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
a vechile having a mass of 500kg is moving with a speed of 10m/s.Sand is dropped into it at the rate of 10kg/min.What force is needed to keep the vechile moving with uniform speed
Answer:
1.67 N
Explanation:
Applying,
F = u(dm/dt)+m(du/dt)................ Equation 1
Where F = force, m = mass of the vehicle, u = speed.
Since u is constant,
Therefore, du/dt = 0
F = u(dm/dt)............... Equation 2
From the question,
Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s
Substitute these values into equation 2
F = 10(10/60)
F = 100/60
F = 1.67 N
Can you think of reasons why the charge on each ball decreases over time and where the charges might go
Answer:
By the principle of corona discharge.
Explanation:
The charge on each ball will decreases over time due to the electrical discharge in air.
According to the principle of corona discharge, when the curvature is small, the discharge of the charge takes placed form the pointed ends.
If one lawn mower causes an 80-dB sound level at a point nearby, four lawnmowers together would cause a sound level of ____________ at that point. a.92 dB b.84 dB c.86 dB d.none of the above
Answer:
The intensity of 4 lawn movers is 86 dB.
Explanation:
Intensity of one lawnmower = 80 dB
Let the intensity is I.
Use the formula of intensity
[tex]dB = 10 log\left ( \frac{I}{Io} \right )\\\\80=10log\left ( \frac{I}{Io} \right )\\\\10^8 = \frac{I}{10^{-12}}\\\\I = 10^{-4} W/m^2[/tex]
Now the intensity of 4 lawn movers is
[tex]dB = 10 log\left ( \frac{4I}{Io} \right )\\\\dB=10log\left ( \frac{4\times10^{-4}}{10^{-12}} \right )\\\\dB = 86 dB\\[/tex]
Cold air rises because it is denser than water, is this true?
Answer:
true
Explanation:
im not sure please dont attack me
Could you show detailed steps in how to solve this problem please
Answer: See attached pic. Hope this helps.
Explanation:
Explain why the flow from the battery increases when the switch is closed. Give the label of the concept(s) that you use from the model of electricity. [
Answer:
Due to the applied filed the electrons move in a particular direction.
Explanation:
Initially when the switch is off, the free electrons move here and there in any random directions in the conductor with the random speeds called thermal velocity. So, tat the net flow is almost zero.
When the battery is connected is switch is ON, the random motion of the electrons aligned in a particular direction due to the force applied by the electric filed, so the net flow is not zero it increases and thus the current flow.
Solids diffuse because the particles cannot move.
A. Can
B. Not enough info
C. Cannot
D. Sometimes will
Solids cannot diffuse.
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
1. Which one of the following is not an organic compound? Why? CH4 C2H6O CaO
2. Fill in the chart below to identify and describe the functional groups associated with organic chemistry. Name General Structure Properties/Uses Alcohol Aldehyde Ketone Fatty acid Ether
3. Explain why carbon is called “the backbone” molecule of organic chemistry and why organic molecules couldn't easily be based on H or O instead.
Answer:
1. CaO is not an organic compound because it doesn’t contain a carbon molecule.
2.
Name General Structure Properties/Uses
Alcohol R-OH (contains a hydroxyl group) Can be poisonous, can be made from fermentation or distillation
Aldehyde R-COH (contains a carbon atom double-bonded to an oxygen and single-bonded to a hydrogen) Makes up formaldehyde and acetaldehyde
Ketone R-CO-R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds) Makes up acetone
Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters
Ether R-O-R (contains double carbon chains connected to an oxygen atom through single bonds) Ethyl ether is very volatile and flammable, used in veterinary medicine
3. Carbon is able to make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.
Explanation:
pf
CaO is not an organic compound because it doesn’t contain a carbon molecule.
Name General Structure Properties/Uses(which contains a hydroxyl group) Can be poisonous, can be made from fermentation or distillation
Aldehyde R-COH (contains a carbon atom double-bonded to oxygen and single-bonded to hydrogen) Makes up formaldehyde and acetaldehyde
Ketone R-CO-1R (contains a carbon atom double-bonded to an oxygen atom and then connected to carbon chains through the other two single bonds) Makes up acetone
Fatty acid R-COOH (contains a carbon atom double-bonded to an oxygen atom, single-bonded to a hydroxyl, and single-bonded to the carbon chain) Makes up fatty acids like acetic acid and stearic acid; used to form esters11
Ether -O-R (contains double carbon chains connected to an oxygen atom through single bonds) Ethyl ether is very volatile and flammable, used in veterinary medicine
Carbon can make four covalent bonds with other elements. This gives it a lot of diversity and the ability to form differently shaped molecules that perform specific functions or fit specific cell receptors in the body. H can form only one bond, and oxygen forms only two bonds, so they don't have as much potential to form a good starting point for organic molecules.
Learn more about organic molecules.
https://brainly.com/question/24225576
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Hi can someon help me how to answer this?
Btw I'm from Philippines
Answer:
Test 1
1.True
2.True
3.True
4.False
5.True
6.True
7.False
8.True
9.True
10.True
yung iba nasa pic
A person with a near point of 85 cm, but excellent distant vision, normally wears corrective glasses. But he loses them while traveling. Fortunately, he has his old pair as a spare.
(a) If the lenses of the old pair have a power of +2.25 diopters, what is his near point (measured from his eye) when he is wearing the old glasses if they rest 2.0 cm in front of his eye?
(b) What would his near point be if his old glasses were contact lenses instead?
Answer:
a) p = 95.66 cm, b) p = 93.13 cm
Explanation:
For this problem we use the constructor equation
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]
where f is the focal length, p and q are the distances to the object and the image, respectively
the power of the lens is
P = 1 / f
f = 1 / P
f = 1 / 2.25
f = 0.4444 m
the distance to the object is
[tex]\frac{1}{p} = \frac{1}{f} -\frac{1}{q}[/tex]
the distance to the image is
q = 85 -2
q = 83 cm
we must have all the magnitudes in the same units
f = 0.4444 m = 44.44 cm
we calculate
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{83}[/tex]
1 / p = 0.010454
p = 95.66 cm
b) if they were contact lenses
q = 85 cm
[tex]\frac{1}{p} = \frac{1}{44.44} - \frac{1}{85}[/tex]
1 / p = 0.107375
p = 93.13 cm
4. Consider a 1 kg block is on a 45° slope of ice. It is connected to a 0.4 kg block by a cable
and pulley. Does the 1 kg block move or down the slope? What is the net force on it and
its acceleration? (8 pts)
If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• m = 1 kg:
∑ F (parallel) = mg sin(45°) - T = ma … … … [1]
∑ F (perpendicular) = n - mg cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• m = 0.4 kg:
∑ F (vertical) = T - mg = ma … … … [2]
Adding equations [1] and [2] eliminates T, so that
((1 kg) g sin(45°) - T ) + (T - (0.4 kg) g) = (1 kg + 0.4 kg) a
(1 kg) g sin(45°) - (0.4 kg) g = (1.4 kg) a
==> a ≈ 2.15 m/s²
The fact that a is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is a ≈ 2.15 m/s², which means the net force on the block would be ∑ F = ma ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
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 tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 59.4 m/s and returns the shot with the ball traveling horizontally at 37.2 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction).
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball?
5 9 . 4
- 3 7 . 2
2 2 . 2
Explanation:
Use the algorithm method.
5 9 . 4
- 3 7 . 2
2 2 . 2
2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.
22.2
22.2
A car is moving at a speed of 60 mi/hr (88 ft/sec) on a straight road when the driver steps on the brake pedal and begins decelerating at a constant rate of 10ft/s2 for 3 seconds. How far did the car go during this 3 second interval?
Answer:
219 ft
Explanation:
Here we can define the value t = 0s as the moment when the car starts decelerating.
At this point, the acceleration of the car is given by the equation:
A(t) = -10 ft/s^2
Where the negative sign is because the car is decelerating.
To get the velocity equation of the car, we integrate over time, to get:
V(t) = (-10 ft/s^2)*t + V0
Where V0 is the initial velocity of the car, we know that this is 88 ft/s
Then the velocity equation is:
V(t) = (-10 ft/s^2)*t + 88ft/s
To get the position equation we need to integrate again, this time we get:
P(t) = (1/2)*(-10 ft/s^2)*t^2 + (88ft/s)*t + P0
Where P0 is the initial position of the car, we do not know this, but it does not matter for now.
We want to find the total distance that the car traveled in a 3 seconds interval.
This will be equal to the difference in the position at t = 3s and the position at t = 0s
distance = P(3s) - P(0s)
= ( (1/2)*(-10 ft/s^2)*(3s)^2+ (88ft/s)*3s + P0) - ( (1/2)*(-10 ft/s^2)*(0s)^2 + (88ft/s)*0s + P0)
= ( (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s + P0) - ( P0)
= (1/2)*(-10 ft/s^2)*9s^2+ (88ft/s)*3s = 219ft
The car advanced a distance of 219 ft in the 3 seconds interval.
For a research project, a student needs a solenoid that produces an interior magnetic field of 0.0100 T. She decides to use a current of 1.00 A and a wire 0.500 mm in diameter. She winds the solenoid in layers on an insulating form 1.00 cm in diameter and 20.0 cm long.
Determine the number of layers of wire needed. (Round your answer up to the nearest integer.)
Determine the total length of the wire. (Use the integer number of layers and the average layer diameter.)
Answer:
[tex]n=3.8[/tex]
Explanation:
From the question we are told that:
Magnetic Field [tex]B=0.01T[/tex]
Current [tex]I=1.00[/tex]
Wire Diameter [tex]d_w=0.5*10^3m[/tex]
Layers Diameter [tex]d_l=1*10^2m[/tex]
Length [tex]l=0.2m[/tex]
Generally the equation for number of layers is mathematically given by
[tex]n=\frac{Bd_w}{\mu_o I}[/tex]
Where
[tex]Vacuum\ permeability=\mu_0[/tex]
[tex]n= \frac{0.01*0.5*10^3m}{4 \pi *10^{-7}*1 }[/tex]
[tex]n=3.8[/tex]
g As they reach higher temperatures, most semiconductors... Selected Answer: have an increased resistance. Answers: have a constant resistance. have an increased resistance. have a decreased resistance.
Answer:
have an increased resistance
Part of your electrical load is a 60-W light that is on continuously. By what percentage can your energy consumption be reduced by turning this light off
Answer:
Following are the solution to the given question:
Explanation:
Please find the complete question in the attached file.
The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.
[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]
Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]
The electricity used is continuously 694W over 30 days.
If just resistor loads (no reagents) were assumed,
[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]
Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]
This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.
Explain whether the unit of work is a fundamental or derived unit
Answer:
answer here
Explanation:
the unit of work is fundamental unit because it doesn't depend on other units.
__________________
Thx
Answer:
The SI unit of work is joule (J)
Explanation:
Joule is a derived unit. ∴ unit of work is a derived unit
In the following calculations, be sure to express the answer in standard scientific notation with the appropriate number of
significant figures.
3.88 x 1079 - 4.701 x 1059
x 10
g
Answer:
-45,597.07
Explanation:
if not in scientific calculator and yung answer nung sa scientific sa comment na lang dinadownload ko ka eh
A mass attached to the end of a spring is oscillating with a period of 2.25 s on a horizontal frictionless surface. The mass was released from rest at
t = 0
from the position
x = 0.0480 m.
Determine the location of the mass at
t = 5.85 s?
Answer:
[tex]X=0.0389m[/tex]
Explanation:
From the question we are told that:
Period of spring [tex]T_s=2.25s[/tex]
Initial Position of Mass [tex]x=0.0480m[/tex]
Final Mass period [tex]T_f=5.85s[/tex]
Generally the equation for the Mass location is mathematically given by
[tex]X=xcos*\frac{2\pi T_s}{T_f}[/tex]
[tex]X=0.048*cos*\frac{2\pi 5.85}{2.25}[/tex]
[tex]X=0.0389m[/tex]
Starting with the Ideal Gas Law, show that the relationship between volume and temperature in an adiabatic process is the one given by :
TfVf^γ^-1 = TiVi^γ-1 = Constant
Answer:
hope it helps
explanation:
An object is 2.0 cm from a double convex lens with a focal length of 1.5 cm. Calculate the image distance
Answer:
0.857 cm
Explanation:
We are given that:
The focal length for a convex lens to be (f) = 1.5cm
The object distance (u) = - 2.0 cm
We are to determine the image distance (v) = ??? cm
By applying the lens formula:
[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]
By rearrangement and making (v) the subject of the above formula:
[tex]v = \dfrac{uf}{u-f}[/tex]
replacing the given values:
[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]
[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]
v = 0.857 cm
Two electrons are passing 20.0 mm apart. What is the electric repulsive force that they exert on each other
Answer:
0.5766422350752*10^-24 N
Explanation:
Couloumb's law states that states that there is an electrical force acting on 2 static charges. The magnitude is directly proportional to the product of the 2 charges.
Strength of electrons : q1 = q2 = 1.602 x 10-19. C
Substitute and solve:
F = (9*10^9)(1.602 x 10-19)(1.602 x 10-19) / (0.02)^2
Done.
How do the magnitudes of the currents through the full circuits compare for Parts I-III of this exercise, in which resistors are combined in series, in parallel, and in combination
Answer: hello tables and data related to your question is missing attached below are the missing data
answer:
a) I = I₁ = I₂ = I₃ = 0.484 mA
b) I₁ = 0.016 amps
I₂ = 0.0016 amps
I₃ = 7.27 * 10^-4 amps
c) I₁ = 1.43 * 10^-3 amp
I₂ = 0.65 * 10^-3 amps
Explanation:
A) magnitude of current for Part 1
Resistors are connected in series
Req = r1 + r2 + r3
= 3300 Ω ( value gotten from table 1 ) ,
V = 1.6 V ( value gotten from table )
hence I ( current ) = V / Req = 1.6 / 3300 = 0.484 mA
The magnitude of current is the same in the circuit
Vi = I * Ri
B) magnitude of current for part 2
Resistors are connected in parallel
V = 1.6 volts
Req = [ ( R1 * R2 / R1 + R2 ) * R3 / ( R1 * R2 / R1 + R2 ) + R3 ]
= [ ( 100 * 1000 / 100 + 1000) * 2200 / ( 100 * 1000 / 100 + 1000 ) + 2200]
= 87.30 Ω
For a parallel circuit the current flow through each resistor is different
hence the magnitude of the currents are
I₁ = V / R1 = 1.6 / 100 = 0.016 amps
I₂ = V / R2 = 1.6 / 1000 = 0.0016 amps
I₃ = V / R3 = 1.6 / 2200 = 7.27 * 10^-4 amps
C) magnitude of current for part 3
Resistors are connected in combination
V = 1.6 volts
Req = R1 + ( R2 * R3 / R2 + R3 )
= 766.66 Ω
Total current ( I ) = V / Req = 1.6 / 766.66 = 2.08 * 10^-3 amps
magnitude of currents
I₁ = ( I * R3 ) / ( R2 + R3 ) = 1.43 * 10^-3 amps
I₂ = ( I * R2 ) / ( R2 + R3 ) = 0.65 * 10^-3 amps
A submarine has a "crush depth" (that is, the depth at which
water pressure will crush the submarine) of 400 m. What is
the approximate pressure (water plus atmospheric) at this
depth? (Recall that the density of seawater is 1025 kg/m3, g=
9.81 m/s2, and 1 kg/(m-s2) = 1 Pa = 9.8692 x 10-6 atm.)
Answer:
P =40.69 atm
Explanation:
We need to find the approximate pressure at a depth of 400 m.
It can be calculated as follows :
P = Patm + ρgh
Put all the values,
[tex]P=1\ atm+1025 \times 9.81\times 400\times 9.8692\times 10^{-6}\ atm/Pa\\\\P=40.69\ atm[/tex]
So, the approximate pressure is equal to 40.69 atm.
Light with a wavelength of 5.0 · 10-7 m strikes a surface that requires 2.0 ev to eject an electron. Calculate the energy, in joules, of one incident photon at this frequency. _____ joules 4.0 x 10 -19 4.0 x 10 -49 9.9 x 10 -32 1.1 x 10 -48
Answer:
pretty sure its 6.2 x 10^-13
Explanation:
I looked it up I'm not a bigbrain but want to help
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
When a golfer tees off, the head of her golf club which has a mass of 158 g is traveling 48.2 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.
Answer:
v₂ = 53.23 m/s
Explanation:
Given that,
The mass of a golf club, m₁ = 158 g = 0.158 kg
The initial speed of a golf club, u₁ = 48.2 m/s
The mass of a golf ball, m₂ = 46 g = 0.046 kg
It was at rest, u₂ = 0
Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 32.7 m/s, v₁ = 32.7 m/s
We use the conservation of energy to find the speed of the golf ball just after impact as follows :
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\\v_2=\dfrac{m_1u_1-m_1v_1}{m_2}\\\\v_2=\dfrac{0.158(48.2)-0.158(32.7)}{0.046}\\\\=53.23\ m/s[/tex]
So, the speed of the golf ball just after the impact is equal to 53.23 m/s.
An object with mass m is located halfway between an object of mass M and an object of mass 3M that are separated by a distance d. What is the magnitude of the force on the object with mass m?A) 8GMm/d^2B) GMm/(4d^2)C) 4GMm/d^2D) GMm/(2d^2)E) 3GMm/2d^2
Answer:
A) 8GMm/d^2
Explanation:
We are given that
[tex]m_1=M[/tex]
[tex]m_2=3M[/tex]
[tex]m_3=m[/tex]
Distance between m1 and m2=d
Distance of object of mass m from m1 and m2=d/2
Gravitational force formula
[tex]F=\frac{Gm_1m_2}{r^2}[/tex]
Using the formula
Force acting between m and M is given by
[tex]F_1=\frac{GmM}{d^2/4}[/tex]
Force acting between m and 3M is given by
[tex]F_2=\frac{Gm(3M)}{d^2/4}[/tex]
Now, net force acting on object of mass is given by
[tex]F=F_2-F_1[/tex]
[tex]F=\frac{Gm(3M)}{d^2/4}-\frac{GmM}{d^2/4}[/tex]
[tex]F=\frac{12GmM}{d^2}-\frac{4GmM}{d^2}[/tex]
[tex]F=\frac{12GmM-4GmM}{d^2}[/tex]
[tex]F=\frac{8GmM}{d^2}[/tex]
Hence, the magnitude of the force on the object with mass m=[tex]\frac{8GmM}{d^2}[/tex]
Option A is correct.
