.
e. is greator than that observed inside the bus.
Answer:
d. may be either greater, smaller, or equal to that observed inside the bus.
Explanation:
The bus is moving at a constant speed. The ball tossed and received by the ball is inside the bus at a speed equal to the speed of the ball. Therefore the speed of the bus becomes zero with respect to the observer inside the bus. Now the observer inside the bus noticed the ball from the inside of the bus, so he threw the ball back and forth from the ball with the speed v relative to the observer. Now the observer outside the bus could see the bus moving at speed relative to its reference point and also throwing the ball from front to back. The speed of the ball to the observer outside the bus The speed of the bus to the observer outside the bus is minus the speed of the ball to the observer inside the bus. Therefore, the ball speed = (u-v) relative to the observer outside the bus.4. The capacitance of a capacitor is increased by a factor of 1.5 when it is completely filled
with a certain dielectric material. Find the dielectric constant of the material and its
electric susceptibility
Answer:
a. Dielectric constant, ε = 1.5 b. Electric susceptibility, χ = 0.5
Explanation:
a. Dielectric constant
Since the capacitance of the capacitor is increased by a factor of 1.5. Let its initial capacitance be C and its final capacitance after adding the material be C'.
Since C' = εC where ε = relative permittivity,
Also, C' = 1.5C
Comparing both equations for C', ε = 1.5.
Since ε = relative permittivity = dielectric constant,
dielectric constant = 1.5
So, the dielectric constant = 1.5
b. Electric susceptibility
The electric susceptibility χ is given by
χ = ε - 1 where ε = dielectric constant
Since ε = 1.5,
χ = ε - 1
χ = 1.5 - 1
χ = 0.5
So the electric susceptibility χ = 0.5
A rubber ball is attached to a string and whirled around in a circle. If the string is 1.0 m long (measured from the center of the baseball to the far end of the string) and the ball’s speed is 10 m/s, what is the ball’s centripetal acceleration?
Centripetal acceleration = (speed squared) / (radius)
Centripetal acceleration = (10 m/s)² / (1.0 m)
Centripetal acceleration = (100 m²/s²) / (1.0 m)
Centripetal acceleration = 100 m/s²
Besides the gravitational force, a 2.80-kg object is subjected to one other constant force. The objectstarts from rest and in 1.20 s experiences a displacement of (4.20 i - 3.30 j) m, where the direction of jis the upward vertical direction. Determine the other force.
Answer:
the other force= (16.3i + 14.6j)N
EXPLANATION:
Given:
Mass=2.80-kg
t= 1.2s
Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.
We will need to calculate the magnitude of the displacement using the below formula;
d = (1/2)at2 + v0t + d0
But note that
d0 = 0,( initial position)
v0 = 0( initial position)
a is the net acceleration
d = √[4.202 + (-3.30)2] m = 5.34 m
Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)
d = (1/2)at2
a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2
the net acceleration is 7.42 m/s2
Acceleration in terms of the vector can be calculated as
a=2(ri - r0)/t^2
Where t =1.2s which is the time
a= 2(4.2i - 3.30j)/ 1.2^2
a=( 5.83i - 4.58j)m/s
now the net force can now be calculated since we have known the value of acceleration, using the formula below;
F(x) = ma - mg
Where a = 5.83i - 4.58j)m/s and g= 9.8m/s
2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2
Therefore, the other force= (16.3i + 14.6j)N