Answer:
82.2 mL
Explanation:
The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;
C1V1=C2V2
Where;
C1= concentration of stock solution
V1= volume of stock solution
C2= concentration of dilute solution
V2= volume of dilute solution
V2= C1V1/C2
V2= 1.48 × 55.6/ 1.0
V2= 82.2 mL
When a mercury thermometer is heated, the mercury expands and rises in the thin tube of glass. What does this indicate about the relative rates of expansion for mercury and glass
Answer:
This means that mercury has a higher or faster expansion rate than glass
Explanation:
This is because When a container expands, the reservoir in the glass expands at the same rate as the glass. Thus, if there is something in a glass and both expand at the same rate, they have no change - but if the contents expand faster, they will fill the container to a higher level, and if the contents expand slower, they will fill the container to a lower level (relative to the new size of the container).
Difference between matter and energy
Answer:
Energy is the strength and vitality required for sustained physical or mental activity.
Matter occupies space and possesses rest mass, especially as distinct from energy.
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In the summer of 2010 a huge piece of ice roughly four times the area of Manhattan and 500 m thick caved off the Greenland mainland.
Required:
a. How much heat would be required to melt this iceberg (assumed to be at 0°C) into liquid water at 0°C?
b. The annual U.S. energy consumption is 1.2 x 10^20 J. If all the U.S. energy was used to melt the ice, how many days would it take to do so?
Answer:
a
[tex]Q = 5.34 *10^{19} \ J[/tex]
b
[tex]T = 0.445 * 365 = 162. 413 \ days[/tex]
Explanation:
From the question we are told that
The area of Manhattan is [tex]a_k = 87.46 *10^{6} \ m^2[/tex]
The area of the ice is [tex]a_i = 4* 87.46 *10^{6 } = 3.498 *10^{8}\ m^2[/tex]
The thickness is [tex]t = 500 \ m \\[/tex]
Generally the volume of the ice is mathematically represented is
[tex]V = a_i * t[/tex]
substituting value
[tex]V = 500 * 3.498*10^{8}[/tex]
[tex]V = 1.75 *10^{11}\ m^3[/tex]
Generally the mass of the ice is
[tex]m_i = \rho_i * V[/tex]
Here [tex]\rho_i[/tex] is the density of ice the value is [tex]\rho _i = 916.7 \ kg/m^3[/tex]
=> [tex]m_i = 916.7 * 1.75*10^{11}[/tex]
=> [tex]m_i = 1.60 *10^{14} \ kg[/tex]
Generally the energy needed for the ice to melt is mathematically represented as
[tex]Q = m _i * H_f[/tex]
Where [tex]H_f[/tex] is the latent heat of fusion of ice and the value is [tex]H_f = 3.33*10^{5} \ J/kg[/tex]
=> [tex]Q = 1.60 *10^{14} * 3.33*10^{5}[/tex]
=> [tex]Q = 5.34 *10^{19} \ J[/tex]
Considering part b
We are told that the annual energy consumption is [tex]G = 1.2*10^{20 } \ J / year[/tex]
So the time taken to melt the ice is
[tex]T = \frac{ 5.34 *10^{19}}{ 1.2 *10^{20}}[/tex]
[tex]T = 0.445 \ years[/tex]
converting to days
[tex]T = 0.445 * 365 = 162. 413 \ days[/tex]
How many heartbeats in a typical human lifetime? Enter your answer as a number (NOT as a power of ten) and in one significant figure.
Answer:
20,000,000,000Explanation:
As we've seen, humans have on average a heart rate of around 60 to 70 beats per minute, give or take. We live roughly 70 or so years, giving us just over 2 billion beats all up.Apr
A river is 87. meters wide and its current flows northward at 6 meters per second. A boat is launched with a velocity of 1.0 meters per second eastward from the west bank of the river. Determine the magnitude and direction of the boat’s resultant velocity as it crosses the river.
Answer:
explained
Explanation:
If a person rows a boat across a rapidly flowing river and tries to head directly for the other shore, the boat instead moves diagonally relative to the shore, as in Figure 1. The boat does not move in the direction in which it is pointed. The reason, of course, is that the river carries the boat downstream. Similarly, if a small airplane flies overhead in a strong crosswind, you can sometimes see that the plane is not moving in the direction in which it is pointed, as illustrated in Figure 2. The plane is moving straight ahead relative to the air, but the movement of the air mass relative to the ground carries it sideways.
A boat is trying to cross a river. Due to the velocity of river the path traveled by boat is diagonal. The velocity of boat v boat is in positive y direction. The velocity of river v river is in positive x direction. The resultant diagonal velocity v total which makes an angle of theta with the horizontal x axis is towards north east direction.
Figure 1. A boat trying to head straight across a river will actually move diagonally relative to the shore as shown. Its total velocity (solid arrow) relative to the shore is the sum of its velocity relative to the river plus the velocity of the river relative to the shore.
An airplane is trying to fly straight north with velocity v sub p. Due to wind velocity v sub w in south west direction making an angle theta with the horizontal axis, the plane’s total velocity is thirty eight point 0 meters per seconds oriented twenty degrees west of north.
Figure 2. An airplane heading straight north is instead carried to the west and slowed down by wind. The plane does not move relative to the ground in the direction it points; rather, it moves in the direction of its total velocity (solid arrow).
In each of these situations, an object has a velocity relative to a medium (such as a river) and that medium has a velocity relative to an observer on solid ground. The velocity of the object relative to the observer is the sum of these velocity vectors, as indicated in Figure 1 and Figure 2. These situations are only two of many in which it is useful to add velocities. In this module, we first re-examine how to add velocities and then consider certain aspects of what relative velocity means.
How do we add velocities? Velocity is a vector (it has both magnitude and direction); the rules of vector addition discussed in Chapter 3.2 Vector Addition and Subtraction: Graphical Methods and Chapter 3.3 Vector Addition and Subtraction: Analytical Methods apply to the addition of velocities, just as they do for any other vectors. In one-dimensional motion, the addition of velocities is simple—they add like ordinary numbers. For example, if a field hockey player is moving at 5 m/s
straight toward the goal and drives the ball in the same direction with a velocity of 30 m/s
relative to her body, then the velocity of the ball is 35 m/s
relative to the stationary, profusely sweating goalkeeper standing in front of the goal.
In two-dimensional motion, either graphical or analytical techniques can be used to add velocities. We will concentrate on analytical techniques. The following equations give the relationships between the magnitude and direction of velocity (
The figure shows components of velocity v in horizontal vx and in vertical y axis v y. The angle between the velocity vector v and the horizontal axis is theta.
Figure 3. The velocity, v, of an object traveling at an angle θ to the horizontal axis is the sum of component vectors and
These equations are valid for any vectors and are adapted specifically for velocity. The first two equations are used to find the components of a velocity when its magnitude and direction are known. The last two are used to find the magnitude and direction of velocity when its components are known.
The equivalent resistance of two resistors connected in series is always greater than the equivalent resistance of the same two resistors connected in parallel. True False
Answer:
True
Explanation:
Because the resistors in series is the sum of the two resistors given as
R= R1+R2
While that of resistors in parallel is the sum of the reciprocal of the resistance given as
1/R = 1/ R1+ 1/R2
So that of series connection will be greater
When an ideal gas undergoes a slow isothermal expansion, A : the work done by the environment is the same as the energy absorbed as heat. B : the increase in internal energy is the same as the work done by the environment. C : the work done by the gas is the same as the energy absorbed as heat. D : the increase in internal energy is the same as the heat absorbed. E : the increase in internal energy is the same as the work done by the gas.
Explanation:
When an ideal gas undergoes a slow isothermal expansion, following phenomenon occur
1. Work done bu the gas = Energy absorbed as heat.
2. Work done by environment = Energy absorbed as heat.
3. Increase in internal energy= Heat absorbed= work done by gas = work done by environment.
Hence all option are correct.
Increase in internal energy is equal to the heat absorbed or work done by gas or environment. All the statements are correct.
If an ideal gas undergoes a slow isothermal expansion,
Work done by the gas is directly proportional energy absorbed as heat.
Work done by environment is directly proportional energy absorbed as heat.
Increase in internal energy is equal to the heat absorbed or work done by gas or environment.
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A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled
Answer:
The period of the motion will still be equal to T.
Explanation:
for a system with mass = M
attached to a massless spring.
If the system is set in motion with an amplitude (distance from equilibrium position) A
and has period T
The equation for the period T is given as
[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]
where k is the spring constant
If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.
Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.
If two identical wires carrying a certain current in the same direction are placed parallel to each other, they will experience a force of repulsion.
a) true
b) false
Answer:
The answer is B. falseExplanation:
Current in the same direction
When current flow through to parallel conductors of a given length, when the current flows in the same direction
1. A force of attraction between the wires occurs and this tends to draw the wires inward
2. A magnetic field in the same direction is produced.
Current in opposite direction
when the current is in opposite direction
1. Force of repulsion between the two wires occurs, draws the wire outward
2. A magnetic field in opposite direction occurs
A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. If the mass is set in motion in a medium that imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position x (t ), of the mass at a function of time, t is
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
A sinusoidal voltage Δv = (100 V) sin (170t) is applied to a series RLC circuit with L = 40 mH, C = 130 μF, and R = 50 Ω.
Required:
a. What is the impedance of the circuit?
b. What is the maximum current in the circuit?
Answer:
See attached file
Explanation:
You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction between the box and the inclined plane is 0.680. a) Determine the static frictional force which holds the box in place. b) You slowly raise one end of the track, slowly increasing the incline of the angle. Determine the maximum angle that the incline can make with the horizontal so that the box just remains at rest. Ms 680 u Fgsin 281 Ffg Mgm r 680 55 4 8
Answer:
[tex]\theta=34 \textdegree[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=55kg[/tex]
Angle [tex]\theta =28.0[/tex]
Coefficient of static friction [tex]\alpha =0.680[/tex]
Generally, the equation for Newtons second Law is mathematically given by
For
[tex]\sum_y=0[/tex]
[tex]N=mgcos \theta[/tex]
for
[tex]\sum_x=0[/tex]
[tex]F_{s}=mgsin\theta[/tex]
Where
[tex]F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta[/tex]
[tex]F_{s}=0.68*55*9.8*cos 28[/tex]
[tex]F_{s}=323.62N[/tex]
Therefore
[tex]\alpha mgcos \theta=mg sin \theta[/tex]
[tex]\theta=tan^{-1}(0.68)[/tex]
[tex]\theta=34 \textdegree[/tex]
(a) The static frictional force which holds the box in place is 323.62 N.
(b) The maximum angle that the incline can make with the horizontal is 34.2⁰.
Net forceThe net force applied to keep the box at rest must be zero in order for the box to remain in equilibrium position. Apply Newton's second law of motion to determine the net force.
∑F = 0
Static frictional forceThe static frictional force is calculated as follows;
Fs = μFncosθ
Fs = 0.68 x (55 x 9.8) x cos28
Fs = 323.62 N
Maximum angle the incline can makeFn(sinθ) - μFn(cosθ) = 0
mg(sinθ) - μmg(cosθ) = 0
μmg(cosθ) = mg(sinθ)
μ(cosθ) = (sinθ)
μ = sinθ/cosθ
μ = tanθ
θ = tan⁻¹(μ)
θ = tan⁻¹(0.68)
θ = 34.2⁰
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help... Please help!!!!!!!!!!!
Answer:
a) 6.8--5.10 thats equal 11.9
b) m=ris/run +10 equal 0.06/8 =7.5*10^-3
A point source emits sound waves with a power output of 100 watts. What is the sound level (in dB) at a distance of 10 m
Answer:
[tex]L = 109.01 db[/tex]
Explanation:
Given
Power, P = 100 W
Distance, d = 10 m
Required
Determine the Sound Level
First, the sound intensity as to be calculated; This is done, as follows;
[tex]I = \frac{P}{4\pi d^2}[/tex]
Substitute for P, d and take π as 3.14
[tex]I = \frac{100}{4 * 3.14 * 10^2}[/tex]
[tex]I = \frac{100}{4 * 3.14 * 100}[/tex]
[tex]I = \frac{100}{1256}[/tex]
[tex]I = 0.0796Wm^{-2}[/tex] --- Approximated
Next is to calculate the Sound Level, as follows
[tex]L = 10 * Log(\frac{I}{I_o})[/tex]
Where [tex]I_o = 10^{-12} Wm^{-2}[/tex]
Substitute for I and Io
[tex]L = 10 * Log(\frac{0.0796}{10^{-12}})[/tex]
[tex]L = 10 * Log(0.0796*10^{12)[/tex]
[tex]L = 10 * Log(0.0796*10^{12)[/tex]
[tex]L = 10 * 10.901[/tex]
[tex]L = 109.01 db[/tex]
Hence, the sound level is 109.01 decibels
If an electron is accelerated from rest through a potential difference of 1.60 x 102V, what is its de Broglie wavelength
Answer:
0.09 x10^-10m
Explanation:
Using wavelength=( 12.27 A)/√V
= 12.27 x 10^-10/ √1.6x10^2
= 0.09x10^-10m
Can you come up with a mathematical relationship, based on your data that shows the relationship between distance from the charges and electric field strength?
Answer:
Explanation:
This question appears incomplete because of the absence of the data been talked about in the question. However, there is a general ruling here and it can be applied to the data at hand.
If an increase in the distance of charges (let's denote with "d") causes the electric field strength (let's denote with"E") to increase, then the mathematical representation can be illustrated as d ∝ E (meaning distance of charge is directly proportional to electric field strength).
But if an increase in the distance of the charges causes the electric field strength to decrease, then the mathematical representation can be illustrated as d ∝ 1/E (meaning distance of charge is inversely proportional to electric field strength).
A scatterplot can also be used to determine this. If there is a positive correlation (correlation value is greater than zero but less than or equal to 1) on the graph, then it is illustrated as "d ∝ E" BUT if there is a negative correlation (correlation value is less than zero but greater than or equal to -1), then it can be illustrated as "d ∝ 1/E".
"Light traveling in a medium with a refractive index 1.11 is incident on a plate of another medium with index of refraction 1.66. At what angle of incidence is the reflected light fully polarized?"
Answer:
56°
Explanation:
Brewsters angle can be simply derived from
n1sin theta1= n2sintheta2= n2costheta1
because the reflected light will be 100% polarized if it is reflected at an angle 90o to the refracted light. Hence, Brewsters angle is
Tan theta= n2/n1
1.66/1.11= 1.495
Theta = 56°
Explanation:
If mirror M2 in a Michelson interferometer is moved through 0.233 mm, a shift of 792 bright fringes occurs. What is the wavelength of the light producing the fringe pattern?
Answer:
The wavelength is [tex]\lambda = 589 nm[/tex]
Explanation:
From the question we are told that
The distance of the mirror shift is [tex]k = 0.233 \ mm = 0.233*10^{-3} \ m[/tex]
The number of fringe shift is n = 792
Generally the wavelength producing this fringes is mathematically represented as
[tex]\lambda = \frac{ 2 * k }{ n }[/tex]
substituting values
[tex]\lambda = \frac{ 2 * 0.233*10^{-3} }{ 792 }[/tex]
[tex]\lambda = 5.885 *10^{-7} \ m[/tex]
[tex]\lambda = 589 nm[/tex]
An object is inside a room that has a constant temperature of 289 K. Via radiation, the object emits three times as much power as it absorbs from the room. What is the temperature (in kelvins) of the object
Answer:
T_object = 380.35 K
Explanation:
From Stefan–Boltzmann law, the power output is given by the formula:
P = σAT⁴
where;
σ is Stefan-Boltzmann constant
A is area of the radiating surface.
T is temperature of the body
Now, we are told that the power the object emitted is 3 times the power absorbed from the room.
Thus, we have;
P_e = 3P_a
Where P_e is power emitted and P_a is power absorbed.
So, we have;
σA(T_object)⁴ = 3σA (T_room)⁴
σA will cancel out to give;
(T_object)⁴ = 3(T_room)⁴
We are given T_room = 289 K
Thus;
(T_object)⁴ = 3 × 289⁴
(T_object) = ∜(3 × 289⁴)
T_object = 380.35 K
A small glass bead charged to 5.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm-long glass rod and is 4.0 cm from the rod's center. The bead is repelled from the rod with a force of 910 N. What is the total charge on the rod?
Answer:
Explanation:
Let B= bead
Q = rod
the electric field at the glass bead pocation is
(Gauss theorem)
E = Q / (2 π d L εo)
the force is
F = q E = q Q / (2 π d L εo)
then
Q = 2 π d L εo F / q
Q = 2*3.14*4x10^-2*10^-1*8.85x10^-12*910x10^-4 / 5x10^-9 = 2.87x10^-8 C = 40.5 nC
two object A and B vertically thrown upward with velocities of 80m/s and 100m/s at two seconds interview where and when will the two object meet.
Answer:
THIS IS YOUR ANSWER:
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14. What's one of the two requirements electric current?
A. There must be an electric potential between two bodies
B. There must be no valence electrons that make their element unstable
C. There must be a carbon element present in the electric current
D. There must be a magnetic force between two bodies
Marko
One of the two requirements of electric current is there must be an electric potential between two bodies
For electric current to flow, there must be an electric potential between two bodies.
This is because electric charge flows from a higher electric potential to a lower electric potential just as, water flows from a higher gravitational potential to a lower gravitational potential.
The difference between the electric potential between the two bodies causes the electric charge to flow between the two bodies.
This flow of electric charge constitutes electric current and electric current will only flow when there is an electric potential between two bodies.
So, one of the two requirements of electric current is there must be an electric potential between two bodies.
So, the answer is A
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Magnetic fields exist
A. Near a magnet
B. Farther away from a magnet
C. Within a magnet
D. All of the above
Answer:
Magnetic fields exist near a magnet, farther away from a magnet, and within a magnet.
So, the answer is D. All of the above.
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A car moving at 30 m/s slows uniformly to a speed of 10 m/s in a time of 5 s. Determine 1. The acceleration of the car. 2. The distance it moves in the third second.
Answer:
Explanation:
Initial velocity , u = 30 m/s
final velocity , v = 10 m/s
time , t = 5 seconds
1. Acceleration = v - u / t
= 10 - 30 / 5
= -20 / 5
= - 4 m/s
what is defect of vision
Answer:
The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resultant magnetic field at the point y
Complete question is;
One long wire carries a current of 30 A along the entire x axis. A second long wire carries a current of 40 A perpendicular to the xy plane and passes through the point (0, 4, 0) m. What is the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis?
Answer:
B_net = 50 × 10^(-7) T
Explanation:
We are told that the 30 A wire lies on the x-plane while the 40 A wire is perpendicular to the xy plane and passes through the point (0,4,0).
This means that the second wire is 4 m in length on the positive y-axis.
Now, we are told to find the magnitude of the resulting magnetic field at the point y = 2.0 m on the y axis.
This means that the position we want to find is half the length of the second wire.
Thus, at this point the net magnetic field is given by;
B_net = √[(B1)² + (B2)²]
Where B1 is the magnetic field due to the first wire and B2 is the magnetic field due to the second wire.
Now, formula for magnetic field due to very long wire is;
B = (μ_o•I)/(2πR)
Thus;
B1 = (μ_o•I_1)/(2πR_1)
Also, B2 = (μ_o•I_2)/(2πR_2)
Now, putting the equation of B1 and B2 into the B_net equation, we have;
B_net = √[((μ_o•I_1)/(2πR_1))² + ((μ_o•I_2)/(2πR_2))²]
Now, factorizing out some common terms, we have;
B_net = (μ_o/2π)√[((I_1)/R_1))² + ((I_2)/R_2))²]
Now,
μ_o is a constant and has a value of 4π × 10^(−7) H/m
I_1 = 30 A
I_2 = 40 A
Now, as earlier stated, the point we are looking for is 2 metres each from wire 2 end and wire 1.
Thus;
R_1 = 2 m
R_2 = 2 m
So, let's calculate B_net.
B_net = ((4π × 10^(−7))/2π)√[(30/2)² + (40/2)²]
B_net = 50 × 10^(-7) T
g In the atmosphere, the shortest wavelength electromagnetic waves are called A. infrared waves. B. ultraviolet waves. C. X-rays. D. gamma rays. E.
Answer:gamma ray
Explanation:
White light containing wavelengths from 410 nm to 750 nm falls on a grating with 7800 slits/cm. Part APart complete How wide is the first-order spectrum on a screen 3.20 m away
Answer:
1.227 m
Explanation:
Given that
Minimum wavelength is 410 nm
Maximum wavelength is 750 nm
Grating is 7800 slits/cm
Distance is 3.2 m
To solve this question, we would use the formula
sin θ = λ/d
sin θ = (410*10^-9) / (0.01/7800)
Sin θ = 410*10^-9 / 1.282*10^-6
Sin θ = 0.32 and θ = 18.67 degrees
For the second wavelength = 750 nm
sin θ = [(0.32x750)/410]
sin θ = (240 / 410)
sin θ = 0.5853 or
θ = 35.8 degrees
And finally, the width of spectrum would be
3.2[tan 35.8 - tan 18.67]
3.2 * 0.3833
= 1.227 m
What is the average value of the magnitude of the Poynting vector (intensity) at 1 meter from a 100-watt light bulb radiating in all directions
Answer:
I = 7.96 W / m²
Explanation:
The light bulb emits a power of P = 100W, this power is distributed over the surface of a sphere, thus the emission is in all directions.
Intensity is defined by power per unit area
I = P / A
The area of a sphere is
A = 4π r²
we substitute
I = P / (4π r²)
in this case it tells us that the distance is r = 1 m
let's calculate
I = 100 / (4π 1²)
I = 7.96 W / m²
ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rises feom 20c to 85c in 10 minutes calculate the s.h.c of metal cylinder
Answer:
692 J/kg/°C
Explanation:
Electric energy added = amount of heat
Power × time = mass × SHC × increase in temperature
Pt = mCΔT
(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)
C = 692 J/kg/°C