Answer:
Explanation:
In a reversible reaction which has reached dynamic equilibrium , rate of forward reaction is equal to rate of backward reaction .
Following is a reversible chemical reaction .
A + B = C + D
Rate of forward reaction = k₁ x [ A ] x [ B ]
Rate of backward reaction = k₂ x [ C ] x [ D ]
k₁ x [ A ] x [ B ] = k₂ x [ C ] x [ D ]
[ A ] x [ B ] = k₂ / k₁ [ C ] x [ D ]
[ A ] x [ B ] = k [ C ] x [ D ]
The products of the forward and backward reactions remain
constant at equilibrium.
Hence option A and D are correct statement .
how can the starch be removed from the leaves of potted plants
Answer:
Explanation:
There are two main ways to de-starch leaves of a plant - the 'Light Exclusion' Method and the 'Carbon Dioxide Deprivation' Method. The 'Light Exclusion' method is a simpler procedure and is used often. Leaves can be destarched by depriving them of light for an extended period of time, usually 24-48 hours.
2 coplas o pregones inventadas por ti relacionadas con la región caribe.
A laser beam enters one of the sloping faces of the equilateral glass prism (n=1.42) and refracts through the prism. Within the prism the light travels horizontally. What is the angle between the direction of the incident ray and the direction of the outgoing ray?
Answer:
30.5°
Explanation:
Since the light travels horizontally through the prism, it undergoes minimum deviation. So, the angle between the direction of the incident ray and that of the outgoing ray D is gotten from
n = [sin(D + α)/2]/sin(α/2) where n = refractive index of prism = 1.42 and α = angle of prism = 60° (since it is a n equilateral glass prism).
Making D subject of the formula, we have
n = [sin(D + α)/2]/sin(α/2)
nsin(α/2) = [sin(D + α)/2]
(D + α)/2 = sin⁻¹[nsin(α/2)]
D + α = 2sin⁻¹[nsin(α/2)]
D = 2sin⁻¹[nsin(α/2)] - α
So, substituting the values of the variables into the equation, we have
D = 2sin⁻¹[nsin(α/2)] - α
D = 2sin⁻¹[1.42sin(60°/2)] - 60°
D = 2sin⁻¹[1.42sin(30°)] - 60°
D = 2sin⁻¹[1.42 × 0.5] - 60°
D = 2sin⁻¹[0.71] - 60°
D = 2(45.23°) - 60°
D = 90.46° - 60°
D = 30.46°
D ≅ 30.5°
what does Newton's third law ? Describe.
Newton's third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A. In other words, forces result from interactions.
Hope it helps
Answer:
Newton's third law state that every action there is equal but opposite reactions
hope this will help you more
Need an answer in hurry u can make the pic big
1. A box contains 10 blue chips. 5 red chips, and 15 yellow chips. Find the odds of choosing the
following:
blue chip
b. yellow chip
c. yellow chip
Answer:
Explanation:
Blue: 10/30
Red: 5/30
Yellow: 15/30
The probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
What is the probability?
The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.
Given that
Blue chips =10
Red chips = 5
Yellow chips = 15
Number of the total samples =10+15+5=30
Probability of choosing Blue chips = [tex]\dfrac{10}{30}= \dfrac{1}{3}[/tex]
Probability of Red chips =[tex]\dfrac{5}{30}=\dfrac{1}{6}[/tex]
Probability of Yellow chips =[tex]\dfrac{15}{30}=\dfrac{1}{2}[/tex]
Thus the probability of finding the blue red and yellow chips is 1/3, 1/6 and 1/2
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What happens if you move a magnet near a coil of wire?
A) current is induced
B)power is consumed
C)the coil becomes magnetized
D) the magnets field is reduced
Need in hurry important please
Answer:
I don't see anything on your question?
You should extending your throwing hand straight up to the sky to follow-through.
O True
O False
False
It's not straight up
What is binding energy?
A.' The attractive forces between the protons in the nucleus and the
electrons
B. The energy required to force two nuclei to undergo nuclear fusion
C. The amount of energy stored in the strong nuclear forces of the
nucleus
D. The amount of energy required to overcome an activation energy
barrier
Please help me out.
Answer:
the answer is B i hope it helps :)
[tex]\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}[/tex]
B. The energy required to force two nuclei to undergo nuclear fusion. ✅
They are usually expressed in terms of [tex]\sf\purple{kJ/mole}[/tex] of nuclei or [tex]\sf\pink{MeV's/nucleon}[/tex].[tex]\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}[/tex]
according to the law of conservation of vhange , what must always be true in a nuclear reaction?
Answer:
The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same
Explanation:
Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.
I NEED THE ANSWER QUICK PLEASEE
which of the following is the correct description of momentum?
-the product of mass and acceleration -the product of mass and velocity
-velocity divided by mass
-acceleration divided by mass
Answer:
The product of mass and velocity is the correct answer
Explanation:
Momentum is defined as mass × velocity
p = mv
Answer:
The product of mass and velocity
Explanation:
I just did it and got it right with a 100%
Posted 1/3/23
Một vô lăng sau khi bắt đầu quay được một phút thì thu được vận tốc 700
vòng/phút. Tính gia tốc góc của vô lăng
reviews the general principles in this problem. A projectile is launched from ground level at an angle of 13.0 ° above the horizontal. It returns to ground level. To what value should the launch angle be adjusted, without changing the launch speed, so that the range doubles?
Answer: θ would equal approximately 28.7°
This is a kinematics problem, where one is only given the theta value 13.0° in regards to the range; thus, the problem is testing one's understanding of the relationships between the variables.
Range (aka x) = (v₀ sin (2θ₀))/g, where θ₀ = 13.0°
Now if we multiply the range by 2, we get:
2x = 2((v₀ sin (2θ₀))/g) → to verbalize, if range equates to (v₀ sin (2θ₀))/g, and doubling the range equals twice the product value, then:
2θ = sin⁻¹(2sin(2(13.0° )) = sin⁻¹(2(0.76255845048)) = sin⁻¹ (1.52511690096) = 57.35560850015109°/2 = θ
Thus, θ = 28.67780425
It's been awhile since I did this; though I hope it helped!
The fastest pitched baseball was clocked at 47 m/s. Assume that the pitcher exerted his force (assumed to be horizontal and constant) over a distance of 1.0 mm, and a baseball has a mass of 145 g.
Required:
a. Draw a free-body diagram of the ball during the pitch.
b. What force did the pitcher exert on the ball during this record-setting pitch?
c. Estimate the force in part b as a fraction of the pitcher's weight.
Answer:
Following are the solution to the given points:
Explanation:
For point a:
Find the schematic of the empty body and in attachment. Upon on ball during the pitch only two forces act:
The strength of the pitcher F is applied that operates horizontally. Its gravity force acting on an object is termed weight, which value is where m denotes mass, and g the acceleration of gravity.
For point b:
[tex]160.2\ N[/tex]
First, they must find that ball's acceleration. You can use the SUVAT equation to achieve that
where
[tex]v = 47\ \frac{m}{s} \\\\u = 0 \\\\a =?\\\\d = 1.0 \ m \\\\[/tex]
Solving for a,
[tex]a=\frac{v^2-u^2}{2d}=\frac{47^2-0}{2(1.0)}=1104.5 \ \frac{m}{s^2}[/tex]
Calculating the mass:
[tex]m = 145 g = 0.145 kg[/tex]
Calculating the force:
[tex]F=ma=0.145 \times 1104.5= 160.2 \ N[/tex]
For point c:
0.195 times the pitcher's weight
[tex]m = 84 \ kg \\\\g = 9.8\ \frac{m}{s^2}\\\\[/tex]
Solving for W:
[tex]W=84 \times 9.8= 823.2 \ N[/tex]
Now the force of Part B could be defined as the fraction of the mass of the pitcher:
[tex]\frac{F}{W}=\frac{160.2}{823.3}=0.195[/tex]
A certain microscope is provided with objectives that have focal lengths of 20 mm , 4 mm , and 1.4 mm and with eyepieces that have angular magnifications of 5.00 × and 15.0 × . Each objective forms an image 120 mm beyond its second focal point.
Answer:
Explanation:
Given that:
Focal length for the objective lens = 20 mm, 4 mm, 1.4 mm
For objective lens of focal length f₁ = 20 mm
s₁' = 120 mm + 20 mm = 140 mm
∴
Magnification [tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{140}{20}[/tex]
[tex]m_1 = 7 \ m[/tex]
For objective lens of focal length f₁ = 4 mm
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{124}{4}[/tex]
[tex]m_1 = 31 \ m[/tex]
For objective lens of focal length f₁ = 1.4 mm
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1 = \dfrac{s'_1}{f_1}[/tex]
[tex]m_1 = \dfrac{121.4}{1.4}[/tex]
[tex]m_1 = 86.71 \ m[/tex]
The magnification of the eyepiece is given as:
[tex]m_e = 5X \ and \ m_e = 15X[/tex]
Thus, the largest angular magnification when [tex]m_1 \ and \ m_e \ are \ large \ is:[/tex]
[tex]M_{large}= (m_1)_{large} \times (m_e)_{large}[/tex]
= 86.71 × 15
= 1300.65
The smallest angular magnification derived when [tex]m_1 \ and \ m_e \ are \ small \ is:[/tex]
[tex]M_{small}= (m_1)_{small} \times (m_e)_{small}[/tex]
= 7 × 5
= 35
The largest magnification will be 1300.65 and the smallest magnification will be 35.
What is magnification?Magnification is defined as the ratio of the size of the image of an object to the actual size of the object.
Now for objective lens and eyepieces, it is defined as the ratio of the focal length of the objective lens to the focal length of the eyepiece.
It is given in the question:
Focal lengths for the objective lens is = 20 mm, 4 mm, 1.4 mm
now we will calculate the magnification for all three focal lengths of the objective lens.
Also, each objective forms an image 120 mm beyond its second focal point.
(1) For an objective lens of focal length [tex]f_1=20 \ mm[/tex]
[tex]s_1'=120\ mm +20 \ mm =140\ mm[/tex]
Magnification will be calculated as
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{140}{20} =7[/tex]
(2) For an objective lens of focal length [tex]f_1= \ 4 \ mm[/tex]
s₁' = 120 mm + 4 mm = 124 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{124}{4} =31[/tex]
(3) For an objective lens of focal length [tex]f_1=1.4\ mm[/tex]
s₁' = 120 mm + 1.4 mm = 121.4 mm
[tex]m_1=\dfrac{s_1'}{f_1} =\dfrac{121.4}{1.4} =86.71[/tex]
Now the magnification of the eyepiece is given as:
[tex]m_e=5x\ \ \ & \ \ m_e=15x[/tex]
Thus, the largest angular magnification when
[tex]m_1 = 86.17\ \ \ \ m_e=15x[/tex]
[tex]m_{large}= (m_1)_{large}\times (m_e)_{large}[/tex]
[tex]m_{large}=86.71\times 15=1300.65[/tex]
The smallest angular magnification derived when
[tex]m_1=7\ \ \ \ m_e=5[/tex]
[tex]m_{small}=(m_1)_{small}\times (m_e)_{small}[/tex]
[tex]m_{small}=7\times 5=35[/tex]
Thus the largest magnification will be 1300.65 and the smallest magnification will be 35.
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A circular ice rink is 20 m in diameter and is to be temporarily enclosed in a hemispherical dome of the same diameter. The ice is maintained at 270 K. On a particular day the inner surface of the dome is measured to be 290 K. Estimate the radiant heat transfer from the dome to the rink if both surfaces can be taken as blackbody.
570k or 1050k
270k+290k= 570k
270k•290k = 1050k
PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?
Answer:
[tex]\lambda=1.39\times 10^{-4}\ s^{-1}[/tex]
Explanation:
Given that,
The half-life of Barium-139 is [tex]4.96\times 10^3[/tex]
A sample contains [tex]3.21\times 10^{17}[/tex] nuclei.
We need to find the decay constant for this decay. The formula for half life is given by :
[tex]T_{1/2}=\dfrac{0.693}{\lambda}\\\\\lambda=\dfrac{0.693}{T_{1/2}}[/tex]
Put all the values,
[tex]\lambda=\dfrac{0.693}{4.96\times 10^3}\\\\=1.39\times 10^{-4}\ s^{-1}[/tex]
So, the decay constant is [tex]1.39\times 10^{-4}\ s^{-1}[/tex].
A body starts from rest and moves with a uniform acceleration of 2 m/s2 in a
straight line. What is the velocity after 5s, How far has it travelled in this time and When will it be 100m from its starting point?
Answer:
Final velocity = 10 m/s
Time taken to travel 100 meter = 8.16 second (Approx.)
Explanation:
Given:
Acceleration = 2 m/s²
Initial velocity = 0 m/s
Find:
Velocity after 5 seconds
Time taken to travel 100 meter
Computation:
Using first equation of motion
v = u + at
v = 0 + (2)(5)
Final velocity = 10 m/s
Using Second equation of motion
s = ut + (1/2)(a)(t²)
100 = (0)(t) + (1/2)(3)(t²)
100 = (1/2)(2)(t²)
100 = (t²)
t = 10
Time taken to travel 100 meter = 8.16 second (Approx.)
A race car starts from rest on a circular track of radius 378 m. The car's speed increases at the constant rate of 0.580 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, find the following.
a. The speed of the race car
b. The distance traveled
c. the elapsed time
Answer:
a) [tex]V=14.904m/s[/tex]
b) [tex]d = 191.49 m[/tex]
c) [tex]t= 25.696 s[/tex]
Explanation:
From the question we are told that:
Radius [tex]r =378m[/tex]
Acceleration [tex]a=0.580[/tex]
a)
Generally the equation for speed of the car is mathematically given by
[tex]a=\frac{v^2}{r}[/tex]
[tex]V=\sqrt{a*r}[/tex]
[tex]V=\sqrt{0.58*383}[/tex]
[tex]V=14.904m/s[/tex]
b)
Generally the equation for distance traveled of the car is mathematically given by
[tex]V^2=u^2+2ad[/tex]
[tex]d=\frac{V^2}{2a}[/tex]
[tex]d=\frac{14.904^2}{2*0.58}[/tex]
[tex]d = 191.49 m[/tex]
c)
Generally the equation for time of the car is mathematically given by
[tex]V=u+at[/tex]
[tex]t=\frac{V}{a}[/tex]
[tex]t=\frac{14.904}{0.58}[/tex]
[tex]t= 25.696 s[/tex]
Which pair of magnets has the strongest attraction between them?
P5. A bullet with an initial velocity of 280 m/s in the x-direction penetrates an initially stationary block of mass 11 kg and emerges on the other side with a final velocity of 70 m/s in the x-direction. The velocity of the block after the collision is 0.2 m/s, also in the x-direction. Assume the block slides on a horizontal frictionless surface. What is the mass of the bullet
Answer:
the mass of the bullet is 10.5 g
Explanation:
Given;
initial velocity, u₁ = 280 m/s
final velocity of the bullet, v₁ = 70 m/s
final velocity of the block, v₂ = 0.2 m/s
mass of the block, m₂ = 11 kg
initial velocity of the block, u₂ = 0
let the mass of the bullet = m₁
Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
280m₁ + 11(0) = 70m₁ + 11 x 0.2
280m₁ = 70m₁ + 2.2
280m₁ - 70m₁ = 2.2
210m₁ = 2.2
m₁ = 2.2/210
m₁ = 0.0105 kg
m₁ = 10.5 g
Therefore, the mass of the bullet is 10.5 g
A toddler weighs 10 kg and raises herself onto tiptoe (on both feet). Her feet are 8 cm long with each ankle joint being located 4.5 cm from the point at which her feet contact the floor. While standing on tip toe:
(a) what is the upward normal force exerted by the floor at the point at which one of the toddler's feet contacts the floor?
(b) what is the tension force in one of her Achilles tendons? (c) what is the downward force exerted on one of the toddler's
ankle joints?
Answer:
a.49 n
b. 63 n
c. 112 n
Explanation:
a.10 times 9.8 from gravity/2 = 49 n
b. 49n times 4.5/8-4.5 = 63 n
c 49n + 63 n = 112 n
A tennis ball of mass of 0.06 kg is initially traveling at an angle of 47o to the horizontal at a speed of 45 m/s. It then was shot by the tennis player and return horizontally at a speed of 35 m/s. Find the impulse delivered to the ball.
Answer:
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
Explanation:
By Impulse Theorem, the motion of the tennis ball is modelled after the following expression:
[tex]Imp = m\cdot (\vec v_{f} - \vec v_{o})[/tex] (1)
Where:
[tex]m[/tex] - Mass of the ball, in kilograms.
[tex]\vec v_{o}[/tex] - Vector of the initial velocity, in meters per second.
[tex]\vec v_{f}[/tex] - Vector of the final velocity, in meters per second.
[tex]Imp[/tex] - Impulse, in meters per second.
If we know that [tex]m = 0.06\,kg[/tex], [tex]\vec v_{o} = \left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})[/tex] and [tex]\vec v_{f} = \left(35\,\frac{m}{s} \right)\cdot (-1, 0)[/tex], then the impulse delivered to the ball is:
[tex]Imp = (0.06\,kg)\cdot \left[\left(35\,\frac{m}{s} \right)\cdot (-1,0) -\left(45\,\frac{m}{s} \right)\cdot (\cos 47^{\circ}, \sin 47^{\circ})\right][/tex]
[tex]Imp = (0.06\,kg)\cdot (-65.670, -32.911)\,\left[\frac{m}{s} \right][/tex]
[tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex]
The impulse delivered to the ball is [tex]Imp = \left(-3.941, 1.975\right)\,\left[\frac{kg\cdot m}{s} \right][/tex].
A 45-kg skydiver jumps out of an airplane and falls 450 m, reaching a maximum speed of 51 m/s before opening her parachute. How much work, in joules, did air resistance do on the skydiver before she opened her parachute
Answer:
The work done by the friction force is - 139927.5 J.
Explanation:
mass of diver, m = 45 kg
distance falls, h = 450 m
initial speed, u = 0 m/s
final speed = 51 m/s
According to the work energy theorem,
Work done by the gravity + work done by the friction force = change in kinetic energy
[tex]m g h + W' = 0.5 m ()v^2 - u^2)\\\\45\times 9.8\times 450 + W' = 0.5\times 45\times (51^2 - 0)\\\\198450 + W' = 58522.5\\\\W' = - 139927.5 J[/tex]
What happens in the gray zone between solid and liquid?-,-
The two most prominent wavelengths in the light emitted by a hydrogen discharge lamp are 656 nm (red) and 486 nm (blue). Light from a hydrogen lamp illuminates a diffraction grating with 550 lines/mm , and the light is observed on a screen 1.7 m behind the grating.
What is the distance between the first-order red and blue fringes?
Express your answer to two significant figures and include the appropriate units.
Answer:
Δd = 7.22 10⁻² m
Explanation:
For this exercise we must use the dispersion relationship of a diffraction grating
d sin θ = m λ
let's use trigonometry
tan θ = y / L
how the angles are small
tant θ = sinθ /cos θ = sin θ
we substitute
sin θ = y / L
d y / L = m λ
y = m λ L / d
let's use direct ruler rule to find the distance between two slits
If there are 500 lines in 1 me, what distance is there between two lines
d = 2/500
d = 0.004 me = 4 10⁻⁶ m
diffraction gratings are built so that most of the energy is in the first order of diffraction m = 1
let's calculate for each wavelength
λ = 656 nm = 656 10⁻⁹ m
d₁ = 1 656 10⁻⁹ 1.7 / 4 10⁻⁶
d₁ = 2.788 10⁻¹ m
λ = 486 nm = 486 10⁻⁹ m
d₂ = 1 486 10⁻⁹ 1.7 / 4 10⁻⁶
d₂ = 2.066 10⁻¹ m
the distance between the two lines is
Δd = d1 -d2
Δd = (2,788 - 2,066) 10⁻¹
Δd = 7.22 10⁻² m
A farmhand pushes a 23 kg bale of hay 3.9 m across the floor of a barn. If she exerts a horizontal force of 91 N on the hay, how much work has she done?
Answer:
W = 354.9 J
Explanation:
Given that,
The mass of a bale of hay, m = 23 kg
The displacement, d = 3.9 m
The horizontal force exerted on the hay, F = 91 N
We need to find the work done. We know that,
We know that,
Work done, W = Fd
So,
W = 91 N × 3.9 m
W = 354.9 J
So, the required work done is 354.9 J.
A 0.25 kg mass is placed on a vertically oriented spring that is stretched 0.56 meters from its equilibrium position. If the spring constant is 105 N/m, how fast will the mass be moving when it reaches the equilibrium position? Hint: You cannot ignore the change in gravitational potential energy in this problem. Please give your answer in units of m/s, however, do not explicitly include units when typing your answer into the answer box.
Answer:
The speed of the ball when it reaches equilibrium position is 3.31 m/s
Explanation:
Given;
mass of the object, m = 0.25 kg
initial displacement of the object, h₁ = 0.56 m
spring constant, k = 105 N/m
displacement at equilibrium position, h₂ = 0
initial velocity of the object, v₁ = 0
velocity of the object at equilibrium position = v₂
The change in gravitational potential energy at the equilibrium position is given as;
ΔP.E = mg(h₂ - h₁)
The change in kinetic energy of the object at the equilibrium position is given as;
ΔK.E = ¹/₂m(v₂² - v₁²)
Apply the principle of conservation of mechanical energy;
ΔK.E + ΔP.E = 0
¹/₂m(v₂² - v₁²) + mg(h₂ - h₁) = 0
¹/₂m(v₂² - 0) + mg(0 - h₁) = 0
¹/₂mv₂² - mgh₁ = 0
¹/₂mv₂² = mgh
¹/₂v₂² = gh
v₂² = 2gh
v₂ = √2gh
v₂ = √(2 x 9.8 x 0.56)
v₂ = 3.31 m/s
Therefore, the speed of the ball when it reaches equilibrium position is 3.31 m/s