Answer:
1
Explanation:
Electrolysis is the passing of an current through a conducting solution, when the occurs, a chemical reaction takes place.
Heating a chemical will always cause a chemical reaction, which is why 3 is also correct
Some information as to why 2 is NOT correct.
2 is NOT a chemical reaction, but rather a process of physical separation. It uses selective boiling and condensation, but is not considered a chemical reaction.
as with 3, heating is not considered a chemical reaction, but rather a physical temperature change. This is always what it is considered to be (e.g boiling water is a physical temperature change, not a chemical reaction)
Hope this helps.
Hope this helps.
A cylindrical disk of wood weighing 45.0 N and having a diameter of 30.0 cm floats on a cylinder of oil of density 0.850 g>cm3 (Fig. E12.19). The cylinder of oil is 75.0 cm deep and has a diameter the same as that of the wood. (a) What is the gauge pressure at the top of the oil column
Answer:
665.25 Pa
Explanation:
Given data :
Weight of the disk, w = 45 N
Diameter, d = 30 cm
= 0.30 m
Therefore, radius of the disk,
[tex]$r=\frac{d}{2}$[/tex]
[tex]$r=\frac{0.30}{2}$[/tex]
= 0.15 m
Now, area of the cylindrical disk,
[tex]$A=\pi r^2$[/tex]
[tex]$A=3.14 \times (0.15)^2$[/tex]
[tex]$=0.07065 \ m^2$[/tex]
∴ The gauge pressure at the top of the oil column is :
[tex]$p=\frac{w}{A}$[/tex]
[tex]$p=\frac{47}{0.07065}$[/tex]
= 665.25 Pa
Therefore, the gauge pressure is 665.25 Pa.
The definition of pressure allows to find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa
The pressure is defined by the relationship between perpendicular force and area.
[tex]P = \frac{F}{A}[/tex]
where P is pressure, F is force, and A is area.
They indicate that the wooden cylinder weighs W = 45.0 N and has a diameter of d = 30 cm = 0.30 m.
The area is:
A = π r² = [tex]\pi \frac{d^2}{4}[/tex]
In the attachment we see a diagram of the forces, where the weight of the cylinder and the thrust are equal.
B-W = 0
B = W
The force applied to the liquid is the weights of the cylinder. Let's replace.
[tex]P= \frac{W}{A} \\P = W \frac{4}{\pi d^2 }[/tex]
Let's calculate.
[tex]P = \frac{45 \ 4 }{\pi \ 0.30^2 }[/tex] P = 45 4 / pi 0.30²
P = 636.6 Pa
In conclusion using the definition of pressure we can find the result for the pressure at the top of the oil cylinder is:
The pressure is: P = 636.6 Pa.
Learn more about pressure here: brainly.com/question/17467912
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration
Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;
[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]
Therefore, the frequency of this mode of vibration is 138.87 Hz
A 49.5-turn circular coil of radius 5.10 cm can be oriented in any direction in a uniform magnetic field having a magnitude of 0.535 T. If the coil carries a current of 26.5 mA, find the magnitude of the maximum possible torque exerted on the coil.
Answer:
The magnitude of the maximum possible torque exerted on the coil is 5.73 x 10⁻³ Nm
Explanation:
Given;
number of turns of the circular coil, N = 49.5 turns
radius of the coil, r = 5.10 cm = 0.051 m
magnitude of the magnetic field, B = 0.535 T
current in the coil, I = 26.5 mA = 0.0265 A
The magnitude of the maximum possible torque exerted on the coil is calculated as;
τ = NIAB
where;
A is the area of the coil
A = πr² = π(0.051)² = 0.00817 m²
Substitute the given values and solve for the maximum torque
τ = (49.5) x (0.0265) x (0.00817) x (0.535)
τ = 0.00573 Nm
τ = 5.73 x 10⁻³ Nm
Part C – RC Circuits in AC Mode 1. Derive Equation 5-6 from Equation 5-5. 2. Using the τ’s you calculated and your measured resistance: a. Calculate the capacitances of the capacitors. b. Compare your calculated and measured values via percent error.
Answer: hi your question is incomplete attached below is the complete question
1) attached below
2) a) 31 Ω, 302.9 Ω
b) 17.3 Ω , 26.4 Ω
Explanation:
1) Deriving Eqn 5-6 from Eqn 5-5
attached below
2) using τ’s calculated and measured resistance
use given data
Tm1 = Rm1 * Cm1
= 329.3 * 333 * 10^-9 = 109.65 μs
Tm2 = Rm2 * Cm2
= 329.3 * 200 * 10^-9 = 658.6 μs
a) Capacitance of capacitors
For Cm1
t₁₂ = 72 μs = R₁' Cm1 log²
∴ R₁' = ( 72 * 10^-6 ) / ( 333 * 10^-9 log² )
= 31 Ω
For Cm2
t₁₂ = 40 μs , ∴ R₂' = 302.9 Ω
b) comparing calculated and measured values via percent error
errors
Rm1 - R₁' = 17.3 Ω
Rm2 - R₂' = 26.4 Ω
true or false A permanent magnet and a coil of wire carrying a current both produce magnetic fields
Answer:
True. A permanent magnet like the earth produces its own B field due to movement of the iron core. The earths magnetic field is the reason why we have an atmosphere and it also is the only defense against solar flares. A coil of wire or solenoid that has current have so much moving charge that the motion of the electrical charge can create a significant G b-field
why is it wrong to leave our light on
Answer:
you will get huge electricity bills ............
Choose the force diagram that best represents a ball thrown upward by Peter, at the
top of its path.
Diagram A
Diagram B
Diagram C
Diagram D
Answer:Diagram A
Explanation:
Since the air resistance is to be neglected, only the gravitational force acts on the ball ( and has acted all the way from the throw upward). Diagram A reflects this fact correctly indicating the gravity acting on the ball downward.
An audience of 2250 fills a concert hall of volume 32000 m^3. If there were no ventilation, by how much would the temperature of the air rise over a period of 2.0 h due to the metabolism of the people (70 W/person)?
A room has dimensions of 15 ft by 15 ft by 20 ft contains air with a density of 0.0724 pounds-mass per cubic feet. The weight of air in the room in pounds-force is
Answer:
the weight of the air in pound-force (lb-f) is 325.8 lbf
Explanation:
Given;
dimension of the room, = 15 ft by 15 ft by 20 ft
density of air in the room, ρ = 0.0724 lbm/ft³
The volume of air in the room is calculated as;
Volume = 15 ft x 15 ft x 20 ft = 4,500 ft³
The mass of the air is calculated as;
mass = density x volume
mass = 0.0724 lbm/ft³ x 4,500 ft³
mass = 325.8 lb-m
The weight of the air is calculated as;
Weight = mass x gravity
Weight = 325.8 lb-m x 32.174 ft/s²
Weight = 10482.29 lbm.ft/s²
The weight of the air in pound-force (lb-f) is calculated as;
1 lbf = 32.174 lbm.ft/s²
[tex]Weight =10,482.29\ lbm.ft/s^2\times \frac{1 \ lbf}{32.174 \ lbm.ft/s^2} \\\\Weight = 325.8 \ lbf[/tex]
Therefore, the weight of the air in pound-force (lb-f) is 325.8 lbf
how do you calculate voltage drop
Answer:
Multiply current in amperes by the length of the circuit in feet to get ampere-feet. Circuit length is the distance from the point of origin to the load end of the circuit.
Divide by 100.
Multiply by proper voltage drop value in tables. The result is voltage drop.
Explanation:
Help pleaseeeeeeeeeeeeee
Answer:
hello the answer is 47m/s
a) Viewers of Star Trek hear of an antimatter drive on the Starship Enterprise. One possibility for such a futuristic energy source is to store antimatter charged particles in a vacuum chamber, circulating in a magnetic field, and then extract them as needed. Antimatter annihilates with normal matter, producing pure energy. What strength magnetic field is needed to hold antiprotons, moving at 5.00 x10^7 m/s in a circular path 2.00m in radius? Antiprotons have the same mass as protons but the opposite (negative) charge.b) Is this field strength obtainable with today's technology or is it a futuristic possibility?
measurement is essential in our life.justify the statement.
Answer:
Measurements allow people to find their way to new places. Measurements such as miles or kilometers are used by GPS systems to give directions. Time measurements help to create schedules so tasks get done on time. Measurements are used in food as well. Ingredients in recipes have to be measured to make the dish correctly. Serving sizes are a measurement that keep people healthy by showing how much of each food you should eat.
A flat coil of wire is placed in a uniform magnetic field that is in the y-direction.
The magnetic flux through the coil is maximum if the coil is:_________.
(a) in the XY plane
(b) in either the XY or the YZ plane
(c) in the XZ plane
(d) in any orientation, because it is constant.
Answer:
The correct answer is c
Explanation:
Flow is defined by
Ф = B . A
bold letters indicate vectors.
The magnetic field is directed to the y axis, The area of the coil is represented by a vector normal to the plane of the coil, so to have a flux
i.i = j.j = k.k = 1
and the tori scalar products are zero
a) If the coil must be in the xy plane so that its normal vector is in the Z axis, so there is no flux
b) if the coil is in the plane yz the normal veto is in the x axis, so the flux is zero
C) If the coil is in XZ, the normal vector points in the y direction, usually the scalar product is one and there is a flux in this configuration
The correct answer is c
Four equal-value resistors are in series with a 5 V battery, and 2.23 mA are measured. What isthe value of each resistor
Answer:
560.54 Ω
Explanation:
Applying,
V = IR'............... Equation 1
Where V = Voltage of the battery, I = currrent, R' = Total resistance of the resistors
make R' the subject of the equation
R' = V/I............ Equation 2
From the question,
Given: V = 5 V, I = 2.23 mA = 2.23×10⁻³ A
Substitute these values into equation 2
R' = 5/(2.23×10⁻³ )
R' = 2242.15 Ω
Since the fours resistor are connected in series and they are equal,
Therefore the values of each resistor is
R = R'/4
R = 2242.15/4
R = 560.54 Ω
Two blocks in contact with each other are pushed to the right across a rough horizontal surface by the two forces shown. If the coefficient of kinetic friction between each of the blocks and the surface is 0.30, determine the magnitude of the force exerted on the 2.0-kg block by the 3.0-kg block.
I assume the blocks are pushed together at constant speed, and it's not so important but I'll also assume it's the smaller block being pushed up against the larger one. (The opposite arrangement works out much the same way.)
Consider the forces acting on either block. Let the direction in which the blocks are being pushed by the positive direction.
The 2.0-kg block feels
• the downward pull of its own weight, (2.0 kg) g
• the upward normal force of the surface, magnitude n₁
• kinetic friction, mag. f₁ = 0.30n₁, pointing in the negative horizontal direction
• the contact force of the larger block, mag. c₁, also pointing in the negative horizontal direction
• the applied force, mag. F, pointing in the positive horizontal direction
Meanwhile the 3.0-kg block feels
• its own weight, (3.0 kg) g, pointing downward
• normal force, mag. n₂, pointing upward
• kinetic friction, mag. f₂ = 0.30n₂, pointing in the negative horizontal direction
• contact force from the smaller block, mag. c₂, pointing in the positive horizontal direction (this is the force that is causing the larger block to move)
Notice the contact forces form an action-reaction pair, so that c₁ = c₂, so we only need to find one of these, and we can get it right away from the net forces acting on the 3.0-kg block in the vertical and horizontal directions:
• net vertical force:
n₂ - (3.0 kg) g = 0 ==> n₂ = (3.0 kg) g ==> f₂ = 0.30 (3.0 kg) g
• net horizontal force:
c₂ - f₂ = 0 ==> c₂ = 0.30 (3.0 kg) g ≈ 8.8 N
The mass of the moon is 7.2 × 10^22 kg and its radius is 1.7×10^6 m.What will be the gravity of the moon to a body of the mass 1 kg on the surface of the moon.
Answer:
1.66 N
Explanation:
The force of gravity of the moon on the body is given by
F = GMm/R² where G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.2 × 10²² kg, m = mass of body = 1 kg and R = radius of moon = 1.7 × 10⁶ m
Substituting the values of the variables into the equation, we have
F = GMm/R²
F = 6.67 × 10⁻¹¹ Nm²/kg² × 7.2 × 10²² kg × 1 kg/(1.7 × 10⁶ m)²
F = 48.024 × 10¹¹ Nm²/2.89 × 10¹² m²
F = 16.62 × 10⁻¹ N
F = 1.662 N
F ≅ 1.66 N
So, the gravity on the moon is 1.66 N
IS ANYONE THERE..??!
Answer:
hmmmmmmmmmmmmmmmmmmmmmmmm y
In 1.0 second, a battery charger moves 0.50 C of charge from the negative terminal to the positive terminal of a 1.5 V AA battery.
Part A:
How much work does the charger do? Answer is 0.75 J
Part B:
What is the power output of the charger in watts?
Answer:
W = Q * V work done on charge Q
A. W = .5 C * 1.5 V = .75 Joules
B. P = W / t = .75 J / 1 sec = .75 Watts
need help pleaseee,question is in the pic
Explanation:
For engine 1,
Energy removed = 239 J
Energy added = 567 J
[tex]\eta_1=\dfrac{239}{567}\cdot100=42.15\%[/tex]
For engine 2,
Energy removed = 457 J
Energy added = 789 J
[tex]\eta_2=\dfrac{457}{789}\cdot100=57.92\%[/tex]
For engine 3,
Energy removed = 422 J
Energy added = 1038 J
[tex]\eta_3=\dfrac{422}{1038}\cdot100=40.65\%[/tex]
So, the engine 2 has the highest thermal efficiency.
as the ball rises the vertical component of it's velocity_____. explain
Answer:
Decreases
Explanation:
because its moving against gravitational attraction and at maximum height its velocity will be and it will decrease until it reaches maximum height and the start to increase again
During the same Olympics, Bolt also set the world record in the 200-m dash with a time of 19.30 s. Using the same assumptions as for the 100-m dash, what was his maximum speed for this race
Answer:
The maximum speed of Bolt for the 100 m race is 14.66 m/s
Explanation:
Given;
initial distance covered by Bolt, d = 200 m
time of this motion, t = 19.3 s
The second distance covered by Bolt, = 100 m
Assuming Bolt maintained the same acceleration for both races.
His acceleration can be determined from the 200 m race.
d = ut + ¹/₂at²
where;
u is his initial velocity = 0
d = ¹/₂at²
[tex]at^2 = 2d\\\\a = \frac{2d}{t^2} \\\\a = \frac{2\times 200}{19.3^2} \\\\a = 1.074 \ m/s^2[/tex]
Let the final or maximum velocity for the 100 m race = v
v² = u² + 2ad₂
v² = 2 x 1.074 x 100
v² = 214.8
v = √214.8
v = 14.66 m/s
The maximum speed of Bolt for the 100 m race is 14.66 m/s
which vector best represents the net force acting on the +3 C charge
20. How much charge will flow through a 2002 galvanometer
connected to a 40092 circular coil of 1000 turns on a wooden
stick 2 cm in diameter? If a magnetic field B=0.011 T parallel to
the axis of the stick is decreased suddenly to zero?
Answer:
5.76 μC
Explanation:
The induce emf, ε = -ΔΦ/Δt where ΔΦ = change in magnetic flux = NAΔB where N = number of turns of coil = 1000, A = cross-sectional area of coil = πd²/4 where d = diameter of coil = 2 cm = 2 × 10⁻² m and ΔB = change in magnetic field strength = B' - B where B' = final magnetic field = 0 T and B = initial magnetic field strength = 0.011 T. So, ΔB = 0 T - 0.011 T = -0.011 T
So, ε = -ΔΦ/Δt
ε = -NAΔB/Δt
ε = -NAΔB/Δt
Also ε = iR where i = current and R = combined resistance of circular coil and galvanometer = 200 Ω + 400 Ω = 600 Ω (since they are in series)
So, iR = -NAΔB/Δt
iΔt = -NAΔB/R
Δq = -NAΔB/R where Δq = charge = iΔt
substituting the values of the variables into the equation, we have
Δq = -1000 × π(2 × 10⁻² m)²/4 × -0.011 T/600 Ω
Δq = -1000 × 4π × 10⁻⁴ m²/4 × -0.011 T/600 Ω
Δq = 0.011π × 10⁻¹ m²T/600 Ω
Δq = 0.03456 × 10⁻¹ m²T/600 Ω
Δq = 5.76 × 10⁻⁶ C
Δq = 5.76 μC
When you take your 1900-kg car out for a spin, you go around a corner of radius 55 m with a speed of 15 m/s. The coefficient of static friction between the car and the road is 0.88. Assuming your car doesn't skid, what is the force exerted on it by static friction?
Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
The image of the object formed by the lens is real, enlarged and inverted. What is the kind of lens ?
Answer:
Converging (convex) lens.
Explanation:
A lens can be defined as a transparent optical instrument that refracts rays of light to produce a real image.
Basically, there are two (2) main types of lens and these includes;
I. Diverging (concave) lens.
II. Converging (convex) lens.
A converging (convex) lens refers to a type of lens that typically causes parallel rays of light with respect to its principal axis to come to a focus (converge) and form a real image. Thus, this type of lens is usually thin at the lower and upper edges and thick across the middle.
Basically, the image of the object formed by a converging (convex) lens. lens is real, enlarged and inverted.
A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the merry-go-round as a solid cylinder and determine the net work needed for this acceleration.
Solution :
Given data :
Mass of the merry-go-round, m= 1640 kg
Radius of the merry-go-round, r = 7.50 m
Angular speed, [tex]$\omega = \frac{1}{8}$[/tex] rev/sec
[tex]$=\frac{2 \pi \times 7.5}{8}$[/tex] rad/sec
= 5.89 rad/sec
Therefore, force required,
[tex]$F=m.\omega^2.r$[/tex]
[tex]$$=1640 \times (5.89)^2 \times 7.5[/tex]
= 427126.9 N
Thus, the net work done for the acceleration is given by :
W = F x r
= 427126.9 x 7.5
= 3,203,451.75 J
This diagram shows the magnetic field lines near the ends of two magnets. There is an error in the diagram.
Two bar magnet with the north pole of one near the south pole of the second. field lines are leaving the north pole and bent away from the south pole of the other. Field lines are leaving the south pole of one and bending away from the north pole of the other.
Which change will correct the error in the diagram?
a)changing the N to S
b)reversing the arrows on the left to point toward the N
c)changing the S to N
d)reversing the arrows on the right to point toward the S
Answer:
changing the N to S. that's how the error will be corrected
Answer:
C is the correct answer
Explanation:
i took the test
A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?
Answer:
a n c
Explanation:
In a new scenario, the block only makes it (exactly) half-way through the rough spot. How far was the spring compressed from its unstretched length
Answer: hello below is the missing part of your question
A mass m = 10 kg rests on a frictionless table and accelerated by a spring with spring constant k = 5029 N/m. The floor is frictionless except for a rough patch. For this rough path, the coefficient of friction is μk = 0.49. The mass leaves the spring at a speed v = 3.4 m/s.
answer
x = 0.0962 m
Explanation:
First step :
Determine the length of the rough patch/spot
F = Uₓ (mg)
and w = F.d = Uₓ (mg) * d
hence;
d( length of rough patch) = w / Uₓ (mg) = 46.55 / (0.49 * 10 * 9.8) = 0.9694 m
next :
work done on unstretched spring length
Given that block travels halfway i.e. d = 0.9694 / 2 = 0.4847 m
w' = Uₓ (mg) * d
= 0.49 * 10 * 9.81 * 0.4847 = 23.27 J
also given that the Elastic energy of spring = work done ( w')
1/2 * kx^2 = 23.27 J
x = [tex]\sqrt{\frac{2*23.27}{5029} }[/tex] = 0.0962 m