Answer:
Explanation:
We shall find electric field at origin due to two given charges sitting on the either side of origin .
Total field will add up due to their same direction .
Field due to a charge Q
= 9 x 10⁹ x Q / R² ; R is distance of point , Q is charge
Field due to first charge
= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴
= 90 x 10¹⁰ N/C
Field due to second charge
= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴
= 112.5 x 10¹⁰ N/C
Total field
= 202.5 x 10¹⁰ N/C
Force on given charge at origin
= charge x field
= 4 x 10⁻³ x 202.5 x 10¹⁰
= 810 x 10⁷ N .
A spherical capacitor contains a charge of 3.40 nC when connected to a potential difference of 240.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 4.10 cm.
Calculate:
a. The capacitance
b. The radius of the inner sphere.
c. The electric field just outside the surface of the inner sphere.
Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
You stand near the edge of a swimming pooland observe through the water an object lying on the bottom of thepool. Which of the following statements correctly describes whatyou see?
a. The apparent depth of the object is less than thereal depth.
b. The apparent depth of the object is greater thanthe real depth.
c. There is no difference between the apparent depth and the actual depth of the object.
Answer:
a
Explanation:
The correct answer would be that the apparent depth of the object is less than the real depth.
The refractive property of light as it passes from air to water would make the depth of the pool appear less shallow than the actual depth to an observed. Hence, an object placed at the bottom of the pool will have an apparent depth that is shallower than its actual depth.
Due to the difference in the density of air and that of water, as the ray of light from an observer standing at the edge of a swimming pool travels from air into the water, it becomes refracted by bending away from the original traveling angle.
The same refraction occurs when light rays from an object inside the pool travel from water into the air. Hence, due to the refraction of the ray of light coming from the object at the bottom of the pool, the depth appears shallower than the actual depth.
Correct option: a
The first step to merging is entering the ramp and _____.
A. honking to indicate your location
B. matching your speed
C. signaling your intent
D. telling your passengers where you're going
Answer:
B. matching your speed
Explanation:
To merge safely, you must identify a gap in traffic and match your speed to the speed of the gap. Before you make your move to fill the gap, you should signal your intent.*
_____
* At least one resource says "The first step ... is to make sure you're traveling at the same speed ..." Then it goes on to say "Use your indicator. Do it early ...." The accompanying animation shows blinkers being activated on the ramp before the merge lane is entered. Apparently, "the first step" is not necessarily the first thing you do.
Answer:
It's C "signaling your intent"
Explanation:
The key thing to look at is they are asking the rest of the first step and that;s C
A rectangular loop of wire carries current I in the clockwise direction. The loop is in a uniform magnetic field B that is parallel to the plane of the loop, in the direction toward the bottom of the page. The length of the rectangle is b and the width is a. What is the net force on the loop by the magnetic field
Answer:
Explanation:
Area of the loop = a b
current = I
magnetic moment of the loop M = area x current
= ab I
Torque on the loop = MB sinθ
here θ = 90
Torque = MB
= abIB
In this case net force on the loop will be zero because here torque is created by two equal and opposite force acting on two opposite sides of the loop so net force will be zero .
Three identical resistors are connected in series to a battery. If the current of 12 A flows from the battery, how much current flows through any one of the resistors
Answer:
Current that flows through any one of the resistors has a value of 12 amperes.
Explanation:
When a group of resistors are connected in series, the same current flows through each resistor. According to the Ohm's Law, the circuit can be represented as follows:
[tex]V_{batt} = i\cdot (R_{1}+R_{2}+R_{3})[/tex]
[tex]i = \frac{V_{batt}}{R_{1}+R_{2}+R_{3}}[/tex]
Where:
[tex]V_{batt}[/tex] - Voltage of the battery, measured in volts.
[tex]R_{1}[/tex], [tex]R_{2}[/tex], [tex]R_{3}[/tex] - Electric resistance of the first, second and third resistors, measured in ohms.
[tex]i[/tex] - Current, measured in amperes.
If [tex]R_{1} = R_{2} = R_{3} = R[/tex], then:
[tex]i = \frac{V_{batt}}{3\cdot R}[/tex]
Current that flows through any one of the resistors has a value of 12 amperes.
The current flows via any of the resistors should have a value of 12 amperes.
Ohm law:At the time When a group of resistors are linked in series, so there is a similar current flow via each resistor.
Here the circuit should be
vbatt = i.(R1 + R2+ R3)
i = Vbatt/R1 + R2 + R3
here
Vbatt means the voltage of the battery
R1,R2, and R3 means the resistance of the first, second and third resistors
I means the current
So, in the case when
R1 = R2 = R3 = R
So,
i = Vbatt/3.R
Learn more about current here: https://brainly.com/question/14956680
A solid cylinder has a diameter of 17.4 mm and a length of 50.3mm. It's mass is 49g . What is its density of the cylinder in metric tonnes per cubic metre? Give your answer to 1 significant figure.
Answer:
4 tonne/m³
Explanation:
ρ = m / V
ρ = 49 g / (π (17.4 mm / 2)² (50.3 mm))
ρ = 0.0041 g/mm³
Converting to tonnes/m³:
ρ = 0.0041 g/mm³ (1 kg / 1000 g) (1 tonne / 1000 kg) (1000 mm / m)³
ρ = 4.1 tonne/m³
Rounding to one significant figure, the density is 4 tonne/m³.
There are fiber optic telephone cables connecting North America and Europe, lying on the bottom of the Atlantic ocean. The wire is 4,500 km long how long and has an index of refraction of 1.5. How long will it take for the signal to cross the ocean? Give your answer in milliseconds.
Answer:
The time taken is [tex]t = 0.0225 \ s[/tex]
Explanation:
From the question we are told that
The length of the wire is [tex]l = 4500 \ km = 4500000 \ m[/tex]
The refractive index is [tex]n_f = 1.5[/tex]
The velocity of the signal is mathematically represented as
[tex]v = \frac{c}{n_f }[/tex]
Where c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
substituting values
[tex]v = \frac{3.0 *10^{8}}{1.5}[/tex]
[tex]v = 2.0*10^{8} \ m/s[/tex]
The time taken is mathematically evaluated as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{4500000}{2.0 *10^{8}}[/tex]
[tex]t = 0.0225 \ s[/tex]
The voltage and power ratings of a particular light bulb, which are its normal operating values, are 110 V and 60 W. Assume the resistance of the filament of the bulb is constant and is independent of operating conditions. If the light bulb is operated at a reduced voltage and the power drawn by the bulb is 36 W. What is the operating voltage of the bulb?
a. 78 V
b. 72 V
c. 66 V
d. 90 V
e. 85 V
Answer:
c. 66 V
Explanation:
p =IV
I =P/V
P1/V1=P2/V2
60/110=36/V2
0.55 = 36/V2
V2 =36/0.55 = 65.5V
V2 = 66V
The time constant of an RC circuit is 2.7 s. How much time t is required for the capacitor (uncharged initially) to gain 0.63 of its full equilibrium charge
Answer:
2.7s
Explanation:
The solution of time required is shown below:-
In the RC circuit condenser charge 63 percent of the full charge from initial time to constant time
Now, the
63% that is equal to 0.63 which is full equilibrium charge
Therefore, the time required to maintain will be Equal to time (t) constant that is 2.7s
So, the correct answer is 2.7s
A particle with mass m = 700 g is found to be moving with velocity v vector (-3.50i cap + 2.90j cap) m/s. From the definition of the scalar product, v^2 = v vector. v vector.
a. What is the particle's kinetic energy at this time? J If the particle's velocity changes to v vector = (6.00i cap - 5.00j cap) m/s,
b. What is the net work done on the particle? J
Answer:
Explanation:
v₁² = v₁ . v₁
= ( - 3.5 i + 2.9 j ).( - 3.5 i + 2.9 j )
= 12.25 + 8.41
= 20.66 m /s
a ) kinetic energy = 1/2 m v₁²
= 1/2 x .7 x 20.66
= 7.23 J
b )
changed velocity v₂ = v₂.v₂
= (6i - 5 j ) . (6i - 5 j )
= 36 + 25
= 61 m /s
kinetic energy = 1/2 m v₂²
= 1/2 x .7 x 61
= 21.35 J
Work done = change in energy
= 21.35 - 7.23
= 14.12 J .
calculate the upthrust aciting on a body if its
true weight is 550 N and apparent weight
lis 490 N
Answer:
As a body moving upward
T=real weight + apparent weight
T=550+490
T=1040
hope u will get the answer:)
Explanation:
A device called an insolation meter is used to measure the intensity of sunlight. It has an area of 100 cm2 and registers 6.50 W. What is the intensity in W/m2
Answer:
650W/m²Explanation:
Intensity of the sunlight is expressed as I = Power/cross sectional area. It is measured in W/m²
Given parameters
Power rating = 6.50Watts
Cross sectional area = 100cm²
Before we calculate the intensity, we need to convert the area to m² first.
100cm² = 10cm * 10cm
SInce 100cm = 1m
10cm = (10/100)m
10cm = 0.1m
100cm² = 0.1m * 0.1m = 0.01m²
Area (in m²) = 0.01m²
Required
Intensity of the sunlight I
I = P/A
I = 6.5/0.01
I = 650W/m²
Hence, the intensity of the sunlight in W/m² is 650W/m²
Two point charges of +2.0 μC and -6.0 μC are located on the x-axis at x = -1.0 cm and x 12) = +2.0 cm respectively. Where should a third charge of +3.0-μC be placed on the +x-axis so that the potential at the origin is equal to zero?
Answer:
x = 0.006 m
Explanation:
The potential at one point is given by
V = k ∑ [tex]q_{i} / r_{i}[/tex]
remember that the potential is to scale, let's apply to our case
V = k (q₁ / x₁ + q₂ / x₂ + q₃ / x)
in this case they indicate that the potential is zero
0 = k (2 10⁻⁶ / (- 1 10⁻²) + (-6 10⁻⁶) / 2 10⁻² + 3 10⁻⁶ / x)
3 / x = + 2 / 10⁻² + 3 / 10⁻²
3 / x = 500
x = 3/500
x = 0.006 m
If the speed of a "cheetah" is 150 m / s. How long does it take to cover 800 m?
Answer:
5.33333... seconds
Explanation:
800 divided by 150 is equal to 5.33333... because it is per second that the cheetah moves at 150miles, the answer is 5.3333.....
explain why cups of soup at a take away kiosk are often sold in white polystrene cups with a lid to stop spillage
Answer:
polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.
the cup helps it become more insulated
which of the following is a physical change?
A. a newspaper burns when placed in a fire.
B.an iron chair rusts when left outside
C.a sample of water boils and releases gas.
D.a plant changes carbon dioxide and water into sugar
Which one of the conditions can cause a particle to move with uniform circular motion in a uniform magnetic field
Given that,
A particle to move with uniform circular motion in a uniform magnetic field.
Suppose, The conditions are,
(I). The charged particle has to be positive and it should be moving in a direction opposite to a uniform magnetic field.
(II). The charged particle should be moving parallel to the magnetic force and perpendicular to the magnetic field.
(III). The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.
We know that,
An particle to move with uniform circular motion.
Here, electric force is perpendicular to velocity of particle.
The electric field is defined as,
[tex]F_{c}=\dfrac{mv^2}{r}[/tex].....(I)
Suppose, there is magnetic field, if a charge moving with velocity and the magnetic field exerts a field.
The magnetic force is defined as,
[tex]F_{m}=qvB[/tex].....(II)
We need to find the magnetic field
Using equation (I) and (II)
[tex]F_{c}=F_{m}[/tex]
[tex]\dfrac{mv^2}{r}=qvB[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
Hence, The magnetic field should be uniform and charge particle should be moving perpendicular to the magnetic field.
(III) is correct option.
Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.
A wedge of glass of refractive index 1.64 has a silver coating on the bottom, as shown in the image attached below.
Determine the smallest distance x to a position where 450-nm light reflected from the top surface of the glass interferes constructively with light reflected from the silver coating on the bottom. The light changes phase when reflected at the silver coating.
Answer:
the smallest distance x = 2.74 × 10⁻³ m or 2.74 mm
Explanation:
From the given information:
The net phase change is zero because both the light ray reflecting from the air-glass surface and silver plate undergo a phase change of [tex]\dfrac{\lambda}{2}[/tex] , as such the condition for the constructive interference is:
nΔy = mλ
where;
n = refractive index
Δy = path length (inside the glass)
So, from the diagram;
[tex]\dfrac{y}{x}=\dfrac{10^{-5} \ m}{0.2 \ m}[/tex]
[tex]\dfrac{y}{x} = 5 \times 10^{-5}[/tex]
[tex]y = 5 \times 10^{-5} x[/tex]
Now;
Δy can now be = 2 ( 5 × 10⁻⁵ [tex]x[/tex])
Δy =1 × 10⁻⁴[tex]x[/tex]
From nΔy = mλ
n( 1 × 10⁻⁴[tex]x[/tex] ) = mλ
[tex]x = \dfrac{m \lambda}{n \times 1 \times 10^{-4} }[/tex]
when the thickness is minimum then m = 1
Thus;
[tex]x = \dfrac{1 \times 450 \times 10^{-9} \ m}{1.64 \times 1 \times 10^{-4} }[/tex]
x = 0.00274 m
x = 2.74 × 10⁻³ m or 2.74 mm
Answer: B. The surface of the coating is rough, so light that shines on it gets scattered in many directions.
Explanation: On Edge!!!!!!!!!!!!!!!!!!!!
An electron initially at rest is accelerated over a distance of 0.210 m in 33.3 ns. Assuming its acceleration is constant, what voltage was used to accelerate it
Answer:
V = 451.47 volts
Explanation:
Given that,
Distance, d = 0.21 m
Initial speed, u = 0
Time, t = 33.3 ns
Let v is the final velocity. Using second equation of motion as :
[tex]d=ut+\dfrac{1}{2}at^2[/tex]
a is acceleration, [tex]a=\dfrac{v-u}{t}[/tex] and u = 0
So,
[tex]d=\dfrac{1}{2}(v-u)t[/tex]
[tex]v=\dfrac{2d}{t}\\\\v=\dfrac{2\times 0.21}{33.3\times 10^{-9}}\\\\v=1.26\times 10^7\ m/s[/tex]
Now applying the conservation of energy i.e.
[tex]\dfrac{1}{2}mv^2=qV[/tex]
V is voltage
[tex]V=\dfrac{mv^2}{2q}\\\\V=\dfrac{9.1\times 10^{-31}\times (1.26\times 10^7)^2}{2\times 1.6\times 10^{-19}}\\\\V=451.47\ V[/tex]
So, the voltage is 451.47 V.
A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius 5 mm , and the outer one a radius 11 mm . The common length of the cylinders is 160 m . What is the potential energy stored in this capacitor when a potential difference 6 V is
Answer:
The potential energy is [tex]PE = 2.031 *10^{-7} \ J[/tex]
Explanation:
From the question we are told that
The inner radius is [tex]r_i = 5 \ mm = 0.005 \ m[/tex]
The outer radius is [tex]r_o = 11 \ mm = 0.011 \ m[/tex]
The common length is [tex]l = 160 \ m[/tex]
The potential difference is [tex]V = 6 \ V[/tex]
Generally the capacitance of the cylindrical capacitor is mathematically represented as
[tex]C = \frac{2 \pi * k * \epsilon_o }{ ln \frac{ r_o }{r_i} } * l[/tex]
Where [tex]\epsilon _o[/tex] is the permitivity of free space with the values [tex]\epsilon _o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]
and k is the dielectric constant of the dielectric material here the dielectric material is free space so k = 1
Substituting values
[tex]C = \frac{2* 3.142 * 1 * 8.85*10^{-12} }{ ln \frac{ 0.011}{0.005} } * 160[/tex]
[tex]C = 1.129 *10^{-8} \ F[/tex]
The potential energy stored is mathematically represented as
[tex]PE = \frac{1}{2} * C * V ^2[/tex]
substituting values
[tex]PE = 0.5 * 1.129 *10^{-8} * (6)^2[/tex]
[tex]PE = 2.031 *10^{-7} \ J[/tex]
Suppose a proton moves to the right and enters a uniform magnetic field into the page. It follows trajectory B with radius rp. An alpha particle (twice the charge and 4 times the mass) enters the same magnetic field in the same way and with the same velocity as the proton. Which path best represents the alpha particle’s trajectory?
Answer:
R = r_protón / 2
Explanation:
The alpha particle when entering the magnetic field experiences a force and with Newton's second law we can describe its movement
F = m a
Since the magnetic force is perpendicular, the acceleration is centripetal.
a = v² / R
the magnetic force is
F = q v x B = q v B sin θ
the field and the speed are perpendicular so the sin 90 = 1
we substitute
qv B = m v² / R
R = q v B / m v²
in the exercise they indicate
the charge q = 2 e
the mass m = 4 m_protón
R = 2e v B / 4m_protón v²
we refer the result to the movement of the proton
R = (e v B / m_proton) 1/2
the data in parentheses correspond to the radius of the proton's orbit
R = r_protón / 2
action and reaction are equal in magnitude and opposite in direction.Then why do they not balance each other.
Explanation:
Newton's third law of motion states that every action has an equal and opposite reaction. This means that forces always act in pairs. Action and reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel out.
Based on the graph below, what prediction can we make about the acceleration when the force is 0 newtons? A. It will be 0 meters per second per second. B. It will be 5 meters per second per second. C. It will be 10 meters per second per second. D. It will be 15 meters per second per second.
Answer:
Option A
Explanation:
From the graph, we came to know that Force and acceleration are in direct relationship.
Also,
Force = 0 when Acceleration = 0
Because Both are 0 at the origin.
Answer:
A. It will be 0 meters per second per second.
Explanation:
The force and acceleration is in a proportional relationship, that means the line goes through the origin.
On the graph, when the force is at 0, the acceleration is 0. The line passes through the origin.
Two charges, +9 µC and +16 µC, are fixed 1 m apart, with the second one to the right. Find the magnitude and direction of the net force (in N) on a −7 nC charge when placed at the following locations.
(a) halfway between the two
(b) half a meter to the left of the +9 µC charge
(c) half a meter above the +16 µC charge in a direction perpendicular to the line joining the two fixed charges (Assume this line is the x-axis with the +x-direction toward the right. Indicate the direction of the force in degrees counterclockwise from the +x-axis.)
Answer:
A) 1.76U×10⁻³N
B) 2.716×10⁻³N
C) 264.5⁰
Explanation:
See detailed workings for (a), (b), (c) attached.
A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm light from a He-Ne laser. Because the lecture hall is very large, the interference pattern will be projected on a wall that is 4.0 m from the slits. For easy viewing by all students in the class, the professor wants the distance between the m
What slit separation is required in order to produce the desired interference pattern?
d=________m
Note: if the professor wants the distance between the m = 0 and m = 1 maxima to be 25 cm
Answer:
d = 1.0128×10⁻⁵m
Explanation:
given:
length L = 4.0m
maximum distance between m = 0 and m = 1 , y = 25cm = 0.25m
wavelength λ = 633nm = 633×10⁻⁹m
note:
dsinθ = mλ (constructive interference)
where d is slit seperation, θ is angle of seperation , m is order of interference , and λ is wavelength
for small angle
sinθ ≈ tanθ
[tex]d (\frac{y}{L}) =[/tex] mλ
[tex]d (\frac{y}{L}) = (1)(633nm)[/tex]
[tex]d(\frac{0.25}{4} ) = (1)(633nm)[/tex]
d = 1.0128×10⁻⁵m
Which of the following statements about stages of nuclear burning (i.e., first-stage hydrogen burning, second-stage helium burning, etc.) in a massive star is not true?
A) As each stage ends, the core shrinks further.
B) Each successive stage of fusion requires higher temperatures than the previous stages.
C) Each successive stage lasts for approximately the same amount of time.
D) Each successive stage creates an element with a higher atomic weight.
Answer:
C) Each successive stage lasts for approximately the same amount of time.
Explanation:
Nuclear burning is a series of nuclear processes through which a star gets its energy. The energy within a star is due to nuclear fusion of lighter elements (hydrogen) into more massive element (helium), with a release of a large amount of energy due to the conversion of some of the mass into energy. Each stage leads to a loss of some of the mass which is converted into energy (option A is valid).
The fusion of four hydrogen atoms into one helium atom means that there is a creation of element with a higher atomic weight (option D is valid), and the energy output of each stage exceeds its energy input, meaning that each stage will require a higher temperature than its previous stages (option B is valid).
A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic
friction 0.50 and a coefficient of static friction 0.65. How much force is required to
get the block moving?
Answer:
The force is [tex]F = 172 \ N[/tex]
Explanation:
From the question we are told that
The mass of the block is [tex]m_b = 27.0 \ kg[/tex]
The coefficient of static friction is [tex]\mu_s = 0.65[/tex]
The coefficient of kinetic friction is [tex]\mu_k = 0.50[/tex]
The normal force acting on the block is
[tex]N = m * g[/tex]
substituting values
[tex]N = 27 * 9.8[/tex]
[tex]N = 294.6 \ N[/tex]
Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as
[tex]F_f = \mu_s * N[/tex]
substituting values
[tex]F_f = 0.65 * 264.6[/tex]
[tex]F_f = 172 \ N[/tex]
Now for this block to move the force require is equal to [tex]F_f[/tex] i.e
[tex]F= F_f[/tex]
=> [tex]F = 172 \ N[/tex]
A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns through an angle of 80 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the 4.0-s interval
Answer:
The time interval is [tex]t = 3 \ s[/tex]
Explanation:
From the question we are told that
The angular acceleration is [tex]\alpha = 4.0 \ rad/s^2[/tex]
The time taken is [tex]t = 4.0 \ s[/tex]
The angular displacement is [tex]\theta = 80 \ radians[/tex]
The angular displacement can be represented by the second equation of motion as shown below
[tex]\theta = w_i t + \frac{1}{2} \alpha t^2[/tex]
where [tex]w_i[/tex] is the initial velocity at the start of the 4 second interval
So substituting values
[tex]80 = w_i * 4 + 0.5 * 4.0 * (4^2)[/tex]
=> [tex]w_i = 12 \ rad/s[/tex]
Now considering this motion starting from the start point (that is rest ) we have
[tex]w__{4.0 }} = w__{0}} + \alpha * t[/tex]
Where [tex]w__{0}}[/tex] is the angular velocity at rest which is zero and [tex]w__{4}}[/tex] is the angular velocity after 4.0 second which is calculated as 12 rad/s s
[tex]12 = 0 + 4 t[/tex]
=> [tex]t = 3 \ s[/tex]
Following are the response to the given question:
Given:
[tex]\to \alpha = 4.0 \ \frac{rad}{s^2}\\\\[/tex]
[tex]\to \theta= 80\ radians\\\\\to t= 4.0 \ s\\\\ \to \theta_0=0\\[/tex]
To find:
[tex]\to \omega=?\\\\\to t=?\\\\[/tex]
Solution:
Using formula:
[tex]\to \theta- \theta_0 = w_{0} t+ \frac{1}{2} \alpha t^2\\\\ \to 80-0= \omega_{0}(4) + \frac{1}{2} (4)(4^2)\\\\ \to 80= \omega_{0}(4) + \frac{1}{2} (4)(16)\\\\\\to 80= \omega_{0}(4) + (4)(8)\\\\\to 80= \omega_{0}(4) + 32\\\\\to 80-32 = \omega_{0}(4) \\\\\to \omega_{0}(4)= 48 \\\\\to \omega_{0}= \frac{48}{4} \\\\ \to \omega_{0} = 12 \frac{rad}{ s} \\\\[/tex]
It would be the angle for rotation at the start of the 4-second interval.
This duration can be estimated by leveraging the fact that the wheel begins from rest.
[tex]\to \omega = \omega_{0} + \alpha t\\\\\to 12 = 0 +4(t) \\\\\to 12 = 4(t) \\\\ \to t=\frac{12}{4}\\\\\to t= 3\ s[/tex]
Therefore, the answer is "[tex]12\ \frac{rad}{s}[/tex] and [tex]3 \ s[/tex]".
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brainly.com/question/7464119
What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
c. The incident light must have at least as much energy as the electron work function
A scientist is testing the seismometer in his lab and has created an apparatus that mimics the motion of the earthquake felt in part (a) by attaching the test mass to a spring. If the test mass weighs 13 N, what should be the spring constant of the spring the scientist use to simulate the relative motion of the test mass and the ground from part (a)?
Complete Question
The complete question is shown on the first uploaded image
Answer:
a
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
b
[tex]k =722.2 \ N/m[/tex]
Explanation:
From the question we are told that
The amplitude is [tex]A = 1.8 \ cm = 0.018 \ m[/tex]
The period is [tex]T = 17 \ s[/tex]
The test weight is [tex]W = 13 \ N[/tex]
Generally the radial acceleration is mathematically represented as
[tex]a = w^2 r[/tex]
at maximum angular acceleration
[tex]r = A[/tex]
So
[tex]a_{max} = w^2 A[/tex]
Now [tex]w[/tex] is the angular velocity which is mathematically represented as
[tex]w = \frac{2 * \pi }{T}[/tex]
Therefore
[tex]a_{max} = [\frac{2 * \pi}{T} ]^2 * A[/tex]
substituting values
[tex]a_{max} = [\frac{2 * 3.142}{17} ]^2 * 0.018[/tex]
[tex]a_{max} = 0.00246 \ m/s^2[/tex]
Generally this test weight is mathematically represented as
[tex]W = k * A[/tex]
Where k is the spring constant
Therefore
[tex]k = \frac{W}{A}[/tex]
substituting values
[tex]k = \frac{13}{0.018}[/tex]
[tex]k =722.2 \ N/m[/tex]