Answer:
In sof the friction with the nylon is very small and the current with the line e is largeummary
Explanation:
When we rub the shoes against the carpet, static electricity is produced, when you touch the metal knob you close the circuit and the current can circulate to three of the body, the value of this current is of the order of micro volts, for which a small discharge, the power that circulates through the body is very small of the order of 0.005 A
When you touch the power line, the voltage may be small, but the amount of current that can generate them is of the order of tens of amps, the electric shock is much greater per location.
In general there is a rule that if the body resumes more than P = 4000W the discharge could be fatal.
In sof the friction with the nylon is very small and the current with the line e is largeummary, the difference is that the current at the stop , so the paper that passes through the body is large and can be dangerous.
In contact with metal doorknob, get a mild shock while with power line of same voltage, fatal the body as the amount of current is more.
What is charging by friction?When the two materials are rubbed each other, then the electric charged generated between them.
This charging of materials, due to the rubbing of two materials against each other, is called the charging by friction.
It is easy to produce a potential difference of several thousand volts between the body and the floor by scuffing your shoes across a nylon carpet.
In this case, the potential difference may be higher, but the value of current is very low. Thus, when the body touches a metal doorknob, it will get a mild shock.
Now, in another case, the contact with a power line of comparable voltage would probably be fatal. This is because in the power line the amount of current is much higher.
Hence, in contact with metal doorknob, get a mild shock while with power line of same voltage, fatal the body as the amount of current is more.
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A man is pulling a 20 kg box with a rope that makes an angle of 60 with the horizontal.If he applies a force of 150 N and a frictional force of 15 N is present, calculate the acceleration of the box.
∑ F (horizontal) = (150 N) cos(60°) - 15 N = (20 kg) a
==> a = ((150 N) cos(60°) - 15 N)/(20 kg) = 3 m/s²
To calculate the acceleration of the box, we need to consider the net force acting on it. So, the acceleration of the box is 3 m/s².
The net force is the vector sum of the applied force and the force of friction. First, let's find the horizontal and vertical components of the applied force:
Horizontal component of the applied force (F[tex]_{horizontal}[/tex]) = F[tex]_{applied}[/tex] × cos(θ)
F[tex]_{horizontal}[/tex] = 150 N × cos(60°)
F[tex]_{horizontal}[/tex] = 150 N × 0.5
F[tex]_{horizontal}[/tex] = 75 N
Vertical component of the applied force (F[tex]_{vertical}[/tex]) = F[tex]_{applied}[/tex] × sin(θ)
F[tex]_{vertical}[/tex] = 150 N × sin(60°)
F[tex]_{vertical}[/tex] = 150 N × (√3 / 2)
F[tex]_{vertical}[/tex] ≈ 129.9 N
Now, let's calculate the net force in the horizontal direction:
Net Force in the horizontal direction (F[tex]_{net horizontal}[/tex]) = F[tex]_{horizontal}[/tex] - F[tex]_{friction}[/tex]
F[tex]_{net horizontal}[/tex] = 75 N - 15 N
F[tex]_{net horizontal}[/tex] = 60 N
Now, we can calculate the acceleration (a) using Newton's second law of motion, F = ma:
F[tex]_{net horizontal}[/tex] = m × a
60 N = 20 kg × a
Now, solve for acceleration (a):
a = 60 N / 20 kg
a = 3 m/s²
So, the acceleration of the box is 3 m/s².
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You have a 1 W light bulb in your lab. It puts out light of only 1 frequency. The wavelength of this light is 500nm. you set up a detector with a surface area of 1 square centimeter facing the light source at a distance of 100m.
Required:
a. Find the number of photons hitting the detector every second.
b. What is the maximum E field of the E M wave hitting the detector?
c. What is the maximum value of the B field of this E M wave?
d. How far away would you have to place the detector to only receive 1 photon per second from the light bulb?
Answer:
a) # _photon = 2.5 10¹⁸ photons / s, b) E = 10⁻² N / C, c) B = 3 10⁻¹¹ T
d) r= 2 10⁹ m
Explanation:
a) Let's solve this exercise in part, let's start by finding the energy of each photon using the Planck relation
E₀ = h f
c = λ f
E₀ = h c /λ
E₀ = 6.63 10⁻³³⁴ 3 10⁸/500 10⁻⁹
E₀ = 3.978 10⁻⁻¹⁹ J
Let's use a direct ratio rule to find the number of photons
#_foton = E / Eo
#_fototn = 1 / 3.978 10⁻¹⁹
# _photon = 2.5 10¹⁸ photons / s
b) The intensity received by the detector is related to the electric field
I = E²
Let's look for the intensity that the detector receives, suppose that the emission is shapeless throughout the space
I = P / A
P = I A
Let's use index 1 for the point on the bulb and index 2 for the point on the detector.
The area of a sphere is
A = 4π r²
P = I₁ A₁ = I₂ A₂
I₁ r₁² = I₂ r₂²
I₂ = I₁ r₁²/r₂²
I₂ = I₁ 1 / 100²
I₂ = I₁ 10⁻⁴
we must know the intensity at the output of the bulb suppose that I₁ = 1 J
I₂ = 10⁻⁴ J
let's look for the electric field
E =√I
E = √10⁻⁴
E = 10⁻² N / C
c) for the calculation of the magnetic field we use that the field is in phase
E / B = c
B = E / c
B = 10⁻² / 3 10⁸
B = 3 10⁻¹¹ T
d) Let's use a direct proportions rule if we fear 2.5 10¹⁸ photons in an area A = 4π R² where R = 100 m how many photons are there in the area of the detector r = 1 cm, A’= 10⁻⁴ m²
#_photons = 2.5 10¹⁸ A_detector / A_sphere
#_photons = 2.5 1018 10-4 / 4π 10⁴
#_photons = 2 10⁹ photons in the detector area
for the number of photons to decrease to 1, the radius of the sphere must be 2 10⁹ m
An observer on Earth sees Planet X to be stationary and also sees a rocket traveling toward Planet X at 0.5c. The rocket emits a pulse of light that travels outward in all directions. According to an observer on Planet X, how fast is the light pulse traveling toward them?
a) 2c/3
b) c/2
c) 2c/3
d) 5c/6
e) c
(E) c
Explanation:
The speed of light is always equal to c regardless of the relative motion of the light source.
As the speed of a particle approaches the speed of light, the momentum of the particle Group of answer choices approaches zero. decreases. approaches infinity. remains the same. increases.
Answer:
approaches infinity
Explanation:
There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.
The relativistic momentum can be written as:
[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
where
u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.
m = mass of the object
c = speed of light.
So, as u tends to c, we will have:
[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.
Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.
A transparent. dielectric coating is applied to glass (εr = 4.μr=1, σ= 0) to eliminate the reflection of red light (wavelength in air of 750 nm).
a. What is the required dielectric constant and minimum thickness of the coating?
b. If violet light (wavelength in air of 420 nm) is shone onto this glass coating (6-0). what percentage of the incident power will be reflected?
Answer:
a) Dielectric constant ( λ ) = 750 * 10^-9 m
minimum thickness of coating ( d ) = 187.5 nm
b) 3.6%
Explanation:
Given data:
wavelength of red light in air = 750 nm
εr = 4
μr = 1, σ = 0
a) Determine the required dielectric constant and min thickness of coating used
Refractive index of coating ( n ) = √εr * μr = √4*1 = 2
the refractive index of glass( ng) = 1.5 which is < 2
λ = 750 * 10^-9 m
Dielectric constant ( λ ) = 750 * 10^-9 m
To determine the minimum thickness we will apply the formula below
d = m λ/2n ; where m = 1
∴ d = 750 nm / 2 ( 2 )
= 187.5 nm
minimum thickness of coating ( d ) = 187.5 nm
b) Determine the percentage of the incident power that will be reflected
R = [ ( n-1 / n + 1 ) - ( n - ng / ng + n ) ]^2
= [ ( 2 - 1 / 2 + 1 ) - ( 2 - 1.5 / 1.5 + 2 ) ]^2
= 0.03628 = 3.6%
Current is a measure of…
A parallel-plate capacitor consists of two plates, each with an area of 29 cm2cm2 separated by 3.0 mmmm. The charge on the capacitor is 7.8 nCnC . A proton is released from rest next to the positive plate. Part A How long does it take for the proton to reach the negative plate
Answer:
t = 2.09 10⁻³ s
Explanation:
We must solve this problem in parts, first we look for the acceleration of the electron and then the time to travel the distance
let's start with Newton's second law
∑ F = m a
the force is electric
F = q E
we substitute
q E = m a
a = [tex]\frac{q}{m} \ E[/tex]
a = [tex]\frac{1.6 \ 10^{-19}}{ 9.1 \ 10^{-31} } \ 7.8 \ 10^{-9}[/tex]
a = 1.37 10³ m / s²
now we can use kinematics
x = v₀ t + ½ a t²
indicate that rest starts v₀ = 0
x = 0 + ½ a t²
t = [tex]\sqrt{\frac{2x}{a} }[/tex]
t = [tex]\sqrt{\frac {2 \ 3 \ 10^{-3}}{ 1.37 \ 10^3} }[/tex]
t = 2.09 10⁻³ s
A 2000 kg truck has put its front bumper against the rear bumper of a 2500 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N.
What is the maximum possible acceleration the truck can give the SUV?
At this acceleration, what is the force of the SUV's bumper on the truck's bumper?
Answer:
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
Explanation:
Concepts and reason
The concept required to solve this problem is Newton’s second law of motion.
Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.
Fundamentals
According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:
F = maF=ma
Here, m is mass and a is the acceleration.
(a)
Rearrange the equation F = maF=ma for a.
a = \frac{F}{m}a=
m
F
Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=
m
F
.
\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}
a=
(2300kg+2400kg)
18,000N
=
(4700kg)
18,000N
=3.83m/s
2
(b)
Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
for a and 2400 kg for m in the equation F = maF=ma .
\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}
F=(2400kg)(3.83m/s
2
)
=9120N
Ans: Part a
The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s
2
.
Part b
The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.
The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².
The force of the SUV's bumper on the truck's bumper is 10000N
What is acceleration?Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.
According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.
F = ma
And, a = F/m
Given, the mass of the ruck , m = 2000 Kg
The mass of the SUV, M = 2500 Kg
The total mass of the both = 2000 + 2500 = 4500 Kg
The maximum force on the trick , F = 18000 N
The maximum acceleration of the truck can give the SUV:
[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]
a = 18000/4500
a = 4 m/s²
The force of the SUV's bumper on the truck's bumper will be:
[tex]F_{max} -f= ma_{max}[/tex]
[tex]f= 18000-2000\times 4[/tex]
[tex]f =10000N[/tex]
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An observer on Earth sees spaceship 1 fly by at 0.80c. 6.0 years later, the observer on Earth sees spaceship 2 fly by at 0.80c, traveling in the same direction as the first. Both spaceships continue to travel with constant velocities. An observer in spaceship 1 observes Earth to pass spaceship 2 ____ years after Earth passed spaceship 1.
Answer:
[tex]t_2=10[/tex]
Explanation:
From the question we are told that:
Velocity of both spaceships [tex]V=0.8c[/tex]
Time [tex]t_1=6[/tex]
Generally the equation for time of spaceship 2 through earth is mathematically given by
[tex]t_2=\frac{t_1}{\sqrt{1-v^2}}[/tex]
[tex]t_2=\frac{6}{\sqrt{1-0.8^2}}[/tex]
[tex]t_2=10[/tex]
Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect. If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be
The question is incomplete, the complete question is;
Light of frequency f falls on a metal surface and ejects electrons of maximum kinetic energy K by the photoelectric effect.
Part A If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be If the frequency of this light is doubled, the maximum kinetic energy of the emitted electrons will be
K/2.
K.
2K.
greater than 2K.
Answer:
2K
Explanation:
Given that the kinetic energy of photo electrons is given by;
K= E -Wo
Where;
K = kinetic energy
E= energy of incident photon
Wo = work function
But;
E= hf
Wo = fo
h= Plank's constant
f= frequency of incident photon
fo= Threshold frequency
So:
K= hf - hfo
Where the frequency of incident light is doubled;
K= 2hf - hfo
Hence, maximum kinetic energy of the emitted electrons in this case will be 2K
System A consists of a mass m attached to a spring with a force constant k; system B has a mass 2m attached to a spring with a force constant k; system C has a mass 3m attached to a spring with a force constant 6k; and system D has a mass m attached to a spring with a force constant 4k. Rank these systems in order of decreasing period of oscillation.
Answer:
T₂ > T₁ > T₃ >T₄
Explanation:
In a simple harmonic motion the angular velocity is
w = [tex]\sqrt{\frac{k}{m} }[/tex]
angular velocity and period are related
w = 2π / T
we substitute
T = [tex]2 \pi \ \sqrt{\frac{m}{k} }[/tex]
let's find the period for each case
a) mass m
spring constant k
T₁ = 2π [tex]\sqrt{\frac{m}{k} }[/tex]
b) mass 2m
spring constant k
T₂ = 2π [tex]\sqrt{\frac{2m}{k} }[/tex]
T₂ = T₁ √2
T₂ = T₁ 1.41
c) masses 3m
spring constant 6k
T₃ = 2π [tex]\sqrt{\frac{3m}{6k} }[/tex]
T₃ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.5}[/tex]
T₃ = T₁ 0.707
d) mass m
spring constant 4k
T₄ = 2π [tex]\sqrt{ \frac{m}{4k} }[/tex]
T₄ = 2π [tex]\sqrt{\frac{m}{k} } \ \sqrt{0.25}[/tex]
T₄ = T₁ 0.5
now let's order the periods in decreasing order
T₂ > T₁ > T₃ >T₄
A rocket explodes into two fragments, one 25 times heavier than the other. The magnitude of the momentum change of the lighter fragment is A) 25 times as great as the momentum change of the heavier fragment. B) The same as the momentum change of the heavier fragment. C) 1/25 as great as the momentum change of the heavier fragment. D) 5 times as great as the momentum change of the heavier fragment. E) 1/4 as great as the momentum change of the heavier fragment.
Answer:
B) The same as the momentum change of the heavier fragment.
Explanation:
Since the initial momentum of the system is zero, we have
0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.
0 = p + p'
p = -p'
Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0 = p
The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'
Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'
So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.
So, option B is the answer
A uniform electric field of strength E points to the right. An electron is fired with a velocity v0 to the right and travels a distance d before coming to a stop. An second electron is then fired upwards through the same field at a velocity of v0. After the electron moving vertical has traveled vertically upwards a distance d, how far will it have moved horizontally?
Answer:
[tex]D_l=d[/tex]
Explanation:
From the question we are told that:
The Electric field of strength direction =Right
The Velocity of The First Electron=V_0
The Velocity of The Second Electron=V_0
Therefore
[tex]V_{e1}=V_{e2}[/tex]
Generally, the equation for the Horizontal Displacement of electron is mathematically given by
[tex]D=\frac{at^2}{2}[/tex]
Where
Acceleration is given as
[tex]a=\frac{V_o}{2d}[/tex]
And
Time
[tex]T=\frac{d}{v_0}[/tex]
Therefore horizontal displacement towards the left is
[tex]D_l=\frac{(\frac{V_o}{2d})(\frac{d}{v_0})^2}{2}[/tex]
[tex]D_l=d[/tex]
a cheetah running at a velocity of 18m/s accelerates at 1m/s² for 5sec what is the final velocity of the cheetah
Two circular coils are concentric and lie in the same plane.The inner coil contains 120 turns of wire, has a radius of 0.012m,and carries a current of 6.0A. The outer coil contains 150turns and has a radius of 0.017 m. What must be the magnitudeand direction (relative to the current in the inner coil) ofthe current in the outer coil, such that the net magnetic field atthe common center of the two coils is zero?
Answer:
[tex]I_2=6.8A[/tex]
Explanation:
From the question we are told that:
Turns of inner coil [tex]N_1=120[/tex]
Radius of inner coil [tex]r_1=0.012m[/tex]
Current of inner coil [tex]I_1=6.0A[/tex]
Turns of Outer coil [tex]N_2=150[/tex]
Radius of Outer coil [tex]r_2=0.017m[/tex]
Generally the equation for Magnetic Field is mathematically given by
[tex]B =\frac{ \mu N I}{2R}[/tex]
Therefore
Condition for the net Magnetic field to be zero
[tex]\frac{N_1* I_1}{( 2 * r_1 )}=\frac{N_2 * I_2}{2 * r_2}[/tex]
[tex]I_2=\frac{(N_1* I_1)*(( 2 * r_2)}{( 2 * r_1)*N_2}[/tex]
[tex]I_2=\frac{(120*6.0)*(( 2 * 0.017)}{( 2 * 0.012)*150}[/tex]
[tex]I_2=6.8A[/tex]
17- How much work is needed for a climber in order to climb 45 m height, where his weight is 70 kg. also, calculate the power required to climb the height in 30 minutes ? g= 9,8 m.sec
Answer:
Work Done= 3150J
Power= 1.75W
Explanation:
Work Done= Force x the distance travelled in the direction of the force (W= f x d)
Weight is a force, i think the qn. stated it wrongly, it should be 70N not 70kg.
Work Done= 70 x 45
=3150J
Power= Work Done/Time
=3150/(30x60)
*convert minutes to seconds since the S.I. unit of Power is joules/seconds(J/s) or watts(W)
=1.75W
According to Newton's law of universal gravitation, the force F between two bodies of constant mass m and M is given by the formula F = G m M d 2 , where G is the gravitational constant and d is the distance between the bodies. a. Suppose that are constants. Find the rate of change of force F with respect to distance d .
Answer:
One can write F = K d^-2 where K = G M m
So dF/dd = -2 K d^-3 = -2 K / d^3 (As d increases F decreases - it is opposite to the direction of F)
Why is oiling done time and again in a sewing machine?
Answer:
to prevent friction on the surfaces
Answer:
Explanation:
Oiling reduces friction between parts with relative motion between them.
Repeated oiling is needed as the film of oil reducing the friction becomes thinner with time as some of the oil gets pushed off of the areas of motion where it can no longer be useful.
Oil also becomes oxidized which reduces the oil's ability to decrease friction.
Oil can also be fouled by dirt, lint or other material. This added material becomes coated in oil and typically gets sequestered away from the moving parts reducing the oil available for lubrication purposes.
A physics instructor wants to project a spectrum of visible-light colors from 400 nm to 700 nm as part of a classroom demonstration. She shines a beam of white light through a diffraction grating that has 600 lines per mm, projecting a pattern on a screen 2.9 m behind the grating.
Required:
How wide is the spectrum corresponding to m=1?
Answer:
Dr = 263 10⁻⁶ m
Explanation:
The diffraction pattern for constructive interference is described by
a sin θ = m λ
in this it indicates that the order of diffraction is m = 1
Let's use a direct proportion rule to find the separation of two slits. If there are 600 lines in 1 me, what is the distance between 2 slits
a = 2 lines 1/600
a = 2/600
a = 3.33 10⁻³ mm = 3.33 10⁻⁴ cm
let's use trigonometry
tan θ = y / L
as the measured angles are small
tan θ = sin θ / cos θ sin θ
sin θ = y / L
we substitute
a y/L = λ
y = λ L / a
for λ = 400 10-9 m
I = 400 10⁻⁹ 2.9 / 3.33 10⁻³
i = 346.89 10⁻⁶ m
f
or λ = 700 nm
y_f = 700 10⁻⁻⁹ 2.9 / 3.33 10⁻³
y_f = 609.609 10⁻⁶ m
the separation of this spectrum
Δr = v_f - i
Dr = (609.609 - 346) 10 ⁻⁶
Dr = 263 10⁻⁶ m
A planet of mass m = 4.25 x 1024 kg orbits a star of mass M = 6.75 x 1029 kg in a circular path. The radius of the orbits R = 8.85 x 107 km. What is the orbital period Tplanet of the planet in Earth days?
285.3 days
Explanation:
The centripetal force [tex]F_c[/tex] experienced by the planet is the same as the gravitational force [tex]F_G[/tex] so we can write
[tex]F_c = F_G[/tex]
or
[tex]m\dfrac{v^2}{R} = G\dfrac{mM}{R^2}[/tex]
where M is the mass of the star and R is the orbital radius around the star. We know that
[tex]v = \dfrac{C}{T} = \dfrac{2\pi R}{T}[/tex]
where C is the orbital circumference and T is orbital period. We can then write
[tex]\dfrac{4\pi^2R}{T^2} = G\dfrac{M}{R^2}[/tex]
Isolating [tex]T^2[/tex], we get
[tex]T^2 = \dfrac{4\pi^2R^3}{GM}[/tex]
Taking the square root of the expression above, we get
[tex]T = 2\pi \sqrt{\dfrac{R^3}{GM}}[/tex]
which turns out to be [tex]T = 2.47×10^7\:\text{s}[/tex]. We can convert this into earth days as
[tex]T = 2.47×10^7\:\text{s}×\dfrac{1\:\text{hr}}{3600\:\text{s}}×\dfrac{1\:\text{day}}{24\:\text{hr}}[/tex]
[tex]\:\:\:\:\:= 285.3\:\text{days}[/tex]
which characteristic of nuclear fission makes it hazardous?
Answer:The radioactive waste
Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei
calculate the force on an object with mass of 50kg and gravity of 10
A nylon string on a tennis racket is under a tension of 285 N . If its diameter is 1.10 mm , by how much is it lengthened from its untensioned length of 29.0 cm ? Use ENylon=5.00×109N/m2.
Answer:
1.74×10⁻³ m
Explanation:
Applying,
ε = Stress/strain............. Equation 1
Where ε = Young's modulus
But,
Stress = F/A.............. Equation 2
Where F = Force, A = Area
Strain = e/L.............. Equation 3
e = extension, L = Length.
Substitute equation 2 and 3 into equation 1
ε = (F/A)/(e/L) = FL/eA............. Equation 4
From the question,
Given: F = 285 N, L = 29 cm = 0.29 m, ε = 5.00×10⁹ N/m²,
A = πd²/4 = 3.14(0.0011²)/4 = 9.4985×10⁻⁶ m²
Substitute these values into equation 4
5.00×10⁹ = (285×0.29)/(9.4985×10⁻⁶×e)
Solve for e
e = (285×0.29)/(5.00×10⁹×9.4985×10⁻⁶)
e = 82.65/4.74925×10⁴
e = 1.74×10⁻³ m
what are the two main types of sound like soundwave
Answer:
acoustic energy and mechanical energy
Explanation:
each type of sounds has to be tackled in their own way.
Q.3. The equivalent resistance across AB is:
(a)1
(c)2
(b)3
(d)4
Answer:
1 ohm
Explanation:
First of all, the equivalent resistance for two resistors (r₁ and r₂) in parallel is given by:
1 / Eq = (1 / r₁) + (1 / r₂)
The equivalent resistance for resistance for two resistors (r₁ and r₂) in series is given by:
Eq = r₁ + r₂
Hence as we can see from the circuit diagram, 2Ω // 2Ω, and 2Ω // 2Ω, hence:
1/E₁ = 1/2 + 1/2
1/E₁ = 1
E₁ = 1Ω
1/E₂ = 1/2 + 1/2
1/E₂ = 1
E₂ = 1Ω
This then leads to E₁ being in series with E₂, hence the equivalent resistance (E₃) of E₁ and E₂ is:
E₃ = E₁ + E₂ = 1 + 1 = 2Ω
The equivalent resistance (Eq) across AB is the parallel combination of E₃ and the 2Ω resistor, therefore:
1/Eq = 1/E₃ + 1/2
1/Eq = 1/2 + 1/2
1/Eq = 1
Eq = 1Ω
An object undergoing simple harmonic motion takes 0.40 s to travel from one point of zero velocity to the next such point. The distance between those points is 50 cm. Calculate (a) the period, (b) the frequency, and (c) the amplitude of the motion.
Answer:
a) [tex]P=0.80[/tex]
b) [tex]1.25Hz[/tex]
c) [tex]A=25cm[/tex]
Explanation:
From the question we are told that:
Travel Time [tex]T=0.40s[/tex]
Distance [tex]d=50cm[/tex]
a)
Period
Time taken to complete one oscillation
Therefore
[tex]P=2*T\\\\P=2*0.40[/tex]
[tex]P=0.80[/tex]
b)
Frequency is
[tex]F=\frac{1}{T}\\\\F=\frac{1}{0.80}[/tex]
[tex]1.25Hz[/tex]
c)
Amplitude:the distance between the mean and extreme position
[tex]A=\frac{50}{2}[/tex]
[tex]A=25cm[/tex]
A resistor is submerged in an insulated container of water. A voltage of 12 V is applied to the resistor resulting in a current of 1.2 A. If this voltage and current are maintained for 5 minutes, how much electrical energy is dissipated by the resistor
Explanation:
Given:
[tex]\Delta t = 5\:\text{min} = 300\:\text{s}[/tex]
[tex]V = 12 V[/tex]
[tex]I = 1.2 A[/tex]
Recall that power P is given by
[tex]P = VI[/tex]
so the amount of energy dissipated [tex]\Delta E[/tex] is given by
[tex]\Delta E = VI\Delta t = (12\:\text{V})(1.2\:\text{A})(300\:\text{s})[/tex]
[tex]\:\:\:\:\:\:\:= 4320\:\text{W} = 4.32\:\text{kW}[/tex]
friction always opposes the _____
Answer:
Friction always opposes the motion
I HOPE ITS RIGHT IF NOT THEN SORRYHAVE A GREAT DAY :)
If a car drives 10 mph South, this is an example of a:
A. Displacement
B. Velocity
C. Speed
D. Distance
Answer:
杰杰伊杜杜杜伊格富尔杰迪耶赫分离福音
Explanation:
莱德利 · 赫耶尔伊 3uritievrirjrirhruebwkwieheoo2hfjcbvi3hd
Answer:
B velocity
Explanation:
What do you understand by moment of inertia and torque?
Word limit 50-60
Please don't copy from any sources. You can rewrite. Plagiarism will be check. Thank you.
Answer:
Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i.e., the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). The axis may be internal or external and may or may not be fixed.
