Answer:
1
[tex]R = 1.692*10^{6} \Omega[/tex]
2
[tex]C = 2.837 *10^{-6} \ F[/tex]
Explanation:
From the question we are told that
The voltage is [tex]E = 110 \ V[/tex]
The current is [tex]I = 6.5 *10^{-5} \ A[/tex]
The time constant is [tex]\tau = 4.8 \ s[/tex]
The resistance of resistor is mathematically evaluated as
[tex]R = \frac{E}{I}[/tex]
substituting values
[tex]R = \frac{ 110 }{ 6.5*10^{-5}}[/tex]
[tex]R = 1.692*10^{6} \Omega[/tex]
The capacitance of the capacitor is mathematically represented as
[tex]C = \frac{\tau}{R}[/tex]
substituting values
[tex]C = \frac{ 4.8}{ 1.692*10^{6}}[/tex]
[tex]C = 2.837 *10^{-6} \ F[/tex]
Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)a. Their electrical potential energy keeps decreasing.b. Their acceleration keeps decreasing.c. Their kinetic energy keeps increasing.d. Their kinetic energy keeps decreasing.e. Their electric potential energy keeps increasing.
Answer:
Explanation:
correct options
a ) Their electrical potential energy keeps decreasing
Actually as they move apart , their electrical potential energy decreases due to increase of distance between them and kinetic energy increases
so a ) option is correct
b ) Their acceleration keeps decreasing
As they move apart , their mutual force of repulsion decreases due to increase of distance between them so the acceleration decreases .
c ) c. Their kinetic energy keeps increasing
Their kinetic energy increases because their electrical potential energy decreases . Conservation of energy law will apply .
The moving apart should be true statements:
a. The electrical potential energy should be reduced.
b. The acceleration should be reduced.
c. The kinetic energy should be increased.
True statements related to moving apart:At the time when the moving part, there is the reduction of the electric potential energy because there is a rise in the distance due to which the increment of the kinetic energy. The reduction of the mutual force of repulsion because of increment in the distance due to this the acceleration should be reduced. There is the increase in the kinetic energy due to the reduction of the electrical potential energy. here the law of conversation of energy should be applied.
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if the current in the circuit decreases, what does that mean about the rate at which the charge(and voltage) change in a capacitor?
2. the exponent of the exponential function contains RC for the given circuit. who's is a constant. use units R and C to find units of RC. write ohms in terms of volts and amps and write farads in terms of volts and coulombs. Simplify
units of RC are__________
Answer:
`1. charge Q, on the capacitor increases, while the current will decrease
2. τ = t = secs
Explanation:
1. consider RC of a circuit to be am external source
voltage across the circuit is given as
v =v₀(1 - [tex]e^{\frac{t}{τ} }[/tex])
where v = voltage
v₀ = peak voltage
t = time taken
τ= time constant
as the charge across the capacitor increases, current decreases
the charge across the circuit is given as
Q= Q₀(1 - [tex]e^{\frac{t}{τ} }[/tex])
charge Q is inversely proportional to the current I
hence the charge across the circuit increases
2. τ = RC
unit of time constant, τ,
= Ω × F
=[tex]\frac{V}{I}[/tex] ˣ [tex]\frac{C}{V}[/tex]
=[tex]\frac{C}{A}[/tex]
=[tex]\frac{C}{C/t}[/tex]
τ = t = secs
A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 15 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.3 m/s^2 , determine the coefficient of static friction between the trunk and the turntable
Answer:
μ = 0.03
Explanation:
In order for the trunk not to slide the frictional force between the turntable and the trunk must be equal to the unbalanced force applied on the trunk by the motion of the turntable. Therefore,
Unbalanced Force = Frictional Force
but,
Unbalanced Force = ma (Newton's second law of motion)
Frictional Force = μN = μW = μmg
Therefore,
ma = μmg
a = μg
μ = a/g
where,
μ = coefficient of static friction between the trunk and the turntable = ?
a = tangential acceleration of trunk = 0.3 m/s²
g = 9.8 m/s²
Therefore,
μ = (0.3 m/s²)/(9.8 m/s²)
μ = 0.03
The tune-up specifications of a car call for the spark plugs to be tightened to a torque of 38N⋅m38N⋅m. You plan to tighten the plugs by pulling on the end of a 25-cm-long wrench. Because of the cramped space under the hood, you'll need to put at an angle of 120∘with respect to the wrench shaft. With what force must you pull?
Answer:
F= 175.5N
Explanation:
Given:
Torque which can also be called moment is defined as rotational equivalent of linear force. It is the product of the external force and perpendicular distance
torque of 38N⋅m
angle of 120∘
Torque(τ): 38Nm
position r relative to its axis of rotation: 25cm , if we convert to metre for consistency we have 0.25m
Angle: 120°
To find the Force, the torque equation will be required which is expressed below
τ = Frsinθ
We need to solve for F, if we rearrange the equation, we have the expression below
F= τ/rsinθ
Note: the torque is maximum when the angle is 90 degrees
But θ= 180-120=60
F= 38/0.25( sin(60) )
F= 175.5N
An astronaut out on a spacewalk to construct a new section of the International Space Station walks with a constant velocity of 2.30 m/s on a flat sheet of metal placed on a flat, frictionless, horizontal honeycomb surface linking the two parts of the station. The mass of the astronaut is 71.0 kg, and the mass of the sheet of metal is 230 kg. (Assume that the given velocity is relative to the flat sheet.)
Required:
a. What is the velocity of the metal sheet relative to the honeycomb surface?
b. What is the speed of the astronaut relative to the honeycomb surface?
Answer:
Explanation:
Let the velocity of astronaut be u and the velocity of flat sheet of metal plate be v . They will move in opposite direction , so their relative velocity
= u + v = 2.3 m /s ( given )
We shall apply conservation of momentum law for the movement of astronaut and metal plate
mu = M v where m is mass of astronaut , M is mass of metal plate
71 u = 230 x v
71 ( 2.3 - v ) = 230 v
163.3 = 301 v
v = .54 m / s
u = 1.76 m / s
honeycomb will be at rest because honeycomb surface is frictionless . Plate will slip over it . Over plate astronaut is walking .
a ) velocity of metal sheet relative to honeycomb will be - 1.76 m /s
b ) velocity of astronaut relative to honeycomb will be + .54 m /s
Here + ve direction is assumed to be the direction of astronaut .
g Question 11 pts Consider two masses connected by a string hanging over a pulley. The pulley is a uniform cylinder of mass 3.0 kg. Initially m1 is on the ground and m2 rests 2.9 m above the ground. After the system is released, what is the speed of m2 just before it hits the ground? m1= 30 kg and m2= 35 kg Group of answer choices 2.1 m/s 1.4 m/s 9.8 m/s 4.3 m/s 1.9 m/s
Answer:
The speed of m2 just before it hits the ground is 2.1 m/s
Explanation:
mass on the ground m1 = 30 kg
mass oat rest at the above the ground m2 = 35 kg
height of m2 above the ground =2.9 m
Let the tension on the string be taken as T
for the mass m2 to reach the ground, its force equation is given as
[tex]m_{2} g - T = m_{2}a[/tex] ....equ 1
where g is acceleration due to gravity = 9.81 m/s^2
and a is the acceleration with which it moves down
For mass m1 to move up, its force equation is
[tex]T - m_{1} g = m_{1} a[/tex]
[tex]T = m_{1}a + m_{1}g[/tex]
[tex]T = m_{1}(a + g)[/tex] ....equ 2
substituting T in equ 1, we have
[tex]m_{2} g - m_{1}(a+g) = m_{2}a[/tex]
imputing values, we have
[tex](35*9.81) - 30(a+9.81) = 35a[/tex]
[tex]343.35 - 30a-294.3 = 35a[/tex]
[tex]343.35 -294.3 = 35a+ 30a[/tex]
[tex]49.05 = 65a[/tex]
a = 49.05/65 = 0.755 m/s^2
The initial velocity of mass m2 = u = 0
acceleration of mass m2 = a = 0.755 m/s^2
distance to the ground = d = 2.9 m
final velocity = v = ?
using Newton's equation of motion
[tex]v^{2}= u^{2} + 2ad[/tex]
substituting values, we have
[tex]v^{2}= 0^{2} + 2*0.755*2.9[/tex]
[tex]v^{2}= 2*0.755*2.9 = 4.379\\v = \sqrt{4.379}[/tex]
v = 2.1 m/s
The headlights of a car are 1.4 m apart. What is the maximum distance (in km) at which the eye can resolve these two headlights? Take the pupil diameter to be 0.30 cm. (Assume the average wavelength of visible light is 555 nm.)
Answer:
5.4x10^4km
Explanation:
See attached file
At what frequency f, in hertz, would you have to move the comb up and down to produce red light, of wavelength 600 nm
Answer:
f = 500 x 10^12Hz
Explanation:
E=hc/wavelength
E=hf
hc/wavelength =hf
c/wavelength =f
f = 3 x 10^8 / 600 x 10^-9 = 500 x 10^12Hz
The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at speed c toward the space vehicle. What is the speed of the light signal as viewed from the space vehicle
Answer:
The speed of the light signal as viewed from the observer is c.
Explanation:
Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.
Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.
Answer:
A) V = -136.36 V , B) V = 4.85 10³ V , C) V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
V = k ∑ [tex]q_{i} / r_{i}[/tex]
where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.
A) We apply this equation to our case
V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
V = 9 10⁹ (1.1 10⁻⁸ / 1.65 + (-3.6 10⁻⁸) / 1.65)
V = 9 10¹ / 1.65 (1.10 - 3.60)
V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
r₂ = 3.30 -0.02 = 3.28 m
r₁ = 0.02 m
we calculate
V = 9 10⁹ (1.1 10⁻⁸ / 0.02 - 3.6 10⁻⁸ / 3.28)
V = 9 10¹ (55 - 1,098)
V = 4.85 10³ V
C) The potential on the surface of sphere B
r₂ = 0.02 m
r₁ = 3.3 -0.02 = 3.28 m
V = 9 10⁹ (1.10 10⁻⁸ / 3.28 - 3.6 10⁻⁸ / 0.02)
V = 9 10¹ (0.335 - 180)
V = 1.62 10⁴ V
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
Two forces act at a point in the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of and newtons, forming an angle of
Answer:
F = 44.22 N
Explanation:
Let force 1, [tex]F_1=19\ N[/tex]
Force 2, [tex]F_2=32\ N[/tex]
The angle between forces, [tex]\theta=118^{\circ}[/tex]
We need to find the magnitude of the resultant force. It is based on the law of cosines. The formula is given by :
[tex]F^2=F_1^2+F_2^2-2AB\cos\theta\\\\F^2=(19)^2+(32)^2-2\times 19\times 32\times \cos(118)\\\\F=\sqrt{(19)^{2}+(32)^{2}-2\times19\times32\times\cos(118)}\\\\F=44.22\ N[/tex]
So, the magnitude of resultant force is 44.22 N.
Correct question: Two forces act on a point on the plane. The angle between the two forces is given. Find the magnitude of the resultant force. forces of 19 and 32 newtons, forming an angle of 118 degrees.
the magnitude of the resultant force is 28.53 N.
The resultant of the two vectors can be calculated using parallelogram theorem.
parallelogram theorem states that if two vectors are represented by the adjacent side of a parallelogram, the resultant of the vectors is the diagonal of the parallelogram drawn from the point of intersection of the vectors.
This can be expressed mathematically as
R² = P²+Q²-2PQcos(180-∅).............. Equation 1
Where R = resultant of the vectors, P and Q = the two vectors respectively, ∅ = angle between the vectors.
From the question,
Given: P = 19 N, Q = 32 N, ∅ = 118°
Substitute these values into equation 2
R² = 19²+32²-2×19×32cos(180-118)
R² = 361+1024-1216cos62°
R² = 1385-1216(0.4695)
R² = 1385-570.878
R² = 814.122
R = √(814.122)
R = 28.53 N
Hence, the magnitude of the resultant force is 28.53 N
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Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)
A. Their electric potential energy keeps decreasing.
B. Their acceleration keeps decreasing.
C. Their kinetic energy keeps increasing.
D. Their kinetic energy keeps decreasing.
E. Their electric potential energy keeps increasing.
Answer:
(A)
Explanation:
We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.
Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.
A light ray in air strikes water at an angle of incidence equal to 40°. If the index of refraction for water is 1.33, what is the angle of refraction?
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°
Four equal masses m are located at the corners of a square of side L, connected by essentially massless rods. Find the rotational inertia of this system about an axis (a) that coincides with one side and (b) that bisects two opposite sides.
Answer:
Explanation:
a )
Moment of inertial of four masses about axis that coincides with one side :
Out of four masses . location of two masses will lie on the axis so their moment of inertia will be zero .
Moment of inertia of the two remaining masses
= m L² + m L²
= 2 mL²
b )
Axis that bisects two opposite sides
Each of the four masses will lie at a distance of L / 2 from this axis so moment of inertia of the four masses
= 4 x m x ( L/2 )²
= 4 x mL² / 4
= m L² .
The metal bar on the conducting rails is moving to the left. The magnitude of the uniform magnetic field is 0.80 Tesla. The magnetic field is directed out of the page. The length L is 0.95 m. The resistance R of the circuit is 3.60 Ω. (a) If the induced current needs to be 0.54 A, what should be the induced emf
Answer:
Explanation:
induced emf = ?
induced emf = induced current x resistance of the circuit
= .54 x 3.6
= 1.944 V
induced emf required = 1.944 V .
A current carrying loop of wire lies flat on a table top. When viewed from above, the current moves around the loop in a counterclockwise sense.
(a) For points OUTSIDE the loop, the magnetic field caused by this current:________.
a. points straight up.
b. circles the loop in a clockwise direction.
c. circles the loop in a counterclockwise direction.
d. points straight down.
e. is zero.
(b) For points INSIDE the loop, the magnetic field caused by this current:________.
a. circles the loop in a counterclockwise direction.
b. points straight up.
c. points straight down.
d. circles the loop in a clockwise direction.
e. is zero
Answer:
D &B
Explanation:
Using Fleming right hand rule that States that if the fore-finger, middle finger and the thumb of left hand are stretched mutually perpendicular to each other, such that fore-finger points in the direction of magnetic field, the middle finger points in the direction of the motion of positive charge, then the thumb points to the direction of the force
Consider a block of mass equal to 10kg sliding on an inclined plane of 30°, as shown in the figure below. The coefficient of kinetic friction between the block and the plane surface is c = 0.4 (a) Determine the value of the horizontal and vertical acceleration of the block. (b) If the block starts from rest in t=0s and when it is in the X=0 and Y=5m position, calculate what its horizontal and vertical position will be at the instant t=1s. (C) How long does the LM block take to reach the base of the tilted plane?
Answer:
(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²
(b) x = 0.665 m and y = 4.62 m
(c) 3.61 s
Explanation:
(a) There are two ways we can solve this. The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).
The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a. Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.
Let's try the first method. Sum of forces in the +y direction:
∑F = ma
N cos 30° + Nμ sin 30° − mg = maᵧ
N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°
Sum of forces in the +x direction:
∑F = ma
N sin 30° − Nμ cos 30° = maₓ
Substituting:
N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°
N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°
N cos 30° − mg = -N sin 30° tan 30°
N (cos 30° + sin 30° tan 30°) = mg
N = mg / (cos 30° + sin 30° tan 30°)
N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)
N = 86.6 N
Now, solving for the accelerations:
N sin 30° − Nμ cos 30° = maₓ
aₓ = N (sin 30° − μ cos 30°) / m
aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg
aₓ = 1.33 m/s²
N cos 30° + Nμ sin 30° − mg = maᵧ
aᵧ = N (cos 30° + μ sin 30°) / m − g
aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²
aᵧ = -0.770 m/s²
Now let's try the second method.
Sum of forces in the perpendicular direction:
∑F = ma
N − mg cos 30° = 0
N = mg cos 30°
Sum of forces in the parallel direction:
∑F = ma
mg sin 30° − Nμ = ma
mg sin 30° − mgμ cos 30° = ma
a = g (sin 30° − μ cos 30°)
a = (10 m/s²) (sin 30° − 0.4 cos 30°)
a = 1.536 m/s²
Solving for the accelerations:
aₓ = a cos 30°
aₓ = 1.33 m/s²
aᵧ = -a sin 30°
aᵧ = -0.770 m/s²
As you can see, the second method is faster and easier, but both methods will give you the same answer.
(b) In the x direction:
Given:
x₀ = 0 m
v₀ = 0 m/s
aₓ = 1.33 m/s²
t = 1 s
Find: x
x = x₀ + v₀ t + ½ at²
x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²
x = 0.665 m
In the y direction:
Given:
y₀ = 5 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
t = 1 s
Find: y
y = y₀ + v₀ t + ½ at²
y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²
y = 4.62 m
(c) In the y direction:
Given:
y₀ = 5 m
y = 0 m
v₀ = 0 m/s
aᵧ = -0.770 m/s²
Find: t
y = y₀ + v₀ t + ½ at²
0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²
t = 3.61 s
If the current flowing through a circuit of constant resistance is doubled, the power dissipated by that circuit will
As light shines from air to another medium, i = 26.0 º. The light bends toward the normal and refracts at 32.0 º. What is the index of refraction? A. 1.06 B. 0.944 C. 0.827 D. 1.21
Explanation:
It is given that,
Angle of incidence from air to another medium, i = 26°
The angle of reflection, r = 32°
We need to find the refractive index of the medium. The ratio of sine of angle of incidence to the sine of angle of reflection is called refractive index. It can be given by :
[tex]n=\dfrac{\sin i}{\sin r}\\\\n=\dfrac{\sin (26)}{\sin (32)}\\\\n=0.82[/tex]
So, the index of refraction is 0.82. Hence, the correct option is C.
lock of mass m2 is attached to a spring of force constant k and m1 . m2. If the system is released from rest, and the spring is initially not stretched or com- pressed, find an expres- sion for the maximum displacement d of m2
Answer:
The maximum displacement of the mass m₂ [tex]= \frac{2(m_1-m_2)g}{k}[/tex]
Explanation:
Kinetic Energy (K) = 1/2mv²
Potential Energy (P) = mgh
Law of Conservation of energy states that total energy of the system remains constant.
i.e; Total energy before collision = Total energy after collision
This implies that: the gravitational potential energy lost by m₁ must be equal to sum of gravitational energy gained by m₂ and the elastic potential energy stored in the spring.
[tex]m_1gd = m_2gd+\frac{1}{2}kd^2\\\\m_1g = m_2g+\frac{1}{2}kd\\\\d = \frac{2(m_1-m_2)g}{k}[/tex]
d = maximum displacement of the mass m₂
A dentist using a dental drill brings it from rest to maximum operating speed of 391,000 rpm in 2.8 s. Assume that the drill accelerates at a constant rate during this time.
(a) What is the angular acceleration of the drill in rev/s2?
rev/s2
(b) Find the number of revolutions the drill bit makes during the 2.8 s time interval.
rev
Answer:
a
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
b
[tex]\theta = 9124.5 \ rev[/tex]
Explanation:
From the question we are told that
The maximum angular speed is [tex]w_{max} = 391000 \ rpm = \frac{2 \pi * 391000}{60} = 40950.73 \ rad/s[/tex]
The time taken is [tex]t = 2.8 \ s[/tex]
The minimum angular speed is [tex]w_{min}= 0 \ rad/s[/tex] this is because it started from rest
Apply the first equation of motion to solve for acceleration we have that
[tex]w_{max} = w_{mini} + \alpha * t[/tex]
=> [tex]\alpha = \frac{ w_{max}}{t}[/tex]
substituting values
[tex]\alpha = \frac{40950.73}{2.8}[/tex]
[tex]\alpha = 14625 .3 \ rad/s^2[/tex]
converting to [tex]rev/s^2[/tex]
We have
[tex]\alpha = 14625 .3 * 0.159155 \ rev/s^2[/tex]
[tex]\alpha = 2327.7 \ rev/s^2[/tex]
According to the first equation of motion the angular displacement is mathematically represented as
[tex]\theta = w_{min} * t + \frac{1}{2} * \alpha * t^2[/tex]
substituting values
[tex]\theta = 0 * 2.8 + 0.5 * 14625.3 * 2.8^2[/tex]
[tex]\theta = 57331.2 \ radian[/tex]
converting to revolutions
[tex]revolution = 57331.2 * 0.159155[/tex]
[tex]\theta = 9124.5 \ rev[/tex]
An air-filled parallel-plate capacitor has plates of area 2.30 cm2 2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. Find the value of its capacitance.
Answer:
[tex]1.357\times 10^{-12}[/tex]
Explanation:
Relevant Data provided
Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2
Distance which indicates d = 1.50 x 10^-3 m
Voltage which indicates V = 12 V
According to the requirement, the computation of value of its capacitance is shown below:-
[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]
[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]
= [tex]1.357\times 10^{-12}[/tex]
Therefore for computing the capacitance we simply applied the above formula.
In the circuit shown, the galvanometer shows zero current. The value of resistance R is :
A) 1 W
B) 2 W
C) 4 W
D) 9 W
Answer:
its supposed to be (a) 1W
A 46-ton monolith is transported on a causeway that is 3500 feet long and has a slope of about 3.7. How much force parallel to the incline would be required to hold the monolith on this causeway?
Answer:
2.9tons
Explanation:
Note that On an incline of angle a from horizontal, the parallel and perpendicular components of a downward force F are:
parallel ("tangential"): F_t = F sin a
perpendicular ("normal"): F_n = F cos a
At a=3.7 degrees, sin a is about 0.064 and with F = 46tons:
F sin a ~~ (46 tons)*0.064 ~~ 2.9tons
Also see attached file
The required force parallel to the incline to hold the monolith on this causeway will be "2.9 tons".
Angle and ForceAccording to the question,
Angle, a = 3.7 degrees or,
Sin a = 0.064
Force, F = 46 tons
We know the relation,
Parallel (tangential), [tex]F_t[/tex] = F Sin a
By substituting the values,
= 46 × 0.064
= 2.9 tons
Thus the response above is appropriate answer.
Find out more information about Force here:
https://brainly.com/question/25239010
A parallel-plate capacitor in air has a plate separation of 1.31 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator.
(a) Determine the charge on the plates before and after immersion.
before pC
after pC
(b) Determine the capacitance and potential difference after immersion.
Cf = F
ΔVf = V
(c) Determine the change in energy of the capacitor.
[ ] nJ
Answer:
a) before immersion
C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F
q = CV = (1.68e-12)(255) = 4.28e-10 C
b) after immersion
q = 4.28e-10 C
Because the capacitor was disconnected before it was immersed, the charge remains the same.
c)*at 20° C
C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F
V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V
e)
U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J
U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J
ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J
Which has more mass electron or ion?
A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g and a volume of 4,064 cm3. The density of seawater is 1.03 g/mL.
a. What is the weight of the statuette?
b. What is the mass of displaced water?
c. What is the weight of displaced water?
d. What is the buoyant force on the statuette?
e. What is the net force on the statuette?
f. How much force would be required to lift the statuette?
Answer:
A) W = 103.55 N
B) mass of displaced water = 4186 g
C) W_displaced water = 41.06 N
D) Buoyant force = 41.06 N.
E) ZERO
F) 62.54 N
Explanation:
We are given;
mass of statuette;m = 10,566 g = 10.566 kg
volume = 4,064 cm³
Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³
A) The dry weight of the statuette can be calculated as;
W = mg
So;
W = 10.556 × 9.81
W = 103.55 N
B) Mass of displaced water is calculated from;
Density = mass/volume
So, mass = Density × Volume
m = 1.03 × 4,064 = 4186 g
C) Weight of displaced water is given by;
W_displaced water = (m_displaced water) × g
W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N
D) The buoyant force is the same as the weight of the displaced water.
Thus, Buoyant force = 41.06 N.
E) The apparent weight of the statuette is calculated from;
Apparent weight = Dry weight - Weight of displaced water
Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.
F. From E above, The Force required to lift the statuette = 62.54 N
The core of an optical fiber has an index of refraction of 1.35 , while the index of refraction of the cladding surrounding the core is 1.21 . What is the critical angle θc for total internal reflection at the core‑cladding interface?
Answer:
The critical angle is [tex]\theta_c = \ 63.68^o[/tex]
Explanation:
From the question we are told that
The refractive index of the core is [tex]n_c = 1.35[/tex]
The refractive index of the cladding is [tex]n_s = 1.21[/tex]
Generally according to Snell's law
[tex]\frac{sin i }{sin r } = \frac{n_s}{n_c }[/tex]
Here for total internal reflection the refractive angle is [tex]r = 90^o[/tex] and the critical angle is equal to the critical angle so [tex]i = \theta_c[/tex]
[tex]\frac{sin \theta_c }{sin (90) } = \frac{n_s}{n_c }[/tex]
substituting values
[tex]\frac{sin \theta_c }{sin (90) } = \frac{1.21}{1.35 }[/tex]
[tex]\theta_c = sin^{-1} [\frac{1.21}{1.35} ][/tex]
[tex]\theta_c = \ 63.68^o[/tex]
A car starts from Hither, goes 50 km in a straight line to Yon, immediately turns around, and returns to Hither. The time for this round trip is 2 hours. The magnitude of the average velocity of the car for this round trip is:
A. 0
B. 50 km/hr
C. 100 km/hr
D. 200 km/hr
E. cannot be calculated without knowing the acceleration
Answer:
The average velocity for this trip is 0 km/hr
Explanation:
We know that average velocity = total displacement/total time.
Now, its displacement is d = final position - initial position.
Since the car starts and ends at its initial position at Hither, if we assume its initial position is 0 km, then its final position is also 0 km.
So, its displacement is d = 0 km - 0 km = 0 km.
Since the total time for the round trip is 2 hours, the average velocity is
total displacement/ total time = 0 km/2 hr = 0 km/hr.
So the average velocity for this trip is 0 km/hr