Answer:
The angle of refraction is 28.68°Explanation:
Given data
angle of incidence, i=40°
angle of refraction,r = ?
index of refraction u= 1.33
applying the formula
[tex]u=\frac{ sin i}{ sin r}[/tex]
According to Snell's law, the incident, normal and refracted rays all act on the same point.
Substituting and solving for r we have.
[tex]1.33= \frac{sin (40)}{sin (r)} \\\\sin(40)= 1.33* sin(r)\\\0.642= 1.33* sin(r)[/tex]
divide both sides by 1.33 we have
[tex]sin(r)= \frac{0.642}{1.33} \\\sin(r)= 0.48\\\r= sin^-^10.48\\\r= 28.68[/tex]
r= 28.68°
A solenoid used to produce magnetic fields for research purposes is 2.2 mm long, with an inner radius of 30 cmcm and 1200 turns of wire. When running, the solenoid produced a field of 1.4 TT in the center. Given this, how large a current does it carry?
Answer:
The current is [tex]I = 2042\ A[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 2.2 \ m[/tex]
The radius is [tex]r_i = 30 \ cm = 0.30 \ m[/tex]
The number of turn is [tex]N = 1200 \ turns[/tex]
The magnetic field is [tex]B = 1.4 \ T[/tex]
The magnetic field produced is mathematically represented as
[tex]B = \frac{\mu_o * N * I }{l }[/tex]
making [tex]I[/tex] the subject
[tex]I = \frac{B * l}{\mu_o * N }[/tex]
Where [tex]\mu_o[/tex] is the permeability of free space with values [tex]\mu_o = 4\pi *10^{-7} N/A^2[/tex]
substituting values
[tex]I = \frac{1.4 * 2.2 }{4\pi *10^{-7} * 1200 }[/tex]
[tex]I = 2042\ A[/tex]
A student builds a rocket-propelled cart for a science project. Its acceleration is not quite high enough to win a prize, so he uses a larger rocket engine that provides 39% more thrust, although doing so increases the mass of the cart by 13%. By what percentage does the cart's acceleration increase?
Answer:
Explanation:
a = F / m
where a is acceleration , F is thrust and m is mass
taking log and differentiating
da / a = dF / F - dm / m
(da / a)x 100 = (dF / F)x100 - (dm / m) x100
percentage increase in a = percentage increase in F - percentage increase in m
= percentage increase in acceleration a = 39 - 13 = 26 %
required increase = 26 %.
A typical home uses approximately 1600 kWh of energy per month. If the energy came from a nuclear reaction, what mass would have to be converted to energy per year to meet the energy needs of the home
Answer:
7.68×10^25kg
Explanation:
The formula for energy used per year is calculated as
Energy used per year =12 x Energy used per month
By substituting Energy used per month in the above formula, we get
Energy used per year =12 x 1600kWh
= 19200kWh
Conversion:
From kWh to J:
1 kWh=3.6 x 10^6 J
Therefore, it is converted to J as
19200 kWh =19200 x 3.6 x 10^6 J
= 6.912×10^10 J
Hence, energy used per year is 6.912×10^10 J
To find the mass that is converted to energy per year.
E = MC^2 ............1
E is the energy used per year
C is the speed of light = 3.0× 10^8m/s
Where E= 6.912×10^10 J
Substituting the values into equation 1
6.912×10^10 J = M × 3.0× 10^8m/s
M = 6.912×10^10 J / (3.0× 10^8m/s)^2
M = 6.912×10^10 J/9×10^16
M = 7.68×10^25kg
Hence the mass to be converted is
7.68×10^25kg
In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum
Complete Question
In a double-slit arrangement the slits are separated by a distance equal to 150 times the wavelength of the light passing through the slits. (a) What is the angular separation between the central maximum and an adjacent maximum? (b) What is the distance between these maxima on a screen 57.9 cm from the slits?
Answer:
a
[tex]\theta = 0.3819^o[/tex]
b
[tex]y = 0.00386 \ m[/tex]
Explanation:
From the question we are told that
The slit separation is [tex]d = 150 \lambda[/tex]
The distance from the screen is [tex]D = 57.9 \ cm = 0.579 \ m[/tex]
Generally the condition for constructive interference is mathematically represented as
[tex]dsin (\theta ) = n * \lambda[/tex]
=> [tex]\theta = sin ^{-1} [\frac{n * \lambda }{ d } ][/tex]
where n is the order of the maxima and value is 1 because we are considering the central maximum and an adjacent maximum
and [tex]\lambda[/tex] is the wavelength of the light
So
[tex]\theta = sin ^{-1} [\frac{ 1 * \lambda }{ 150 \lambda } ][/tex]
[tex]\theta = 0.3819^o[/tex]
Generally the distance between the maxima is mathematically represented as
[tex]y = D tan (\theta )[/tex]
=> [tex]y = 0.579 tan (0.3819 )[/tex]
=> [tex]y = 0.00386 \ m[/tex]
An elastic circular bar is fixed at one end and attached to a rubber grommet at the other end. The grommet functions as a torsional spring with spring constant k. If a concentrated torque of magnitude Ta is applied in the center of the bar, what is the rotation at the end of the bar, φ(L)? Assume a constant shear modulus G and polar moment of inertia J.
Answer:
2.1 rad(anticlockwise).
Explanation:
So, we are given the following data or parameters or information in the question above:
=> "The torsional stiffness of the spring support is k = 50 N m/rad. "
=> "If a concentrated torque of mag- nitude Ta = 500 Nm is applied in the center of the bar"
=> "L = 300 mm Assume a shear modu- lus G = 10 kN/mm2 and polar monnent of inertia J = 2000 mln"
Hence;
G × J = 10 kN/mm2 × 2000 mln = 20 Nm^2.
Also, L/2 = 300 mm /2 = 0.15 m (converted to metre).
==> 0.15/20 (V - w) + θ = 0.
==> 0.15/20 (V - w ) = -θ.
Where V = k = 50 N m/rad
w = 183.3 θ.
Therefore, w + Vθ = 500 Nm.
==> 183.3 + 50 θ = 500 Nm.
= 6.3
Anticlockwise,
θ = 2.1 rad.
A skater of mass 45.0 kg standing on ice throws a stone of mass 7.65 kg with a speed of 20.9 m/s in a horizontal direction. Find:
a. The speed of the skater after throwing the stone.
b. The distance over which the skater will move in the opposite direction if the coefficient of kinetic friction between his skates and the ice is 0.03.
Answer:
Explanation:
know that there is no external force on skater and the stone so the total momentum of the system will remains constant
so we will have
here we have
so the skater will move back with above speed
now the deceleration of the skater is due to friction given as
Answer:
(a) 3.553 m/s
(b) 21.46 m
Explanation:
(a) Applying the law of of momentum,
Total momentum before collision = Total momentum after collision
mu+m'u' = mv+m'v'.................. Equation 1
Where m and m' are the mass of skater and stone respectively, u and u' are the initial velocity of skater and stone respectively, v and v' are the final velocity of the skater and the stone respectively.
Note, u = 0 m/s, u' = 0 m/s
Therefore,
0 = mv+m'v'
-mv = m'v'................ Equation 2
make v the subject of the equation
v = -m'v'/m............. Equation 3
Given: m = 45 kg, m' = 7.65 kg, v' = 20.9 m/s
Substitute into equation 3
v = 7.65(20.9)/45
v = -3.553 m/s
Hence the speed of the skater = 3.553 m/s
(b) F = mgμ..............Equation 4
But F = ma
Therefore,
ma = mgμ
a = gμ............... Equation 5
Where a = acceleration of the skater, g = acceleration due to gravity, μ = coefficient of kinetic friction
Given: μ = 0.03, g = 9.8 m/s²
Substitute into equation 5
a = 0.03(9.8)
a = 0.294 m/s²
Using the equation of motion,
v² = u²+2as............. Equation 6
Where s = distance moved by the skater.
note that u = 0 m/s.
therefore,
v² = 2as
s = v²/2a................ Equation 7
Given: v = 3.553 m/s, a = 0.294
Substitute into equation 7
s = 3.553²/(2×0.294)
s = 12.62/0.588
s = 21.46 m
(a) Find the speed of waves on a violin string of mass 717 mg and length 24.3 cm if the fundamental frequency is 980 Hz. (b) What is the tension in the string? For the fundamental, what is the wavelength of (c) the waves on the string and (d) the sound waves emitted by the string? (Take the speed of sound in air to be 343 m/s.)
Answer:
a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c) λ = 0.486 m , d) λ = 0.35 m
Explanation:
a) The speed of a wave on a string is
v = √T /μ
also all the waves fulfill the relationship
v = λ f
they indicate that the fundamental frequency is f = 980 Hz.
The wavelength that is fixed at its ends and has a maximum in the center
L = λ / 2
λ = 2L
we substitute
v = 2 L f
let's calculate
v = 2 0.243 980
v = 476.28 m / s
b) The tension of the rope
T = v² μ
the density of the string is
μ = m / L
T = v² m / L
T = 476.28² 0.717 / 0.243
T = 6.69 10⁵ N
c) λ = 2L
λ = 2 0.243
λ = 0.486 m
d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air
v = λ f
λ= v / f
λ = 343/980
λ = 0.35 m
Consider an electromagnetic wave where the electric field of an electromagnetic wave is oscillating along the z-axis and the magnetic field is oscillating along the x-axis.
In what directions is it possible that the wave is traveling?
A. The-z direction.
B. The ty direction
C. The +x direction.
D. The -y direction
E. The -x direction.
F. The +z direction.
Answer:
The wave will be travelling in the y-axis
Explanation:
An e-m wave has a spatially varying electric field that is always associated with a magnetic field that changes over time and vice versa. The electric field and the magnetic field oscillates perpendicularly to each other, and together form a wave that travels in a perpendicular direction to the magnetic and the electric field in space. The movement of the e-m wave through space is usually away from the source where it is generated. So, if the electric field travels in the z-axis, and the magnetic field travels through along the x-axis, then the e-m wave generated will travel in the y-axis direction.
What is the wavelength λλlambda of the wave described in the problem introduction? Express the wavelength in terms of the other given variables and constants
Complete Question
The complete question is shown on the first uploaded image
Answer:
The wavelength is [tex]\lambda= \frac{2 \pi }{k}[/tex]
Explanation:
From the question we are told that
The electric field is [tex]\= E = E_o sin (kx - wt )\r j[/tex]
The magnetic field is [tex]\= B = B_0 sin (kx -wt) \r k[/tex]
From the above equation
and k is the wave number which is mathematically represented as
[tex]k = \frac{2 \pi }{\lambda }[/tex]
=> [tex]\lambda= \frac{2 \pi }{k}[/tex]
Where [tex]\lambda[/tex] is the wavelength
"On a movie set, an alien spacecraft is to be lifted to a height of 32.0 m for use in a scene. The 260.0-kg spacecraft is attached by ropes to a massless pulley on a crane, and four members of the film's construction crew lift the prop at constant speed by delivering 135 W of power each. If 18.0% of the mechanical energy delivered to the pulley is lost to friction, what is the time interval required to lift the spacecraft to the specified height?"
Answer:
The time interval required to lift the spacecraft to this specified height is 123.94 seconds
Explanation:
Height through which the spacecraft is to be lifted = 32.0 m
Mass of the spacecraft = 260.0 kg
Four crew member each pull with a power of 135 W
18.0% of the mechanical energy is lost to friction.
work done in this situation is proportional to the mechanical energy used to move the spacecraft up
work done = (weight of spacecraft) x (the height through which it is lifted)
but the weight of spacecraft = mg
where m is the mass,
and g is acceleration due to gravity 9.81 m/s
weight of spacecraft = 260 x 9.81 = 2550.6 N
work done on the space craft = weight x height
==> work = 2550.6 x 32 = 81619.2 J
this is equal to the mechanical energy delivered to the system
18.0% of this mechanical energy delivered to the pulley is lost to friction.
this means that
0.18 x 81619.2 = 14691.456 J is lost to friction.
Total useful mechanical energy = 81619.2 J - 14691.456 J = 66927.74 J
Total power delivered by the crew to do this work = 135 x 4 = 540 W
But we know tat power is the rate at which work is done i.e
[tex]P = \frac{w}{t}[/tex]
where p is the power
where w is the useful work done
t is the time taken to do this work
imputing values, we'll have
540 = 66927.74/t
t = 66927.74/540
time taken t = 123.94 seconds
A 4g bullet, travelling at 589m/s embeds itself in a 2.3kg block of wood that is initially at rest, and together they travel at the same velocity. Calculate the percentage of the kinetic energy that is left in the system after collision to that before.
Answer:
The percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
Explanation:
Given;
mass of bullet, m₁ = 4g = 0.004kg
initial velocity of bullet, u₁ = 589 m/s
mass of block of wood, m₂ = 2.3 kg
initial velocity of the block of wood, u₂ = 0
let the final velocity of the system after collision = v
Apply the principle of conservation of linear momentum
m₁u₁ + m₂u₂ = v(m₁+m₂)
0.004(589) + 2.3(0) = v(0.004 + 2.3)
2.356 = 2.304v
v = 2.356 / 2.304
v = 1.0226 m/s
Initial kinetic energy of the system
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(0.004)(589)² = 693.842 J
Final kinetic energy of the system
K.E₂ = ¹/₂v²(m₁ + m₂)
K.E₂ = ¹/₂ x 1.0226² x (0.004 + 2.3)
K.E₂ = 1.209 J
The kinetic energy left in the system = final kinetic energy of the system
The percentage of the kinetic energy that is left in the system after collision to that before = (K.E₂ / K.E₁) x 100%
= (1.209 / 693.842) x 100%
= 0.174 %
Therefore, the percentage of the kinetic energy that is left in the system after collision to that before is 0.174 %
A body of mass 2.5kg is raised 4.0m above the ground.Calculate the potential energy if g=10m/s squared
Answer:
100 joulesExplanation:
[tex]mass = 2.5kg\\height = 4.0\\Acceleration \: due \:to\:gravity = 10m/s^2\\\\P.E = mgh\\P.E = 2.5kg\times10\times4\\\\Potential \: Energy = 100 joules[/tex]
The potential energy if g = 10m/s² is 98 J
What is Potential Energy ?Potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.
Formula for potential energy :
P.E. = mgh
m = mass (kg)g = gravity (m/s²)h = height/distance (m)Given :
The book is held from the ground of a distance = 4.0 m,
so
h = 4.0 m
we know that the book weighs,
2.5 kg
so
m = 2.5 kg.
Now we just put it in the formula ;
PE = (2.5kg) × (9.8 m/s²) × ( 4.0 m)
P E = 98 J
Therefore, The potential energy if g = 10m/s² is 98 J
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4. The Richter scale describes how much energy an earthquake releases. With every increase of 1.0 on the scale, 32 times more energy is released. How many times more energy would be released by a quake measuring 2.0 more units on the Richter scale?
Answer:
64 times
Explanation:
if increase of 1 gives you 32
then increase of 2 will give you its double
64
If you increase one, you get 32 then multiplying by 2 will give you 64, which is its double.
What is Earthquake?An earthquake is a sudden energy released in the Earth's lithosphere that causes shock wave, which cause the Earth's surface to shake. Earthquakes can range in strength from ones that are so small that no one can feel them to quakes that are so powerful that they uproot entire cities, launch individuals and objects into the air, and harm vital infrastructure.
The frequency, kind, and intensity of earthquakes observed over a specific time period are considered to be the seismic activity of an area.
The average rate of earthquake energy output per unit volume determines the basicity of a certain area of the Earth. The non-earthquake seismic rumbling is also alluded to as a tremor.
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Question 8
A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The
damping factor for the spring is determined to be 0.4 and the spring oscillates 12 times each
second. Find an equation for the displacement, D(t), of the spring from equilibrium in terms of
seconds, t.
D(t) =
Can someone please help me ASAP?!!!!
Answer: D(t) = [tex]8.e^{-0.4t}.cos(\frac{\pi }{6}.t )[/tex]
Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:
y = [tex]a.sin(\omega.t)[/tex] or y = [tex]a.cos(\omega.t)[/tex]
where:
|a| is initil displacement
[tex]\frac{2.\pi}{\omega}[/tex] is period
For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:
[tex]y=a.e^{-ct}.cos(\omega.t)[/tex] or [tex]y=a.e^{-ct}.sin(\omega.t)[/tex]
For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:
[tex]y=a.e^{-ct}.cos(\omega.t)[/tex]
period = [tex]\frac{2.\pi}{\omega}[/tex]
12 = [tex]\frac{2.\pi}{\omega}[/tex]
ω = [tex]\frac{\pi}{6}[/tex]
Replacing values:
[tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex]
The equation of displacement, D(t), of a spring with damping factor is [tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex].
Sally who weighs 450 N, stands on a skate board while roger pushes it forward 13.0 m at constant velocity on a level straight street. He applies a constant 100 N force.
Work done on the skateboard
a. Rodger Work= 0J
b. Rodger work= 1300J
c. sally work= 1300J
d. sally work= 5850J
e. rodger work= 5850J
Answer:
b. Rodger work = 1300 J
Explanation:
Work done: This can be defined as the product of force and distance along the direction of the force.
From the question,
Work is done by Rodger using a force of 100 N in pushing the skateboard through a distance of 13.0 m.
W = F×d............. Equation 1
Where W = work done, F = force, d = distance.
Given: F = 100 N, d = 13 m
Substitute these values into equation 1
W = 100(13)
W = 1300 J.
Hence the right option is b. Rodger work = 1300 J
A motor is designed to operate on 117 V and draws a current of 12.3 A when it first starts up. At its normal operating speed, the motor draws a current of 3.38 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed
Answer:
a) using
R=V/I =117/12.3
R=9.5 ohms
b)
E=V-I*R =117-3.38*9.5
E=84.8Volts
c)
at (1/3)rd of normal speed ,back emf is (1/3) of its maximum
value
E=(1/3)*84.8=28.3Volts
Current drawn
I=V-Eback/R =117-28.3/9.5
I=9.33A
Explanation:
The resistance is = 9.5 ohms
The back emf developed at normal speed is = 84.90 volts
The current drawn at one-third normal speed =9.33 A.
To calculate the resistance of the armature coil this formula is used;
V = IR
make R the subject of formula,
R = V/I
where R = resistance
V = voltage
I = Current
R = 117/12.3
R = 9.5 ohms
To calculate the back emf developed at normal speed, this formula is used;
E = V + Ir ( for normal emf)
But for back emf which is the difference between the supplied voltage and the loss from the current through the resistance, this formula is used;
E = V - Ir
where V = 117v
I = 3.38
r = 9.5
E = 117 - ( 3.38 × 9.5)
= 117 - 32.11
= 84.90 volts
To calculate the current drawn at one-third normal speed;
1/3 of Emf = 1/3 × 84.90
= 28.3volts
Therefore current (I) = V - E/ R
= 117- 28.3/9.5
= 88.7/9.5
= 9.33 A
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EXAMPLE 5 Find the radius of gyration about the x-axis of a homogeneous disk D with density rho(x, y) = rho, center the origin, and radius a. SOLUTION The mass of the disk is m = rhoπa2, so from these equations we have 2 = Ix m = 1 4πrhoa4 rhoπa2 = a2 4 .
Answer:
Radius of gyration = a/2.
Explanation:
So, from the question above I can see that the you are already answering the question and you are stuck up or maybe that's how the problem is set from the start. Do not worry, you are covered in any of the ways. So, from the question we have that;
"The mass of the disk is m = ρπa^2, so from these equations we have y^2 = Ix/m."
(NB: I changed the "rho" word to its symbol).
Thus, the radius of gyration with respect to x-axis = (1/4 πρa^4)/ πρa^2 = a^2/4.
Therefore, the Radius of gyration = a/2.
"A power of 200 kW is delivered by power lines with 48,000 V difference between them. Calculate the current, in amps, in these lines."
Answer:
9.6×10⁹ A
Explanation:
From the question above,
P = VI.................... Equation 1
Where P = Electric power, V = Voltage, I = current.
make I the subject of the equation
I = P/V............. Equation 2
Given: P = 200 kW = 200×10³ W, V = 48000 V.
Substitute these vales into equation 2
I = 200×10³×48000
I = 9.6×10⁹ A.
Hence the current in the line is 9.6×10⁹ A.
An object has an acceleration of 6.0 m/s/s. If the net force was doubled and the mass was one-third the original value, then the new acceleration would be _____ m/s/s.
Hahahahaha. Okay.
So basically , force is equal to mass into acceleration.
F=ma
so when F=ma , we get acceleration=6m/s/s
Force is doubled.
Mass is 1/3 times original.
2F=1/3ma
Now , we rearrange , and we get 6F=ma
So , now for 6 times the original force , we get 6 times the initial acceleration.
So new acceleration = 6*6= 36m/s/s
A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s
With what angular speed is the stick spinning after the collision?
Answer:
63.44 rad/s
Explanation:
mass of bullet = 3.3 g = 0.0033 kg
initial velocity of bullet [tex]v_{1}[/tex] = 250 m/s
final velocity of bullet [tex]v_{2}[/tex] = 140 m/s
loss of kinetic energy of the bullet = [tex]\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})[/tex]
==> [tex]\frac{1}{2}*0.0033*(250^{2} - 140^{2} )[/tex] = 70.785 J
this energy is given to the stick
The stick has mass = 250 g =0.25 kg
its kinetic energy = 70.785 J
from
KE = [tex]\frac{1}{2} mv^{2}[/tex]
70.785 = [tex]\frac{1}{2}*0.25*v^{2}[/tex]
566.28 = [tex]v^{2}[/tex]
[tex]v= \sqrt{566.28}[/tex] = 23.79 m/s
the stick is 1.5 m long
this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m
The angular speed will be
Ω = v/r = 23.79/0.375 = 63.44 rad/s
Determine the slit spacing d. Explain which measurement you made, show your calculation and your result for the slit spacing. There are several measurements you can make.
Answer:
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
Explanation:
To determine the gap spacing we must use the constructive interference equation
d sin θ = m λ
as the angles are small
tan θ = sin θ / cos θ
tan θ = sin θ
and the definition of tangent is
tan θ = y / L
Thus
sin θ = y / L
when replacing
d y / L = m λ
d = m λ L / y
with this equation we can know what parameter should be measured.
The quantities to measure are:
* the distance to the screen
* The distance from the central maximum to each interference
* in order of interference
* wavelength
As a wheel turns, the angle through which it has turned varies with time as β(t)=Ct + Bt3 where C=0.400rad/s and B=0.0120rad/s3. Calculate the angular velocity w(t) as a function of time.
Answer:
ω(t) = 0.4 + 0.036 t²
Explanation:
The angular displacement of the disk is given as the function of time:
β(t) = Ct + B t³
where,
C = 0.4 rad/s
B = 0.012 rad/s³
Therefore,
β(t) = 0.4 t + 0.012 t³
Now, for angular velocity ω(t), we must take derivative of angular displacement with respect to t:
ω(t) = dβ/dt = (d/dt)(0.4 t + 0.012 t³)
ω(t) = 0.4 + 0.036 t²
Huygens claimed that near the surface of the Earth the velocity downwards of an object released from rest, vy, was directly proportional to the square root of the distance it had fallen, . This is true if c is equal to
Answer:
the expression is False
Explanation:
From the kinematics equations we can find the speed of a body in a clean fall
v = v₀ - g t
v² = V₀² - 2 g y
If the body starts from rest, the initial speed is zero (vo = 0)
v= √ (2g y)
In the first equation it gives us the relationship between speed and time.
With the second equation we can find the speed in which the distance works, this is the expression, see that speed is promotional at the height of a delicate body.
Therefore the expression is False
two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister
A horizontal uniform meter stick is supported at the 50.0 cm mark. It has a mass of 0.52 kg, hanging from it at the 20.0 cm mark and a mass of 0.31 kg mass hanging from the 60.0 cm mark. Determine the position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance. Group of answer choices
Answer: 70.5 cm
Explanation:
The position on the meter stick, at which one would hang a third mass of 0.61 kg, to keep the meter stick in balance will be at the side of 0.31kg.
You will use the moment techniques.
That is,
Sum of the clockwise moment = sum of anticlockwise moments
Please find the attached file for the remaining explanation and solution.
Four 50-g point masses are at the corners of a square with 20-cm sides. What is the moment of inertia of this system about an axis perpendicular to the plane of the square and passing through its center
Answer:
moment of inertia I ≈ 4.0 x 10⁻³ kg.m²
Explanation:
given
point masses = 50g = 0.050kg
note: m₁=m₂=m₃=m₄=50g = 0.050kg
distance, r, from masses to eachother = 20cm = 0.20m
the distance, d, of each mass point from the centre of the mass, using pythagoras theorem is given by
= (20√2)/ 2 = 10√2 cm =14.12 x 10⁻² m
moment of inertia is a proportion of the opposition of a body to angular acceleration about a given pivot that is equivalent to the entirety of the products of every component of mass in the body and the square of the component's distance from the center
mathematically,
I = ∑m×d²
remember, a square will have 4 equal points
I = ∑m×d² = 4(m×d²)
I = 4 × 0.050 × (14.12 x 10⁻² m)²
I = 0.20 × 1.96 × 10⁻²
I = 3.92 x 10⁻³ kg.m²
I ≈ 4.0 x 10⁻³ kg.m²
attached is the diagram of the equation
A particle with a charge of 4.0 μC has a mass of 5.0 × 10 -3 kg. What electric field directed upward will exactly balance the weight of the particle?
Answer:
E = 12.25 x 10³ N/C = 12.25 KN/C
Explanation:
In order to balance the weight of the object the electrostatic force due to the electric field must be equal to the weight of the body or charge. Therefore,
Electrostatic Force = Weight
E q = mg
where,
E = Electric Field = ?
m = Mass of the Charge = 5 x 10⁻³ kg
g = acceleration due to gravity = 9.8 m/s²
q = magnitude of charge = 4 μC = 4 x 10⁻⁶ C
Therefore,
E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)
E = 0.049 N/4 x 10⁻⁶ C
E = 12.25 x 10³ N/C = 12.25 KN/C
Suppose that a 117.5 kg football player running at 6.5 m/s catches a 0.43 kg ball moving at a speed of 26.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in opposite directions.
(d) Calculate the change in kinetic energy of the system, in joules, in this case.
Answer:
a) 6.57 m/s
b) 53.75 J
c) 6.37 m/s
d) -98.297 J
Explanation:
mass of player = [tex]m_{p}[/tex] = 117.5 kg
speed of player = [tex]v_{p}[/tex] = 6.5 m/s
mass of ball = [tex]m_{b}[/tex] = 0.43 kg
velocity of ball = [tex]v_{b}[/tex] = 26.5 m/s
Recall that momentum of a body = mass x velocity = mv
initial momentum of the player = mv = 117.5 x 6.5 = 763.75 kg-m/s
initial momentum of the ball = mv = 0.43 x 26.5 = 11.395 kg-m/s
initial kinetic energy of the player = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.5^{2}[/tex] = 2482.187 J
a) according to conservation of momentum, the initial momentum of the system before collision must equate the final momentum of the system.
for this first case that they travel in the same direction, their momenta carry the same sign
[tex]m_{p}[/tex][tex]v_{p}[/tex] + [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
where v is the final velocity of the player.
inserting calculated momenta of ball and player from above, we have
763.75 + 11.395 = (117.5 + 0.43)v
775.145 = 117.93v
v = 775.145/117.93 = 6.57 m/s
b) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.57^{2}[/tex] = 2535.94 J
change in kinetic energy = 2535.94 - 2482.187 = 53.75 J gained
c) if they travel in opposite direction, equation becomes
[tex]m_{p}[/tex][tex]v_{p}[/tex] - [tex]m_{b}[/tex][tex]v_{b}[/tex] = ([tex]m_{p}[/tex] +[tex]m_{b}[/tex])v
763.75 - 11.395 = (117.5 + 0.43)v
752.355 = 117.93v
v = 752.355/117.93 = 6.37 m/s
d) the player's new kinetic energy = [tex]\frac{1}{2} mv^{2}[/tex] = [tex]\frac{1}{2}[/tex] x 117.5 x [tex]6.37^{2}[/tex] = 2383.89 J
change in kinetic energy = 2383.89 - 2482.187 = -98.297 J
that is 98.297 J lost
Solve 3* +5-220t = 0
Answer:
t = 27.5
Explanation:
[tex]3 + 5 -220t = 0[/tex]
Well to solve for t we need to combine like terms and seperate t.
So 3+5= 8
8 - 220t = 0
We do +220 to both sides
8 = 220t
And now we divide 220 by 8 which is 27.5
Hence, t = 27.5
A typical electric oven has two separate heating elements: one on top and one on the bottom. The bottom element is used for baking while the top element is used to broil foods. When only the bottom element is active and glowing red hot, what heat transfer mechanisms carry most of the heat to the food in the oven?
Answer:
Convection and Radiation mechanisms carry most of the heat
Explanation:
This is because Convection proceeds strongy as heated air rises from the hot element while Radiation is also strong, although the material of the cooking pots will how effective it is.