Answer:
The right answer is "0.273 m".
Explanation:
Given:
Power (P),
[tex]\frac{1}{f} = 2D[/tex]
Near point,
u = 0.6 m
As we know,
⇒ [tex]\frac{1}{v} -\frac{1}{u}=\frac{1}{f} = 2[/tex]
By substituting the values, we get
⇒ [tex]\frac{1}{v} -\frac{1}{0.6} =2[/tex]
[tex]\frac{1}{v}=2+\frac{1}{0.6}[/tex]
[tex]\frac{1}{v} =\frac{1.2+1}{0.6}[/tex]
[tex]\frac{1}{v}=\frac{2.2}{0.6}[/tex]
By applying cross-multiplication, we get
[tex]0.6=2.2 \ v[/tex]
[tex]v = \frac{0.6}{2.2}[/tex]
[tex]S_{near} = 0.273 \ m[/tex]
trong cùng một nhiệt độ, lượng năng lượng trên mỗi mol của chất khí nào lớn nhất
a) Khí đơn nguyên tử
b) Khí có từ ba nguyên tử
c) Khí lưỡng nguyên tử
You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. How much force must you exert to keep the pole motionless in a horizontal position?
Answer:
[tex]F=133N[/tex]
Explanation:
From the question we are told that:
Length [tex]l=3.0m[/tex]
Mass [tex]m=24kg[/tex]
Distance from Tip [tex]d=35cm[/tex]
Generally, the equation for Torque Balance is mathematically given by
[tex]mg(l/2)=F(l-d)[/tex]
[tex]2*9.81(3/2)=F(3-35*10^-2)[/tex]
Therefore
[tex]F=133N[/tex]
I need help with this please!!!!
Answer:
1.84 hours
I hope it's helps you
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
14. What's one of the two requirements electric current?
A. There must be an electric potential between two bodies
B. There must be no valence electrons that make their element unstable
C. There must be a carbon element present in the electric current
D. There must be a magnetic force between two bodies
Marko
One of the two requirements of electric current is there must be an electric potential between two bodies
For electric current to flow, there must be an electric potential between two bodies.
This is because electric charge flows from a higher electric potential to a lower electric potential just as, water flows from a higher gravitational potential to a lower gravitational potential.
The difference between the electric potential between the two bodies causes the electric charge to flow between the two bodies.
This flow of electric charge constitutes electric current and electric current will only flow when there is an electric potential between two bodies.
So, one of the two requirements of electric current is there must be an electric potential between two bodies.
So, the answer is A
Learn more about electric current here:
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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.
Air is compressed polytropically from 150 kPa, 5 meter cube to 800 kPa. The polytropic exponent for the process is 1.28. Determine the work per unit mass of air required for the process in kilojoules
a) 1184
b) -1184
c) 678
d) -678
Answer:
wegkwe fhkrbhefdb
Explanation:B
As a skydiver accelerates downward, what force increases? A. Gravity B. Thrust C. Air resistance D. Centripetal
Answer:
(A) Gravity is you're answer.
Explanation:
When an object or human is falling at an increased rate, The force of gravity is taking place.
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
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 bullet is dropped from the same height when another bullet is fired horizontally they will hit the ground
Answer:
simultaneously
Time taken to reach the ground depends on the vertical component of velocity, not horizontal component of velocity.
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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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
What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
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
Convert 385k to temperature of
Answer:
233.33°F
Explanation:
(385K - 273.15) * 9/5 + 32 = 233.33°F
Susan is quite nearsighted; without her glasses, her far point is 34 cm and her near point is 17 cm . Her glasses allow her to view distant objects with her eye relaxed. With her glasses on, what is the closest object on which she can focus?
Answer:
[tex]u=34cm[/tex]
Explanation:
From the question we are told that:
Far point is [tex]V=34 cm[/tex]
Near point is [tex]u=17 cm[/tex]
Therefore
Focal Length
[tex]f=-34cm[/tex]
Generally the equation for the Lens is mathematically given by
[tex]\frac{1}{u}=\frac{1}{f}-\frac{1}{v}[/tex]
[tex]\frac{1}{u}=\frac{1}{-34}-\frac{1}{-17}[/tex]
[tex]u=34cm[/tex]
A 1050 kg car accelerates from 11.3 m/s to 26.2 m/s . What impulse does the engine give?
Answer:
I = 15,645. kg*m/s or 15,645 N*s
Explanation:
I = m(^v)
I = 1050kg((26.2m/s-11.3m/s)
I = 15,645. kg*m/s
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
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 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
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
An object of mass 80 kg is released from rest from a boat into the water and allowed to sink. While gravity is pulling the object down, a buoyancy force of 1/50 times the weight of the object is pushing the object up (weight=mg). If we assume that water resistance exerts a force on the abject that is proportional to the velocity of the object, with proportionality constant 10 N-sec/m, find the equation of motion of the object. After how many seconds will the velocity of the object be 40 m/s? Assume that the acceleration due to gravity is 9.81 m/sec^2.
Answer:
a) Fnet = mg - Fb - Fr
b) 8.67 secs
Explanation:
mass of object = 80 kg
Buoyancy force = 1/50 * weight ( 80 * 9.81 ) = 15.696
Proportionality constant = 10 N-sec/m
a) Calculate equation of motion of the object
Force of resistance on object due to water = Fr ∝ V
= Fr = Kv = 10 V
Given that : Fb( due to buoyancy ) , Fr ( Force of resistance ) acts in the positive y-direction on the object while mg ( weight ) acts in the negative y - direction on the object.
Fnet = mg - Fb - Fr
∴ Equation of motion of the object ( Ma = mg - Fb - Fr )
b) Calculate how long before velocity of the object hits 40 m/s
Ma = mg - Fb - Fr
a = 9.81 - 0.1962 - 0.125 V = 9.6138 - 0.125 V
V = u + at ---- ( 1 )
u = 0
V = 40 m/s
a = 9.6138 - 0.125 V
back to equation 1
40 = 0 + ( 9.6138 - 0.125 (40) ) t
40 = 4.6138 t
∴ t = 40 / 4.6138 = 8.67 secs
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
A train starts from rest (at position zero) and moves with constant acceleration. On the first observation, its velocity is 20m/s and 80seconds later the velocity became 60m/s. At 80s calculate the position, average velocity, and the constant acceleration over the interval.(7-points)
The value of the acceleration is a = 0.5 m/s². The position at 80 s is x = 3200 m and finally the average velocity is v = 40 m/s.
Acceleration:
We can use the fallowing kinematic equation to get the acceleration at 80 s.
[tex]a=\frac{v_{f}-v_{i}}{t}[/tex]
Where:
v(i) is the initial velocity (20 m/s)v(f) is the final velocity (60 m/s)t is the interval (80 s)The, we have:
[tex]\vec{a}=\frac{60-20}{80}[/tex]
[tex]\vec{a}=0.5\: m/s^{2}[/tex]
Position:
Knowing the acceleration we can find the position using the falling equation.
[tex]\vec{x}=v_{i}t+0.5at^{2}[/tex]
[tex]\vec{x}=20*80+0.5*0.5*80^{2}[/tex]
[tex]\vec{x}=3200 m[/tex]
Average velocity:
The definition of the average velocity is:
[tex]\vec{v}=\frac{\Delta x}{t}[/tex]
[tex]\vec{v}=\frac{x_{f}-x_{i}}{t}[/tex]
[tex]\vec{v}=\frac{3200-0}{80}[/tex]
[tex]\vec{v}=40\: m/s[/tex]
Learn more about the kinematic equations here:
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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.
May someone help...please. Pretty please...
If a person is 18 % shorter than average, what is the ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height? Assume that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same.
f(short)/f(avg)=?
We have that the ratio of his walking pace to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
given the assumption and the calculation given below
From the question we are told that:
Consider a person 18\% shorter than average
Let average height of a person be [tex]10m[/tex]
Therefore
The height of an [tex]18\%[/tex] shorter man is mathematically given as
H=10*0.18
H=8.2m
Generally, the equation for velocity is mathematically given by
[tex]v=\frac{1}{2\pi} \sqrt{{g}{l}}[/tex]
Where we have the Assumption that a person's leg swings like a pendulum and that the angular amplitude of everybody's stride is about the same
Therefore
[tex]\frac{V_1}{V_2}=\frac{l_1}{l_2}[/tex]
[tex]\frac{V_1}{V_2}={82}{100}[/tex]
[tex]\frac{V_2}{V_1}=1.10[/tex]
In conclusion
The ratio of his walking pace (that is, the frequency 'f' of his motion) to the walking pace of a person of average height is
[tex]\frac{V_2}{V_1}=1.10[/tex]
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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
