Answer:
113 degrees
1.974 radians
Explanation:
The angle can be calculated using below expression
Angle = arc/r
Where arc= diameter of d = 310 millimeters
r= s = 306 millimeters long.
Since both have the same unit, no need for conversion
Angle = arc/r = 306/(0.5×310 )
= 306/155
= 1.974 radians
Then to find the angle In degrees we have
R x 180/pi = D
Where R= angle in radians= 1.974 radians
1.974 x 180/π
Where π= 22/7
= 113 degrees( approximately)
Therefore, central angle is subtended by one of the brake pads is 1.974 radians
Which one would be felt louder: a sound with a SIL of 60dB and a frequency of 40Hz or a sound with a SIL of 60dB and a frequency of 100Hz ? a) they both will appear to have the same loudness b) sound with a SIL of 60dB c) sound with a SIL of 60dB and a frequency of 1000Hz d) whichever is sounded first e) cannot say and a frequency of 40ã
Answer:
a) they both will appear to have the same loudness
Explanation:
When we talk about how loud a sound is , we describe it in decibels (dB). The value of this unit on measurement in this question is what we are going to build our answer on.
Their sound intensity level is the same at 60db even though each of these separate sounds have different frequencies of 40Hz and 100Hz respectively.
The both of them will therefore have the same loudness actually as their sound intensity level has no difference bat 60db each.
please help! What is the relationship between velocity and acceleration?
Answer:Acceleration implies any change in the velocity of the object with respect to time. Velocity is nothing but the rate of change of displacement. On the other hand, acceleration is the rate of change of velocity with respect to time.
Explanation:
A block of mass 12.2 kg is sliding at an initial velocity of 3.9 m/s in the positive x-direction. The surface has a coefficient of kinetic friction of 0.173. (indicate the direction with signs of your answer) (a) What is the force of kinetic friction in N acting on the block? (b) What is the block's acceleration in /s^2? (c) How Far will it slide (in m) before coming to rest? Plz answer as soon as possible
Answer:
Explanation:
a) Force of friction = μ R where μ is coefficient of kinetic friction and R is reaction force
R = mg where m is mass of the block
Force of friction F = μ x mg
= .173 x 12.2 x 9.8
= 20.68 N
b ) Only force of friction is acting on the body so
deceleration = force / mass = 20.68 / 12.2 = 1.7 m /s²
acceleration = - 1.7 m /s²
c )
v² = u² - 2 a s
v = 0 , u = 3.9 m /s
a = 1.7 m /s
0 = 3.9² - 2 x 1.7 x s
s = 4.47 m
A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?
Answer:
(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰
(a)(ii) The maximum torque is 0.132 Nm
(b) The orientation of the coil is 45⁰
Explanation:
Given;
diameter of the circular wire, d = 8.6 cm = 0.086 m
radius of the wire, r = d /2 = 0.043 m
number of turns, N = 15 turns
magnetic field, B = 0.56 T
The torque on the wire is given by;
τ = NIABsinθ
where;
θ is the orientation of the wire
(a) maximum torque occurs when the orientation of the wire is at 90⁰
The maximum torque is given by;
τ = NIABsin(90⁰)
τ = NIAB
τ = (15)(2.7)(π x 0.043²)(0.56)
τ = 0.132 Nm
(b)
71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm
[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]
2. An ambulance traveling at 20 m/s emits a sound at 500 Hz. What frequency does a person standing on the corner of a street detect?
On the image at right, the two magnets are the same. Which paper clip would be harder to remove?
Answer:B
Explanation: The book is thinner making magnets attraction stronger, making the paper clip harder to move
A penny is dropped from rest from a building 100m tall. what kind of motion is this
A. centripetal
B. Free fall
C. Linear
D. projectile
Answer:
this is a projectile
Answer:
D. projectile
Explanation:
Since it's dropped from a rest that means that it's velocity at the beginning is 0.
(1.5 pts) A woman pushes on a box to the left. If the box is accelerating, what forces are working on the
Question 2:
box? (Draw both y and x forces)
Answer:
Nope
Explanation:
How do lenses and mirrors compare in their interactions with light?
A. Lenses spread apart light; mirrors do not.
B. Lenses reflect light; mirrors do not.
C. Lenses refract light; mirrors do not.
D. Lenses focus light; mirrors do not.
Answer:
C. lenses refract light; mirrors do not
This question involves the concepts of reflection and refraction.
The comparison of lenses and mirrors in their interaction with light is "C. Lenses refract light; mirrors do not.".
LENSES AND MIRRORSWhen it comes to the interaction with light, the key difference between lenses and mirrors is the difference of refraction and reflection. Reflection means the complete rebound of the light rays after striking on a surface without any absorption or transmission. On the other hand, refraction is the bending of light rays, while passing through a medium, without any rebound or absorption.
Lenses are tansparent from both sides, so they refract the light rays. While, mirrors are coated opaque from one side, so they reflect back the light rays.
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A rod is pivoted about its center and oriented horizontally. A 5.0-N force directed upward is applied 4.0 m to the left of the pivot and another upward 5.0-N force is applied 1.5 m to the right of the pivot. What is magnitude of the total torque about the pivot?
Answer:
The total torque is 27.5 Nm
Explanation:
Given;
5.0-N force directed upward is applied 4.0 m to the left of the pivot,
5.0-N force directed upward is applied 1.5 m to the right of the pivot,
Taking the moment about the pivot, the total torque is given by;
τ = Fr
where;
F is the appllied force
r is the radius of the force arm
τ = (5 N x 4 m) + (5 N x 1.5 m)
τ = 27.5 Nm
Therefore, the total torque is 27.5 Nm
In the Bohr model of the hydrogen atom, an electron in the 1st excited state moves at a speed of 2.19 106 m/s in a circular path having a radius of 5.29 10-11 m. What is the effective current associated with this orbiting electron?
Answer:
I = 1.05x10⁻³ A
Explanation:
By definition, an electric current is the rate of charge flow at a given time:
[tex] I = \frac{q}{t} [/tex]
Where:
q: is the electrons charge = 1.602x10⁻¹⁹ C
t: is the time
In a circular motion, the time is given by:
[tex] t = T = \frac{2\pi}{\omega} = \frac{2\pi}{v/r} = \frac{2\pi r}{v} [/tex]
Where:
ω: is the angular speed = v/r
v: is the speed = 2.19x10⁶ m/s
r: is the radius = 5.29x10⁻¹¹ m
[tex] t = \frac{2\pi r}{v} = \frac{2\pi 5.29 \cdot 10^{-11} m}{2.19 \cdot 10^{6} m/s} = 1.52 \cdot 10^{-16} s [/tex]
Now, the effective current is:
[tex] I = \frac{q}{t} = \frac{1.602 \cdot 10^{-19} C}{ 1.52 \cdot 10^{-16} s} = 1.05 \cdot 10^{-3} A [/tex]
Therefore, the effective current associated with this orbiting electron is 1.05x10⁻³ A.
I hope it helps you!
Which is increased when the string of a stringed instrument is tightened?
timbre
pitch
wavelength
loudness
When the string of the instrument is tightened then the length of the string decreases hence the pitch and frequency will increase so, option B is correct.
What is pitch?The frequency at which the sound waves that create a sound vibrate determines its pitch. High-frequency sound waves produce high-pitched noises, and low-frequency sound waves make low-pitched noises. The ability to discern between harsh and flat sounds is known as pitch.
A string will vibrate at a varied frequency depending on its length. Higher frequency and higher pitch are produced by shorter strings.
The pitch rises as the anxiety does as well. A string's length is also crucial. A string vibrates and makes music when it is supported at two points and pulled. The pitch of this string will, however, rise if the length is shortened.
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Which image illustrates the interaction of a light wave with a mirror?
t J
A
с
.
A. A
B. B
C. C
D. D
0
Answer:
I'm pretty sure its A
Explanation:
because its a reflection- Hope you get a good grade!
how does tom and jerry movie character influence your attitude
Answer:
it makes me wish I was a cartoon
Answer:
goofy and stupid and act like a kid
Explanation:
In a certain time, light travels 3.50 km in a vacuum. During the same time, light travels only 2.35 km in a liquid. What is the refractive index of the liquid?
Answer:
1.45
Explanation:
Refractive index of the liquid is given as;
Refractive index = [tex]\frac{speed of light in vacuum}{speed of light in liquid}[/tex]
But,
speed = [tex]\frac{distance}{time}[/tex]
Since a certain light of specific wavelength was used during the same time, let the time be represented by t.
So that;
speed of light in vacuum = [tex]\frac{3500}{t}[/tex]
speed of light in the liquid = [tex]\frac{2350}{t}[/tex]
Refractive index = [tex]\frac{3500}{t}[/tex] ÷ [tex]\frac{2350}{t}[/tex]
= [tex]\frac{3500}{t}[/tex] x [tex]\frac{t}{2350}[/tex]
Refractive index = [tex]\frac{3500}{2350}[/tex]
= 1.4536
= 1.45
The refractive index of the liquid is 1.45.
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
A Ping-Pong ball has a mass of 2.3 g and a terminal speed of 9.3 m/s. The drag force is of the form bv2 What is the value of b?
Answer:
The value is [tex]b = 0.00026 \ kg / m[/tex]
Explanation:
From the question we are told that
The mass of the Ping-Pong is [tex]m = 2.3 \ g = 0.0023 \ kg[/tex]
The terminal speed is [tex]v = 9.3 \ m/s[/tex]
The drag force is [tex]bv^2[/tex]
Generally the resultant force on the Ping- Pong is mathematically represented as
[tex]F = mg - bv^2[/tex]
when terminal velocity is attained , the resultant force is zero so
[tex]0 = mg - bv^2[/tex]
=> [tex]b = \frac{m * g}{v^2}[/tex]
=> [tex]b = \frac{0.0023 * 9.8}{ 9.3 ^2}[/tex]
=> [tex]b = 0.00026 \ kg / m[/tex]
A tall, open container is full of glycerine. At what depth h below the surface of the glycerine is the pressure 2370 Pa greater than atmospheric pressure? The density of glycerine is 1.26X10^3 kg/m^3
Answer:
So, at the depth of 24 cm below the surface of the glycerine the pressure is 2970 Pa. Hence, this is the required solution.
Explanation:
Given that,
Pressure exerted by the surface of glycerine, P = 2970 Pa and it is greater than atmospheric pressure.
The density of glycerine,
We need to find the depth h below the surface of the glycerine. The pressure due to some depth is given by :
h = 0.24 meters
or
h = 24 cm
7) A moving object is rolling on a surface that is 5 m off the ground. The object is moving at a constant speed of 4 m/s. If the object is 3.2 kg, what is the final energy of the ball after rolling for 10 m, assuming friction is negligible?
156.8 J
131.2 J
182.4 J
25.6 J
8) A spreadsheet application is used to create a computational model of the energy experienced by a pendulum. How do the energy values of the pendulum relate?
The sum of the potential energy and the kinetic energy is always constant.
The sum of the potential energy and the kinetic energy is always 0.
The potential energy is always greater than the kinetic energy.
The kinetic energy is always equal to the potential energy.
131.2 J and The last one on number 8
I gave the same answer and it passed.
7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.
8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.
What is energy?Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.
Given:
A moving object is rolling on a surface that is 5 m off the ground,
The speed of the object, v = 4 m/s,
The mass of the object, m = 3.2 kg,
Calculate the kinetic energy after 10 meters as shown below,
KE = 1/2 × 4² × 3.2
KE = 25.6 J
Calculate the potential energy as shown below,
PE = 3.2 × 9.8 × 5
PE = 156.8 J
Thus, total energy = KE + PE
The total energy = 25.6 + 156.8
The total energy = 182.4 J
8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.
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Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
Answer:
a
[tex]\lambda = 3.68 *10^{-36} \ m[/tex]
b
[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 180 \ kg[/tex]
The speed of the person is [tex]v = 1 \ m/s[/tex]
The energy of the proton is [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]
Generally the de Broglie wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m * v }[/tex]
Here h is the Planck constant with the value
[tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
So
[tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]
=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]
Generally the energy of the proton is mathematically represented as
[tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]
Here [tex]m_p[/tex] is the mass of proton with value [tex]m_p = 1.67 *10^{-27} \ kg[/tex]
=> [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]
=> [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]
=> [tex]v = 3.09529 *10^{7} \ m/s[/tex]
So
[tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]
so [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]
=> [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
car driving on a circular test track shows a constant speedometer reading of 100 kph for one lap. a. Describe the car's speed during this time. b.
Answer:
Speed = 100 km/h
Explanation:
Given:
Speedometer reading = 100 kph for one lap
Assume;
Time taken to complete one lap = 1 hour
Computation:
Speed = Distance / Time
Speed = 100 / 1
Speed = 100 km/h
You are driving on the highway at a speed of 40 m/s (which is over the speed limit) when you notice a cop in front of you. To avoid a ticket, you press on the brake and slow to a speed of 30 m/s over the course of 5 seconds. What is the acceleration of the car? WORK=BRAINLIEST
What is your car's initial velocity?
What is your car's final velocity?
How long does it take the car to slow down?
Write the equation you will use to solve this problem.
What is the acceleration of your vehicle?
+ 2.0 m/s^2
- 2.0 m/s^2
+ 8.0 m/s^2
- 6.0 m/s^2
Explanation:
U = 40m/s
V = 30m/s
T = 5 sec
A = ?
[tex]a = \frac{u - v}{t}[/tex]
[tex]a = \frac{40 - 30}{5}[/tex]
[tex]a = \frac{10}{5}[/tex]
[tex]a = 2[/tex]
since it's decreasing in speed, The acceleration will be " - 2.0ms^-2 " or " - 2.0m/s^2 "
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Btw don't mind me answering twice. I want the free points and maybe another brainliest? lol.
Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, what direction and value is his acceleration air resistance is negligible.
A. 9.8 m/s/s west
b. 9.8m/s/s east
C. 9.8m/s's down
d 9.8m/s/s south
Answer:
9.8m/s^2 down (option C)
Explanation:
The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.
As a rough model of the impact of walking/running, consider that half the mass of the body falls from a height of 4.77-cm onto a single foot. (During a typical stride, an adult's center-of-mass moves approximately this distance vertically). Use the kinematic equations to calculate the speed of an object falling from this height at the moment of impact with the ground under the influence of gravity.A. As a rough model of the impact of walking, consider that half of the mass of the entire body strikes the ground with a downward velocity of 1.0 m/s and comes to a full vertical stop over an impact duration of 20 ms. Calculate the force associated with this single step for a person with a mass of 74.2 kg. B. Calculate the stress (solid pressure) of a force of 1880 N applied across the 0.4 cm^2 cross-sectional area of the typical Achilles tendon. For reference, the maximum rupture stress of tendons has been reported in the range of 100-150 MPa.
Answer:
0.967 m/s
1855 N
[tex]46.375\ \text{MPa}[/tex]
Explanation:
v = Final velocity
u = Initial velocity = 0
s = Displacement = 4.77 cm
g = a = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the kinematic equations
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]
The velocity of the object at the moment of impact is 0.967 m/s
Now
[tex]\Delta v[/tex] = Change in velocity = 1 m/s
t = Time taken = 20 ms
m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]
[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]
The force associated with a single step of the person is 1855 N
A = Area = [tex]0.4\ \text{cm}^2[/tex]
Stress is given by
[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]
The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]
The speed of object during falling is 0.967 m/s.
(A) The magnitude of force associated with this single step for a person is 1855 N.
(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
Given data:
The height of fall is, h = 4.77 cm = 0.0477 m.
The magnitude of downward velocity is, v' = 1.0 m/s.
The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].
The mass of person is, m = 74.2 kg.
The magnitude of force is, F' = 1880 N.
The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].
The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,
[tex]v^{2}=u^{2}+2gh[/tex]
Solving as,
[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]
Thus, the speed of object during falling is 0.967 m/s.
(A)
Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.
Apply the expression of average force as,
[tex]F =\dfrac{m'v'}{t}[/tex]
Solving as,
[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]
Thus, the magnitude of force associated with this single step for a person is 1855 N.
(B)
The expression for the stress is given as,
[tex]\sigma = \dfrac{F'}{A}[/tex]
Solving as,
[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]
Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
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A T-shirt cannon mounted at the top of an arena needs to fire a t-shirt into the first row, a horizontal distance of 39 meters away. If the cannon launches t-shirts at 12 m/s, how high is the cannon mounted?
Question 1 options:
3.3 m
16.2 m
53.4 m
8.9 m
Answer:
h = 51.75 m
nearest answer is:
53.4 m
Explanation:
First we analyze the horizontal motion. Since, the air friction is assumed to be negligible. Hence, the horizontal motion shall be uniform. Therefore,
s = V₀ₓ t
where,
s = horizontal distance = 39 m
V₀ₓ = Horizontal Initial Velocity = 12 m/s
t = time = ?
Therefore,
39 m = (12 m/s)t
t = 39 m/12 m/s
t = 3.25 s
Now, we analyze the vertical motion. Applying newton's second equation of motion to vertical motion:
h = V₀y t + (1/2)gt²
where,
h = height of cannon = ?
V₀y = initial vertical velocity = 0 m/s
g = 9.8 m/s²
Therefore,
h = (0 m/s)(3.25 s) + (1/2)(9.8 m/s²)(3.25 s)²
h = 51.75 m
an object of mass 4kg moving with initial velocity if 20m/s accelerates for 10s and attaind a final velocity of 60m/s calculate the acceleration
Answer:
given us,
mass= 4×9.8gm m(9.8) formula
= 39.2
final velocity (v)= 60m/s
initial velocity (u)= 20m/s
time(t)= 10s
acleration(a)=?
now,
accleration(a)= v-u/t=60- 20/10
=40/10
=4m/s
:. the acceleration is 4 m/s
Explanation:
first we have to calculate mass and we can use acceleration formula
An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 and the temperature remains constant?
Answer:
2.22 kPaExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
From the question we have
[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]
We have the final answer as
2.22 kPaHope this helps you
A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is present, given by B = 4.0i + 9.0x2 j , with x in meters and B in mT. Calculate the k-component of the force on the 2 m segment of the conductor that lies between x = 1.0 m and x = 3.0 m.
Answer:
0.546 [tex]\hat k[/tex]
Explanation:
From the given information:
The force on a given current-carrying conductor is:
[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]
where the length usually in negative (x) direction can be computed as
[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]
Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:
[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]
[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]
[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]
where;
current I = 7.0 A
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]
F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]
F = 0.546 [tex]\hat k[/tex]
Two identical bars are conducting heat from a region of higher temperature to one of lower temperature. In arrangement A, the bars conduct 80 J of heat in a certain amount of time. How much heat is conducted in B during the same time
Answer:
Q' = 320 J
Explanation:
The arrangements are given in attachment. The Fourier's Law of Heat Conduction states that:
Q = KAΔT/L
where,
Q = Heat Transferred
K = Constant (Conduction Coefficient)
A = Surface Area of Heat Transfer
ΔT = Difference of Temperature between two surfaces
L = Length between surfaces
For Arrangement AL
Q = 80 J
Therefore,
80 = KAΔT/L ------------- equation (1)
Now, for arrangement B:
A' = 2 A (As, the rods are now connected in parallel with each other)
L' = L/2
Therefore,
Q' = K(2A)ΔT/(L/2)
Q' = 4 KAΔT/L
using equation (1)
Q' = 4(80 J)
Q' = 320 J
A car traveling initially at a speed of 20 m/s accelerates to a speed of 31 m/s over a distance of 45 meters.
What is the magnitude of the car's acceleration?
Answer:abc defg hijk lmnop qrs tuv wx y and z
Explanation: now i know my abc's