Answer:
Y = 5.03 x 10⁻³ m = 5.03 mm
Explanation:
Using Young's Double-slit formula:
[tex]Y = \frac{\lambda L}{d}[/tex]
where,
Y = Fringe Spacing = Width of bright fringe = ?
λ = wavelength = 633 nm = 6.33 x 10⁻⁷ m
L = Screen distance = 3.1 m
d = slit width = 0.39 mm = 3.9 x 10⁻⁴ m
Therefore,
[tex]Y = \frac{(6.33\ x\ 10^{-7}\ m)(3.1\ m)}{3.9\ x\ 10^{-4}\ m}[/tex]
Y = 5.03 x 10⁻³ m = 5.03 mm
. 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
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]
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
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.
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
In an experiment, a student brings up the rotational speed of a piece of laboratory apparatus to 24 rpm. She then allows the apparatus to slow down uniformly on its own, and counts 236 revolutions before the apparatus comes to a stop. The moment of inertia of the apparatus is known to be 0.076 kg m2. What is the magnitude of the torque on the apparatus
Answer:
T = 6.43 x 10⁻⁵ N.m
Explanation:
First, we will calculate the deceleration of the apparatus by using the third equation of motion:
[tex]2\alpha \theta = \omega_f^2-\omega_i^2[/tex]
where,
α = angular decelration = ?
θ = angular displacement = (236 rev)(2π rad/rev) = 1482.83 rad
ωi = initial angular speed = (24 rpm)(2π rad/1 rev)(1 min/ 60 s) = 2.51 rad/s
ωf = final angular speed = 0 rad/s
Therefore,
[tex]2\alpha(1482.83\ rad) = (0\ rad/s)^2-(2.51\ rad/s)^2\\\\\alpha = -\frac{(2.51\ rad/s)^2}{2965.66\ rad} \\\\\alpha = - 8.46\ x\ 10^{-4}\ rad/s^2[/tex]
negative sign shows deceleration
Now, for torque:
T = Iα
where,
T = Torque = ?
I = moment of inertia = 0.076 kg.m²
Therefore,
T = (0.076 kg.m²)(8.46 x 10⁻⁴ N.m)
T = 6.43 x 10⁻⁵ N.m
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.
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 .
Engineers are working on a design for a cylindrical space habitation with a diameter of 7.50 km and length of 29.0 km. The habitation will simulate gravity by rotating along its axis. With what speed (in rad/s) should the habitation rotate so that the acceleration on its inner curved walls equals 8 times Earth's gravity
Answer:
The speed will be "0.144 rad/s".
Explanation:
Given that,
Diameter,
d = 7.50 km
Radius,
R = [tex]\frac{7.5}{2} \ Km[/tex]
Acceleration on inner curve,
= 8 times
Now,
As we know,
⇒ [tex]\omega^2R=8g[/tex]
or,
⇒ [tex]\omega=\sqrt{\frac{8g}{R} }[/tex]
On substituting the values, we get
⇒ [tex]=\sqrt{\frac{8\times 9.8}{\frac{7.5}{2}\times 10^3 } }[/tex]
⇒ [tex]=\sqrt{\frac{78.4}{3750} }[/tex]
⇒ [tex]=\sqrt{0.0209}[/tex]
⇒ [tex]=0.144 \ rad/s[/tex]
a runner completed the 100 yard dash in 10 seconds. what was the runners average speed
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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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 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]
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.
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
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 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
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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pls can anyone solve this
Answer:
3 pls give me brainliest
Explanation:
If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be Group of answer choices 1.0 m/s2. We cannot tell from the information given. 2.2 m/s2. 3.0 m/s2.
Answer:
We cannot tell from the information given
Explanation:
Given;
mass of the box, m = 5 kg
first force, F₁ = 10 N
second force, F₂ = 5 N
(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;
∑Fx = 10 N - 5 N
= 5 N
Apply Newton's second law of motion;
∑Fx = ma
a = ∑Fx/m
a = 5 / 5
a = 1 m/s² in the direction of the 10 N force.
(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;
∑Fx = 10 N + 5 N
∑Fx = 15 N
a = 15 / 5
a = 3 m/s²
Therefore, the information given is not enough to determine the acceleration of the box.
In Case 1, a mass M hangs from a vertical spring having spring constant k and is at rest in its equilibrium position. In Case 2 the mass has been lifted a distance D vertically upward. If we define the potential energy in Case 1 to be zero, what is the potential energy of Case 2
Answer: hello your question is incomplete attached below is the complete question
answer : 1/2 KD^2 ( option A )
Explanation:
P.E ( potential energy ) = mgd
In case 1 P.E = 0 i.e. mgd = 0
Given that in case 2 the Mass M had moved through the Distance D by the compression of the spring
The potential energy of the M in case 2
= P.E of M at rest + P.E of the spring
= 0 + 1/2 KD^2
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?
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]
the organelle involved in cell secretion is
Answer:
Golgi apparatus (bodies).
Explanation:
A cell can be defined as the fundamental or basic functional, structural and smallest unit of life for all living organisms. Some living organisms are unicellular while others are multicellular in nature.
Generally, cells have the ability to independently replicate themselves. In a cell, the "workers" that perform various functions or tasks for the survival of the living organism are referred to as organelles. Some examples of cell organelles found in all living organisms such as trees, birds, and bacteria include; nucleus, cytoplasm, cell membrane, golgi apparatus, mitochondria, lysosomes, ribosomes, chromosomes, endoplasmic reticulum, vesicles, etc.
Golgi apparatus is also referred to as Golgi bodies and it functions as a packaging unit in living organisms, especially eukaryotic cells because it prepares protein and lipid molecules for export by chemically tagging them.
Hence, the Golgi apparatus (bodies) is an organelle involved in cell secretion through the transportation (export) of proteins and lipids out of a cell.
Discuss the chemical bond exist in silicon crystal?
A certain 20-A circuit breaker trips when the current in it equals 20 A. What is the maximum number of 100-W light bulbs you can connect in parallel in an ideal 120-V dc circuit without tripping this circuit breaker
Answer: 28
Explanation:
Given
Circuit breaker current is [tex]I=20\ A[/tex]
Power of the light bulb is [tex]P=100\ W[/tex]
Voltage of the DC-circuit is [tex]V=120\ V[/tex]
If the resistance are connected in parallel, they must have same voltage i.e. 120 V
So, Resistance is given by
[tex]\Rightarrow R=\dfrac{V^2}{P}\\\\\Rightarrow R=\dfrac{120^2}{100}\\\\\Rightarrow R=144\ \Omega[/tex]
For the 20 A current and 120 V battery, net resistance is
[tex]\Rightarrow R_{net}=\dfrac{120}{20}\\\\\Rightarrow R_{net}=6\ \Omega[/tex]
Suppose there are n resistance in the circuit connected in parallel.
[tex]\Rightarrow \dfrac{144}{n}=R_{net}\\\\\Rightarrow n=\dfrac{144}{6}\\\\\Rightarrow n=28.8\approx 28\ \text{for current to be less than 20A}[/tex]
Thus, there can maximum of 28 bulbs.
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 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.
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_____
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°