I NEED HELP PLEASE, THANKS! :)
Three positive charges A, B, and C, and a negative charge D are placed in a line as shown in the diagram. All four charges are of equal magnitude. The distances between A and B, B and C, and C and D are equal.
a. Which charge experiences the greatest net force? Which charge experiences the smallest net force?
b. Find the ratio of the greatest to the smallest net force.
Answers
Answer:
1. Largest force: C; smallest force: B; 2. ratio = 9:1
Explanation:
The formula for the force exerted between two charges is
[tex]F=K\dfrac{ q_{1}q_{2}}{r^{2}}[/tex]
where K is the Coulomb constant.
q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.
For simplicity, let's combine Kq₁q₂ into a single constant, k.
Then, we can write
[tex]F=\dfrac{k}{r^{2}}[/tex]
1. Net force on each particle
Let's
Call the distance between adjacent charges d. Remember that like charges repel and unlike charges attract.Define forces exerted to the right as positive and those to the left as negative.
(a) Force on A
[tex]\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}[/tex]
(b) Force on B
[tex]\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}[/tex]
(C) Force on C
[tex]\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}[/tex]
(d) Force on D
[tex]\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}[/tex]
(e) Relative net forces
In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.
[tex]F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\[/tex]
2. Ratio of largest force to smallest
[tex]\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}[/tex]
Related Questions
2. Fill in the blanks with appropriate words.
a. The force which acts towards the centre is called .......
force.
Answers
Answer:
centripetal force
Explanation:
because this force always compels the body to move in circle
To break a chemical bond, particles need to collide with a certain amount of a]energy. b]atoms. c]molecules. d]chemicals.
Answers
Answer:
Energy
Explanation:
The stronger the force it has the more quickly the bonds will break
As there will be more energy they will vibrate more and the bonds will separate easily..
Q1. Tractors are often used on sloping fields, so stability is important in their design. On the diagram, the center of the X marks the center of the tractor.
a) Explain why the tractor has not toppled over.
b) Give two features of the tractor which affects its stability.
Q2. The diagram shows two buses. Bus A is empty. Bus B contains bags of sand upstairs to represent passengers. Each bus has been tilted as far as it can without falling over.
a) Each bus will topple over if it is tilted any further. Explain why this will happen.
b) What difference does it make to the stability of the bus when the upper deck is full of passengers?
c) Why are the bags of sand in bus B only put upstairs?
Q3. A tractor is being used on a rough ground. What is the safest position for its center of mass? it is an emergency I will give you 20 points
Answers
Answer:
a) The tractor has not toppled over because the the center of mass is acting in between the wheels
b) The features which affect its stability are;
1) The location of the center of gravity of the tractor close to the ground
2) The increased spacing between the left and the right wheels of the tractor
Q2. a) The center of mass wheel act in the region to the right away the wheel span (space in between the wheel) which creates a clockwise effect to, topple the buses over
b) It raises the bus's center of mass higher
c) To raise the center of mass to its highest practical level
Q3. At the bottom middle location within the tractor
Explanation:
a) The tractor has not toppled over because the vertical line from the center of gravity to the ground is still in between the wheel base such that since the tractor will topple over by turning clockwise about the right tire due to its weight, the weight of the tractor is currently acting in the anticlockwise direction to the right end tire of the tractor keeping the tractor's tires in good contact with the ground
b) The features which affect its stability are;
1) The location of the center of gravity of the tractor close to the ground
2) The increased spacing between the left and the right wheels of the tractor
Q2. a) When either of the buses are tilted further, the mass of the buses is acting on the tires and in their tilted condition, the mass is acting mainly on the right tire, whereby as the vertical line from the center of mass to the ground is in between the wheel bases the mass of the buses still serve s the restoring force to keep the bus on ground by providing anticlockwise moment where clockwise lotion is required for the bus to topple over
If the buses are tilted further, the vertical line from the center of mass will cross to the other side of the right tire such that the mass now provides clockwise moment turning the buses clockwise and since there are no opposing anticlockwise moment to balance that of the now clockwise moment of the buses it will continue to turn clockwise with the result that they eventually topple over
b) When the upper deck is full of passengers, there will be an appreciable proportion of the total mass of the bus in the upper deck such that the level of the center of mass will be raised up higher
c) Sand bags are only put upstairs because by allowing for the disproportionate distribution of load such that the majority of the load of the bus is concentrated upstairs, the center of mass will be at its maximum height allowing the bus to be tested for stability in the worst case scenario
Therefore, the sands are only put upstairs to raise the center of mass
Q3. Being that the tractor is being used on a rough ground, the safest position will be at the lowest possible location at the middle of both the front and rear wheels and the left and right tires.
A 1500 kg car on flat ground is moving 5.25 m/s. Its engine creates a 1250 N forward force as the car moves 42.8 m. What is its final velocity?
Answers
Answer:
9.93 m/s
Explanation:
The following data were obtained from the question:
Mass (m) = 1500 Kg
Initial velocity (u) = 5.25 m/s
Force (F) = 1250 N
Distance (s) = 42.8 m
Final velocity (v) =...?
Next, we shall determine the acceleration of car. This can be achieved as shown below:
Force (F) = mass (m) x acceleration (a)
1250 = 1500 x a
Divide both side by 1500
a = 1250/1500
a = 0.83 m/s²
Therefore, the acceleration of the car is 0.83 m/s²
Now, we can determine the final velocity of car as shown below:
Initial velocity (u) = 5.25 m/s
Distance (s) = 42.8 m
Acceleration (a) = 0.83 m/s²
Final velocity (v) =...?
v² = u² + 2as
v² = (5.25)² + 2 x 0.83 x 42.8
v² = 27.5625 + 71.048
v² = 98.6105
Take the square root of both side
v = √98.6105
v = 9.93 m/s
Therefore, the final velocity of the car is 9.93 m/s
Answer:
9.93
Explanation:
too easy, just use a calculator lol
Elements in the same periodic table group have the same___
Answers
Answer:
elements in the same periodic table group have the same valence electrons
what are the fundamental units used in physics?
Answers
Answer:
Metre, kilogram, second, ampere, kelvin, candela, and mole
Explanation:
A base unit (also referred to as a fundamental unit) is a unit adopted for measurement of a base quantity
What are the units of impulse
Answers
Answer:
Newton Second
Explanation:
The SI unit of impulse in Newton Second (N.s)
What is the speed of a bus if it
covers 1.5km in 30 seconds?
a. 50km/Hr
b. 500m/s
c. 50m/s
d. 1500m/s
Answers
[tex]answer \\ = 50 \: m / s \\ solution \\ distance \: travelled\: by \: bus = 1.5 \: km \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1.5 \times 1000 \: m \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 1500 \: m \\ time \: = 30 \: seconds \\ speed = \frac{distance}{time} \\ \: \: \: \: \: \: \: \: \: \: = \frac{1500m}{30s } \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = 50 \: m/s \\ hope \: it \: helps[/tex]
Answer:
C. 50 m/s
Explanation:
Given:
Distance = 1.5 km = 1500 m
Time = 30 sec
Required:
Speed = ?
Formula
Speed = Distance / Time
Solution:
Putting the given in the formula
Speed = 1500/30
Speed = 50 m/s
Hence, the force of gravitation between Earth and Jupiter is 8.16 * 10 N
Enample 5: An apple with the mass of 200 g falls from a tree. What is the acceleration
of the apple towards the earth? What is the acceleration of the earth towards the apple?
(Given: mass of the earth = 6*10kg radius of the earth 6.4 10 m, G=6.67 * 10 " Nm
kg and neglecting height of the tree).
Answers
Answer:
Acceleration of the apple towards the earth is: 9.7705 m/s2.
Acceleration produced on earth towards the apple
[tex]0.325*10^{-24} \ \ m/s2.[/tex]
Explanation:
Given :
[tex]m = 200 g[/tex]
We have to convert into the kg ,
so
[tex]m=0.2 kg[/tex]
M=[tex]6*10^{24}[/tex] kg
[tex]R=6.4*10^{6} m[/tex]
[tex]G= 6.67*10^{-11} \ Nm^{2} kg-2[/tex]
As the gravitational force is existing among the earth and the apple
so
[tex]F=\frac{GMm}{R^{2} }[/tex]
Putting the value of G,M,m and [tex]R^{2}[/tex], we get
[tex]\frac{6.67*10^{-11} *6*10^{24} *0.2 }{(6.4*10^{6} )^{2} } \\\\F= 1.9541\ N[/tex]
Consider a1 and a2 be a acceleration due to the gravitational force of earth attraction created on the apples
F= a1 *m
[tex]a1\ =\ \frac{F}{m}[/tex]
a1=[tex]\frac{1.9541}{0.2}[/tex]
[tex]a1=9.7705 \ m/s2[/tex]
Acceleration of the apple towards the earth is: 9.7705 m/s2.
again
[tex]F= a2 *M[/tex]
a2=[tex]\frac{F}{M}[/tex]
[tex]a2=\frac{ 1.9541}{6**10^{24} } \\a2=0.325* 10^{-24}\ m/s2[/tex]
So, Acceleration produced on earth towards the apple
[tex]0.325*10^{-24} \ \ m/s2.[/tex]
which two elements define a story's setting?
geographical location
character description
weather conditions
plot structure
dialogue and action
Reset
Next
Answers
Answer:
Geographical location, and time and weather conditions
Answer:
Georgraphical location and weather conditions
Explanation:
Setting is where the story takes place so the location would be vital and same with weather conditions because it determines what the character(s) do
The estimated distance of the earth from the sun is 149000000km. The speed of the light is 300000km/s. What is the time taken for the light to travel from the sun to the earth
Answers
Answer:
The time taken for the light to travel from the earth to the sun = 496.67 seconds
Explanation:
Estimated distance of the earth to the sun = 149000000km
Speed of light = 300000km/s
Time taken for the light to travel from the sun to the earth = ?
Speed = Distance/time
time = Distance/speed
time = 149000000/300000
time = 496.67 s
The time taken for the light to travel from the earth to the sun = 496.67 seconds
A transformer has 500 turns of the primary winding and 10 turns of the secondary winding. a) Determine the secondary voltage if the secondary circuit is open and the primary voltage is 120 V
Answers
Answer:
First we use the fact that the ratio of the voltage on the secondary and on the primary will be the same as the ratio of the number of turns of both coils,
u=u^2/U^1 = N^2/N^1
we evaluate the unkown secondary voltage
u^ 2=N^2/N^1 u1
We substitute the given values -
u^2= 10/500 * 120v = 2.4v
Hence, the answer is 2.4v.
You can make a smoothie in a blender with a power of 400 watts and an efficiency of 85 percent. How much energy is actually being used to make the smoothie in 30 seconds?
Answers
Answer:
total energy is 14.12kj
energy at 85% is 12kJ
Explanation:
How much net force is required to keep a 3-kg object moving to the right with a constant speed of
6.0 m/s?
Answers
Answer:
Explanation:
Here,
m = 3 Kg
a = 0 m/s^2 (velocity is constant so the acceleration is zero)
We have,
F = ma
or, F = 3 x 0
or, F = 0 N
Constant velocity can always be associated with a zero net force.
Need Help !
A 1500 kg car on flat ground is
moving 5.25 m/s. Its engine
creates a 1250 N forward force as the car moves 42.8 m. What is its final velocity?
Answers
Answer: 5.96m/s
Explanation:
Given the following :
Mass of car (m) = 1500kg
Velocity (V) = 5.25m/s
Forward force of engine = 1250N
Diatance moved = 4.8m
Final Velocity =?
Final kinetic energy = Initial kinetic energy + work done by engine
Initial kinetic energy = 0.5 × mass × velocity^2
Initial kinetic energy = 0.5 × 1500 × 5.25^2
Initial kinetic energy = 20671.875 J
Work done by engine = Force × distance
Work done by engine = 1250 × 4.8 = 6000J
Final kinetic energy = (20671.875 + 6000) J
= 26671.875 J
From kinetic energy = 0.5mv^2
26671.875 = 1/2 × 1500 × v^2
53343.75 = 1500v^2
v^2 = 35.5625
v = sqrt(35.5625)
v = 5.96m/s
Apart from height above ground, what does the amount of gravitational potential energy that an object has on earth depend on?
Answers
Answer:
Mass
Explanation: the more mass the more GPE
Apart from the height above the ground, the object's mass on which the amount of gravitational potential depends.
What is gravitational potential energy?The law of gravity gives rise to the generic term for gravitational potential energy, which is equal to the effort made to defy gravity in order to move a mass to a specific location in space. It makes logical to choose the gravitational potential energy zero at an infinite distance away because, due to the inverse square nature of the gravity force, the force approaches zero for huge distances. Since gravity performs positive work as a mass approaches a planet, the gravitational potential energy is then negative. When a mass approaches a huge body, it enters a "bound state," where it is trapped until an external source of energy enables it to break free.
There is gravitational potential energy in any object raised above the earth ( or GPE). The mass and height of an object above the surface of the Earth determine how much gravitational potential energy it has.
To get more information about gravitational potential energy :
https://brainly.com/question/13003047
#SPJ2
Un bombero alejado d = 31.0 m de un edificio en llamas dirige un chorro de agua desde una manguera contra incendios a nivel del suelo con un ángulo de θi = 33.0° arriba de la horizontal como se muestra en la figura siguiente. Si la rapidez del chorro cuando sale de la manguera es vi = 40.0 m/s, ¿a qué altura (en m) golpeará el edificio? m
Answers
Answer:
El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.
Explanation:
El chorro de agua exhibe un movimiento parabólico, dado que este tiene una inclinación inicial y la única aceleración es debida a la gravitación terrestre. Las ecuaciones cinemáticas que modelan el fenómeno son:
Distancia horizontal (en metros)
[tex]x = x_{o} + v_{o}\cdot t \cdot \cos \theta[/tex]
Donde:
[tex]x_{o}[/tex] - Posición horizontal inicial, medida en metros.
[tex]t[/tex] - Tiempo, medido en segundos.
[tex]v_{o}[/tex] - Velocidad inicial, medida en metros por segundo.
[tex]\theta[/tex] - Angulo de inclinación del chorro de agua, medido en grados sexagesimales.
Distancia vertical (en metros)
[tex]y = y_{o} + v_{o}\cdot t \cdot \sin \theta + \frac{1}{2}\cdot g \cdot t^{2}[/tex]
Donde:
[tex]y_{o}[/tex] - Posición vertical inicial, medida en metros.
[tex]g[/tex] - Constante gravitacional, medida en metros por segundo al cuadrado.
Partiendo de la primera ecuación, se despeja el tiempo:
[tex]t = \frac{x - x_{o}}{v_{o}\cdot \cos \theta}[/tex]
Si [tex]x = 31\,m[/tex], [tex]x_{o} = 0\,m[/tex], [tex]v_{o} = 40\,\frac{m}{s}[/tex] and [tex]\theta = 33^{\circ}[/tex], entonces:
[tex]t = \frac{31\,m-0\,m}{\left(40\,\frac{m}{s} \right)\cdot \cos 33^{\circ}}[/tex]
[tex]t = 0.924\,s[/tex]
La altura máxima se calcula por sustitución directa de términos en la segunda ecuación. Si [tex]y_{o} = 0\,m[/tex], [tex]v_{o} = 40\,\frac{m}{s}[/tex], [tex]t = 0.924\,s[/tex] y [tex]g = -9.807\,\frac{m}{s^{2}}[/tex], entonces:
[tex]y = 0\,m + \left(40\,\frac{m}{s} \right)\cdot (0.924\,s)\cdot \sin 33^{\circ} + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right) \cdot (0.924\,s)^{2}[/tex]
[tex]y = 15.943\,m[/tex]
El chorro golpea el edificio a una altura de 15.943 metros con respecto al suelo.
A plane mirror is useful for seeing:
Answers
They are used to see round dangerous bends
A 2-kg object is moving 6 m/s in a horizontal direction.
How much net force is required for this object to continue its state of motion?
Α. Ο Ν
B.
2N
C.
3 N
12 N
Answers
Answer: A) O N
Explanation:
An object in motion will maintain its state of motion. The presence of an unbalanced force changes the velocity of the object.
A block of mass 4 kg rests on a horizontal surface without friction, when subjected to the action of 5 horizontal forces as shown in the figure (wait for the figure). Answer: (a) How strong is the resulting force? (b) What is the direction of the resulting force? (c) If the block takes action to come out of inertia, what will be its acceleration acquired?
Answers
Answer:
a) F = 40 N
b) It is acting towards left
c) a = 10 m/s²
Explanation:
As the block of mass and its forces are not shown, the question is incomplete.
First we attach the diagram of the attached question (See below)
We can see in the diagram that there are total of 5 forces acting on that block. Consider the forces acting towards right to be positive and the forces acting towards left be negative (or vice versa, the answer would be same). Let them be:
F₁ = 10 N
F₂ = 17 N
F₃ = 21 N
F₄ = -35 N
F₅ = -53 N
(a)
To find resultant force, we'll add all of the 5 forces.
Resultant Force = F₁ + F₂ + F₃ + F₄ + F₅
Resultant Force = 10 + 17 +21 - 35 - 53
Resultant Force = -40 N
(b)
As the resultant force has a negative sign, and previously we stated that the forces acting towards left are negative, so the Resultant Force is acting in the left direction.
(c)
We know that
F = (m)(a)
Where F is the resultant force we found. Resultant Force was -40N , where negative sign only represents the direction. We'll consider only the magnitude which is 40N, where m is the mass of the block which is 4 kg. Substitute in the equation.
40 = (4)(a)
a = (40)(4)
a = 10 m/s² (in left direction)
Your cardiovascular system gets better when you increase the time and intensity of your workouts. What should the word in italics be replaced with to make this sentence more precise?
Answers
Answer: The word better should be replaces with stronger and efficient.
Explanation:
The cardiovascular system of the body works better if the the body id performing intense workout and the heart becomes more efficient.
The workout sessions provides more oxygen to the heart which makes it stronger and provides more oxygen to the vital organs of the body.
There are less chances of occurring the problems related to the cardio vascular system of the body such as high blood pressure, blockage, et cetera.
Three dogs are fighting over a bone. One is pulling to the left with a force of 20N, another
to the right with a force of 35N and the third upward with a force of 15N. What is the net
force on the bone? This is a multiple-select problem: You must select a magnitude for the
force and an angle for the force.
Answers
Answer:
Net force = 21.12 N
angle for the force α = 45°
Explanation:
In this problem, we will take the left as the negative x-axis, the right as the positive x-axis, and the upward direction as the positive y-axis.
The 20 N force to the left has an x component of -20 N, and y component of 0 N
The 35 N force to the right has a x component of 35 N, and a y component of 0 N
The 15 N upwards has an x component of 0 N, and a y component of 15 N
We resolve the forces into the x and y components.
for the x component Fx
Fx = -20 + 35 + 0 = 15 N
For the y component Fy
Fy = 0 + 0 + 15 = 15 N
Net force Fn = [tex]\sqrt{Fx^{2} + Fy^{2} }[/tex]
[tex]Fn = \sqrt{(15)^{2} + (15^{2} ) }[/tex]
Net force Fn = 21.21 N
for the angle,
[tex]tan\alpha = \frac{Fy}{Fx} = \frac{15}{15}[/tex]
[tex]tan\alpha = 1[/tex]
α = [tex]tan^{-1}[/tex] 1
angle for force α = 45°
Supriya has a pre-paid electricity connection at home. On a sultry Sunday afternoon, she checks that she only has ₹9 left in the balance. She wants to use the air conditioner (AC), rated at 1.2 kW. The price of electricity is ₹10 /unit in her area. How long can she run the AC before her balance is over and the power runs out?
Answers
Answer: 0.75 hour(s) or 45 minutes
Explanation:
1 unit of electricity = 1kwh
Amount of energy consumed by 1kw load in one hour
Price per unit = ₹10
Air conditioner rating(power) = 1.2Kw
Given that balance = ₹9
Therefore, supriya's total energy cost = energy balance = ₹9
Total energy cost = power in kilowatt × time of operation (in hour) × cost per unit
₹9 = 1.2kw × time × ₹10
₹9 = 12 × time
Time = 9 / 12
Time = 0.75 hours
0.75 × 60 = 45 minutes
A block of mass m is placed on a smooth wedge of inclination θ . The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block has a magnitude: (A) mg tan θ (B) mg sin θ (C) mg cos θ (D) mg sec θ
Answers
Answer: mg/Cosθ
Explanation:
Taking horizontal acceleration of wedge as 'a'
FCosΘ = FsinΘ
F = mass(m) × acceleration(a) = ma
For horizontal resolution g = 0
Therefore,
Horizontal = Vertical
maCosΘ = mgSinΘ
aCosΘ = gSinΘ
a = gSinΘ/CosΘ
Recall from trigonometry :
SinΘ/Cosθ = tanΘ
Therefore,
a = gtanΘ
Normal force acing on the wedge:
mgCosΘ + maSinΘ - - - - (y)
Substitute a = gtanΘ into (y)
mgCosΘ + mgtanΘsinΘ
tanΘ = sinΘ/cosΘ
mgCosΘ + mgsinΘ/cosΘsinΘ
mgCosΘ + mgsin^2Θ/cosΘ
Factorizing
mg(Cosθ + sin^2Θ/cosΘ)
Taking the L. C. M
mg[(Cos^2θ + sin^2Θ) /Cosθ]
Recall: Cos^2θ + sin^2Θ = 1
mg[ 1 /Cosθ]
mg/Cosθ
What is the energy of a photon with a frequency of 2.2 x 1016 Hz? Planck's constant is 6.63 x 10-34 Jos.
O 1.5 x 10-17
8.8 x 10-17
O 1.5 10-16
O 8.8 10-16
Answers
Answer: 1.5 × 10^-17
Explanation:
Given the following :
Frequency(f) = 2.2 × 10^16 Hz
Planck's constant(h) = 6.63 × 10^-34
The energy of a photon 'E' is given as the product of frequency and the planck's constant
E = hf
E = (6.63 × 10^-34) × (2.2 × 10^16)
E = 6.63 × 2.2 × 10^(-34 +16)
E = 14.586 × 10^-18
E = 1.4586 × 10^-17
E = 1.5 × 10-17 (2 S. F)
Answer:
C. 1.5 × 10–16 J
Explanation:
Question: A swimmer has a swimming speed of 5.75 m/s in still water. When she
heads directly across the river, she ends up travelling downstream at an angle of 40°
(with respect to a line perpendicular to the shore).
(a) What is the speed of the current?
(b) What heading would the swimmer need to have in order to reach a point directly
across the river?
(c) If the river is 15 m wide, how long would a trip directly across the river take?
Answers
Answer:
(a) 4.82 m/s
(b) The simmer need to head 57.05° upstream of the river to reach a point directly opposite the river
(c) 4.024 seconds
Explanation:
The parameters given are;
The speed of the swimmer = 5.75 m/s
The direction of the resultant motion = 40° downstream
The speed of the current = x m/s
Therefore in a given second, we have that the swimmer moves 5.75 meters across the river while the current moves x meters downstream
Which gives;
[tex]tan(40^{\circ}) = \dfrac{x}{5.75}[/tex]
x = 5.75 × tan(40°) = 4.82 m/s
The speed of the current = 4.82 m/s
(b) If the swimmer swims at a direction θ to reach a point, p, directly opposite the river, we have;
The resultant speed across the river will be the swimmer's swimming speed which is 5.75 m/s
5.75² = y² + 4.82²
y = √(5.75² - 4.82²) = √9.78
tan(θ) = 4.82/(√9.78) = 1.54
θ = tan⁻¹(1.54) = 57.05°
The simmer need to swim 57.05° upstream to reach a point directly opposite the river
(c) If the river is 15 m wide, we have;
Time (t) = Distance/speed
The component of the swimmer's speed directly across the river = 5.75/(tan(57.05°)
The component of the swimmer's speed directly across the river = 3.73 m/s
The time, t, to cross the river is therefore;
t = 15/3.73 = 4.024 seconds
I NEED HELP PLEASE, THANKS! Describe each Newton Law. :)
Answers
Explanation:
Newton's first law: Inertia
"An object in motion stays in motion, and an object at rest stays at rest, until acted upon by an unbalanced force."
This means that an object does not speed up, slow down, or change direction, unless the forces acting on it are unbalanced.
Newton's second law: Net force
∑F = ma
The net force (the sum of the forces acting on an object) is equal to the object's mass times its acceleration.
Newton's third law: Action/reaction
"For every action, there is an opposite and equal reaction."
This means that when one object pushes on another object (action), the second object pushes back with an equal and opposite force (reaction).
A wooden block of 0.60 kg is placed on a rough surface where it is accelerated at the rate of 4 m/s2. If the force applied on the block is 12N,what frictionless force is affecting the motion of block?
Answers
Answer:
Take a look below
Explanation:
Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 4.50 cm . Point a is at the center of the square, and point b is at the empty corner closest to q2. Take the electric potential to be zero at a distance far from both charges.Part AWhat is the electric potential at point a due to q1 and q2?Part BWhat is the electric potential at point b?Part CA point charge q3 = -4.00 μC moves from point a to point b. How much work is done on q3 by the electric forces exerted by q1 and q2?
Answers
Answer:
A) 0 V
B) -117 kV
C) -0.468 J
Explanation:
q1=+2.00μC q2=−2.00μC q3 = -4.00 μC
A) The electric potential at point a due to q1 and q2 ([tex]V_a[/tex]) is given as:
[tex]V_a=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=r_1=d.\ Therefore\\V_a=k(\frac{q_1}{d}+\frac{q_2}{d} )=k(\frac{2}{d}-\frac{2}{d} )=0[/tex]
B) he electric potential at point b due to q1 and q2 ([tex]V_b[/tex]) is given as:
[tex]V_b=k\Sigma \frac{q}{r_i} = k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )\\ but\ r_2=4.5 cm=0.045m,\ r_1=\sqrt{0.045^2+0.045^2}= 0.0636.\ Therefore\\V_b= k(\frac{q_1}{r_1}+\frac{q_2}{r_2} )= 9*10^{9}(\frac{2*10^{-6}}{0.0636}-\frac{2*10^{-6}}{0.045} )=-117\ kV[/tex]
C) The work done on q3 by the electric forces exerted by q1 and q2 (W) is given by:
[tex]W=q_3(V_a - V_b)=-4*10^{-6}(0-(-117*10^3))=-0.468\ J[/tex]
Physics students study a piano being pulled across a room on a rug. They know that when it is at rest, it experiences a gravitational force of 2200 N. When it is being pulled with 1200 N it is sliding across the room in equilibrium, but the students feel like they had to pull harder than 1200 N at first to begin motion.
Answers
Answer:
4
Explanation:
hope this helps
Answer:
The static frictional force is greater than the kinetic frictional force, so the static frictional force is greater than 1200 N.
Explanation:
Edge 2021