Answer:
W = 14700 J
Explanation:
This is an exercise on Newton's second law.
To solve it we must fix a coordinate system, the most common is an axis parallel to the ramp and the other perpendicular axis, we write Newton's second law
Y Axis . Perpendicular to the ramp
N - Wy = 0
X axis. Parallel to the ramp, we assume it is positive when the ramp is going up
F - Wx = m a
in this case F = 1470 N and it is parallel to the plane.
Work is defined by
W = F .d
boldface indicates vectors
W = F d cos θ
let's calculate
W = 1470 10 cos 0
W = 14700 J
A woman is standing at the rim of a nonuniform cylindrical horizontal platform initially at rest. The platform is free to rotate about frictionless orthogonal axle that goes through its center and has 4 m in diameter and moment of inertia of 500 kgm2. The woman then starts walking along the rim in clockwise direction at a constant speed of 1.50 m/s relative to the Earth. If a woman has 60 kg, how much work does she do to set herself and the platform into motion?
Answer:
e
Explanation:
i took it myself and got it right
How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to produce an electric field of 1450 N/C just outside the surface of the sphere
Answer:
1.007 × 10^(10) electron
Explanation:
We are given;
Electric Field;E = 1450 N/C
Diameter;d = 20 cm = 0.2 m
So, Radius: r = 0.2/2 = 0.1 m
Formula for Electric field just outside the sphere is given by the formula;
E = kq/r²
Where;
E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge
k is a constant with a value of 9 x 10^(9) N.m²/C²
Making q the subject, we have;
q = Er²/k
Thus,
q = 1450 × 0.1²/(9 × 10^(9))
q = 1.61 × 10^(-9) C
Now, total charge q is also given by the formula;
q = Ne
Where;
e is charge on electron which is 1.6 × 10^(-19)
N is number of excess electrons
Making N the formula, we have;
N = q/e
N = (1.61 × 10^(-9))/(1.6 × 10^(-19))
N = 1.007 × 10^(10) electron
Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The longer one has length 8L, diameter 4D, and resistance R2. How do the resistances of these two resistors compare
Answer:
the resistance of the longer one is twice as big as the resistance of the shorter one.
Explanation:
Given that :
For the shorter cylindrical resistor
Length = L
Diameter = D
Resistance = R1
For the longer cylindrical resistor
Length = 8L
Diameter = 4D
Resistance = R2
So;
We all know that the resistance of a given material can be determined by using the formula :
[tex]R = \dfrac{\rho L }{A}[/tex]
where;
A = πr²
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
For the shorter cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{ 4 \rho L }{\pi \ D ^2}[/tex]
For the longer cylindrical resistor ; we have:
[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]
since 2 r = D
[tex]R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]
[tex]R = \dfrac{32\rho L }{\pi \ (4 D) ^2}[/tex]
[tex]R = \dfrac{2\rho L }{\pi \ (D) ^2}[/tex]
Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :
[tex]\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}[/tex]
[tex]\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}[/tex]
[tex]\dfrac{R_s}{R_L} =2[/tex]
[tex]{R_s}=2{R_L}[/tex]
Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.
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.
In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.
Answer:
The angular speed is [tex]w = 5.89 \ rad/s[/tex]
Explanation:
From the question we are told that
The time taken is [tex]t = 1.6 s[/tex]
The number of somersaults is n = 1.5
The total angular displacement during the somersault is mathematically represented as
[tex]\theta = n * 2 * \pi[/tex]
substituting values
[tex]\theta = 1.5 * 2 * 3.142[/tex]
[tex]\theta = 9.426 \ rad[/tex]
The angular speed is mathematically represented as
[tex]w = \frac{\theta }{t}[/tex]
substituting values
[tex]w = \frac{9.426}{1.6}[/tex]
[tex]w = 5.89 \ rad/s[/tex]
Refer the attached photo
Answer:
A
Explanation:
since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet
Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What is the mass of the string (in kg)?
Answer:
m = 0.0125 kg
Explanation:
Let us apply the formula for the speed of a wave on a string that is under tension:
[tex]v = \sqrt{\frac{F}{\mu} }[/tex]
where F = tension force
μ = mass per unit length
Mass per unit length is given as:
μ = m / l
where m = mass of the string
l = length of the string
This implies that:
[tex]v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }[/tex]
Let us make mass, m, the subject of the formula:
[tex]v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}[/tex]
From the question:
F = 20 N
l = 4.50 m
v = 85 m/s
Therefore:
[tex]m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg[/tex]
Two long, parallel wires carry currents in the same direction. If I1 = 10 A, and I2 = 20 A, and they are d = 1.0 m apart, what is the magnetic field at a point P midway between them?
Answer:
The magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T
Explanation:
Given;
current in the first wire, I₁ = 10 A
current in the second wire, I₂ = 20 A
distance between the two wires, d = 1.0 m
Magnetic field at mid point between two parallel wires is calculated as;
[tex]B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r}(I_1 +I_2)[/tex]
where;
r is the midpoint between the wires, = 0.5 m
μ₀ is the permeability of free space, = 4π x 10⁻⁷
[tex]B = \frac{\mu_o }{2\pi r}(I_1 +I_2)\\\\B = \frac{4\pi*10^{-7} }{2\pi *0.5}(10 +20)\\\\B = \frac{4\pi*10^{-7} *30}{2\pi *0.5}\\\\B = 1.2 *10^{-5} \ T[/tex]
Therefore, the magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T
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
Find the ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same.
Answer:
The ratio of the new force over the original force is 16
Explanation:
Recall the formula for the gravitational force between two masses M1 and M2 separated a distance D:
[tex]F_G=G\,\frac{M_1\,\,M_2}{D^2}[/tex]
So now, if the masses M1 and M2 are quadrupled and the distance stays the same, the new force becomes:
[tex]F'_G=G\,\frac{4M_1\,\,4M_2}{D^2}=G\,\frac{16\,\,M_1\,\,M_2}{D^2}=16\,\,G\,\frac{M_1\,\,M_2}{D^2}= 16\,\,F_G[/tex]
which is 16 times the original force.
So the ratio of the new force over the original force is 16
The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.
What does Newton's law of gravitation state?Newton's law of gravitation states that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.
The formula for Newton's law of gravitation is:
[tex]F = G \frac{m_1m_2}{r^{2} }[/tex]
where,
F is the gravitational force.G is the gravitational constant.m₁ and m₂ are the masses of both objects.r is the distance between the objects.The initial force between the planets is:
[tex]F_1 = G \frac{m_1m_2}{r^{2} }[/tex]
The force between the planets if the masses of both planets are quadrupled but the distance between them stays the same is:
[tex]F_2 = G \frac{4m_14m_2}{r^{2} } = 16 G \frac{m_1m_2}{r^{2} }[/tex]
The ratio of F₂ to F₁ is:
[tex]\frac{F_2}{F_1} =\frac{16 G \frac{m_1m_2}{r^{2} }}{G \frac{m_1m_2}{r^{2} }} = \frac{16}{1}[/tex]
The ratio of the gravitational force between two planets if the masses of both planets are quadrupled but the distance between them stays the same is 16:1.
Learn more about Newton's gravitational law here: https://brainly.com/question/9373839
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
A vector quantity has direction, a scalar quantity does not.
Explanation:
hope you like then comment plz
A rod has length 0.900 mm and mass 0.500 kgkg and is pivoted at one end. The rod is not uniform; the center of mass of the rod is not at its center but is 0.500 mm from the pivot. The period of the rod's motion as a pendulum is 1.49 ss. What is the moment of inertia of the rod around the pivot
Answer:
The moment of inertia is [tex]I =0.14 \ kg \cdot m^2[/tex]
Explanation:
From the question we are told that
The length of the rod is [tex]l = 0.900 \ m[/tex]
The mass of the rod is [tex]m = 0.500 \ kg[/tex]
The distance of the center of mass from the pivot is [tex]d = 0.500 \ m[/tex]
The period of the rod's motion is [tex]T = 1.49 \ s[/tex]
Generally the period of the motion is mathematically represented as
[tex]T = 2 \pi * \sqrt{\frac{I}{m* g * d} }[/tex]
Where [tex]I[/tex] is the moment of inertia about the pivot so making [tex]I[/tex] the subject of formula
[tex]I = [\frac{T}{2\pi } ]^2 * m * g * d[/tex]
substituting values
[tex]I = [\frac{1.49}{2* 3.142 } ]^2 * 0.5 * 9.8 * 0.5[/tex]
[tex]I =0.14 \ kg \cdot m^2[/tex]
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]
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.
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
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
For more information on length conversions, visit
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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.
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 .
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.
An 100 V/m electric field is directed along the x axis. If the potential at the origin is 300 V, what is potential at the point ( -2m, 0) point
Answer:
200volts
Explanation:
Pls see attached file
Answer:
100 V
Explanation:
Electric field E = 100 V/m
Potential at the origin = 300 V
Potential at point (-2m, 0) i.e 2 m behind the origin = ?
From the equation ΔV = EΔd,
ΔV = [tex]V_{0} - V_{x}[/tex]
where [tex]V_{0}[/tex] is the potential at origin,
and [tex]V_{x}[/tex] is the potential at point (-2, 0)
E = electric field
Δd = 0 - (-2) = 2 m
[tex]V_{0} - V_{x}[/tex] = 300 - [tex]x[/tex]
equating, we have
300 - [tex]x[/tex] = 100 x 2
300 - [tex]x[/tex] = 200
[tex]x[/tex] = 100 V
If the diameter of a radar dish is doubled, what happens to its resolving power, assuming that all other factors remain unchanged? Its resolving power
a. is reduced to one-half its original value.
b. Quadruples.
c. is reduced to one-quartet its original value.
d. Halves.
e. Doubles.
Answer:
e. Doubles.
Explanation:
Resolving power is given by the formula as follows :
[tex]\dfrac{1}{d\theta}=\dfrac{D}{1.22\lambda}[/tex]
Here, [tex]d\theta[/tex] is the angle subtended by two distant objects
D is diameter of the telescope
Here, the diameter of a radar dish is doubled, assuming all other factors remain unchanged, then the resolving power gets doubled. Hence, the correct option is (e).
Which of the following is not considered a behavior?
A. eating
B. anxiety
C. sleeping
D. crying
━━━━━━━☆☆━━━━━━━
▹ Answer
B. Anxiety
▹ Step-by-Step Explanation
Anxiety isn't a behavior since it's a feeling. Behavior and feeling are different things therefore, anxiety is the correct answer.
Hope this helps!
- CloutAnswers ❁
Brainliest is greatly appreciated!
━━━━━━━☆☆━━━━━━━
Eating, sleeping, and crying all are considered as behaviors. However, anxiety cannot be considered as a behavior because it is a feeling. Thus, the correct option is B.
What is Anxiety?Anxiety is an intense feeling of excessive, and persistent worry and the fear about everyday situations. This includes fast heart rate, rapid breathing, sweating, and feeling tired constantly may occur.
Behavior is the range of actions and mannerisms which are made by individuals, organisms, systems or the artificial entities in some environment. These systems can include other systems or organisms as well as the inanimate physical environment. Behaviors include eating, sleeping, and crying. Anxiety is not a behavior, it is a feeling.
Therefore, the correct option is B.
Learn more about Anxiety here:
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I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation
Answer:
The amplitude of the oscillation is 2.82 cm
Explanation:
Given;
mass of attached block, m = 4.1 kg
energy of the stretched spring, E = 3.8 J
period of oscillation, T = 0.13 s
First, determine the spring constant, k;
[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]
where;
T is the period oscillation
m is mass of the spring
k is the spring constant
[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]
Now, determine the amplitude of oscillation, A;
[tex]E = \frac{1}{2} kA^2[/tex]
where;
E is the energy of the spring
k is the spring constant
A is the amplitude of the oscillation
[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]
Therefore, the amplitude of the oscillation is 2.82 cm
In the photoelectric effect, if the intensity of light shone on a metal increases, what will happen?
Answer:
C) There will be more electrons ejected
Explanation:
The number of electrons ejected whenever a photoelectric effect is identified it is proportional to the intensity of the incident light
Nevertheless, the photoelectrons' maximal kinetic energy is independent of their light intensity
Therefore, the maximum speed of the electron ejected doesn't really depend on the light intensity
So, if the intensity rises, only the number of electrons ejected will rised
Therefore the option c is correct
Answer:
C) There will be more electrons ejected
Explanation:
In the photoelectric effect, photons with an energy of E are shone upon a piece of metal, and if the energy of the photons overcome the work function ϕ of the metal, then electrons with will be ejected from the metal with a kinetic energy KE.
E_photon = Φ + KE
Each photon is capable of ejecting one electron from the metal. Therefore, increasing the intensity of the light (the number of photons shone on the metal) will increase the number of electrons ejected from the metal.
A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3
Answer:
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
Explanation:
Density is the mass per unit volume
Density = mass/volume = m/V
Volume of a cylinder V = πr^2 h
Given;
Height h = 48.5mm = 0.0485 m
Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m
Substituting the values;
Volume V = π×(0.0245^2)×0.0485
V = 0.000091458438030 m^3
V = 0.000091458 m^3
The mass is given as;
Mass = 1 kg
So, the density can be calculated as;
Density = 1/0.000091458
Density = 10933.92825785 kg/m^3
Density = 10,933.93 kg/m^3
the density of the material is 10,933.93 kg/m^3
A long straight metal rod has a radius of 2.0 mm and a surface charge of density 0.40 nC/m2. Determine the magnitude of the electric field 3.0 mm from the axis.
Answer:
Explanation:
Gauss Theorem
E2πrL=o2πRL/εo
then
E=oR/(rεo)
E=(0.4*10^-9*2*10^-3) / (3*10^-3*8.85*10^-12)
= 30.13 N/C
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
(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must you give the pail at the highest point of the circle if no water is to spill from it
Answer:
The minimum speed required is 2.62m/s
Explanation:
The value of gravitational acceleration = g = 9.81 m/s^2
Radius of the vertical circle = R = 0.7 m
Given the mass of the pail of water = m
The speed at the highest point of the circle = V
The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.
Below is the calculation:
[tex]\frac{mV^{2}}{R} = mg[/tex]
[tex]V = \sqrt{gR}[/tex]
[tex]V = \sqrt{9.81 \times 0.7}[/tex]
[tex]V = 2.62 m/s[/tex]
Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?
Answer:
17.94 kN/C is the electric field intensity at the origin due to the charges.
Explanation:
From the question, we are told that
The distance of 1 μC from origin = 1 m
The distance of 2 μC from origin = 2 m
The distance of 3 μC from origin = 3 m
The distance of 4 μC from origin = 5 m
Therefore, for us to find the electric field intensity, we'll solve below:
The formula for Electric field intensity = ( k * q ) / ( r * r )
where , r is distance ,
k = 9 * 10^9 ,
and , q is charge .
now ,
electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]
=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C
=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C
=> electric field intensity at the origin = 17.94 kN/C