Answer:
a) for the torque to be maximum, sin should be maximum
i.e (sinФ)maximum = 1
b) therefore the Maximum torque is
Tmax = 0.1838 × 1 = 0.1838 N.m
c) Given the torque is 71.0% of its maximum value; Ф = 45.24⁰ ≈ 45⁰
Explanation:
Given that; Diameter is 8.40 cm,
Radius (R) = D/2 = 8.40/2 = 4.20 cm = 0.042 m
Number of turns (N) = 17
Current in the loop (I) = 3.20 A
Magnetic field (B) = 0.610 T
Let the angle between the loop's area vector A and the magnetic field B be
Now. the area of the loop is;
A = πR²
A = 3.14 ( 0.042 )²
A = 0.005539 m²
Torque on the loop (t) = NIABsinФ
t = 17 × 3.20 ×0.005539 × 0.610 × sinФ
t = 0.1838sinФ N.m
for the torque to be maximum, sin should be maximum
i.e (sinФ)maximum = 1
therefore the Maximum torque is
Tmax = 0.1838 × 1 = 0.1838 N.m
Given the torque is 71.0% of its maximum value
t = 0.71 × tmax
t = 0.71 × 0.1838
t = 0.1305
Now
0.1305 N.m = 0.1838 sinФ N.m
sinФ = 0.1305 / 0.1838
sinФ = 0.71001
Ф = sin⁻¹ 0.71001
Ф = 45.24⁰ ≈ 45⁰
A brass rod is 185 cm long and 1.60 cm in diameter. What force must be applied to each end of the rod to prevent it from contract- ing when it is cooled from 120.0°C to 10.0°C?
Answer:
42000N
Explanation:
First you calculate how much it would contract, and secondly you then calculate the force to stretch it by that amount.
1) linear thermal expansion coef brass 19e-6 /K
∆L = αL∆T = (19e-6)(1.85)(110) = 0.00387 meter or 3.87 mm
Second part involves linear elasticity.
for brass, young's modulus is 15e6 psi or 100 GPa
cross-sectional area of rod is π(0.008)² = 0.0002 m²
F = EA∆L/L
F = (100e9)(0.0002)(0.00387) / (1.85)
F = 42000 or 42 kN
A body is thrown vertically upwards with a speed of 95m / s and after 7s it reaches its maximum height. How fast does it reach its maximum height? What was the maximum height reached?
Explanation:
u = 95 m/sec ( Initial speed)
t = 7 sec ( Time of ascent)
According to Equations of Motion :
[tex]s = ut - \frac{1}{2} g {t}^{2} [/tex]
Max. Height = 95 * 7 - 4.9 * 49 = 424. 9 = 425 m
Answer:
332.5 m
Explanation:
At the maximum height, the velocity is 0.
Given:
v₀ = 95 m/s
v = 0 m/s
t = 7 s
Find: Δy
Δy = ½ (v + v₀) t
Δy = ½ (0 m/s + 95 m/s) (7 s)
Δy = 332.5 m
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for their diffraction patterns to be resolved by Rayleigh's criterion
Answer:
The distance is [tex]D = 0.000712 \ m[/tex]
Explanation:
From the question we are told that
The wavelength of the light source is [tex]\lambda = 700 \ nm = 700 *10^{-9} \ m[/tex]
The distance from a pin hole is [tex]x = 9\ m[/tex]
The diameter of the pin hole is [tex]d = 1.2 \ mm = 0.0012 \ m[/tex]
Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is
mathematically represented as
[tex]D = \frac{1.22 \lambda }{d }[/tex]
substituting values
[tex]D = \frac{1.22 * 700 *10^{-9} }{ 0.0012 }[/tex]
[tex]D = 0.000712 \ m[/tex]
It takes 85N of force to accelerate a ball at a rate of 15 m/s². What is the mass of the ball?
Explanation:
F = ma
85 N = m (15 m/s²)
m ≈ 5.7 kg
A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s
Answer:
a) A = 1.99 μm , b) λ = 0.4134 m , c) v = 57.2 m / s , d) s = - 1,946 nm ,
e) v_max = 1,739 mm / s
Explanation:
A sound wave has the general expression
s = s₀ sin (kx - wt)
where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is
s = 1.99 sin (15.2 x - 869 t)
a) the amplitude of the wave is
A = s₀
A = 1.99 μm
b) wave spectrum is
k = 2π /λ
in the equation k = 15.2 m⁻¹
λ = 2π / k
λ = 2π / 15.2
λ = 0.4134 m
c) the speed of the wave is given by the relation
v = λ f
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 869 / 2π
f = 138.3 Hz
v = 0.4134 138.3
v = 57.2 m / s
d) To find the instantaneous velocity, we substitute the given distance and time into the equation
s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
s = 1.99 sin (0.77368 - 2.55486)
remember that trigonometry functions must be in radians
s = 1.99 (-0.98895)
s = - 1,946 nm
The negative sign indicates that it shifts to the left
e) the speed of the oscillating part is
v = ds / dt)
v = - s₀(-w) cos (kx -wt)
the maximum speed occurs when the cosines is 1
v_maximo = s₀w
v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
let's reduce to mm / s
v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
v_max = 1,739 mm / s
a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s
Calculation of Wavelength
When A sound wave has the general expression is:
Then, s = s₀ sin (kx - wt)
Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is
s is = 1.99 sin (15.2 x - 869 t)
a) When the amplitude of the wave is
A is = s₀
Thus, A = 1.99 μm
b) When the wave spectrum is
k is = 2π /λ
Now, in the equation k = 15.2 m⁻¹
Then, λ = 2π / k
After that, λ = 2π / 15.2
Thus, λ = 0.4134 m
c) When the speed of the wave is given by the relation is:
Then, v = λ f
Now, the angular velocity and frequency are related is:
w is = 2π f
Then, f = w / 2π
After that, f = 869 / 2π
Now, f = 138.3 Hz
Then, v = 0.4134 138.3
Thus, v = 57.2 m / s
d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation
Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
After that, s = 1.99 sin (0.77368 - 2.55486)
Then remember that trigonometry functions must be in radians
After that, s = 1.99 (-0.98895)
Thus, s = - 1,946 nm
When The negative sign indicates that it shifts to the left
e) When the speed of the oscillating part is
Then, v = ds / dt)
Now, v = - s₀(-w) cos (kx -wt)
When the maximum speed occurs when the cosines is 1
Then, v_maximo = s₀w
After that, v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
Now, let's reduce to mm / s
Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
Therefore, v_max = 1,739 mm / s
Finf more informmation about Wavelength here:
https://brainly.com/question/6352445
Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth
Answer:
Vrel= 0.75c
Explanation:
See attached file
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system. The group of boxes accelerates at 1.516 m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387 N. Between the box of mass m2 and the box of mass m3, the force meter reads F23=2304 N. Assume that the ropes and force meters are massless.
The question is incomplete. Here is the complete question.
Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.
(a) What is the total mass of the three boxes?
(b) What is the mass of each box?
Answer: (a) Total mass = 2384.5kg;
(b) m1 = 915kg;
m2 = 605kg;
m3 = 864.5kg;
Explanation: The image of the boxes is described in the picture below.
(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:
[tex]F_{pull}=m_{T}.a[/tex]
[tex]m_{T}=\frac{F_{pull}}{a}[/tex]
[tex]m_{T}=\frac{3615}{1.516}[/tex]
[tex]m_{T}=2384.5[/tex]
Total mass of the system of boxes is 2384.5kg.
(b) For each mass, analyse each box and make them each a free-body diagram.
For [tex]m_{1}[/tex]:
The only force acting On the [tex]m_{1}[/tex] box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.
[tex]m_{1} = \frac{F_{12}}{a}[/tex]
[tex]m_{1} = \frac{1387}{1.516}[/tex]
[tex]m_{1}[/tex] = 915kg
For [tex]m_{2}[/tex]:
There are two forces acting on [tex]m_{2}[/tex]: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:
[tex]m_{2} = \frac{F_{23}-F_{12}}{a}[/tex]
[tex]m_{2} = \frac{2304-1387}{1.516}[/tex]
[tex]m_{2}[/tex] = 605kg
For [tex]m_{3}[/tex]:
[tex]m_{3} = m_{T} - (m_{1}+m_{2})[/tex]
[tex]m_{3} = 2384.5-1520.0[/tex]
[tex]m_{3}[/tex] = 864.5kg
PLEASE HELP FAST Five-gram samples of brick and glass are at room temperature. Both samples receive equal amounts of energy due to heat flow. The specific heat capacity of brick is 0.22 cal/g°C and the specific heat capacity of glass is 0.22 cal/g°C. Which of the following statements is true? 1.The temperature of each sample will increase by the same amount. 2.The temperature of each sample will decrease by the same amount. 3.The brick will get hotter than the glass. 4.The glass will get hotter than the brick.
Answer:
1.The temperature of each sample will increase by the same amount
Explanation:
This is because, since their specific heat capacities are the same and we have the same mass of each substance, and the same amount of energy due to heat flow is supplied to both the glass and brick at room temperature, their temperatures would thereby increase by the same amount.
This is shown by the calculation below
Q = mcΔT
ΔT = Q/mc where ΔT = temperature change, Q = amount of heat, m = mass of substance and c = specific heat capacity of substance.
Since Q, m and c are the same for both substances, thus ΔT will be the same.
So, the temperature of each sample will increase by the same amount
One solenoid is centered inside another. The outer one has a length of 54.0 cm and contains 6750 coils, while the coaxial inner solenoid is 4.00 cm long and 0.170 cm in diameter and contains 21.0 coils. The current in the outer solenoid is changing at 35.0 A/s .What is the mutual inductance of the solenoids?Find the emf induced in the inner solenoid.
Answer:
M₁₂ = 1.01 10⁻⁴ H , Fem = 3.54 10⁻³ V
Explanation:
The mutual inductance between two systems is
M₁₂ = N₂ Ф₁₂ / I₁
where N₂ is the number of turns of the inner solenoid N₂ = 21.0, i₁ the current that flows through the outer solenoid I₁ = 35.0 A / s and fi is the flux of the field of coil1 that passes through coil 2
the magnetic field of the coil1 is
B = μ₀ n I₁ = μ₀ N₁/l I₁
the flow is
Φ = B A₂
the area of the second coil is
A₂ = π d₂ / 4
Φ = μ₀ N₁ I₁ / L π d² / 4
we substitute in the first expression
M₁₂ = N₂ μ₀ N₁ / L π d² / 4
M₁₂ = μ₀ N₁ N₂ π d² / 4L
d = 0.170 cm = 0.00170 m
L = 4.00 cm = 0.00400 m
let's calculate
M₁₂ = 4π 10⁻⁷ 6750 21 π 0.0017²/ (4 0.004)
M₁₂ = π² 0.40966 10⁻⁷ / 0.004
M₁₂ = 1.01 10⁻⁴ H
The electromotive force is
Fem = - M dI₁ / dt
Fem = - 1.01 10⁻⁴ 35.0
Fem = 3.54 10⁻³ V
Determine the final angular velocity of a particle that rotates 4500 ° in 3 seconds and an angular acceleration of 8 Rad / s ^ 2
Answer:
the final angular velocity of the particle is approximately 38.18 Rad/s
Explanation:
To start with, let's make sure that units of angle measure are the same, converting everything into radians:
[tex]4500^o\, \frac{\pi}{180^o}= 25\,\pi[/tex]
And now we can use the kinematic formulas for rotational motion:
[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2[/tex]
Therefore we can find the initial angular velocity [tex]\omega_0[/tex] of the particle:
[tex]\theta-\theta_0=\omega_0\,t+\frac{1}{2} \alpha\,t^2\\25\,\pi=\omega_0\,(3)+\frac{1}{2} (8)\,(3)^2\\25\,\pi-36=\omega_0\,(3)\\\omega_0=\frac{25\,\pi-36}{3} \\\omega_0\approx 14.18\,\,\,rad/s[/tex]
and now we can estimate the final angular velocity using the kinematic equation for angular velocity;
[tex]\omega=\omega_0\,+\alpha\,t\\\omega=14.18+8\,(3)\\\omega=38.18\,\,\,rad/s[/tex]
A 23 cm tall object is placed in front of a concave mirror with a radius of 37 cm. The distance of the object to the mirror is 86 cm. Calculate the focal length of the mirror.
Answer:
18.5 cm
Explanation:
From;
1/u + 1/v = 1/f
Where;
u= object distance = 86cm
image height = 23 cm
Radius of curvature = 37 cm
The radius of curvature (r) is the radius of the sphere of which the mirror forms a part.
Focal length (f) = radius of curvature (r)/2 = 37cm/2 = 18.5 cm
Therefore, the focal length of the mirror is 18.5 cm
A 5.0 kg block hangs from a spring with spring constant 2000 N/m. The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium. a) What is the total mechanical energy of the motion
Answer:
Explanation:i think this would help u
A 25 cm diameter circular saw blade spins at 3500 rpm. How fast would you have to push a straight hand saw to have the teeth move through the wood at the same rate as the circular saw teeth
Answer:
The answer is "45.79 m/s"
Explanation:
Given values:
diameter= 25 cm
w= 3500 rpm
Formula:
[tex]\boxed{v=w \times r} \ \ \ \ \ \ _{where} \ \ \ w = \frac{rad}{s} \ \ \ and \ \ \ r = meters[/tex]
Calculating r:
[tex]r= \frac{diameter}{2}[/tex]
[tex]=\frac{25}{2}\\\\=12.5 \ cm[/tex]
converting value into meters: [tex]12.5 \times 10^{-2} \ \ meter[/tex]
calculating w:
[tex]w= diameter \times \frac{2\pi}{60}\\[/tex]
[tex]= 3500 \times \frac{2\times 3.14}{60}\\\\= 3500 \times \frac{2\times 314}{6000}\\\\= 35 \times \frac{314}{30}\\\\= 35 \times \frac{314}{30}\\\\=\frac{10990}{30}\\\\=\frac{1099}{3}\\\\=366.33[/tex]
w= 366.33 [tex]\ \ \frac{rad}{s}[/tex]
Calculating v:
[tex]v= w\times r\\[/tex]
[tex]= 366.33 \times 12.5 \times 10^{-2}\\\\= 366.33 \times 12.5 \times 10^{-2}\\\\= 4579.125 \times 10^{-2}\\\\\boxed{=45.79 \ \ \frac{m}{s}}[/tex]
A fish is 11.9 cm from the front surface of a fish bowl of radius 33 cm. Where does the fish appear to be to someone in air viewing it from in front of the bowl
Answer:
The fish would appear 42.7 cm on the left side from the front of the bowl.
Explanation:
The fish (object) distance = 11.9 cm, radius of curvature of the bowl = 33 cm. The distance of image of the fish (image distance) can be determined by applying the mirror formula;
[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{u}[/tex] + [tex]\frac{1}{v}[/tex]
where f is the focal length of the reflecting surface, u is the object distance and v is the image distance.
But, f = [tex]\frac{radius of curvature}{2}[/tex]
= [tex]\frac{33}{2}[/tex]
f = 16.5 cm
Substitute f = 16.5 = [tex]\frac{165}{10}[/tex], and u = 11.9 = [tex]\frac{119}{10}[/tex] in equation 1;
[tex]\frac{10}{165}[/tex] = [tex]\frac{10}{119}[/tex] + [tex]\frac{1}{v}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{10}{165}[/tex] - [tex]\frac{10}{119}[/tex]
= [tex]\frac{1190 - 1650}{19635}[/tex]
[tex]\frac{1}{v}[/tex] = [tex]\frac{-460}{19635}[/tex]
⇒ v = [tex]\frac{19635}{-460}[/tex]
= -42.6848
v = 42.7 cm
The fish would appear 42.7 cm on the left side from the front of the bowl.
What is 3/4 of 12 and 24
Answer:
3/4 of 12 = 16
3/4 of 24 = 32
Those are the answers based on how your question sounded
Explanation:
Answer:
27
Explanation:
3/4 of (12 and 24)
of means ×
and means +
therefore,3/4× (12+24)
3/4×(36)
3×9
27
"Light traveling in a medium with a refractive index 1.11 is incident on a plate of another medium with index of refraction 1.66. At what angle of incidence is the reflected light fully polarized?"
Answer:
56°
Explanation:
Brewsters angle can be simply derived from
n1sin theta1= n2sintheta2= n2costheta1
because the reflected light will be 100% polarized if it is reflected at an angle 90o to the refracted light. Hence, Brewsters angle is
Tan theta= n2/n1
1.66/1.11= 1.495
Theta = 56°
Explanation:
Coherent light with wavelength 601 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on the screen?
Answer:
The wavelength is [tex]\lambda = 1805 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the light is [tex]\lambda = 601 \ nm = 601 *10^{-9} \ m[/tex]
The distance of the screen is D = 3.0 m
The fringe width is [tex]y = 4.84 \ mm = 4.84 *10^{-3} \ m[/tex]
Generally the fringe width for a bright fringe is mathematically represented as
[tex]y = \frac{ \lambda * D }{d }[/tex]
=> [tex]d = \frac{ \lambda * D }{ y }[/tex]
=> [tex]d = \frac{ 601 *10^{-9} * 3}{ 4.84 *10^{-3 }}[/tex]
=> [tex]d = 0.000373 \ m[/tex]
Generally the fringe width for a dark fringe is mathematically represented as
[tex]y_d = [m + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
Here m = 0 for first order dark fringe
So
[tex]y_d = [0 + \frac{1}{2} ] * \frac{\lambda D }{d }[/tex]
looking at which we see that [tex]y_d = y[/tex]
[tex]4.84 *10^{-3} = [0 + \frac{1}{2} ] * \frac{\lambda * 3 }{ 0.000373 }[/tex]
=> [tex]\lambda = 1805 *10^{-9} \ m[/tex]
=> [tex]\lambda = 1805 nm[/tex]
Two wires carry current I1 = 73 A and I2 = 31 A in the opposite directions parallel to the x-axis at y1 = 3 cm and y2 = 13 cm. Where on the y-axis (in cm) is the magnetic field zero?
Answer:
The position on the y-axis where the magnetic field is zero is at y = 10 cm
Explanation:
The magnetic field B due to a long straight wire carrying a current, i at a distance R from the wire is given by
B = μ₀i/2πR
Now, let y be the point where the magnetic fields of both wires are equal.
So, the magnetic field due to wire 1 carrying current i₁ = 73 A is
B₁ = μ₀i₁/2π(y - 3) and
the magnetic field due to wire 2 carrying current i₂ = 31 A is
B₂ = μ₀i₂/2π(13 - y)
At the point where the magnetic field is zero, B₁ = B₂. So,
μ₀i₁/2π(y - 3) = μ₀i₂/2π(13 - y)
cancelling out μ₀ and 2π, we have
i₁/(x - y) = i₂/(13 - y)
cross-multiplying, we have
(13 - y)i₁ = (y - 3)i₂
Substituting the values of i₁ and i₂, we have
(13 - y)73 = (y - 3)31
949 - 73y = 31y - 93
Collecting like terms, we have
949 + 93 = 73y + 31y
1042 = 104y
dividing through by 104, we have
y = 1042/104
y = 10.02 cm
y ≅ 10 cm
So, the position on the y-axis where the magnetic field is zero is at y = 10 cm
An elderly sailor is shipwrecked on a desert island but manages to save his eyeglasses. The lens for one eye has a power of 1.28 diopters, and the other lens has a power of 8.50 diopters. What is the magnifying power of the telescope he can construct with these lenses
Heat and thermodynamics Numerical
Answer:
K = 227.04 W/m.°C
Explanation:
First we need to find the heat required to melt the ice:
q = m H
where,
q = heat required = ?
m = mass of the ice = 8.5 g = 8.5 x 10⁻³ kg
H = Latent heat of fusion of ice = 3.34 x 10⁵ J/kg
Therefore,
q = (8.5 x 10⁻³ kg)(3.34 x 10⁵ J/kg)
q = 2839 J
Now, we find the heat transfer rate through rod:
Q = q/t
where,
t = time = (10 min)(60 s/1 min) = 600 s
Q = Heat Transfer Rate = ?
Therefore,
Q = 2839 J/600 s
Q = 4.73 W
From Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
K = Thermal Conductivity = ?
A = cross sectional area = 1.25 cm² = 1.25 x 10⁻⁴ m²
L = Length of rod = 60 cm = 0.6 m
ΔT = Difference in temperature = 100°C - 0°C = 100°C
Therefore,
4.73 W = K(1.25 X 10⁻⁴ m²)(100°C)/0.6 m
K = (4.73 W)/(0.0208 m.°C)
K = 227.04 W/m.°C
A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire
Answer:
The direction of the force is towards the East.
Explanation:
Using the right hand rule, the force on the current carrying conductor is east.
In the right hand rule, if the hand is held with the fingers pointed parallel to the palm representing the magnetic field, and the thumb held at right angle to the rest of the fingers representing the direction of the current, then the palm will push in the direction of the force.
In this case, the thumb is pointing downwards, with the fingers pointing north away from the body in the direction of the earth's magnetic field, the palm will push east.
Proposed Exercises: Strength and Acceleration in Circular Movement In the situation illustrated below, a 7kg sphere is connected to a rope so that it can rotate in a vertical plane around an O axis perpendicular to the plane of the figure. When the sphere is in position A, it has a speed of 3m/s. Determine for this position the modulus of tension on the string and the rate at which the tangential velocity is increased.
Answer:
81 N
7.1 m/s²
Explanation:
Draw a free body diagram of the sphere. There are two forces:
Weight force mg pulling straight down,
and tension force T pulling up along the rope.
Sum of forces in the centripetal direction:
∑F = ma
T − mg sin 45° = m v² / r
T = m (g sin 45° + v² / r)
T = (7 kg) (10 m/s² sin 45° + (3 m/s)² / 2 m)
T = 81 N
Sum of forces in the tangential direction:
mg cos 45° = ma
a = g cos 45°
a = (10 m/s²) cos 45°
a = 7.1 m/s²
A guitar string is 90 cm long and has a mass of 3.5g . The distance from the bridge to the support post is L=62cm, and the string is under a tension of 540N . What are the frequencies of the fundamental and first two overtones? Express your answers using two significant figures. Enter your answers in ascending order separated by commas.
Answer:
[tex]v_1 = 301 Hz[/tex]
[tex]v_2 = 601 \ \ Hz[/tex]
[tex]v_3 = 901 \ Hz[/tex]
Explanation:
From the question we are told that
The length of the string is [tex]l = 90 \ cm = 0.9 \ m[/tex]
The mass of the string is [tex]m_s = 3.5 \ g =0.0035 \ kg[/tex]
The distance from the bridge to the support post [tex]L = 62 \ c m = 0.62 \ m[/tex]
The tension is [tex]T = 540 \ N[/tex]
Generally the frequency is mathematically represented as
[tex]v = \frac{n}{2 * L } [\sqrt{ \frac{T}{\mu} } ][/tex]
Where n is and integer that defines that overtones
i.e n = 1 is for fundamental frequency
n = 2 first overtone
n =3 second overtone
Also [tex]\mu[/tex] is the linear density of the string which is mathematically represented as
[tex]\mu = \frac{m_s}{l}[/tex]
=> [tex]\mu = \frac{0.0035 }{ 0.9 }[/tex]
=> [tex]\mu = 0.003889 \ kg/m[/tex]
So for n = 1
[tex]v_1 = \frac{1}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_1 = 301 \ Hz[/tex]
So for n = 2
[tex]v_2 = \frac{2}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v_2 = 601 \ Hz[/tex]
So for n = 3
[tex]v_3 = \frac{3}{2 * 0.62 } [\sqrt{ \frac{ 540}{0.003889} } ][/tex]
[tex]v =901 \ Hz[/tex]
How could a country benefit from making it into space?
Answer:
space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.
. Two waves that have the same wavelengths and amplitudes are traveling in opposite directions on a string. If each wave has a speed of 10 m/s and there are moments when the string is not moving, what is the wavelength of the waves if the time between each moment that the string is flat is 0.5 s?
Answer:
10m
Explanation:
Since Given frequency f= 1/t
and velocity ν=10 m/s
We know ν=λf
λ= ν/f
= 10/1/0.5
=5m
Since both the waves are similar but moves in opposite direction its total wavelength of the wave will be 10 m
In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect of this change on the impedance of the circuit
Answer:
Impedance increases for frequencies below resonance and decreases for the frequencies above resonance
Explanation:
See attached file
Explanation:
The charger for your electronic devices is a transformer. Suppose a 60 Hz outlet voltage of 120 V needs to be reduced to a device voltage of 3.0 V. The side of the transformer attached to the electronic device has 45 turns of wire.
How many turns are on the side that plugs into the outlet?
Answer:
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
Explanation:
The transformer turns ratio is given by the following equation:
V₁/V₂ = N₁/N₂
where,
V₁ = Voltage of outlet = 120 V
V₂ = Device Voltage = 3 V
N₁ = No. of turns on outlet side = ?
N₂ = No. of turns on side of device = 45
Therefore,
120 V/3 V = N₁/45
N₁ = (40)(45)
N₁ = 1800 turns
So, the side of the transformer that plugs into the outlet has 1800 turns.
A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.
Required:
What is the charge on the dust particle?
Answer:
The charge on the dust particle is [tex]q_d = 6.94 *10^{-13} \ C[/tex]
Explanation:
From the question we are told that
The length is [tex]l = 2.0 \ m[/tex]
The width is [tex]w = 4.0 \ m[/tex]
The charge is [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]
The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]
Generally the electric field on the carpet is mathematically represented as
[tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]
Where [tex]\epsilon _o[/tex] is the permittivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
substituting values
[tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]
[tex]E = -70621.5 \ N/C[/tex]
Generally the electric force keeping the dust particle on the air equal to the force of gravity acting on the particles
[tex]F__{E}} = F__{G}}[/tex]
=> [tex]q_d * E = m * g[/tex]
=> [tex]q_d = \frac{m * g}{E}[/tex]
=> [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]
=> [tex]q_d = 6.94 *10^{-13} \ C[/tex]
During the first part of this lab, we want to determine how the object distance is related to what two quantities
Answer and Explanation:
The computation of the object distance related to two quantities is shown below:
It could find out by using the lens formula which is shown below:
[tex]\frac{1}{v} - \frac{1}{u} = \frac{1}{f}[/tex]
where,
v = image distance
u = object distance
f = focal length
It could be found by applying the above formula i.e considering the image distance, object distance and the focal length
g A solenoid 63.5 cm long has 960 turns and a radius of 2.77 cm. If it carries a current of 2.28 A, find the magnetic field along the axis at its center.Find the magnetic field on the solenoidal axis at the end of the solenoid.
Answer:
The value is [tex]B = 0.0043 \ T[/tex]
Explanation:
From the question we are told that
The length of the solenoid is [tex]l = 63.5 = 0.635 \ m[/tex]
The number of turns is [tex]N = 960 \ turns[/tex]
The current is [tex]I = 2.28 \ A[/tex]
Generally the magnetic field is mathematically represented as
[tex]B = \mu _o * n * I[/tex]
Where n is the number of turn per unit length which is mathematically evaluated as
[tex]n = \frac{N}{l}[/tex]
[tex]n = \frac{960}{0.635}[/tex]
[tex]n = 1512 \ turns /m[/tex]
and [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4\pi * 10^{-7} N/A^2[/tex]
So
[tex]B = 4\pi * 10^{-7} * 1512 * 2.28[/tex]
[tex]B = 0.0043 \ T[/tex]
