Answer:
a
[tex]\theta _1 =0.687 ^o[/tex]
[tex]\theta _2 =0.630 ^o[/tex]
b
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.1 will not be valid
Explanation:
From the question we are told that
The slit grating is [tex]N = 68 \ slits / cm = 6800 \ slits / m[/tex]
The order of spectrum is [tex]n = 4[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 6800}[/tex]
=> [tex]a = 0.000147 \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 0.000147 } ][/tex]
=> [tex]\theta _2 =0.630 ^o[/tex]
Gnerally if grating is [tex]N = 12800 \ slits per cm = 1280000 \ slits / m[/tex]
Generally the width of the slit is mathematically represented as
[tex]a = \frac{1}{ 1280000}[/tex]
=> [tex]a = 7.813 *10^{-7} \ m[/tex]
Generally the condition for constructive interference is
[tex]asin\theta = n * \lambda[/tex]
Now for the first wavelength the angle is evaluated as
[tex]\theta _1 = sin ^{-1} [ \frac{n \lambda_1 }{a} ][/tex]
[tex]\theta _1 = sin ^{-1} [ \frac{4* 434 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _1 = sin ^{-1} [ 2.22][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
=> [tex]\theta _1 =0.687 ^o[/tex]
Now for the second wavelength the angle is evaluated as
[tex]\theta _2 = sin ^{-1} [ \frac{n \lambda_2 }{a} ][/tex]
=> [tex]\theta _2 = sin ^{-1} [ \frac{4* 410 *10^{-9} }{ 7.813*10^{-7} } ][/tex]
=> [tex]\theta _2 = sin ^{-1} [2.1 ][/tex]
Generally given that the domain arcsine function is between -1 and 1 then the arcsine of 2.22 will not be valid
Galaxy B moves away from galaxy A at 0.577 times the speed of light. Galaxy C moves away from galaxy B in the same direction at 0.731 times the speed of light. How fast does galaxy C recede from galaxy A?
Answer:
The value is [tex]p = 0.7556 c[/tex]
Explanation:
From the question we are told that
The speed at which galaxy B moves away from galaxy A is [tex]v = 0.577c[/tex]
Here c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
The speed at which galaxy C moves away from galaxy B is [tex]u = 0.731 c[/tex]
Generally from the equation of relative speed we have that
[tex]u = \frac{p - v}{ 1 - \frac{ p * v}{c^2} }[/tex]
Here p is the velocity at which galaxy C recede from galaxy A so
[tex]0.731c = \frac{p - 0.577c }{ 1 - \frac{ p * 0.577c}{c^2} }[/tex]
=> [tex]0.731c [1 - \frac{ p * 0.577}{c}] = p - 0.577c[/tex]
=> [tex]0.731c - 0.4218 p = p - 0.577c[/tex]
=> [tex]0.731c + 0.577c = p + 0.4218 p[/tex]
=> [tex]1.308 c = 1.731 p[/tex]
=> [tex]p = 0.7556 c[/tex]
Which two things occur when electric charges move through a conductor?
A. The field is perpendicular to the electron flow.
B. The field is parallel to the electron flow.
C. An electric field forms.
D. A magnetic field forms.
The two correct answers are A and D ap3x approved!
A C and D
C because ofc when electric charge flow electric field is produced
Answer:
A & D
Explanation:
A 80 kg parent and a 20 kg child meet at the center of an ice rink. They place their hands together and push. The parent pushes the child with a force of 25 N, what is the force acting on the parent?
Answer:
force acting on the parent = 25 N .
Explanation:
According to third law of Newton , there is equal and opposite reaction to every action . Here force by the parent on child is action and the force by child on parent is reaction . The former is given as 25 N so force by child on parent will also be 25 N .
Answer is 25 N .
Please help It is Anatomy and Phys
Think about the last time you had your temperature taken. Describe the circumstances that led you to have your temperature taken, including the mechanism by which your temperature was measured (oral thermometer, ear thermometer, etc.), so that given the right equipment, you could demonstrate the technique yourself to another person. What would have been considered an abnormal temperature in that situation?
Answer:
the last time i had my tempature taken was at disney prings about a week ago -_- they used one of those gun things that dont touch u and the SHOVED me forward so i guess i was fine
Explanation:
Answer:
thermometer
Explanation:
According to the law of conservation of momentum, in an isolated system:
A. The initial and final velocities remain the same.
B. The total initial momentum is equal to the total final momentum.
C. One of the colliding objects must be at rest.
D. The final momentum is always less than the initial momentum.
Answer:
Explanation:B
What is the Basic SI unit for distance/length
A. Meters
B. Liters
C. Grams
D. Millimeters
A 12 kg coyote runs towards a rabbit in the Ortega Mountains
with a velocity of 8 m/s. Determine the momentum of the
coyote.
What is "given" in the word problem?
A) mass and velocity
B) momentum
C) momentum and mass
D) momentum and velocity
Given :
A 12 kg coyote runs towards a rabbit in the Ortega Mountains with a velocity of 8 m/s.
To Find :
The momentum of the coyote.
Solution :
The given in the question is mass of coyote which is 12 kg and velocity which is 8 m/s .
Momentum is given by :
M = mass × velocity
M = 12 × 8 kg m/s
M = 96 kg m/s
Hence, this is the required solution.
Can someone help with my physics homework? please
A go cart engine applies a force of 888N and moves the cart forward 22m.
a) How much work is done?
b) What is doing the work?
c) If the driver wants to go further will the amount of work increase or decrease? Do you need a bigger engine to go
further?
d) We put on a bigger engine (1111N) but the cart still moves forward 22m. How much work is done now?
e) Why would you put on a bigger engine if you are still moving 22m?
f) Work requires a change in energy, which engine uses more gas to go 22m?
g) Even an empty semi truck uses much more gas than a car. Why?Find the soultions
Answer:
a) 19536 joules of work are done.
b) The work is done by the engine on the structure of the cart.
c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.
d) 24442 joules of work are done.
e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.
f) The bigger engine uses more gas to go 22 meters.
g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.
Explanation:
a) If force applied in the cart is uniform, that is, constant in magnitude and direction and is parallel to distance travelled by the car, the work done on the cart is defined by the following equation:
[tex]W = F\cdot \Delta s[/tex] (1)
Where:
[tex]F[/tex] - Force applied by the cart, measured in newtons.
[tex]\Delta s[/tex] - Distance travelled by the car, measured in meters.
[tex]W[/tex] - Work done on the cart, measured in joules.
If we know that [tex]F = 888\,N[/tex] and [tex]\Delta s = 22\,m[/tex], then the work done on the cart is:
[tex]W =(888\,N)\cdot (22\,m)[/tex]
[tex]W = 19536\,J[/tex]
19536 joules of work are done.
b) The work is done by the engine on the structure of the cart.
c) There are three options: (i) Keeping the engine and changing the travelled distance, (ii) Changing the engine and keeping the travelled distance, (iii) Changing the engine and the travelled distance.
d) If we know that [tex]F = 1111\,N[/tex] and [tex]\Delta s = 22\,m[/tex] , then the work on the cart is:
[tex]W = (1111\,N)\cdot (22\,m)[/tex]
[tex]W = 24442\,N[/tex]
24442 joules of work are done.
e) We may change for a bigger engine if it allows a greater acceleration and higher peak speed.
f) The gas consumption is directly proportional to the square of velocity and mass of the cart and, hence, to the work done on the cart. In consequence, we conclude that the bigger engine uses more gas to go 22 meters.
g) An empty semi truck uses more gas than a car since the first has much more mass than a car and is designed for moves big loads and for being fast.
a. The amount of work done by the go cart engine is 19,536 Nm.
b. Work is being done by the go cart engine i.e the engine installed on the go cart.
c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.
d. The amount of work done when a bigger engine is used is 24,442 Nm.
e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.
f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.
g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.
Given the following data:
Force A = 888 NewtonDistance = 22 meterForce B = 1111 Newtona. To determine the amount of work done by the go cart engine:
Mathematically, the work done by an object is given by the formula;
[tex]Work\;done = Force \times distance[/tex]
Substituting the given parameters into the formula, we have;
[tex]Work\;done = 888 \times 22[/tex]
Work done = 19,536 Nm.
b. Work is being done by the go cart engine i.e the engine installed on the go cart.
c. The amount of work will increase if the driver wants to go further because work done is directly proportional to the distance covered. No, you don't need a bigger engine to go further.
d. To determine the amount of work done when a bigger engine is used:
[tex]Work\;done = Force \times distance[/tex]
[tex]Work\;done = 1111 \times 22[/tex]
Work done = 24,442 Nm.
e. The reason why a bigger engine is used is, so that the engine can easily do more work in the same amount of distance.
f. Since we know that work requires a change in energy, the bigger engine would use more gas to go 22 meters.
g. An empty semi truck uses much more gas than a car because it has more weight (mass) and as such requires more energy, in accordance with the law of inertia.
Read more: https://brainly.com/question/22599382
What is the atomic number for the following Bohr model?
e
P= 8
N= 8
---
a
8
b
2
OOOO
с
6
d
16
Answer:
8
Explanation:
Atomic number in the Bohr model given is 8.
For every atom, the atomic number is the number of protons in such an atom.
The protons are the positively charged particles found in an atom. They are called the atomic number.
Protons are found in the nucleus of an atom. Together with neutrons, the occupy the tiny nucleus of the atom. By so doing and coupled with their mass, the form the mass of atom. The atomic number is used to identify every atom.As the building collapses, the volume of air inside the building decreases, while the mass of the air stays the same. This means that the _____ of the air inside the building will increase.
Answer:
Density
Explanation:
A person standing a certain distance from four identical loudspeakers is hearing a sound level intensity of 125 dB. What sound level intensity would this person hear if all but one are turned off?
Answer:
[tex]\mathbf{\beta = 123.75 \ dB}[/tex]
Explanation:
From the question, using the expression:
[tex]125 \ dB = 10 \ log (\dfrac{I}{I_o})[/tex]
where;
[tex]I_o = 10^{-12} \ W/m^2[/tex]
[tex]I = 10^{12.5} \times 10^{-12} \ W/m^2[/tex]
[tex]I = 3.162 \ W/m^2[/tex]
This is a combined intensity of 4 speakers.
Thus, the intensity of 3 speakers = [tex]\dfrac{3.162\times 3}{4}[/tex]
= 2.372 W/m²
Thus;
[tex]\beta = 10 \ log ( \dfrac{2.372}{10^{-12}} ) \ W/m^2[/tex]
[tex]\mathbf{\beta = 123.75 \ dB}[/tex]
A circuit component that is a composed of a semiconductor layer sandwiched between two other semiconductor layers is a(n)?
Explanation:
It's a transistor. Hope that helps!
Answer:
D. Transistor
Explanation:
Edge 2021
A 1430 kg is moving at 25.6 m/s when a force is applied, in the direction of the cars motion. The car speeds up to 31.3 m/s. If the force is applied for 5.4 s what is the magnitude of the force
The car accelerates with magnitude a such that
31.3 m/s = 25.6 m/s + a (5.4 s)
→ a = (31.3 m/s - 25.6 m/s) / (5.4 s) ≈ 1.056 m/s²
Then the applied force has a magnitude F of
F = (1430 kg) a ≈ 1500 N
if A=3i +2j+3k ,find the magnitude of A+B and A-B
Since vector B was not specified, I'll assume one at random. You can later answer your own question.
Answer:
[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}[/tex]
[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}[/tex]
Explanation:
Given:
[tex]\vec A=3\hat i +2\hat j+3\hat k[/tex]
And (assumed):
[tex]\vec B=-5\hat i +8\hat j-4\hat k[/tex]
Find the magnitude of
[tex]\vec A+\vec B[/tex]
[tex]\vec A-\vec B[/tex]
Given a vector
[tex]\vec P=x\hat i +y\hat j+z\hat k[/tex]
The magnitude of the vector is:
[tex]\mid\mid \vec P\mid \mid=\sqrt{x^2+y^2+z^2}[/tex]
First part:[tex]\vec A+\vec B =3\hat i +2\hat j+3\hat k-5\hat i +8\hat j-4\hat k[/tex]
[tex]\vec A+\vec B =-2\hat i +10\hat j-\hat k[/tex]
The magnitude of the sum is:
[tex]\mid\mid \vec A+\vec B \mid \mid=\sqrt{(-2)^2+10^2+(-1)^2}=\sqrt{4+100+1}[/tex]
[tex]\mathbf{\mid\mid \vec A+\vec B \mid \mid=\sqrt{105}}[/tex]
Second part:[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k-(-5\hat i +8\hat j-4\hat k)[/tex]
[tex]\vec A-\vec B =3\hat i +2\hat j+3\hat k+5\hat i -8\hat j+4\hat k[/tex]
[tex]\vec A-\vec B =8\hat i -6\hat j+7\hat k[/tex]
The magnitude of the difference is:
[tex]\mid\mid \vec A-\vec B \mid \mid=\sqrt{8^2+(-6)^2+7^2}=\sqrt{64+36+49}[/tex]
[tex]\mathbf{\mid\mid \vec A-\vec B \mid \mid=\sqrt{149}}[/tex]
a cannon has a mass 2500. it fires a cannon-Ball during a routine exercise. the cannon is 1000 times heavier than the cannon ball. the cannon ball leaves the barrel at a horizontal velocity of 160 m/s. the cannon comes to rest 2 seconds after the cannon ball was fired. calculate the magnitude of average net force that causes the cannon to rest
Answer:
F = 200 [N]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which can be calculated by means of the following equation.
[tex]P=m*v[/tex]
where:
P = lineal momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
Now we must understand that the momentum is conserved before and after the firing of the cannon. Before firing the cannon we have the mass of the cannon and mass of the cannonball together at rest (speed = 0). After firing the cannon the cannonball moves forward with positive speed, while the cannon moves back (negative), in this way knowing the masses of each one we can determine the speed of the cannon.
[tex](m_{cannon}+m_{ball})*v_{1}=-(m_{cannon}*v_{2})+(m_{ball}*v_{3})[/tex]
where:
m_cannon = 2500 [kg]
m_ball = 2.5 [kg]
v₁ = 0 (velocity of the group before firing)
v₂ = velocity of the cannon after firing [m/s]
v₃ = 160 [m/s] (velocity of the cannonball after firing)
[tex](2500+2.5)*0 = -(2500*v_{2})+(2.5*160)\\v_{2}=0.16[m/s][/tex]
Now using the following equation of kinematics, we can calculate the acceleration.
[tex]v_{f}=v_{o}-a*t[/tex]
where:
Vf = final velocity = 0 (cannon comes to rest)
Vo = initial velocity = 0.16 [m/s]
a = acceleration [m/s²]
t = time = 2 [s]
[tex]0 = 0.16 - a*2\\2*a=0.16\\a = 0.08 [m/s^{2}][/tex]
With the value of acceleration, we can use Newton's second law which tells us that the forces acting on a body is equal to the product of mass by acceleration.
ΣF = m*a
where:
F = force [N] (units of Newtons)
m = mass = 2500 [kg]
a = acceleration = 0.08 [m/s²]
[tex]F = 2500*0.08\\F = 200 [N][/tex]
The force of gravity between any two ordinary objects on the earth is.......
A) stronger when closer to the earth
B) stronger if the objects are closer to each other C) always downward
D) stronger than the force of gravity from the earth
Answer:
c
Explanation: beacuse newtons law of phisics says ''what goes up must come down''
The force of gravity between any two ordinary objects on the earth is stronger if the objects are closer to each other.
The force of gravity between two object on the earth surface is given Newton's law of universal gravitation;
[tex]F= \frac{Gm_1m_2}{r^2}[/tex]
where;
G is gravitational constantm₁ and m₂ are the masses of the two objectsr is the distance between the two objectsThe distance between the object is inversely proportional to the force of gravity between the objects. The smaller the distance between the two objects, the greater the force of gravity and vice versa.
Thus, we can conclude that the force of gravity between any two ordinary objects on the earth is stronger if the objects are closer to each other.
Learn more here:https://brainly.com/question/10609143
name the force that help us to walk
A 12 N net force is applied to an object as it moves a distance of 3.0 m: Use the
Work-Kinetic Energy Theorem to determine the object's change in kinetic energy.
Enter your answer in Joules.
Answer:
4 joules
Explanation:
Justify this statement "Density of the impure water is greater than of pure water.
Explaining the concept:
To explain this, we first need to know what density actually is:
Density is the mass of a given substance per unit volume, which means, that is the amount of mass present in unit volume (1 cm³ , 1m³)
Impure water is water, but with impurities added to it
Why is the density of Impure water is greater:
When we compare the mass of equal volumes of Water and Impure water, we will find that the mass of Impure water will be more, that is because the extra mass is of the impurities
Since impure water has impurities uniformly spread in it, we will find more matter in unit volume of Impure water than in Pure water
That is why the density of Impure water is greater
A 1369.4 kg car is traveling at 28.9 m/s when the driver takes his foot off the gas pedal. It takes 5.1 s for the car to slow down to 20 m/s. How large is the net force slowing the car
Answer:
F = 2389.603 N
Explanation:
Given:
Mass m = 1,369.4 kg
Initial velocity u = 28.9 m/s
Final velocity v = 20 m/s
Time t = 5.1 s
Find:
Net force
Computation:
a = (v - u)/t
a = (20 - 28.9)/5.1
a = -1.745 m/s²
F = ma
F = (1369.4)(1.745)
F = 2389.603 N
How does opening a parachute slow the fall rate of a skydiver?
Answer:
the air is being stopped in the pocket of the parachutes theys why the parachute has a certain shape so that the air that gets caught inside of it as the skydiver goes down slows down the landing
what is the potential energy of a 30kg rock that falls 15 meters
Answer:
4500 JExplanation:
The potential energy of a body can be found by using the formula
PE = mgh
where
m is the mass
h is the height
g is the acceleration due to gravity which is 10 m/s²
From the question we have
PE = 30 × 10 × 15
We have the final answer as
4500 JHope this helps you
If the magnitude of the electric field of an electromagnetic wave is 3x10^3 V/m, what is the the magntude of the magnetic field?
a. 1.1 x 10^-12 T
b. 9 x 10^11 T
c. 10^5 T
d. 10^-5 T
e. 3 x 10^3 T
Answer:
The value of the magnetic field is [tex]B = 1.0*10^{-5} \ T[/tex]
The correct option is d
Explanation:
From the question we are told that
The magnitude of the electric field is [tex]E = 3*10^{3} \ V/m[/tex]
Generally the magnitude of the magnetic field is mathematically represented as
[tex]B = \frac{E}{c}[/tex]
Here c is the speed of light [tex]c = 3.0*10^{8} \ m/s[/tex]
=> [tex]B = \frac{E}{c}[/tex]
=> [tex]B = \frac{3.0*10^{3}}{ 3.0*10^{8}}[/tex]
=> [tex]B = 1.0*10^{-5} \ T[/tex]
The mass and coordinates of three objects are given below: m1 = 6.0 kg at (0.0, 0.0) m, m2 = 1.5 kg at (0.0, 4.1) m, and m3 = 4.0 kg at (1.9, 0.0) m. Determine where we should place a fourth object with a mass m4 = 7.9 kg so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m
Answer:
The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].
Explanation:
Vectorially speaking, the center of gravity with respect to origin ([tex]\vec r_{cg}[/tex]), measured in meters, is defined by the following formula:
[tex]\vec r_{cg} = \frac{m_{1}\cdot \vec r_{1}+m_{2}\cdot \vec r_{2}+m_{3}\cdot \vec r_{3}+m_{4}\cdot \vec r_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}[/tex] (1)
Where:
[tex]m_{1}[/tex], [tex]m_{2}[/tex], [tex]m_{3}[/tex], [tex]m_{4}[/tex] - Masses of the objects, measured in kilograms.
[tex]\vec r_{1}[/tex], [tex]\vec r_{2}[/tex], [tex]\vec r_{3}[/tex], [tex]\vec r_{4}[/tex] - Location of the center of mass of each object with respect to origin, measured in meters.
If we know that [tex]\vec r_{cg} = (0,0)\,[m][/tex], [tex]\vec r_{1} = (0,0)\,[m][/tex], [tex]\vec r_{2} = (0, 4.1)\,[m][/tex], [tex]\vec r_{3} = (1.9,0.0)\,[m][/tex], [tex]m_{1} = 6\,kg[/tex], [tex]m_{2} = 1.5\,kg[/tex], [tex]m_{3} = 4\,kg[/tex] and [tex]m_{4} = 7.9\,kg[/tex], then the equation is reduced into this:
[tex](0,0) = \frac{(6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4.0\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4}}{6\,kg+1.5\,kg+4\,kg+7.9\,kg}[/tex]
[tex](6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4} = (0,0)\,[kg\cdot m][/tex]
[tex](7.9\,kg)\cdot \vec r_{4} = -(6\,kg)\cdot (0,0)\,[m]-(1.5\,kg)\cdot (0,4.1)\,[m]-(4\,kg)\cdot (1.9,0)\,[m][/tex]
[tex]\vec r_{4} = -0.759\cdot (0,0)\,[m]-0.190\cdot (0,4.1)\,[m]-0.506\cdot (1.9,0)\,[m][/tex]
[tex]\vec r_{4} = (0, 0)\,[m] -(0, 0.779)\,[m]-(0.961,0)\,[m][/tex]
[tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex]
The location of the center of gravity of the fourth mass is [tex]\vec r_{4} = (-0.961\,m,-0.779\,m)[/tex].
How long does it take a plane, traveling at a constant speed of 123 m/s, to fly once around a circle whose radius is 4330 m?
Answer:
3.7 minExplanation:
Step one:
given data
speed = 123m/s
radius of circle= 4330m
Step two:
We need to find the circumference of the circle, it represents the distance traveled
C=2πr
C= 2*3.142*4330
C= 27209.72m
Step three:
We know that velocity= distance/time
time= distance/velocity
time= 27209.72/123
time=221.2 seconds
in minute = 221.2/60
time= 3.7 min
E=6.63E-34J-s*4.6E14 hz
An object with a mass of 32 kg has an initial energy of 500). At the end of the experimentthe velocity of the object is recorded as 5.1 m/s . the object travelled 50 m to get to this point, what was the average force of friction on object during the tripAssume no potential energy Show all work
Answer:
F = 1.68 N
Explanation:
Let's solve this exercise in parts.
Let's use the concept of conservation of the mechanical nerve
initial
Em₀ = 500 J
The energy is totally kinetic
Em₀ = K = ½ m v₀²
v₀ = [tex]\sqrt{\frac{2 Em_{o} }{m} }[/tex]
v₀ = √ (2 500/32)
v₀ = 5.59 m / s
now with kinematics we can find a space
v² = v₀² - 2 a x
the negative sign is because the body is stopping
a =[tex]( \frac{v_{o}^{2} - v^{2} }{2x} )[/tex]
let's calculate
a = (5.59² - 5.1²) / 2 50
a = 0.0524 m / s²
Finally let's use Newton's second law
F = ma
F = 32 0.0524
F = 1.68 N
A cicada produces a sound of roughly 5000 Hz frequency. What category would that sound fall into?
Supersonic
Normal range
Infrasonic
Ultrasonic
Help ASAP plz and thx u
Answer:
a). a = F/m
Explanation:
Formula is F=ma
When a neutral atom gains or loses
electrons, it becomes charged
and is called a(n)
Answer:
It is called an ion.
Explanation:
Atoms of elements can lose or gain electrons making them no longer neutral, they become charged. A charged atom is called an ion. When an atom loses electron(s) it will lose some of its negative charge and so becomes positively charged. A positive ion is formed where an atom has more protons than electrons.