In a redox (reduction-oxidation) reaction, the molecule that gains an electron is reduced. Reduction and oxidation always occur together in such reactions. Reduced molecules often act as energy carriers in metabolic pathways.
Explanation:During a redox (reduction-oxidation) reaction, the molecule that gains an electron is referred to as being reduced. This process is accompanied by an energy transfer. When a molecule loses an electron (a process termed as oxidation), energy is released; this energy and the electron are then transferred to another molecule, resulting in reduction of that molecule. These two processes, oxidation and reduction, always occur together in a redox reaction.
For example, considering a metabolic pathway involving Nicotinamide adenine dinucleotide (NAD+), when it accepts a hydride ion (H-) from a hydrogen atom, it gets reduced to form NADH. Here, NAD+ is the molecule that gets reduced in this redox reaction.
In summary, in redox reactions, the molecule that gains an electron and is reduced can act as an energy carrier, an important function in metabolism, facilitating the controlled transfer of energy within the cell. The concepts of oxidation and reduction are integral to understanding energy extraction and utilization in cells.
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In the electron dot symbol shown, what kind of bond connects the two nitrogen atoms in this molecule?
ionic bond
covalent bond
double covalent bond
triple covalent bond
Answer:
triple covalent bond
Explanation:
A triple covalent bond is formed when three pairs of electrons (six electrons) are shared between the two combining atoms. A triple bond is shown by marking three short lines between the two symbols of the atoms. It requires three more electrons to attain the stable octet.
- Hope this helps! If you need a further explanation please let me know.
Final answer:
A triple covalent bond connects the two nitrogen atoms in this molecule N2.
Explanation:
A triple covalent bond connects the two nitrogen atoms in this molecule. Nitrogen atoms have 5 valence electrons and need to share 3 pairs of electrons to achieve a valence octet. The triple bond between the nitrogen atoms allows them to share 3 pairs of electrons and form a stable diatomic molecule, N2.
A thin film of alcohol (n = 1.36) lies on a flat glass plate (n = 1.51). When monochromatic light, whose wavelength λ can be changed, is incident normally, the reflected light is minimum for wavelength λ = 512 nm and a maximum for wavelength = 640 nm. What is the minimum thickness of the film?
To find the minimum thickness of the film, one should understand thin film interference. For minimum reflection, this corresponds to the condition of destructive interference. Using this knowledge, the minimum thickness can be calculated using the provided values.
Explanation:One needs to understand thin film interference to solve this problem. This concept involves understanding how light waves interact when they hit a thin film, and this interaction depends on the film's thickness, the light's wavelength, and the medium's indices of refraction.
Given that minimum reflection is obtained for a wavelength of 512 nm, we know that this condition corresponds to destructive interference. For destructive interference in a thin film like the one we're discussing, the additional distance traveled by the second ray of the light (which reflects off the lower surface of the film) must be odd multiples of λ/2, where λ is the wavelength of the light in the medium of the film.
In this case, we know that the film is made of alcohol and the light in this medium has a different wavelength than in vacuum, given by λ' = λ/n. Here, n is the refractive index of alcohol, so λ' = 512 nm/1.36.
For minimum thickness, and therefore first-order destructive interference, we have that 2t = λ'/2, or t = λ'/4, where t is the thickness of the film, the thing we're looking to find. Substituting in our value for λ', we can calculate that the minimum thickness of the film would be 512 nm/(1.36 x 4).
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The minimum thickness of the film is approximately 470.6 nm.
To find the minimum thickness of the film, we need to consider the conditions for constructive and destructive interference for normally incident light on a thin film.
For destructive interference (minimum reflected light), the optical path difference between the reflected rays from the top and bottom surfaces of the film must be an odd multiple of half the wavelength of the light in the medium of the film. This can be expressed as:
[tex]2n_f_t = (2m + 1)\frac{\lambda_d}{2}[/tex]
where [tex]\( n_f \)[/tex] is the refractive index of the film, (t) is the thickness of the film, [tex]\( \lambda_d \)[/tex] is the wavelength of the light in air for which destructive interference occurs, and (m) is an integer.
For constructive interference (maximum reflected light), the optical path difference must be an even multiple of half the wavelength of the light in the medium of the film. This can be expressed as:
[tex]2n_f_t = m\lambda_c[/tex]
where [tex]\( \lambda_c \)[/tex] is the wavelength of the light in air for which constructive interference occurs.
Given that the destructive interference occurs at [tex]\( \lambda_d = 512 \)[/tex] nm and the constructive interference occurs at [tex]\( \lambda_c = 640 \)[/tex] nm, we can set up two equations using the above conditions:
For destructive interference:
[tex]2 \times 1.36 \times t = (2m + 1)\frac{512}{2}[/tex]
For constructive interference:
2 x 1.36 x t = m x 640
We can solve these two equations simultaneously to find the minimum thickness (t). Let's first express (m) in terms of (t) from both equations:
From destructive interference:
2 x 1.36 x t = (2m + 1) x 256
[tex]t = \frac{(2m + 1) \times 256}{2 \times 1.36}[/tex]
From constructive interference:
2 x 1.36 x t = m x 640
[tex]t = \frac{m \times 640}{2 \times 1.36}[/tex]
Since the thickness (t) must be the same for both conditions, we can equate the two expressions for (t):
[tex]\frac{(2m + 1) \times 256}{2 \times 1.36} = \frac{m \times 640}{2 \times 1.36}[/tex]
2m + 1 x 256 = m x 640
512m + 256 = 640m
256 = 640m - 512m
256 = 128m
m = 2
Now that we have the value of (m), we can substitute it back into either equation to find (t). Using the destructive interference equation:
[tex]t = \frac{(2 \times 2 + 1) \times 256}{2 \times 1.36}[/tex]
[tex]t = \frac{5 \times 256}{2.72}[/tex]
[tex]t = \frac{1280}{2.72}[/tex]
t = 470.6 nm
A sphere of radius R has total charge Q. The volume charge density (C/m3) within the sphere is rho(r)=C/r2, where C is a constant to be determined. Part A The charge within a small volume dV is dq=rhodV. The integral of rhodV over the entire volume of the sphere is the total charge Q. Use this fact to determine the constant C in terms of Q and R. Hint: Let dV be a spherical shell of radius r and thickness dr. What is the volume of such a shell? Express your answer in terms of the variables Q and R.
Answer: C = Q/4πR
Explanation:
Volume(V) of a sphere = 4πr^3
Charge within a small volume 'dV' is given by:
dq = ρ(r)dV
ρ(r) = C/r^2
Volume(V) of a sphere = 4/3(πr^3)
dV/dr = (4/3)×3πr^2
dV = 4πr^2dr
Therefore,
dq = ρ(r)dV ; dq =ρ(r)4πr^2dr
dq = C/r^2[4πr^2dr]
dq = 4Cπdr
FOR TOTAL CHANGE 'Q', we integrate dq
∫dq = ∫4Cπdr at r = R and r = 0
∫4Cπdr = 4Cπr
Q = 4Cπ(R - 0)
Q = 4CπR - 0
Q = 4CπR
C = Q/4πR
The value of C in terms of Q and R is [Q/4πR]
A horizontal 826 N merry-go-round of radius 1.17 m is started from rest by a constant horizontal force of 57.8 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-goround after 3.47 s. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J. 028 10.0
Answer:
The kinetic energy of the merry-go-round is [tex]\bf{475.47~J}[/tex].
Explanation:
Given:
Weight of the merry-go-round, [tex]W_{g} = 826~N[/tex]
Radius of the merry-go-round, [tex]r = 1.17~m[/tex]
the force on the merry-go-round, [tex]F = 57.8~N[/tex]
Acceleration due to gravity, [tex]g= 9.8~m.s^{-2}[/tex]
Time given, [tex]t=3.47~s[/tex]
Mass of the merry-go-round is given by
[tex]m &=& \dfrac{W_{g}}{g}\\~~~~&=& \dfrac{826~N}{9.8~m.s^{-2}}\\~~~~&=& 84.29~Kg[/tex]
Moment of inertial of the merry-go-round is given by
[tex]I &=& \dfrac{1}{2}mr^{2}\\~~~&=& \dfrac{1}{2}(84.29~Kg)(1.17~m)^{2}\\~~~&=& 57.69~Kg.m^{2}[/tex]
Torque on the merry-go-round is given by
[tex]\tau &=& F.r\\~~~&=& (57.8~N)(1.17~m)\\~~~&=& 67.63~N.m[/tex]
The angular acceleration is given by
[tex]\alpha &=& \dfrac{\tau}{I}\\~~~&=& \dfrac{67.63~N.m}{57.69~Kg.m^{2}}\\~~~&=& 1.17~rad.s^{-2}[/tex]
The angular velocity is given by
[tex]\omega &=& \alpha.t\\~~~&=& (1.17~rad.s^{-2})(3.47~s)\\~~~&=& 4.06~rad.s^{-1}[/tex]
The kinetic energy of the merry-go-round is given by
[tex]E &=& \dfrac{1}{2}I\omega^{2}\\~~~&=&\dfrac{1}{2}(57.69~Kg.m^{2})(4.06~rad.s^{-1})^{2}\\~~~&=& 475.47~J[/tex]
How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes
Answer:
7.54 x[tex]10^{-7}[/tex] C
Explanation:
The complete question is:
Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular aluminum plates, each 19 cm in diameter, separated by 1.0 cm.How much charge can be added to each of the plates before a spark jumps between the two plates? For such flat electrodes, assume that value of 3×106N/C of the field causes a spark.
SOLUTION:
As you know that, parallel plate capacitance can be defined as
C=εo A/d
Where,
εo is the permittivity of free space constant, A is area of the capacitor plates, and d is the distance between them
Also, C= Q/V
where,
'Q' defines charge stored on the capacitor and 'V' express the potential difference between the plates.
Equating the equations.
εo A/d = Q/V
Q= (εo A V)/d
for a uniform electric field (such as the one between the plates of a parallel-plate capacitor)
V= Ed ->where 'E' defines magnitude of the electric field.
Therefore,
Q= (εo A E d)/d =>εo A E ->eq(1)
The area of the plates is given by
A= πr² => π(d/2)² =>π(0.19/2)
A=0.0284m²
Given: E= 3 x [tex]10^{6}[/tex] N/C
Substituting all the required values in eq(1)
(1)=> Q= (8.85x[tex]10^{-12}[/tex]) (-.0284)(3 x [tex]10^{6}[/tex] )
Q= 7.54 x[tex]10^{-7}[/tex] C
Therefore,7.54 x[tex]10^{-7}[/tex] C can be added to each of the plates before a spark jumps between the two plates
What must be included when describing the displacement of an object?
Answer:
Measurement and Direction
Explanation:
Displacement is a vector quantity because it has magnitude (the measure o displacement)
Answer:
Measurement and direction.
Explanation:
A wire runs left to right and carries a current in the direction shown.
A horizontal line representing a wire. There is a vector right labeled I. There is a point labeled Z below the line.
What is the direction of the magnetic field at point Z?
Answer (C)
Explanation:HOPE THIS HELPS
The direction of the magnetic field at point Z is into the page.
What is magnetic field?The magnetic field is the region of space where a charged object experiences magnetic force when it is moving.
When a charged particle such as an electron or proton moves inside a conductor, magnetic lines of force rotate around the particle. Since, this relative motion caused the magnetic field to generate.
A wire runs left to right and carries a current in the direction shown (attached diagram).
A horizontal line representing a wire. There is a vector right labeled I. There is a point labeled Z below the line.
According to the Fleming's right hand rule, the direction of magnetic field is into the page.
Thus, the direction of the magnetic field at point Z is into the page.
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a force sets an object in motion. when the force is multiplied by the time of its application we call the quantity impulse which changes the momentum of that object. what do we call the quantity force x (times) distance, and what quantity can this change?
Answer:
Work done
Explanation:
Impulse of an object is given by the product of force and time. It is also equal to the change on momentum of an object. The quantity that is given by the product of force and distance is called work done by an object.Work done also depends on the angle between force and displacement.The work done on an object is given as the product of force and distance traveled by the object.
What is impulse?Impulse is the force applied to an object for a certain period of time
J = Ft (Ns)
The change in the momentum of an object is equal to the impulse experienced by the object.
The work done on an object is given as the product of force and distance traveled by the object.
W = Fd (J)
The work done on an object causes change in the position of the object.
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Two objects collide and bounce apart. Assuming no outside forces act on the system, which best describes the total momentum after the collision?
It is always greater than it was before the collision.
It is often greater than it was before the collision.
It is always the same as it was before the collision.
It is often the same as it was before the collision.
Answer:
It is always the same as it was before the collision.
Answer:
The answer is C on Edge
Explanation:
When laser light shines on a screen after passing through two closely spaced slits, it becomes
Answer:
diffracted into semicircular waves. constructive interference occurs where the waves are crest to crest or trough to trough, destructive interference occurs where they are crest to trough. The light that falls on the screen produces bands of light and dark fringes on the screen as a result of these constructive and destructive interferences. This is called the young's slit experiment.
Laser light passing through two closely spaced slits and shining on a screen will produce an interference pattern of bright and dark fringes due to the wave properties of light, specifically diffraction and interference, which is illustrated through Young's double-slit experiment.
Explanation:When a laser light passes through two closely spaced slits and shines on a screen, it creates an interference pattern known as interference fringes. This phenomenon occurs due to the wave property of light known as diffraction and interference. Light waves passing through the two slits interfere with each other, creating a series of bright and dark lines or fringes on the screen.
The bright lines, called constructive interference, happen when the light waves enhance each other, and the dark lines, called destructive interference, occur when the light waves cancel each other out. The spacing and number of these fringes depend on the wavelength of the laser light and the distance between the slits. This is illustrated through Young's double-slit experiment.
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Three fire hoses are connected to a fire hydrant. Each hose has a radius of 0.017 m. Water enters the hydrant through an underground pipe of radius 0.088 m. In this pipe the water has a speed of 2.7 m/s. (a) How many kilograms of water are poured onto a fire in one hour by all three hoses
Answer:
2.36 x 10^5 kg
Explanation:
radius of hose, r = 0.017 m
radius of underground pipe, R = 0.088 m
number of hoses, n = 3
velocity of water in underground pipe, V = 2.7 m/s
Let v is the velocity of water in each hose.
According to the equation of continuity
A x V = n x a x v
π R² x V = n x π x r² x v
0.088 x 0.088 x 2.7 = 3 x 0.017 x 0.017 x v
v = 24.12 m/s
(a) Amount of water poured onto a fire in one hour by all the three hoses
= n x a x v x density of water x time
= 3 x 3.14 x 0.017 x 0.017 x 24.12 x 1000 x 3600
= 2.36 x 10^5 kg
Thus, the amount of water poured onto the fire in one hour is 2.36 x 10^5 kg.
a) Using the principles of fluid dynamics, specifically the continuity equation, we found that the velocity of water in the hoses is 2.02 m/s. The volume flow rate in the hoses is 0.0055 m³/s. Therefore, the mass of water poured onto a fire in one hour by all three hoses is 19800000 kg.
Explanation:a) To answer this question, we need to apply the principles of fluid dynamics, specifically the concept of continuity equation which states that the mass flow rate (the mass of water passing through a given spot per unit time) through a pipe system is constant.
First, let's calculate the cross sectional area of the hydrant pipe (A1) which is πr² = π(0.088m)² ≈ 0.0243 m².
Then, calculate the area of one fire hose (A2) which is π(0.017m)² ≈ 0.000907 m².
Since there are three hoses, the total area for the three hoses (A3) is 3 * A2 ≈ 0.00272 m².
Then, we apply the continuity equation, which is
A1v1=A3v3. We can solve for v3 (the velocity of the water in the hoses), v3 = A1v1/A3 ≈ 2.02 m/s.
To find the mass of water delivered in one hour, we first find the volume flow rate in the hoses,
Q= A3v3 = 0.00272 m² * 2.02 m/s ≈ 0.0055 m³/s.
The volume of water in one hour is Q * 3600s = 19800 m³.
Then, since water has a density (ρ) of 1000 kg/m³, the mass (m) of the water is ρ*volume which equals
19800 * 1000 kg = 19800000 kg.
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A student lifts a 1.5 kg book from a chair seat 41 cm off the ground to a table that is 71 cm off the ground.
What is the book's change in gravitational potential energy?
Answer:
4.4j
Explanation:
The change in gravitational potential energy of the book is 3.57 J.
Explanation:The change in gravitational potential energy of the book can be calculated using the equation:
ΔPE = mgh
where ΔPE is the change in potential energy, m is the mass of the book, g is the acceleration due to gravity (9.8 m/s²), and h is the change in height.
In this case, the mass of the book is 1.5 kg, the initial height is 41 cm (or 0.41 m), and the final height is 71 cm (or 0.71 m).
Substituting these values into the equation:
ΔPE = (1.5 kg)(9.8 m/s²)(0.71 m - 0.41 m) = 3.57 J
Therefore, the book's change in gravitational potential energy is 3.57 J.
A thermometer reading 10 degrees C is brought into a room with a constant temperature of 38 degrees Upper C. (a) If the thermometer reads 14 degrees C after 4 minutes, what will it read after being in the room for 6 minutes?
Answer:
Explanation:
We shall apply Newton's laws of cooling .
dT / dt = k ( T₁ - T₂ )
dT is change in temperature in time dt , k is constant , T₁ and T₂ are temperature of cold object and hot surrounding respectively.
For the first case of 4 minutes
(14 - 10)/ 4 = k [(14+10) /2 - 38 ] , T₁ = (14+10) /2 , the average temperature
1 = - 26 k
Using the formula for next two minutes
T-14 / 2 = k [(14+T) /2 - 38 ]
T-14 / 2 = 1/26[ 38 - (14+T) /2 ]
T / 2 - 7 = 1.46 - .27 - T / 52
T/2 + T/52 = 7 + 1.46 - .27
= .52 T = 8.19
T = 15.75 degree
10 points
True or False? Charging by friction happens when you rub two objects
together and PROTONS move from one object to the other.
Answer:
true
Explanation:
yes because I said so
Soccer ball is heading east positive direction with momentum of 2kgm/s. It is kicked backwards and a fraction of a second later is moving west negative direction withmomentum of -2.3. What impulse did the kicker put on the ball on kgm/s?
Answer:
-4.3 kgm/s
Explanation:
Impulse which the soccer ball experiences is equal to the change in momentum of a body.
It is given mathematically as:
Impulse = m*Δv = [tex]mv - mu[/tex]
Initial momentum of soccer ball = 2 kgm/s
Final momentum of the soccer ball = -2.3 kgm/s
Therefore, Impulse will be:
Impulse = [tex]-2.3 - 2[/tex]
Impulse = [tex]-4.3 kgm/s[/tex]
The impulse put on the soccer ball by the kicker is -4.3 kgm/s.
The highest waterfall in the world is the Salto Angel in Venezuela. Its longest single falls has a height of 807 m. If water at the top of the falls is at 14.2°C, what is the maximum temperature of the water at the bottom of the falls? Assume all the kinetic energy of the water as it reaches the bottom goes into raising its temperature.
Answer:
Maximum temperature of the water at the bottom of the fall, θ₂ = 16.09°C
Explanation:
The height of the waterfall, H = 807 m
At the top of the fall, [tex]\theta_{1} = 14.2^{0} C[/tex]
For water, the specific heat capacity, C = 4200 J/kg/°C
Potential energy at the top of the waterfall, PE = mgH
According to the principle of conservation of energy,
Kinetic energy at the bottom of the waterfall = Potential energy at the top
Therefore, KE = mgh
Applying the conservation of energy again,
KE at the bottom = Heat Energy due to raise in temperature
Heat energy due to raise in temperature = MCΔθ
MCΔθ = MgH
CΔθ = gH
Δθ = gH/C
Δθ = (9.81*807)/4200
Δθ = 1.89°C
[tex]\triangle \theta = \theta_{2} - \theta_{1}[/tex]
1.89 = θ₂ - 14.2
θ₂ = 14.2 + 1.89
θ₂ = 16.09°C
Putting a marshmallow directly in the fire
A. Conduction
B. Convention
C. Radiation
Answer:
A conduction is the answer
Explanation:
Answer: Radiation
Explanation:
Conduction involves thermal energy between a conductor/insulator, convection is when warm fluids rise and cool fluids sink due to a distinction in the density of their particles. None of these match a marshmallow over fire. Radiation is the correct answer.
A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.25 m/s2 and feels no appreciable air resistance. When it has reached a height of 500 m , its engines suddenly fail so that the only force acting on it is now gravity.a) What is the maximum height this rocket will reach above the launch pad?b) How much time after engine failure will elapse before the rocket comes crashing down to the launch pad?
Answer:
a) 112.5 m
b) 15.81s
Explanation:
a)We can use the following equation of motion to calculate the velocity v of the rocket at s = 500 m at a constant acceleration of a = 2.25 m/s2
[tex]v^2 = 2as[/tex]
[tex]v^2 = 2*2.25*500 = 2250[/tex]
[tex]v = \sqrt{2250} = 47.4 m/s[/tex]
After the engine failure, the rocket is subjected to a constant deceleration of g = -10 m/s2 until it reaches its maximum height where speed is 0. Again if we use the same equation of motion we can calculate the vertical distance h traveled by the rocket after engine failure
[tex]0^2 - v^2 = 2gh[/tex]
[tex]-2250 = 2(-10)h[/tex]
[tex]h = 2250/20 = 112.5 m[/tex]
So the maximum height that the rocket could reach is 112.5 + 500 = 612.5 m
b) Using ground as base 0 reference, we have the following equation of motion in term of time when the rocket loses its engine:
[tex]s + vt + gt^2/2 = 0[/tex]
[tex]500 + 47.4t - 10t^2/2 = 0[/tex]
[tex]5t^2 - 47.4t - 500 = 0[/tex]
[tex]t= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]t= \frac{47.4\pm \sqrt{(-47.4)^2 - 4*(5)*(-500)}}{2*(5)}[/tex]
[tex]t= \frac{47.4\pm110.67}{10}[/tex]
t = 15.81 or t = -6.33
Since t can only be positive we will pick t = 15.81s
Answer:
A) the maximum height this rocket will reach above the launch pad is 614.68 m
B) time elapsed after engine failure before the rocket comes crashing down to the launch pad is 16.025 sec.
Explanation:
Detailed explanation and calculation is shown in the image below
According to the diagram, which pair could both be north poles?
The given diagram of the magnet shows attraction which occurs between opposite poles therefore, A and C could both be the north poles. Thus, the correct option is C.
What is a magnet?A magnet is a material or an object which produces a magnetic field around itself. The magnetic field of this magnet is invisible however it is responsible for the most notable property of a magnet which is a force that is responsible for pulling on other ferromagnetic materials, such as iron, steel, nickel, cobalt, etc. towards it and attracts or repels other magnets as well.
Some important properties of magnet include a magnet has two poles which are north pole and south pole, magnets attract ferromagnetic materials, the similar poles of two magnets repel each other, whereas the opposite poles of two magnets attract each other. A magnet hung up in the air always comes to rest facing the north-south direction. The poles of a magnet are arranged in pairs.
Therefore, the correct option is C.
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A firefighter who weighs 192 lb slides down an infinitely long fire pole that exerts a friction resistive force with magnitude proportional to his speed, with k D 2:5 lb-s/ft. Assuming that he starts from rest, find his velocity as a function of time and find his terminal velocity
Answer:
Explanation:
check the pictures attached to further understand and i hope it works
Answer:
[tex]v = \frac{-384}{5} (1 - \frac{12}{5} e^{ -5t/12})[/tex]
Explanation:
Weight of the firefighter, W = 192 lb
W = mg
g = 32 ft/s²
Mass of the firefighter, m = W/g
m = 192/32
m = 6 slugs
k = 2.5 lb-s/ft
The force, F = ma = kv
2.5v = 6a
a = 2.5v/6
a = 5v/12
The fundamental dynamic equation;
dv/dt + drag + gravity = 0
dv/dt = -g-a
dv/dt = -32-5v/12..............(a)
The motion will attain terminal velocity when dv/dt = 0
-32 - 5v/12 = 0
-32 = 5v/12
-384 = 5v
v₀ = -384/5
v₀ = -384/5
dv/dt = -32 - 5v/12
[tex]\frac{dv}{-32 -5v/12} = dt[/tex]
[tex]-12/5 ln(32 + 5v/12) = t + c\\ln(32 + 5v/12) = -5t/12 + lnc\\ln(32 + 5v/12) - ln c = -5t/12\\ln\frac{32 + 5v/12}{c} = -5t/12[/tex]
Take exponential of both sides
[tex]\frac{32 + 5v/12}{c} =e^{ -5t/12}\\32 + 5v/12 = ce^{ -5t/12}\\5v/12 = -32 + ce^{ -5t/12}\\v = 12/5 (-32 + ce^{ -5t/12})\\[/tex]
c = v₀ = -384/5
[tex]v = 12/5 (-32 - \frac{384}{5} e^{ -5t/12})\\v = \frac{-384}{5} (1 - \frac{12}{5} e^{ -5t/12})[/tex]
A rifle fires a 1.06 x 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by 6.61 x 10-2 m from its unstrained length. The pellet rises to a maximum height of 6.26 m above its position on the compressed spring. Ignoring air resistance, determine the spring constant.
Answer:
k = 2.65 10² N / m
Explanation:
As the pellets shoot up we can use energy conservation
Starting point. Compressed spring rifle
Em₀ = Ke = ½ k x²
Final point. Highest point of the path
[tex]Em_{f}[/tex] = U = m g y
Energy is conserved
Em₀ = Em_{f}
½ k x² = m g y
k = 2 m g y / x²
Let's calculate
k = 2 1.06 10⁻² 9.8 6.26 / (6.61 10⁻²)²
k = 2.65 10² N / m
In the skeletal system, these structures are responsible for connecting bones.
A.
ligaments and tendons
B.
cartilage and calcium
C.
tendons and muscles
D.
bones and ligaments
Answer:
answer is A
Explanation:
Answer:
A.ligaments and tendons
Explanation:
a man lift a load of 20kg through a height of 4m in 10sec . calculate the workdone
Answer:
The answer to your question is Work = 784.8 Joules
Explanation:
Data
mass = 20 kg
height = 4 m
time = 10 s
Mechanic work is defined as the force applied to a body so it moves a distance.
Formula
Work = force x distance
Process
1.- Calculate the force
Force = mass x gravity
-Substitution
Force = 20 x 9.81
-Result
Force = 196.2 N
2.- Calculate the Work
Work = 196.2 x 4
-Result
Work = 784.8 Joules
A cylindrical fishing reel has a mass of 0.7 kg and a radius of 5.24 cm. A friction clutch in the reel exerts a restraining torque of 1.88 N · m if a fish pulls on the line. The fisherman gets a bite, and the reel begins to spin with an angular acceleration of 65.7 rad/s 2 . What force does the fish exert on the line? Answer in units of N.
Final answer:
The force exerted by the fish on the line is 35.87 N.
Explanation:
The force exerted by the fish on the line can be calculated using the equation:
Torque = Moment of Inertia x Angular Acceleration
In this case, the moment of inertia of the fishing reel can be calculated using the formula:
Moment of Inertia = [tex]0.5 x Mass x Radius^2[/tex]
Substituting the given values, we can calculate the moment of inertia and then solve for the force exerted by the fish:
Moment of Inertia =[tex]0.5 x 0.7 kg x (5.24 cm)^2[/tex]
Angular Acceleration = [tex]65.7 rad/s^2[/tex]
Using the equation Torque = Moment of Inertia x Angular Acceleration, we can rearrange the equation to solve for the force:
Torque = Force x Radius
Solving for the force:
Force = Torque / Radius
Substituting the given values, we can calculate the force exerted by the fish:
Force = 1.88 N.m / (5.24 cm)
Converting the radius to meters:
Force = 1.88 N.m / (0.0524 m) = 35.87 N
The force exerted by the fish on the line is calculated using the reel's moment of inertia, angular acceleration, and the restraining torque applied by the clutch. The resulting force is found to be 37.08 N.
To find the force that the fish exerts on the line, we need to consider the torque and angular acceleration.
First, let's find the moment of inertia (I) for the cylindrical reel. The moment of inertia for a cylinder rotating around its central axis is given by:I = 0.5 * m * r²
m = 0.7 kg (mass of the reel)
r = 0.0524 m (radius of the reel, converted from cm to meters)
Therefore:
I = 0.5 * 0.7 kg * (0.0524 m)² = 9.62 * 10⁻⁴ kg·m²
We are given the angular acceleration (α) as:α = 65.7 rad/s²
Using Newton's second law for rotation, we find the net torque (τ_net) on the reel:
τ_net = I * α
τ_net = 9.62 * 10⁻⁴ kg·m² * 65.7 rad/s² = 0.0632 N·m
The friction clutch exerts a restraining torque (τ_friction) of 1.88 N·m. The total torque exerted by the fish includes overcoming this friction:τ_fish = τ_net + τ_friction
τ_fish = 0.0632 N·m + 1.88 N·m = 1.9432 N·m
The force (F) exerted by the fish on the line can be related to the torque using:τ_fish = F * r
Therefore:F = τ_fish / r
F = 1.9432 N·m / 0.0524 m = 37.08 N
Thus, the force exerted by the fish on the line is 37.08 N.
Complete Question: A cylindrical fishing reel has a mass of 0.7 kg and a radius of 5.24 cm. A friction clutch in the reel exerts a restraining torque of 1.88 N · m if a fish pulls on the line. The fisherman gets a bite, and the reel begins to spin with an angular acceleration of 65.7 rad/s 2 . What force does the fish exert on the line? Answer in units of N.
Which example illustrates an engineering improvement that has led to greater knowledge about
space?
analyzing rock samples from the moon to determine the composition of the moon
science experiments on the International Space Station that study how crystals grow in
space
development of the space shuttle to allow humans to explore the moon
Answer:
D.
Explanation:
development of the space shuttle to allow humans to explore the moon
Label the parts that make up the human heart. Drag the items on the left to the correct location on the right.
A long string carries a wave; a 7.00-m segment of the string contains five complete wavelengths and has a mass of 180 g. The string vibrates sinusoidally with a frequency of 55.0 Hz and a peak-to-valley displacement of 20.0 cm. (The "peak-to-valley" distance is the vertical distance from the farthest positive position to the farthest negative position.)(a) Write the function that describes this wave traveling in the positive x direction. (Use the following as necessary: x and t. x is in meters and t is in seconds.)y =
Answer:
y = 0.1sin(4.5x - 345.6t)
Explanation:
Parameters given:
Length of 5 wavelengths = 7 m
Length of one wavelength, λ = 7/5 = 1.4 m
Frequency of wave, f = 55 Hz
Peak to Valley displacement = 20 cm = 0.2 m
Amplitude is half of the Peak to Valley displacement = 0.1 m
The wave function of a wave traveling in the positive x direction is given generally as:
y = Asin(kx - wt)
Where A = amplitude
k = Wave factor = 2π/λ
x = displacement on the x axis
w = angular frequency = 2πf
t = time taken
=> Wave factor, k = 2π/λ = 2π/(1.4) = 4.5 m^(-1)
=> Angular frequency, w = 2πf = 2π * 55 = 345.6 Hz
Therefore, inserting all the necessary parameters, we get that the wave function is:
y = 0.1sin(4.5x - 345.6t)
A spring is compressed by 50 cm from it's equilibrium length by an unattached block of mass 5.0 kg (the spring constant is 400 N/m). The box is released from rest, slides along a friction-less surface for a time, and then ascends a ramp of length 10 m, which is inclined at an angle of 20 degrees above the horizontal. What is the maximum distance that the box travels up the ramp?
Answer:
L = 2.98 m
Explanation:
In this exercise we are told that there is no friction, so we can use energy conservation
Starting point. Compressed spring
Em₀ = Ke = ½ k x²
Final point. At the highest point of the ramp
[tex]Em_{f}[/tex] = U = mg h
As there is no friction the energy is conserved
Emo = Em_{f}
½ k x² = mg h
h = ½ k x² / mg
h = ½ 400 0.50² / (5.0 9.8)
h = 1.02 m
This is the height that the body reaches, to calculate the distance traveled on the ramp let's use trigonometry
sin θ = h / L
L = h / sin θ
L = 1.02 / sin 20
L = 2.98 m
Which of the following situations would produce an average velocity of zero?
A) a round-trip to school and back
B) a horse galloping in a field
C) the criss-crossing path of a flying bug
D) a trip, first to the moon, then onto Mars
A solution that contains the maximum amount of solute for a given amount of solvent at a particular temperature is called
Answer:
saturated solution is the correct answer.
Explanation:
A saturated solution is a chemical solution, comprised of solute and solvent and in which no more further solute can dissolve in the solvent.A saturated solution contains the highest concentration of solute dissolved in a solvent.factors affecting the saturation solution are: the solution's temperature and pressure and also chemical composition of the substance addAn example of a saturated solution is carbonated water.