Answer:
y = 10.2 m
Explanation:
It is given that,
Charge, [tex]q_1=-3\ nC[/tex]
It is placed at a distance of 9 cm at x axis
Charge, [tex]q_2=+4\ nC[/tex]
It is placed at a distance of 16 cm at x axis
We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,
[tex]\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0[/tex]
Here,
[tex]r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}[/tex]
So,
[tex]\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}[/tex]
Squaring both sides,
[tex]3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m[/tex]
So, at a distance of 10.2 m on the y axis the electric potential equals 0.
According to the question,
Charge,
[tex]q_1 = -3 \ nC[/tex] (9 cm at x-axis)[tex]q_2 = +4 \ nC[/tex] (16 cm at x-axis)Now,
→ [tex]\frac{kq_1}{r_1} +\frac{kq_2}{r_2} =0[/tex]
or,
→ [tex]\frac{kq_1}{r_1} =-\frac{kq_2}{r_2}[/tex]
→ [tex]\frac{q_1}{r_1} = \frac{q_2}{r_2}[/tex]
here,
[tex]r_1 = \sqrt{y^2+81}[/tex]
[tex]r^2 = \sqrt{y^2+225}[/tex]
By substituting the values,
→ [tex]\frac{-3 }{\sqrt{y^2+225} } = -\frac{4}{\sqrt{y^2+225} }[/tex]
By applying cross-multiplication,
[tex]3\times \sqrt{y^2+225} = 4\times \sqrt{y^2+81}[/tex]
By squaring both sides, we get
→ [tex]9(y^2+225) = 16(y^2+81)[/tex]
[tex]9y^2+2025 = 16 y^2+1296[/tex]
[tex]2025-1296=7y^2[/tex]
[tex]7y^2=729[/tex]
[tex]y = 10.2 \ m[/tex]
Thus the solution above is correct.
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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).
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.
Determine the position in the oscillation where an object in simple harmonic motion: (Be very specific, and give some reasoning to your answer.) has the greatest speed has the greatest acceleration experiences the greatest restoring force experiences zero restoring force g
Answer:
Explanation:
The greatest speed is attained at middle point or equilibrium point or where displacement from equilibrium point is zero .
When the object remains at one of the extreme point it experiences greatest acceleration but at that point velocity is zero . Due to acceleration , its velocity goes on increasing till it come to equilibrium point . At this point acceleration becomes zero . After that its velocity starts decreasing because of negative acceleration . Hence at middle point velocity is maximum .
The greatest acceleration is attained at maximum displacement or at one of the two extreme end .
Greatest restoring force too will be at position where acceleration is maximum because acceleration is produced by restoring force .
Restoring force is proportional to displacement or extension against restoring force . So it will be maximum when displacement is maximum .
Zero restoring force exists at equilibrium position or middle point or at point where displacement is zero . It is so because acceleration at that point is zero .
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]
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]
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
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.....
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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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
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]
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.
You set two parallel slits 0.1 mm apart at a distance of 2 m from a screen and illuminate them with light of wavelength 450 nm. The distance between a bright spot in the interference pattern and the dark spot adjacent to it is Group of answer choices
Answer:
Δx = 9 x 10⁻³ m = 9 mm
Explanation:
The formula for fringe spacing in Young's Double Slit Experiment is given as follows:
Δx = λL/d
where,
Δx = fringe spacing = ?
λ = wavelength of light = 450 nm = 450 x 10⁻⁹ m
L = Distance between slits and screen = 2 m
d = distance between slits = 0.1 mm = 0.1 x 10⁻³ m
Therefore,
Δx = (450 x 10⁻⁹ m)(2 m)/(0.1 x 10⁻³ m)
Δx = 9 x 10⁻³ m = 9 mm
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 inventor claims to have developed a food freezer that, in steady-state conditions, requires a power input of 0.25 kW to extract energy by heat transfer at a rate of 3050 J/s from the freezer contents, which are at a temperature of 270 K. Determine if this claim is real considering an ambient temperature of 293 K. (a) Can the freezer operation in such conditions
Answer:
The inventors claim is not real
a) No the the freezer cannot operate in such conditions
Explanation:
From the question we are told that
The power input is [tex]P_i = 0.25 kW = 0.25 *10^{3} \ W[/tex]
The rate of heat transfer [tex]J = 3050 J/s[/tex]
The temperature of the freezer content is [tex]T = 270 \ K[/tex]
The ambient temperature is [tex]T_a = 293 \ K[/tex]
Generally the coefficient of performance of a refrigerator at idea conditions is mathematically represented as
[tex]COP = \frac{T }{Ta - T}[/tex]
substituting values
[tex]COP = \frac{270 }{293 - 270}[/tex]
[tex]COP =11.7[/tex]
Generally the coefficient of performance of a refrigerator at real conditions is mathematically represented as
[tex]COP = \frac{J}{P_i}[/tex]
substituting values
[tex]COP = \frac{3050}{0.25 *10^{3}}[/tex]
[tex]COP = 12.2[/tex]
Now given that the COP of an ideal refrigerator is less that that of a real refrigerator then the claims of the inventor is rejected
This is because the there are loss in the real refrigerator cycle that are suppose to reduce the COP compared to an ideal refrigerator cycle where there no loss that will reduce the COP
A skater spins at 3rev/s when she stretches her arms outward. If she keeps her fists on her chest she can spin at 4.5rev/s and her body inertia is 3kg.m2. What is her body inertia when she stretches her arms outward?
Answer:
Body inertia I = 4.5 kg/m^2
Explanation:
Here, we want to calculate the body inertia when the arms are stretched outwards.
We know from the question that angular momentum is conserved
Thus;
I * 3 = 4.5 * 3
I = 4.5 kg/m^2
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
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 .
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
A car is moving along a road at 28.0 m/s with an engine that exerts a force of
2,300.0 N on the car to balance the drag and friction so that the car maintains a
constant speed. What is the power output of the engine?
Answer:
Power = Force × Distance/time
Power = Force × Velocity
Power = 2,300.0 N × 28.0 m/s²
Power = 64400 Nm/s
Explanation:
First show the formula of Power
Re-arrange formula and used to work out Power
Pretty simple stuff!
Hope this Helps!!
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]
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
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
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.
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
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
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
The 2-Mg truck is traveling at 15 m/s when the brakes on all its wheels are applied, causing it to skid for 10 m before coming to rest. The total mass of the boat and trailer is 1 Mg. Determine the constant horizontal force developed in the coupling C, and the friction force developed between the tires of the truck and the road during this time.
Answer:
constant horizontal force developed in the coupling C = 11.25KN
the friction force developed between the tires of the truck and the road during this time is 33.75KN
Explanation:
See attached file
The friction force between the tires of the truck and the road is 22500 N.
Calculating the friction force:It is given that a 2 Mg truck ( m = 2000 Kg) is initially moving with a speed of u = 15 m/s.
Distance traveled before coming to rest, s = 10m
The final velocity of the truck will be zero, v = 0
When the breaks are applied, only the frictional force is acting on the truck and it is opposite to the motion of the truck.
The frictional force is given by:
f = -ma
the acceleration of the truck = -a
The negative sign indicates that the acceleration is opposite to the motion.
Applying the third equation of motion we get:
v² = u² -2as
0 = 15² - 2×a×10
225 = 20a
a = 11.25 m/s²
So the magnitude of frictional force is:
f = ma = 2000 × 11.25 N
f = 22500 N
Learn more about friction force:
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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 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
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.