Over the first 16.4 m, the person performs
W = (60.0 N) (16.4 m) = 984 J
of work.
Over the remaining 6.88 m, they perform a varying amount of work according to
F(x) ≈ 60.0 N + (-8.72 N/m) x
where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :
W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J
(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)
So the total work performed by the person on the box is
984 J + 206.4 J = 1190.4 J ≈ 1190 J
When you shine a beam of light, which is composed of just two different colors, red and green, onto a diffraction grating which color gets diffracted more
Answer:
The diffraction grating separates light into colors as the light passes through the many fine slits of the grating. This is a transmission grating. ... The prism separates light into colors because each color passes through the prism at a different speed and angle.
A wire, 0.60 m in length, is carrying a current of 2.0 A and is placed at a certain angle with respect to the magnetic field of strength 0.30 T. If the wire experiences a force of 0.18 N, what angle does the wire make with respect to the magnetic field
Answer:
[tex]\theta=30 \textdegree[/tex]
Explanation:
From the question we are told that:
Current [tex]I=2.0A[/tex]
Length [tex]L=0.60m[/tex]
Magnetic field [tex]B=0.30T[/tex]
Force [tex]F=0.18N[/tex]
Generally the equation for Force is mathematically given by
[tex]F = BIL sin\theta[/tex]
[tex]sin\theta=\frac{F}{BIL}[/tex]
[tex]\theta=sin^{-1}\frac{0.18}{0.3*2*0.6}[/tex]
[tex]\theta=30 \textdegree[/tex]
how will be electric lines of force where intensity of electric field is maximum ?
a. wider
b. +ve to -ve
c. narrow
d. -ve to +ve
i'm pretty sure the answer is A wider
Electric lines of force where intensity of electric field is maximum when its wider.
What is Electric field?The physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field. It can also refer to a system of charged particles' physical field.
Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.
Electrical technology makes use of electric fields, which are significant in many branches of physics. For instance, in atomic physics and chemistry, the electric field acts as an attracting force to hold atoms' atomic nuclei and electrons together.
Therefore, Electric lines of force where intensity of electric field is maximum when its wider.
To learn more about electric field, refer to the link:
https://brainly.com/question/1443103
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Si un resorte de constante elástica 1300 n/m se comprime 12 cm ¿Cuanta energía almacena? Y si estira 12cm ¿Cuanta energía almacena?
La energía que almacena el resorte cuando se comprime y estira 12 cm es 9,4 J.
La energía potencial elástica del resorte se puede calcular con la siguiente ecuación:
[tex] E_{p} = \frac{1}{2}kx^{2} [/tex]
En donde:
k: es la constante del resorte = 1300 N/m
x: es la distancia de compresión o de elongación = 12 cm = 0,12 m
Dado que la energía es proporcional al cuadrado de la distancia recorrida por el resorte (x), la energía almacenada por el resorte durante la compresión será la misma que la energía almacenada por la elongación.
Por lo tanto, la energía almacenada es:
[tex]E_{p} = \frac{1}{2}kx^{2} = \frac{1}{2}1300 N/m*(0,12 m)^{2} = 9,4 J[/tex]
Entonces, la energía del resorte cuando se comprime y cuando se estira es la misma, a saber 9,4 J.
Para saber más sobre energía potencial visita este link: https://brainly.com/question/156316?referrer=searchResults
Espero que te sea de utilidad!
Answer:
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
Explanation:
La Energía Potencial Elástica almacenada por el resorte ([tex]U_{e}[/tex]), en joules, se calcula a partir de la Ley de Hooke, la definición de Trabajo y el Teorema del Trabajo y la Energía, cuya expresión se presenta abajo:
[tex]U_{e} = \frac{1}{2}\cdot k\cdot (x_{f}^{2}-x_{o}^{2})[/tex] (1)
Donde:
[tex]k[/tex] - Constante elástica del resorte, en newtons por metro.
[tex]x_{o}[/tex] - Posición inicial del resorte, en metros.
[tex]x_{f}[/tex] - Posición final del resorte, en metros.
Nótese que el resorte sin deformar tiene una posición de cero, la tensión tiene un valor positivo y la compresión, negativo.
Asumiendo que en ambos casos el resorte se encuentra inicialmente sin deformar, se reduce (1) a una forma de función par, es decir, una función que cumple con la propiedad de que [tex]f(x) = f(-x)[/tex], se encuentra que al comprimirse o estirarse en la misma medida almacena la misma cantidad de energía.
La cantidad de energía a almacenar es:
[tex]U_{e} = \frac{1}{2}\cdot \left(1300\,\frac{N}{m} \right)\cdot (0,12\,m)^{2}[/tex]
[tex]U_{e} = 9,360\,J[/tex]
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
As the speed of a particle approaches the speed of light, the momentum of the particle Group of answer choices approaches zero. decreases. approaches infinity. remains the same. increases.
Answer:
approaches infinity
Explanation:
There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.
The relativistic momentum can be written as:
[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
where
u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.
m = mass of the object
c = speed of light.
So, as u tends to c, we will have:
[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]
Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.
Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.
Trình bày những hiểu biết của em về đại lượng vận tốc dài, vận tốc góc(định nghĩa, công thức, ý nghĩa, đơn vị, loại đại lượng).
which characteristic of nuclear fission makes it hazardous?
Answer:The radioactive waste
Explanation:Fission is the splitting of a heavy unstable nucleus into two Lighter nuclei
A regulation soccer field for international play is a rectangle with a length between 100 m and a width between 64 m and 75 m. What are the smallest and largest areas that the field could be?
Answer:
The smallest and largest areas could be 6400 m and 7500 m, respectively.
Explanation:
The area of a rectangle is given by:
[tex] A = l*w [/tex]
Where:
l: is the length = 100 m
w: is the width
We can calculate the smallest area with the lower value of the width.
[tex] A_{s} = 100 m*64 m = 6400 m^{2} [/tex]
And the largest area is:
[tex] A_{l} = 100 m*75 m = 7500 m^{2} [/tex]
Therefore, the smallest and largest areas could be 6400 m and 7500 m, respectively.
I hope it helps you!
Answer:
the largest areas that the field could be is [tex]A_l[/tex]=7587.75 m
the smallest areas that the field could be is [tex]A_s[/tex]=6318.25 m
Explanation:
to the find the largest and the smallest area of the field measurement error is to be considered.
we have to find the greatest possible error, since the measurement was made nearest whole mile, the greatest possible error is half of 1 mile and that is 0.5m.
therefore to find the largest possible area we add the error in the mix of the formular for finding the perimeter with the largest width as shown below:
[tex]A_l[/tex]= (L+0.5)(W+0.5)
(100+0.5)(75+0.5) = (100.5)(75.5) = 7587.75 m
To find the smallest length we will have to subtract instead of adding the error factor value of 0.5 as shown below:
[tex]A_s[/tex]= (L-0.5)(W-0.5)
(100-0.5)(64-0.5) = (99.5)(63.5) = 6318.25 m
Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
A car is driving towards an intersection when the light turns red. The brakes apply a constant force of 1,398 newtons to bring the car to a complete stop in 25 meters. If the weight of the car is 4,729 newtons, how fast was the car going initially
Answer:
the initial velocity of the car is 12.04 m/s
Explanation:
Given;
force applied by the break, f = 1,398 N
distance moved by the car before stopping, d = 25 m
weight of the car, W = 4,729 N
The mass of the car is calculated as;
W = mg
m = W/g
m = (4,729) / (9.81)
m = 482.06 kg
The deceleration of the car when the force was applied;
-F = ma
a = -F/m
a = -1,398 / 482.06
a = -2.9 m/s²
The initial velocity of the car is calculated as;
v² = u² + 2ad
where;
v is the final velocity of the car at the point it stops = 0
u is the initial velocity of the car before the break was applied
0 = u² + 2(-a)d
0 = u² - 2ad
u² = 2ad
u = √2ad
u = √(2 x 2.9 x 25)
u =√(145)
u = 12.04 m/s
Therefore, the initial velocity of the car is 12.04 m/s
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
state the laws of reflection
Answer:
Explanation:
The law of reflection says that the reflected angle (measured from a vertical line to the surface called the normal) is equal to the reflected angle measured from the same normal line.
All other properties of reflection flow from this one statement.
Parallel Wires: Two long, parallel wires carry currents of different magnitudes. If the current in one of the wires is doubled and the current in the other wire is halved, what happens to the magnitude of the magnetic force that each wire exerts on the other?
Answer:
Explanation:
Given force between 2 currents carrying
wires = F₀
Magnetic force between the2 wires =F₀= (μ₀/4π) x ( 2 (μ₀/4π) x ( 2I₁I₂ / μ) x L
where I₁=Current in wire 1
I₂= Current in wire 2
L= Length of the wire
when one current is doubled and the other is halved
I₁= 2 I₁
I₂= I₂/2
F₀ = (μ₀/4π) x ( 2× (2I₁) (I₂/2) / μ) x L
If a boy lifts a mass of 6kg to a height of 10m and travels horizontally with a constant velocity of 4.2m/s, calculate the work done? Explain your answer.
Answer:
W = 641.52 J
Explanation:
The work done here will be the sum of potential energy and the kinetic energy of the boy. Here potential energy accounts for vertical motion part while the kinetic energy accounts for the horizontal motion part:
[tex]Work\ Done = Kinetic\ Energy + Potential\ Energy\\\\W = K.E +P.E\\\\W = \frac{1}{2}mv^2+mgh\\\\[/tex]
where,
W = Work Done = ?
m = mass = 6 kg
v = speed = 4.2 m/s
g = acceleration dueto gravity = 9.81 m/s²
h = height = 10 m
Therefore,
[tex]W = \frac{1}{2}(6\ kg)(4.2\ m/s)^2+(6\ kg)(9.81\ m/s^2)(10\ m)[/tex]
W = 52.92 J + 588.6 J
W = 641.52 J
What Are the type's of Tidal turbines?
Answer:
Types of tidal turbines
Axial turbines.
Crossflow turbines.
Flow augmented turbines.
Oscillating devices.
Venturi effect.
Tidal kite turbines.
Turbine power.
Resource assessment.
Answer:
Axial turbines
Crossflow turbines
flow augmented turbines
I need help with this please!!!!
Answer:
1.84 hours
I hope it's helps you
How do you know that a liquid exerts pressure?
Answer:
The pressure of water progressively increases as the depth of the water increases. The pressure increases as the depth of a point in a liquid increases. The walls of the vessel in which liquids are held are likewise subjected to pressure. The sideways pressure exerted by liquids increases as the liquid depth increases.
If you dive underwater, you notice an uncomfortable pressure on your eardrums due to the increased pressure. The human eardrum has an area of about 70 mm217 * 10-5 m22, and it can sustain a force of about 7 N without rupturing. If your body had no means of balancing the extra pressure (which, in reality, it does), what would be the maximum depth you could dive without rupturing your eardrum
Answer:
[tex]h=10m[/tex]
Explanation:
From the question we are told that:
Area [tex]a=70 x 10^{-6}[/tex]
Force [tex]F=7N[/tex]
Generally the equation for Pressure is mathematically given by
Pressure = Force/Area
[tex]P=\frac{F}{A}[/tex]
[tex]P=\frac{ 7}{(70 * 10^{-6})}[/tex]
[tex]P= 1*10^{5} Pa[/tex]
Generally the equation for Pressure is also mathematically given by
[tex]P=hpg[/tex]
Therefore
[tex]h=\frac{P}{hg}[/tex]
[tex]h=\frac{10000}{1000*9.8}[/tex]
[tex]h=10m[/tex]
When the lightbulbs were used as the resistors, you observed only a flash of light, as opposed to a continuous glow. Explain why that behavior is expected. After all, the light bulb is directly connected to the power supply.
Solution :
Whenever the lightbulbs are used as resistors, we throw the switch to the left. This allows the current to flow through the circuit which causes the bulb to glow and also the capacitor gets charged. When the capacitor gets fully charged, the electric field becomes constant between its two plates. Now there is no displacement current induced in the plates of the capacitor. The capacitor works as an open switch and the bulb gets switched off.
And thus the bulb flashes for the moment as opposed to continuous glow.
How much amount of water can be decomposed
through electrolysis by passing 2 F charge?
Answer:
So, with 2 Faraday of electricity, we can decompose (2/4 × 2) = 1 mole of water. So 18 grams of water is decomposed.
2- A student ran 135 meters in 15 seconds. What was the student's velocity?
*
7.5 m/s
9 m/s
12 m/s
15 m/s
Answer:
9 Brainly hahaha ............huh
1 Poin Question 4 A 85-kg man stands in an elevator that has a downward acceleration of 2 m/s2. The force exerted by him on the floor is about: (Assume g = 9.8 m/s2) А ON B 663 N C) 833 N D) 1003 N
Answer:
D) 1003 N
Explanation:
Given the following data;
Mass of man = 85 kg
Acceleration of elevator = 2 m/s²
Acceleration due to gravity, g = 9.8 m/s²
To find the force exerted by the man on the floor;
Force = mg + ma
An ideal double slit interference experiment is performed with light of wavelength 640 nm. A bright spot is observed at the center of the resulting pattern as expected. For the 2n dark spot away from the center, it is known that light passing through the more distant slit travels the closer slit.
a) 480 nm
b) 600 nm
c) 720 nm
d) 840 nm
e) 960 nm
Answer:
960 nm
Explanation:
Given that:
wavelength = 640 nm
For the second (2nd) dark spot; the order of interference m = 1
Thus, the path length difference is expressed by the formula:
[tex]d sin \theta = (m + \dfrac{1}{2}) \lambda[/tex]
[tex]d sin \theta = (1 + \dfrac{1}{2}) 640[/tex]
[tex]d sin \theta = ( \dfrac{3}{2}) 640[/tex]
dsinθ = 960 nm
A car accelerates from 0 m/s to 25 m/s in 5 seconds. What is the average acceleration of the car.
Answer:
5 m/s I hope it will help you
Explanation:
mark me as a brainlist answer
A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length [tex]l[/tex] = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance [tex]R_{sol[/tex] = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
[tex]R_o[/tex] = 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) = [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]
so
ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )
ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]
ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
Answer:
1.28 *10^-5 A
Explanation:
Same work as above answer. Needs to be more precise
Explain why liquid particles at a high pressure would need more
energy to change to a gas than liquid particles at a low pressure.
Answer:
Liquids evaporate faster as they heat up and more particles have enough energy to break away. The particles need energy to overcome the attractions between them. ... At this point the liquid is boiling and turning to gas. The particles in the gas are the same as they were in the liquid they just have more energy.
Mass A, 2.0 kg, is moving with an initial velocity of 15 m/s in the x-direction, and it collides with mass M, 4.0 kg, initially moving at 7.0 m/s in the x-direction. After the collision, the two objects stick together and move as one. What is the change in kinetic energy of the system as a result of the collision, in joules
Answer:
the change in the kinetic energy of the system is -42.47 J
Explanation:
Given;
mass A, Ma = 2 kg
initial velocity of mass A, Ua = 15 m/s
Mass M, Mm = 4 kg
initial velocity of mass M, Um = 7 m/s
Let the common velocity of the two masses after collision = V
Apply the principle of conservation of linear momentum, to determine the final velocity of the two masses;
[tex]M_aU_a + M_mU_m = V(M_a + M_m)\\\\(2\times 15 )+ (4\times 7) = V(2+4)\\\\58 = 6V\\\\V = \frac{58}{6} = 9.67 \ m/s[/tex]
The initial kinetic of the two masses;
[tex]K.E_i = \frac{1}{2} M_aU_a^2 \ + \ \frac{1}{2} M_mU_m^2\\\\K.E_i = (0.5 \times 2\times 15^2) \ + \ (0.5 \times 4\times 7^2)\\\\K.E_i = 323 \ J[/tex]
The final kinetic energy of the two masses;
[tex]K.E_f = \frac{1}{2} M_aV^2 \ + \ \frac{1}{2} M_mV^2\\\\K.E_f = \frac{1}{2} V^2(M_a + M_m)\\\\K.E_f = \frac{1}{2} \times 9.67^2(2+ 4)\\\\K.E_f = 280.53 \ J[/tex]
The change in kinetic energy is calculated as;
[tex]\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 280.53 \ J \ - \ 323 \ J\\\\\Delta K.E = -42.47 \ J[/tex]
Therefore, the change in the kinetic energy of the system is -42.47 J
1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of 2 m above the elevator floor when the elevator floor when the elevator is 28 m above the ground.
a. What's the maximum height?
b. How long does it take for the ball to return to the elevator floor?
(a) The maximum height reached by the ball from the ground level is 75.87m
(b) The time taken for the ball to return to the elevator floor is 2.21 s
The given parameters include:
constant velocity of the elevator, u₁ = 10 m/sinitial velocity of the ball, u₂ = 20 m/sheight of the boy above the elevator floor, h₁ = 2 mheight of the elevator above the ground, h₂ = 28 mTo calculate:
(a) the maximum height of the projectile
total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)
at maximum height the final velocity of the projectile (ball), v = 0
Apply the following kinematic equation to determine the maximum height of the projectile.
[tex]v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m[/tex]
The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection
h = h₁ + h₂ + h₃
h = 28 m + 2 m + 45.87 m
h = 75.87 m
(b) The time taken for the ball to return to the elevator floor
Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m
Apply the following kinematic equation to determine the time to return to the elevator floor.
[tex]h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s[/tex]
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the 2kg block slids down a firctionless curved ramp starting from rest at heiht of 3m what is the speed of the block at the bottemvof the ramp
A
Explanation:
1qdeeeeeeeeeeehhhhhhhhhwilffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffffff.
What is utilization of energy
Explanation:
Energy utilization focuses on technologies that can lead to new and potentially more efficient ways of using electricity in residential, commercial and industrial settings—as well as in the transportation sector
