Answer:
0.069 N, in the X directionExplanation:
According to Flemming's left hand rule, it sates that if the first three fingers of the left hand are held mutually at right angles to one another, the fore finger will point in the direction of magnetic field, the middle finger will point in direction of current, while the thumb will point to the direction of force.
Mathematically the law is stated as
F= BIL
given data
Magnetic field B= 0.43T
Current I= 4.9 A
length of conductor L= 3.3cm to meter , 3.3/100= 0.033 m
Applying the formula the force is calculated as
F= 0.43*4.9* 0.033= 0.069 N
According to Flemming's rule the direction of all parameters are mutually perpendicular to one another, then the Force is in the X direction
Determine the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
Answer:
T = 3.14 hours
Explanation:
We need to find the orbital period (in hours) of an observation satellite in a circular orbit 1,787 km above Mars.
We know that the radius of Mars is 3,389.5 km.
So, r = 1,787 + 3,389.5 = 5176.5 km
Using Kepler's law,
[tex]T^2=\dfrac{4\pi ^2}{GM}r^3[/tex]
M is mass of Mars, [tex]M=6.39\times 10^{23}\ kg[/tex]
So,
[tex]T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 6.39\times 10^{23}}\times (5176.5 \times 10^3)^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times6.39\times10^{23}}\times(5176.5\times10^{3})^{3}}\\\\T=11334.98\ s[/tex]
or
T = 3.14 hours
So, the orbital period is 3.14 hours
You shine unpolarized light with intensity 52.0 W/m2 on an ideal polarizer, and then the light that emerges from this polarizer falls on a second ideal polarizer. The light that emerges from the second polarizer has intensity 15.0 W/m2. Find the intensity of the light that emerges from the first polarizer.
Answer:
The intensity of light from the first polarizer is [tex]I_1 = 26 W/m^2[/tex]
Explanation:
The intensity of the unpolarized light is [tex]I_o = 52.0 \ W/m^2[/tex]
Generally the intensity of light that emerges from the first polarized light is
[tex]I_1 = \frac{I_o}{2 }[/tex]
substituting values
[tex]I_1 = \frac{52. 0}{2 }[/tex]
[tex]I_1 = 26 W/m^2[/tex]
The A block, with negligible dimensions and weight P, is supported by the coordinate point (1.1/2) of the parabolic fixed grounded surface, from equation y = x^2/2 If the block is about to slide, what is the coefficient of friction between it and the surface; determine the force F tangent to the surface, which must be applied to the block to start the upward movement.
Answer:
μ = 1
F = P√2
Explanation:
The parabola equation is: y = ½ x².
The slope of the tangent is dy/dx = x.
The angle between the tangent and the x-axis is θ = tan⁻¹(x).
At x = 1, θ = 45°.
Draw a free body diagram of the block. There are three forces:
Weight force P pulling down,
Normal force N pushing perpendicular to the surface,
and friction force Nμ pushing up tangential to the surface.
Sum of forces in the perpendicular direction:
∑F = ma
N − P cos 45° = 0
N = P cos 45°
Sum of forces in the tangential direction:
∑F = ma
Nμ − P sin 45° = 0
Nμ = P sin 45°
μ = P sin 45° / N
μ = tan 45°
μ = 1
Draw a new free body diagram. This time, friction force points down tangential to the surface, and applied force F pushes up tangential to the surface.
Sum of forces in the tangential direction:
∑F = ma
F − Nμ − P sin 45° = 0
F = Nμ + P sin 45°
F = (P cos 45°) μ + P sin 45°
F = P√2
A certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied. In approximately what distance would the car stop had it been going 66.0mph
Answer: 156.02 metre.
Explanation:
Give that a certain car traveling 33.0mph skids to a stop in 39m from the point where the brakes were applied.
Let us use third equation of motion,
V^2 = U^2 + 2as
Since the car is decelerating, V = 0
And acceleration a will be negative.
U = 33 mph
S = 39 m
Substitute both into the formula
0 = 33^2 - 2 × a × 39
0 = 1089 - 78a
78a = 1089
a = 1089 / 78
a = 13.96 m/h^2
If we assume that the car decelerate at the same rate.
the distance the car will stop had it been going 66.0mph will be achieved by using the same formula
V^2 = U^2 + 2as
0 = 66^2 - 2 × 13.96 × S
4356 = 27.92S
S = 4356 / 27.92
S = 156.02 m
Therefore, the car would stop at
156.02 m
A simple pendulum is 3.00 m long. (a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2? s (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2? s (c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2? s
Answer:
a,)3.042s
b)4.173s
c)3.281s
Explanation:
For a some pendulum the period in seconds T can be calculated using below formula
T=2π√(L/G)
Where L = length of pendulum in meters
G = gravitational acceleration = 9.8 m/s²
Then we are told to calculate
(a) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating upward at 3.00 m/s2?
Since oscillations for this pendulum is located in the elevator that is accelerating upward at 3.00 then
use G = 9.8 + 3.0 = 12.8 m/s²
Period T=2π√(L/G)
T= 2π√(3/12.8)
T=3.042s
b) (b) What is the period of small oscillations for this pendulum if it is located in an elevator accelerating downward at 3.00 m/s2?
G = 9.8 – 3.0 = 6.8 m/s²
T= 2π√(3/6.8)
T=4.173s
C)(c) What is the period of this pendulum if it is placed in a truck that is accelerating horizontally at 3.00 m/s2?
Net acceleration is
g'= √(g² + a²)
=√(9² + 3²)
Then period is
T=2π√(3/11)
T=3.281s
An electron is trapped between two large parallel charged plates of a capacitive system. The plates are separated by a distance of 1 cm and there is vacuum in the region between the plates. The electron is initially found midway between the plates with a kinetic energy of 11.2 eV and with its velocity directed toward the negative plate. How close to the negative plate will the electron get if the potential difference between the plates is 100 V? (1 eV = 1.6 x 10-19 J)
Answer:
The electron will get at about 0.388 cm (about 4 mm) from the negative plate before stopping.
Explanation:
Recall that the Electric field is constant inside the parallel plates, and therefore the acceleration the electron feels is constant everywhere inside the parallel plates, so we can examine its motion using kinematics of a constantly accelerated particle. This constant acceleration is (based on Newton's 2nd Law:
[tex]F=m\,a\\q\,E=m\,a\\a=\frac{q\,E}{m}[/tex]
and since the electric field E in between parallel plates separated a distance d and under a potential difference [tex]\Delta V[/tex], is given by:
[tex]E=\frac{\Delta\,V}{d}[/tex]
then :
[tex]a=\frac{q\,\Delta V}{m\,d}[/tex]
We want to find when the particle reaches velocity zero via kinematics:
[tex]v=v_0-a\,t\\0=v_0-a\,t\\t=v_0/a[/tex]
We replace this time (t) in the kinematic equation for the particle displacement:
[tex]\Delta y=v_0\,(t)-\frac{1}{2} a\,t^2\\\Delta y=v_0\,(\frac{v_0}{a} )-\frac{a}{2} (\frac{v_0}{a} )^2\\\Delta y=\frac{1}{2} \frac{v_0^2}{a}[/tex]
Replacing the values with the information given, converting the distance d into meters (0.01 m), using [tex]\Delta V=100\,V[/tex], and the electron's kinetic energy:
[tex]\frac{1}{2} \,m\,v_0^2= (11.2)\,\, 1.6\,\,10^{-19}\,\,J[/tex]
we get:
[tex]\Delta\,y= \frac{1}{2} v_0^2\,\frac{m (0.01)}{q\,(100)} =11.2 (1.6\,\,10^{-19})\,\frac{0.01}{(1.6\,\,10^{-19})\,(100)}=\frac{11.2}{10000} \,meters=0.00112\,\,meters[/tex]Therefore, since the electron was initially at 0.5 cm (0.005 m) from the negative plate, the closest it gets to this plate is:
0.005 - 0.00112 m = 0.00388 m [or 0.388 cm]
Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?
Answer:
Explanation:
This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.
If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).
But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).
A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".
A hot cup of coffee is placed on a table. Which will happen because of conduction? Answer options with 4 options A. The temperature of the coffee will decrease while the temperature of the table decreases. B. The temperature of the coffee will increase while the temperature of the table increases. C. The temperature of the coffee will decrease while the temperature of the table increases. D. The temperature of the coffee will increase while the temperature of the table decreases.
Answer:
C.Explanation:
C. The temperature of the coffee will decrease while the temperature of the table increases.
A solid block is attached to a spring scale. When the block is suspended in air, the scale reads 21.2 N; when it is completely immersed in water, the scale reads 18. 2 N. What are the volume and density of the block?
Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
Answer:
The volume of the block is 306 cm³
The density of the block is 7.07 g/cm³
Explanation:
Given;
weight of block in air, [tex]W_a[/tex] = 21.2 N
Weight of block in water, [tex]W_w[/tex] = 18.2 N
Mass of the block in air;
[tex]W_a = mg[/tex]
21.2 = m x 9.8
m = 21.2 / 9.8
m = 2.163 kg
mass of the block in water;
[tex]W_w = mg[/tex]
18.2 = m x 9.8
m = 18.2 / 9.8
m = 1.857 kg
Apply Archimedes principle
Mass of object in air - mass of object in water = density of water x volume of object
2.163 kg - 1.857 kg = 1000 kg/m³ x Volume of block
0.306 kg = 1000 kg/m³ x Volume of block
Volume of the block = [tex]\frac{0.306 \ kg}{1000 \ kg/m^3}[/tex]
Volume of the block = 3.06 x 10⁻⁴ m³
Volume of the block = 306 cm³
Determine the density of the block
[tex]Density = \frac{mass}{volume} \\\\Density =\frac{2163 \ g}{306 \ cm^3} \\\\Density = 7.07 \ g/cm^3[/tex]
You are performing an experiment that requires the highest-possible magnetic energy density in the interior of a very long current-carrying solenoid. Which of the following adjustments increases the energy density?a. Increasing only the length of the solenold while keeping the turns per unit lengh flxed. b. Increasing the number of turns per unit length on the solenold. c. Increasing the cross-sectional area of the solenoid. d. None of these. e. Increasing the current in the solenoid.
Answer:
The correct choice is B & E.
Explanation:
A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.
Cheers!
A system of four particles moves along a dimension. The center of mass is at rest, and the particles do not interact with any objects outside of the system. Find the velocity of v4 at t=2.83 seconds given the details for the motion of particles 1,2,3
Answer:
v = - 14.08 m / s
Explanation:
The definition of center of mass is
[tex]x_{cm}[/tex] = 1 /M ∑sun [tex]x_{i} m_{i}[/tex]
where M is the total mass of the system and [tex]x_{i}[/tex] and [tex]m_{i}[/tex] are the position and mass of each component.
The velocity of the center of mass can be found by deriving this expression with respect to time
[tex]v_{cm}[/tex] = 1 / M ∑ m_{i} [tex]v_{i}[/tex] vi
let's find the total mass
M = m₁ + m₂ + m₃ + m₄
M = 1.45 + 2.81 +3.89 + 5.03
m = 13.18 kg
let us substitute in the velocity of the center of mass [tex]v_{cm}[/tex] = 0
0 = 13.18 (m₁ v₁ + m₂ v₂ + m₃v₃ + m₄v₄)
v₄ = - (m₁ v₁ + m₂ v₂ + m₃v₃) / m₄
let's substitute the given values
v₄ = -[1.45 (6.09 +0.299 t) +2.81 (7.83 + 0.357t) +3.89 (8.09 + 0.405 t)] / 5.03
They ask us for the calculations for a time t = 2.83 s
v₄ = - [8.8305 + 1.227 + 22.00 + 2.839 + 31.47 +4.4585] / 5.03
v = - 14.08 m / s
The velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.
The given parameters;
[tex]m_1 = 1.45 \ kg, \ \ v_1(t) = (6.09 \ m/s) + (0.299 \ m/s^2)\times t\\\\m_2 = 2.81 \ kg, \ \ v_2(t) = (7.83 \ m/s) + (0.357 \ m/s^2)\times t \\\\m_3 = 3.89 \ kg, \ \ v_3(t) = (8.09 \ m/s) + (0.405 \ m/s^2)\times t\\\\m_4 = 5.03 \ kg[/tex]
The velocity of the center mass of the particles is calculated as;
[tex]M_{cm}V_{cm} = m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4\\\\V_{cm} = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}} \\\\0 = \frac{m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4}{M_{cm}}\\\\m_1v_1 + m_2 v_2 + m_3v_3 + m_4v_4 = 0\\\\m_4v_4 = -(m_1v_1 + m_2 v_2 + m_3v_3)\\\\v_4 = \frac{-(m_1v_1 + m_2 v_2 + m_3v_3)}{m_4}[/tex]
The velocity of particle 1 at time, t = 2.83 s;
[tex]v_1 = 6.09 \ + \ 0.299\times 2.83\\\\v_1 = 6.94 \ m/s[/tex]
The velocity of particle 2 at time, t = 2.83 s;
[tex]v_2 = 7.83\ + \ 0.357\times 2.83\\\\v_2 = 8.84 \ m/s[/tex]
The velocity of particle 3 at time, t = 2.83 s;
[tex]v_3 = 8.09\ + \ 0.405 \times 2.83\\\\v_3 = 9.24 \ m/s[/tex]
The velocity of the particle 4 at time, t = 2.83 s;
[tex]v_4 = \frac{-(m_1v_1 + m_2v_2 + m_3v_3)}{m_4} \\\\v_4 = \frac{-(1.45\times 6.94\ + \ 2.81\times 8.84\ + \ 3.89 \times 9.24)}{5.03} \\\\v_4 = -14 .1 \ m/s[/tex]
Thus, the velocity of the particle 4 at time, t = 2.83 s, is -14.1 m/s.
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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]
A charged particle moving through a magnetic field at right angles to the field with a speed of 25.7 m/s experiences a magnetic force of 2.98 10-4 N. Determine the magnetic force on an identical particle when it travels through the same magnetic field with a speed of 4.64 m/s at an angle of 29.2° relative to the magnetic field.
Answer:
The magnetic force would be:
[tex]F\approx 2.625\,\,10^{-5}\,\,N[/tex]
Explanation:
Recall that the magnetic force on a charged particle (of charge q) moving with velocity (v) in a magnetic field B, is given by the vector product:
F = q v x B
(where the bold represents vectors)
the vector product involves the sine of the angle ([tex]\theta[/tex]) between the vectors, so we can write the relationship between the magnitudes of these quantities as:
[tex]F=q\,v\,B\,sin(\theta)[/tex]
Therefore replacing the known quantities for the first case:
[tex]F=q\,v\,B\,sin(\theta)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\,sin(90^o)\\2.98\,\,10^{-4} \,\,N=q\,(25.7\,\,m/s)\,B\\q\,\,B=\frac{2.98\,\,10^{-4} }{25.7} \,\frac{N\,\,s}{m}[/tex]
Now, for the second case, we can find the force by using this expression for the product of the particle's charge times the magnetic field, and the new velocity and angle:
[tex]F=q\,v\,B\,sin(\theta)\\F=q\,(4.64\,\,m/s)\,B\,sin(29.2^o)\\F=q\,B(4.64\,\,m/s)\,\,sin(29.2^o)\\F=\frac{2.98\,\,10^{-4} }{25.7} \,(4.64\,\,m/s)\,\,sin(29.2^o)\\F\approx 2.625\,\,10^{-5}\,\,N[/tex]
: A spaceship is traveling at the speed 2t 2 1 km/s (t is time in seconds). It is pointing directly away from earth and at time t 0 it is 1000 kilometers from earth. How far from earth is it at one minute from time t 0
Answer:
145060km
Explanation: Given that
speed = dx/dt = 2t^2 +1
integrate
x = 2/3t^3 + t + c (c is constant, x is in km, t is in second)
given that at t=0, x = 1000
so 1000 = 2/3 X (0)^3 + 0 + c
or c = 1000
So x = 2/3t^3 + t + 1000
for t = 1 min = 60s
x = 2/3 X 60^3 + 60 + 1000
x = 2/3×216000+ 1060
x = 144000+1060
= 145060km
At one minute, it will be 145060km far from the earth
You have two capacitors and want to connect them across a voltage source (battery) to store the maximum amount of energy. Should they be connected in series or in parallel?
Answer:
In parallel
Explanation:
Ctotal = C1 + C2 + ... + Cn
If a transformer has 50 turns in the primary winding and 10 turns on the secondary winding, what is the reflected resistance in the primary if the secondary load resistance is 250 W?
Answer:
The reflected resistance in the primary winding is 6250 Ω
Explanation:
Given;
number of turns in the primary winding, [tex]N_P[/tex] = 50 turns
number of turns in the secondary winding, [tex]N_S[/tex] = 10 turns
the secondary load resistance, [tex]R_S[/tex] = 250 Ω
Determine the turns ratio;
[tex]K = \frac{N_P}{N_S} \\\\K = \frac{50}{10} \\\\K = 5[/tex]
Now, determine the reflected resistance in the primary winding;
[tex]\frac{R_P}{R_S} = K^2\\\\R_P = R_SK^2\\\\R_P = 250(5)^2\\\\R_P = 6250 \ Ohms[/tex]
Therefore, the reflected resistance in the primary winding is 6250 Ω
Six automobiles are initially traveling at the indicated velocities. The automobiles have different masses and velocities. The drivers step on the brakes and all automobiles are brought to rest.Automobile 1: 500kg, 10m/sAutomobile 2: 2000kg, 5m/sAutomobile 3: 500kg, 20m/sAutomobile 4: 1000kg, 20m/sAutomobile 5: 1000kg, 10m/sAutomobile 6: 4000kg, 5m/sRequired:a. Rank these automobiles based on the magnitude of their momentum before the brakes are applied, from largest to smallest.b. Rank these automobiles based on the magnitude of the impulse needed to stop them, from largest to smallest.c. Rank the automobiles based on the magnitude of the force needed to stop them, from largest to smallest.
Answer:
A. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
B. largest: (4000 kg, 5 m/s; 1000 kg, 20 m/s)
medium: (2000 kg, 5 m/s; 500 kg, 20 m/s; 1000 kg, 10 m/s)
smallest: (500 kg, 10 m/s)
C. You can't say anything about the forces required until we know about the time frames required for each one to stop. So If they all stopped in the same time interval, then the rankings are the same.
"Can we consider light wave as a single frequency wave? Either Yes or No, explain the reason of your answer. "
Answer:
Well, yes.
We can have an isolated light wave that is defined by only one frequency (and one wavelenght). But this is not a really common situation, most of the light that we can see in nature, is actually a composition of different waves with different frequencies.
Even if we have, for example, a red laser, the actual frequency of the light that comes from the laser may be in a range of frequencies, so the actual wave is a composition of different waves with really close frequencies.
An example of a light wave defined by only one frequency can be, for example, the photon that comes out of a change in energy of an electron.
Here we have a single photon, with a single frequency, that is modeled as a single frequency wave.
What is the displacement of the object after 3 seconds?
Answer:
3 meters
Explanation:
Distinguish between physical and chemical changes. Include examples in your explanations.
Answer:
Chemical changes are recognized when a substance changes its properties permanently and it cannot be the same substance as before.
Instead the physical changes implies that if you can return to the same substance through a reverse process.
Explanation:
A chemical change is, by example, a combustion, if a sheet of paper burns, its result is ashes, the ashes cannot go back to being a sheet of paper because its properties changed, heat energy was generated that changed matter permanently.
A physical change, by example, is that of freezing water, the water becomes ice, but this can easily become water again if the temperature is increased, its properties do not change and the chemistry of the substance does not change.
A chemical change takes place when a chemical reaction takes place, while when a matter changes forms but not the chemical identity then a physical change takes place.
• A product or a new compound formation takes place from a chemical change as the rearrangement of atoms takes place to produce novel chemical bonds.
• In a chemical change always a chemical reaction takes place.
• Some of the chemical changes examples are souring milk, burning wood, digesting food, mixing acid and base, cooking food, etc.
• In a physical change no new chemical species form.
• The changing of the state of a pure substance between liquid, gas, or solid is a physical change as there is no change in the identity of the matter.
• Some of the physical changes are melting of ice, tempering of steel, breaking a bottle, crumpling a sheet of aluminum foil, boiling water, and shredding paper.
Thus, a new substance is formed during a chemical change, while a physical change does not give rise to a new substance.
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Each proton-proton cycle generates 26.7 MeV of energy. If 9.9 Watts are generated via by the proton-proton cycle, how many billion neutrinos are produced
Answer:
4.635 *10^12 Neutrinos
Explanation:
Here in this question, we are to determine the number of neutrinos in billions produced, given the power generated by the proton-proton cycle.
We proceed as follows;
In proton-proton cycle generates 26.7 MeV of energy and in this cycle two neutrinos are produced.
From the question, we are given that
Power P = 9.9 watts = 9.9 J/s
Watts is same as J/s
The number of proton-proton cycles required to generate E energy is N = E / E '
Where E ' = Energy generated in proton-proton cycle which is given as 26.7 Mev in the question
Converting Mev to J, we have
= 26.7 x1.6 x10 -13 J
To get the number N which is the number of proton-proton cycle required, we have;
N = 9.9 /(26.7 x1.6 x10^-13) = 2.32 * 10^12
Since we have two proton cycles( proton-proton), it automatically means 2 neutrinos will be produced.
Therefore number of neutrions produced = 2 x Number of proton-proton cycles = 2 * 2.32 * 10^12 = 4.635 * 10^12 neutrinos
which category would a person who has an IQ of 84 belong ?
Equal currents of magnitude I travel into the page in wire M and out of the page in wire N. The direction of the magnetic field at point P which is at the same distance from both wires is
Answer:
The direction of the magnetic field on point P, equidistant from both wires, and having equal magnitude of current flowing through them will be pointed perpendicularly away from the direction of the wires.
Explanation:
Using the right hand grip, the direction of the magnet field on the wire M is counterclockwise, and the direction of the magnetic field on wire N is clockwise. Using this ideas, we can see that the magnetic flux of both field due to the currents of the same magnitude through both wires, acting on a particle P equidistant from both wires will act in a direction perpendicularly away from both wires.
How long will it take a spacecraft travelling at 99% the speed of light (gamma = 7) to reach
the star Sirius which is 8.6 light-years away according to people on Earth ? How long will it
take according to the crew of the ship?
Answer:
The time taken is [tex]t = 2.739 *10^{8} \ s[/tex]
Explanation:
From the question we are told that
The speed of the spacecraft is [tex]v = 0.99c[/tex]
where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]
=> [tex]v = 0.99 * 3.0 *10^{8 } = 2.97*10^{8}\ m/s[/tex]
The distance of Sirius is [tex]d = 8.6 \ light-years = 8.6 * 9.461*10^{15}= 8.135*10^{16} \ m[/tex]
Generally the time taken is mathematically represented as
[tex]t = \frac{d}{v}[/tex]
substituting values
[tex]t = \frac{8.136 *10^{16}}{2.97 *10^{8}}[/tex]
[tex]t = 2.739 *10^{8} \ s[/tex]
If you stood on a planet having a mass four times higher than Earth's mass, and a radius two times 70) lon longer than Earth's radius, you would weigh:________
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth.
CHECK COMPLETE QUESTION BELOW
you stood on a planet having a mass four times that of earth mass and a radius two times of earth radius , you would weigh?
A) four times more than you do on Earth.
B) two times less than you do on Earth.
C) the same as you do on Earth
D) two times more than you do on Earth
Answer:
OPTION C is correct
The same as you do on Earth
Explanation :
According to law of gravitation :
F=GMm/R^2......(a)
F= mg.....(b)
M= mass of earth
m = mass of the person
R = radius of the earth
From law of motion
Put equation b into equation a
mg=GMm/R^2
g=GMm/R^2
g=GM/R^2
We know from question a planet having a mass four times that of earth mass and a radius two times of earth radius if we substitute we have
m= 4M
r=(2R)^2=4R^2
g= G4M/4R^2
Then, 4in the denominator will cancel out the numerator we have
g= GM/R^2
Therefore, g remain the same
A cyclist moves effortlessly at a constant speed of 12 m / s, but enters a muddy area where the coefficient of kinetic friction is 0.60. Will the rider leave the muddy area without having to pedal if the mud extends 11m? If so, how fast will it emerge?
Answer:
3.5 m/s
Explanation:
There are 3 forces on the cyclist:
Weight force mg pulling down,
Normal force N pushing up,
and friction force Nμ pushing left.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
∑F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
a = -(10 m/s²)(0.60)
a = -6 m/s²
Velocity reached at end of 11 m:
v² = v₀² + 2aΔx
v² = (12 m/s)² + 2 (-6 m/s²) (11 m)
v = √12 m/s
v ≈ 3.5 m/s
Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (Assume the small-angle approximation is valid here.)
(a) If the distance between the central bright fringe and either of the adjacent bright fringes is 1.97 cm, find the wavelength of the incident light.
(b) At what angle does the next set of bright fringes appear?
Answer:
a
[tex]\lambda = 1.667 nm[/tex]
b
[tex]\theta = 0.8681^o[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 0.220 \ mm = 0.00022 \ m[/tex]
The is distance of the screen from the slit is [tex]D = 2.60 \ m[/tex]
The distance between the central bright fringe and either of the adjacent bright [tex]y = 1.97 cm = 1.97 *10^{-2}\ m[/tex]
Generally the condition for constructive interference is
[tex]d sin \tha(\theta ) = n \lambda[/tex]
From the question we are told that small-angle approximation is valid here.
So [tex]sin (\theta ) = \theta[/tex]
=> [tex]d \theta = n \lambda[/tex]
=> [tex]\theta = \frac{n * \lambda }{d }[/tex]
Here n is the order of maxima and the value is n = 1 because we are considering the central bright fringe and either of the adjacent bright fringes
Generally the distance between the central bright fringe and either of the adjacent bright is mathematically represented as
[tex]y = D * sin (\theta )[/tex]
From the question we are told that small-angle approximation is valid here.
So
[tex]y = D * \theta[/tex]
=> [tex]\theta = \frac{ y}{D}[/tex]
So
[tex]\frac{n * \lambda }{d } = \frac{y}{D}[/tex]
[tex]\lambda =\frac{d * y }{n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }[/tex]
[tex]\lambda = 1.667 *10^{-6}[/tex]
[tex]\lambda = 1.667 nm[/tex]
In the b part of the question we are considering the next set of bright fringe so n= 2
Hence
[tex]dsin (\theta ) = n \lambda[/tex]
[tex]\theta = sin^{-1}[\frac{ n * \lambda }{d} ][/tex]
[tex]\theta = sin^{-1}[\frac{ 2 * 1667 *10^{-9}}{ 0.00022} ][/tex]
[tex]\theta = 0.8681^o[/tex]
A large reflecting telescope has an objective mirror with a 14.0 m radius of curvature. What angular magnification in multiples does it produce when a 3.25 m focal length eyepiece is used? ✕
Answer:
The magnification is [tex]m = -2.15[/tex]
Explanation:
From the question we are told that
The radius is [tex]r = 14.0 \ m[/tex]
The focal length eyepiece is [tex]f_e = 3.25 \ m[/tex]
Generally the objective focal length is mathematically represented as
[tex]f_o = \frac{r}{2}[/tex]
=> [tex]f_o = \frac{14}{2}[/tex]
=> [tex]f_o = 7 \ m[/tex]
The magnification is mathematically represented as
[tex]m = - \frac{f_o }{f_e }[/tex]
=> [tex]m = - \frac{7 }{ 3.25 }[/tex]
=> [tex]m = -2.15[/tex]
The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter and 1.2 m long. The starter motor runs for 0.760 s until the car engine starts.Required:a. How much charge passes through the starter motor? b. How far does an electron travel along the wire while the starter motor is on?(mm)
Answer:
(a)106.4C
b)0.5676mm
Explanation:
(a)To get the charge that have passed through the starter then The current will be multiplied by the duration
I= current
t= time taken
Q= required charge
Q= I*t = 140*0.760 = 106.C
(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)
diameter of the conductor is 4.20 mm
But Radius= diameter/2= 4.20/2=
The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then
radius of the conductor is 0.0021m.
We can now calculate the area of the conductor which is
A = π*r^2
= π*(0.0021)^2 = 13.85*10^-6 m^2
We can proceed to calculate the current density below
J = 140/13.85*10^-6 = 10108303A/m
According to the listed reference:
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s
Therefore , the distance traveled is:
x = v*t = 0.7468 * 0.760 = 0.5676mm
(a) The charge passes through the starter motor is 106.4C.
(b) An electron travel along the wire while the starter motor is on 0.5676mm.
ElectronAnswer (a)
I= current
t= time taken
Q= required charge
Q= I*t
Q= 140*0.760
Q= 106.C
Answer (b)
The n electron travel along the wire while the starter motor is on:
Diameter of the conductor is 4.20 mm
Radius= diameter/2= 4.20/2
Radius =2.1mm
Radius of the conductor is 0.0021m.
A = π*r^2
A= π*(0.0021)^2
A= 13.85*10^-6 m^2
Where e= 1.6*10^-19
n= 8.46*10^28
Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 )
Vd =0.0007468m/s
Vd =0 .7468 mm/s
The distance traveled is:
x = v*t
x= 0.7468 * 0.760
x = 0.5676mm
Learn more about "Electron":
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An astronomer is measuring the electromagnetic radiation emitted by two stars, both of which are assumed to be perfect blackbody emitters. For each star she makes a plot of the radiation intensity per unit wavelength as a function of wavelength. She notices that the curve for star A has a maximum that occurs at a shorter wavelength than does the curve for star B. What can she conclude about the surface temperatures of the two stars
Answer:
Star A has a higher surface temperature than star B.
Explanation:
The effective temperature of a star can be determined by means of its spectrum and Wien's displacement law:
[tex]T = \frac{2.898x10^{-3} m. K}{\lambda max}[/tex] (1)
Where T is the effective temperature of the star and [tex]\lambda_{max}[/tex] is the maximum peak of emission.
A body that is hot enough emits light as a consequence of its temperature. For example, if an iron bar is put in contact with fire, it will start to change colors as the temperature increase, until it gets to a blue color, that scenario is known as Wien's displacement law. Which establishes that the peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase and higher wavelengths as the temperature decreases.
Therefore, star A has a higher surface temperature than star B, as it is shown in equation 1 since T and [tex]\lambda max[/tex] are inversely proportional.
