Answer:
F = 23.1 10⁻¹⁴ N, points inside the page
Explanation:
The magnetic force is given by the expression
F = q v x B
bold letters indicate vectors, the magnitude of this formula is
F = q v B sin θ
in this case the magnetic field is in the east direction and the particle moves north, so the angle is 90 and the sine 90 = 1
F = q v B
F = 2 1.6 10⁻¹⁹ 3.8 10⁵ 1.9
F = 23.1 10⁻¹⁴ N
the direction of the force is given by the right hand rule, for a positive charge
the thumb points in the directional of the speed, in this case towards the North
the other fingers extended in the direction of the magnetic field, in this case to the East
the palm points in the direction of the force, consequently it points inside the page,
why doping method is used to design a diode circuit
Answer:
To increase the conductivity of the material.
Explanation:
Generally , the group 4 elements are non conductor but in certain conditions, such as doping or the increase in temperature, they becomes conductor.
The doping is the process of mixing of pentavalent or the trivalent material into tetra valent material in the very small amount, so that the material becomes conductor.
In making a diode we need two types of the materials, n type semiconductor and p type semi conductor.
When the trivalent impurity is added in the tetra valent element, the semiconductor becomes n type because an electron is left for the conduction.
When the pentavalent impurity is added in the tetra valent element, the semiconductor becomes p type because a hole is left for the conduction.
Discuss the chemical bond exist in silicon crystal?
Air in a thundercloud expands as it rises. If its initial temperature is 292 K and no energy is lost by thermal conduction on expansion, what is its temperature when the initial volume has tripled
Answer:
Explanation:
It is a case of adiabatic expansion .
[tex]T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}[/tex]
T₁ , T₂ are initial and final temperature , V₁ and V₂ are initial and final volume.
Given ,
V₂ = 3 V₁ and T₁ = 292 . γ for air is 1.4 .
[tex]( 3 )^{\gamma-1}= \frac{292}{ T_2}[/tex]
[tex]( 3 )^{1.4-1}= \frac{292}{ T_2}[/tex]
1.552 = 292 / T₂
T₂ = 188 K .
One coulomb represents how many electrons?
a. 1 electron
b. 100 electrons
C. 6.25 quintillion electrons
d. 6.25 million-million electrons
e, none of the above
Answer:
6.24 x 1018 electrons.
Explanation:
So I think C
A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B particles with a combined energy of 720 units are added to the container, the system is allowed to come to thermal equilibrium.Part A) At equilibrium, how many energy units does each A particle have?Part B) At equilibrium, how many energy units does each B particle have?
Answer:
"8 units" is the appropriate answer.
Explanation:
According to the question,
Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.
Now,
The total energy will be:
= [tex]480+720[/tex]
= [tex]1200 \ units[/tex]
The total number of particles will be:
= [tex]50+100[/tex]
= [tex]150[/tex]
hence,
Energy of each A particle or each B particle will be:
= [tex]\frac{1200}{150}[/tex]
= [tex]8 \ units[/tex]
pls can anyone solve this
Answer:
3 pls give me brainliest
Explanation:
On the way home from school, Mr. X drives the first 10 miles at 55 mi/hr, the next 20 miles at 70 mi/hr, and the last 5 miles at 35 mi/hr. How far does Mr. X live from school?
Given :
On the way home from school, Mr. X drives the first 10 miles at 55 mi/hr, the next 20 miles at 70 mi/hr, and the last 5 miles at 35 mi/hr.
To Find :
How far does Mr. X live from school.
Solution :
To find the distance between Mr. X residence from school is simply given by summing all the distance he travelled .
So, distance = 10 + 20 + 5 miles
distance = 35 miles.
Hence, this is the required solution.
A bullet of m=30g reaches 900m/s in a 550mm long gun barrel. What total energy does the bullet have upon exiting the gun? *
A)12150000J
B)810000J
C)12150J
D)14850000J
show your work please
Answer:
C)12150J
Explanation:
KE = (1/2)(m)(v²)
KE = .5*30g(1kg/1000g)*(900m/s)²
KE = 12,150J
What is the period of revolution of a satellite with mass m that orbits the earth in a circular path of radius 7880 km (about 1500 km above the surface of the earth)
Answer:
Explanation:
For time period of revolution , the expression is as follows .
T² = 4π² R³ /GM , M is mass of the earth.
Putting the values
T² = 4π² (7880 x 10³)³ /(6.67 x 10⁻¹¹ )( 5.97 x 10²⁴ )
T² = 4.846 x 10⁷ s
T = 6.961 x 10³ s
= 6961 s
= 116 minutes .
which statement regarding the idealized model of motion called free fall is true?
a. the effect of air resistance is factored in the equation of motion in the idealized model called free fall.
b. free fall only models motion for objects that do not have an initial velocity in the upward direction.
c. the idealized model of the motion called free fall applies in cases where distance of the fall is large compared with the radius of the astronomical body on which the fall occurs.
d. a freely falling object has a constant acceleration due to gravity.
what type of material is good at transferring heat?
A. A thermal insulator, such as air
B. A thermal conductor, such as water
C. A thermal insulator, such as water
D, A thermal conductor, such as air
A thermal conductor, such as water is the type of material that is good at transferring heat. Option B is correct.
What is a heat conductor?Heat conductors are substances that effectively conduct thermal energy.
The kind of substance that is effective in transferring heat is a thermal conductor, such as water.
Metals have excellent heat conductivity. Heat insulators are substances that poorly conduct thermal energy. Thermal insulators include gases like air and materials like plastic and wood.
Water conducts heat 24.17 times more quickly than air. The kind of substance that is effective in transferring heat is a thermal conductor, such as water.
Hence, option B is correct.
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When you press on a circular table of radius 0.5 m, with a force of 20N, what will be exerted pressure?
50pa
25pa
40N/m
10N.m
Answer and I will give you brainiliest
Answer:
Note the word "circular".
Meaning the area would be found using that of a circle = πr²
Recall
Pressure = Force/Area
Area = π x 0.5² = 0.785m²
Pressure = 20/0.785
= 25pa ( to the nearest whole number)
Pressure = Force/Area
From the question,we have force ryt? so now let's find the area.As the table is circular, we have to use the formula of area of circle;
= πr²
= 3.14 × 0.5 × 0.5
= 3.14 × 0.25
= 0.785 m²
Put the values of force and area in formula;
= 20/0.785
= 20000/785
= 25.45 Pa
= 25 Pa (Approx)
Ferdinand the frog is hopping from lily pad to lily pad in search of a good fly
for lunch. If the lily pads are spaced 2.4 m apart, and Ferdinand jumps with a
speed of 5.4m/s, taking 0.60 s to go from lily pad to lily pad, at what angle
must Ferdinand make each of his jumps?
Answer:
θ = 33°
Explanation:
Here, we can use the formula for the total time of flight of a projectile to calculate the launch angle of frog:
[tex]T = \frac{2\ u\ Sin\theta}{g} \\\\Sin\theta = \frac{Tg}{2u}[/tex]
where,
θ = launch angle = ?
T = Total time of flight = 0.6 s
g = acceleration due to gravity = 9.81 m/s²
u = launch speed = 5.4 m/s
Therefore,
[tex]Sin\theta = \frac{(0.6\ s)(9.81\ m/s^2)}{(2)(5.4\ m/s)}\\\\\theta = Sin^{-1}(0.545)[/tex]
θ = 33°
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
a). How fast is the plane moving at takeoff?
b). How long does ot take the plane to travel down the runway?
A plane mirror produces images of objects that have an orientation that is _____, a size that is _______ (compared to that of the object) and a type that is _____
Answer:
RIGHT, SAME SIZE, VIRTUAL
Explanation:
Plane mirrors comply with the law of reflection where the angle of incidence is equal to the angle of reflection
therefore to complete the sentences:
A plane mirror produces images of objects that have an orientation that is RIGHT __, a size that is _SAME SIZE____ (compared to that of the object) and a type that is VIRTUAL_____
What is the primary evidence used to determine how the Moon formed? O A. Moon craters and Earth craters were caused by the same asteroid
strike.
B. Moon rocks and Earth rocks are made up of many of the same
materials.
O C. The Moon and Earth are exactly the same age.
D. The Moon and Earth have similar atmospheres.
What is the primary evidence used to determine how the Moon formed?
[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]
A. Moon craters and Earth craters were caused by the same asteroid strike. ✅
[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]
a runner completed the 100 yard dash in 10 seconds. what was the runners average speed
Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating about a perpendicular axis at one end is ml2/3. Write your answer with one decimal place.
Answer:
2.2 s
Explanation:
Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point = mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)
So, T = 2π√(I/mgh)
T = 2π√(mL²/3 /mgL/2)
T = 2π√(2L/3g)
substituting the values of the variables into the equation, we have
T = 2π√(2L/3g)
T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))
T = 2π√(3.66 m/(29.4 m/s² ))
T = 2π√(0.1245 s² ))
T = 2π(0.353 s)
T = 2.22 s
T ≅ 2.2 s
So, the period of the man's leg is 2.2 s
Planet K2-116b has an Average orbital radius of 7.18x10^9 m around the star K2-116. It has a mass of about 0.257 times the mass of the earth and an orbital period of 2.7 days.
What is the orbital speed of the planet?
Determine the mass of the star.
a) v = 1.94 × 10^5 m/s
b) Ms = 2.09 × 10^24 kg
Explanation:
Given:
m = 0.257M (M = mass of earth = 5.972×10^24 kg)
= 1.535×10^24 kg
r = 7.18×10^9 m
T = 2.7 days × (24 hr/1 day) × (3600 s/1 hr)
= 2.3328×10^5 s
a) To find the orbital speed of the planet, we need to find the circumference of the planet's orbit first:
C = 2×(pi)×r
= 2(3.14)(7.18×10^9m)
= 4.51×10^10 m
The orbital speed v is then given by
v = C/T
= (4.51×10^10 m)/(2.33×10^5 s)
= 1.94 × 10^5 m/s
b) We know that centripetal force Fc is given by
Fc = mv^2/r
where v = orbital speed
r = average orbital radius
m = mass of planet
We also know that the gravitational force FG between the star K2-116 and the planet is given by
FG = GmMs/r^2
where m = mass of planet
Ms = mass of star K2-116
r. = average orbital radius
G = universal gravitational constant
= 6.67 × 10^-11 m^3/kg-s^2
Equating Fc and FG together, we get
Fc = FG
mv^2/r = GmMs/r^2
Note that m and one of the r's get cancelled out so we are left with
v^2 = GMs/r
Solving for the mass of the star Ms, we get
Ms = rv^2/G
=(7.18 × 10^9 m)(1.94 × 10^5 m/s)^2/(6.67 × 10^-11 m^3/kg-^2)
= 2.09 × 10^24 kg
A circular loop of wire with radius 10.0 cm is located in the xy-plane in a region of uniform magnetic field. A field of 2 T is directed in the z-direction, which is upward. (a) What is the magnetic flux through the loop
Answer:
[tex]\phi=628.3[/tex]
Explanation:
From the question we are told that
Radius [tex]r=10.0 cm[/tex]
Magnetic field[tex]B=2T[/tex]
Generally the equation for area of circular path is mathematically given by
[tex]Area=\pi r^2[/tex]
[tex]A=\pi 10^2[/tex]
[tex]A=314.15m^2[/tex]
Generally the equation for Magnetic flux is mathematically given by
[tex]\phi=BA[/tex]
[tex]\phi=2*314.15[/tex]
[tex]\phi=628.3[/tex]
A car sitting at rest begins
accelerating at 2.40 m/s2 for
15.0 seconds. How far has the
car gone?
The distance covered by the car for the speed of 2.40 m/s² for 15 seconds is 36 meters
To find the distance the given datas are:
Speed = 2.40 m/s²
Time = 15 seconds.
What is distance?Distance is the total movement of an object without any regard to direction.Distance can be evaluated how much an object moves from starting point to the end point.The distance completely depends upon the speed and time, i,e., the object covering some area with some particular time interval with the particular speed.Formula of distance,Distance = Speed × Time.
Distance will be measured in meter, kilometer, etc..Distance is a Scalar quantity.Substituting the given datas in the formula,
Distance = 2.40 × 15
= 36 m
The Car went at the distance of about 36 meters.
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What is the momentum of a lab cart with a mass of 0.60 [kg] and a speed of 2.2 [m/s]?
-3.67 [kg m/s)
-0.27 [kg m/s]
-1.32 [kg m/s)
-0.82 [kg m/s)
Answer:
1.32kgm/s
Explanation:
use the formula: p=mv
The momentum of a lab cart = -1.32 kg m/s
What is momentum?It is measure of the inertia of a body/ object .It can be calculated by multiplying mass with velocity .
General formula for momentum = M = m * v
given
mass = 0.60 kg
speed = 2.2 m/s
velocity = - 2.2 m/s ( answer is in negative , since mass is a scaler quantity but velocity is a vector quantity hence , velocity can be negative )
momentum = mass * velocity
= 0.60 * (-2.2 ) = -1.32 kg m/s
The momentum of a lab cart =c) -1.32 kg m/s
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A seagull flying horizontally over the ocean at a constant speed of 2.60 m/s carries a small fish in its mouth. It accidentally lets go of the fish, and 2.50 s after the bird lets go, the fish lands in the ocean. (a) Just before reaching the ocean, what is the horizontal component of the fish's velocity in m/s
Answer:
Horizontal Component of Fish's Velocity = 2.6 m/s
Explanation:
In this scenario, we will neglect the effects of the air resistance on the small fish. Since there is no resisting force available in the horizontal direction. Therefore, the horizontal component of the velocity of the fish will remain equal to the horizontal component of the velocity of the seagull and it will remain the same throughout the whole motion.
Horizontal Component of Fish's Velocity = Constant Horizontal Speed of Seagull
Horizontal Component of Fish's Velocity = 2.6 m/s
A pair of glasses uses a nonreflective coating of index of refraction 1.4 to minimize reflection of light with wavelength 500nm. If the index of refraction of the glass is 1.5, what is the minimum non-zero thickness of the coating
Answer:
d = 178.57 10⁻⁹ m
Explanation:
For this exercise we must find the thickness to minimize the reflection, so the interference for the reflection must be destructive.
To find the expression we must take into account, two things:
* When the light goes from an index mordant medium to one with a higher refractive incoe, it undergoes a phase change of 180 (pi radians)
* within the film the wavelength of light is modulated by the index of refraction
λₙ = λ₀/ n
In this case the light passes from the air to the reflective layer and undergoes a phase change of ∫π rad, then it is reflected in the film-glass layer where it undergoes another phase change of π rad, therefore the total change of phase is 2π radians, this change is the or changes its value
period of the trigonometric functions, therefore its value does not change
the expression for destructive interference is
d sin θ = (me + ½) λₙ
d sin θ = (m + ½) λ₀ / n
the minimum thickness occurs for m = 0 and if we take perpendicular incidence the sine = 1
d = λ₀ /2 n
l
et's calculate
d = 500 10⁻⁹ /( 2 1.4)
d = 178.57 10⁻⁹ m
Q5: An ice skater moving at 12 m/s coasts
to a halt in 95m on an ice surface. What is the coefficient
of (kinetic) friction between ice and skates?
u = 0.077
Explanation:
Work done by friction is
Wf = ∆KE + ∆PE
-umgx = ∆KE,. ∆PE =0 (level ice surface)
-umgx = KEf - KEi = -(1/2)mv^2
Solving for u,
u = v^2/2gx
= (12 m/s)^2/2(9.8 m/s^2)(95 m)
= 0.077
Kinetic friction is the ratio of the friction force to the normal force experienced by a body in moving state.The coefficient of kinetic friction between the ice and skates is 0.077.
Given-
velocity of the ice skater is 12 m/ sec.
Work done by the friction is the sum of the change of the kinetic energy and the change in potential energy.
[tex]W_{f}=\bigtriangleup KE +\bigtriangleup PE[/tex]
The value for the potential energy will be equal to Zero in this case. Therefore the work done by the friction is,
[tex]W_{f}=\bigtriangleup KE +0[/tex]
Kinetic energy is directly proportional to the mass of the object and to the square of its velocity and work done can be given as,
[tex]W_{f} =u_{f} mgx[/tex]
Here, [tex]u_{f}[/tex] is friction force, [tex]m[/tex] is mass, [tex]g[/tex] is gravity and x is the distance .
Equate the value of kinetic energy and work done of friction for further result, we get,
[tex]u_{f} mgx=\dfrac{1}{2} \times mv^2[/tex]
[tex]u_{f} =\dfrac{1}{2gx} \times v^2[/tex]
[tex]u_{f} =\dfrac{1}{9.8\times 95} \times 12^2[/tex]
[tex]u_{f} =0.077[/tex]
Hence, the coefficient of kinetic friction between the ice and skates is 0.077.
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1./ The upward net force on the space shuttle at launch is 10,000,000 N. What is the least amount of charge you could move from its nose to the launch pad, 60 m below, and thereby prevent it from lifting off
Answer:
[tex]Q=2C[/tex]
Explanation:
From the question we are told that:
Force [tex]F=10000000N[/tex]
Distance [tex]d=60m[/tex]
Where
[tex]Q_1=Q_2[/tex]
Generally the equation for Force is mathematically given by
[tex]F=\frac{KQ^2}{r^2}[/tex]
Where
[tex]K=9*10^9[/tex]
Therefore
[tex]Q^2=\frac{Fr^2}{K}[/tex]
[tex]Q^2=\frac{10000000*60^2}{98*10^9}[/tex]
[tex]Q=\sqrt{\frac{10000000*60^2}{9*10^9}}[/tex]
[tex]Q=2C[/tex]
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface. The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position. At what location are the kinetic energy and the potential energy the same
Given :
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.
The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
To Find :
At what location are the kinetic energy and the potential energy the same.
Solution :
Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.
So,
[tex]P.E = K.E\\\\\dfrac{kx^2}{2} = \dfrac{mv^2}{2}\\\\x = v\sqrt{\dfrac{m}{k}}\\\\x = 1.5 \times \sqrt{\dfrac{0.5}{20}}\ m\\\\x = 0.238 \ m[/tex]
Hence, this is the required solution.
What do the spheres in this model represent?
A. Molecules
B. electrons
C. Planets and the sun
D. Atoms
Answer:
Explanation:
the spheres cant be electrons as they will repel each other.
so the ans is D. Atoms
Answer:
D. Atoms
Explanation:
The spheres in this model represents the atoms. So, option (D) is correct answer.
If a cat can exert 2000N Of Force to move a trailer 50m is 20 seconds how much power did the car use ?
им putins брат, почему вы обманываете нашу систему образования, Это теперь запрещено в России.
. A stationary mass explodes into two parts of mass 4 kg and 40 kg . If the K.E of larger mass is 100 J . The K.E of small mass will be
Answer:
10J
Explanation:
KE = (1/2)mv²
100J = (.5)(40kg)v²
v²=(100J)/(20kg)
v²= 5
KE = 5(.5)(4kg)
KE = 10J