Answer:
[tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
Explanation:
Given that
Wavelength [tex]\lambda=588 \mathrm{nm}[/tex]
slit separation [tex]\mathrm{d}=2464 \mathrm{nm}[/tex]
slit screen distance [tex]\mathrm{D}=11 \mathrm{cm}[/tex]
We know that for double slit the maxima condition is that
[tex]\operatorname{dsin} \theta=m \lambda[/tex]
[tex]\sin \theta=\frac{m \lambda}{d}[/tex]
[tex]\theta=\sin ^{-1}\left(\frac{\mathrm{m} \lambda}{\mathrm{d}}\right)[/tex]
For small angle approximation, [tex]\sin \theta \approx \tan \theta \approx \theta[/tex]
[tex]\tan \theta=\frac{y_{m}}{D}[/tex]
[tex]y_{m}=D \times \tan \left[\sin ^{-1}\left(\frac{m \lambda}{d}\right)\right][/tex]
Now [tex]y_{4}[/tex] [tex]y_{4}=D \times \tan \left[\sin ^{-1}\left(\frac{4 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{4 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=35.22 \mathrm{cm}[/tex]
Again [tex]y_{3}=D \times \tan \left[\sin ^{-1}\left(\frac{3 \lambda}{d}\right)\right]=11 \times \tan \left[\sin ^{-1}\left(\frac{3 \times 588 \mathrm{nm}}{2464 \mathrm{nm}}\right)\right]=11.27 \mathrm{cm}[/tex]
Hence [tex]y_{4}-y_{3}=35.22-11.27=23.95 \mathrm{cm}[/tex]
Three masses are located in the x-y plane as follows: a mass of 6 kg is located at (0 m, 0 m), a mass of 4 kg is located at (3 m, 0 m), and a mass of 2 kg is located at (0 m, 3 m). Where is the center of mass of the system?
A. (1 m, 2 m)
B. (2 m, 1 m)
C. (1 m, 1 m)
D. (1 m, 0.5 m)
E. (0.5 m, 1 m)
D. (1m, 0.5m)
Explanation:
The center of mass (or center of gravity) of a system of particles is the point where the weight acts when the individual particles are replaced by a single particle of equivalent mass. For the three masses, the coordinates of the center of mass C(x, y) is given by;
x = (m₁x₁ + m₂x₂ + m₃x₃) / M ----------------(i)
y = (m₁y₁ + m₂y₂ + m₃y₃) / M ----------------(ii)
Where;
M = sum of the masses
m₁ and x₁ = mass and position of first mass in the x direction.
m₂ and x₂ = mass and position of second mass in the x direction.
m₃ and x₃ = mass and position of third mass in the x direction.
y₁ , y₂ and y₃ = positions of the first, second and third masses respectively in the y direction.
From the question;
m₁ = 6kg
m₂ = 4kg
m₃ = 2kg
x₁ = 0m
x₂ = 3m
x₃ = 0m
y₁ = 0m
y₂ = 0m
y₃ = 3m
M = m₁ + m₂ + m₃ = 6 + 4 + 2 = 12kg
Substitute these values into equations (i) and (ii) as follows;
x = ((6x0) + (4x3) + (2x0)) / 12
x = 12 / 12
x = 1 m
y = (6x0) + (4x0) + (2x3)) / 12
y = 6 / 12
y = 0.5m
Therefore, the center of mass of the system is at (1m, 0.5m)
A Nearsighted Eye. A certain very nearsighted person cannot focus on anything farther than 36.0 cm from the eye. Consider the simplified model of the eye. In a simplified model of the human eye, the aqueous and vitreous humors and the lens all have a refractive index of 1.40, and all the refraction occurs at the cornea, whose vertex is 2.60 cm from the retina.
Required:
a. If the radius of curvature of the cornea is 0.65 cm when the eye is focusing on an object 36.0 cm from the cornea vertex and the indexes of refraction are as described before, what is the distance from the cornea vertex to the retina?
b. What does this tell you about the shape of the nearsighted eye?
1. This distance is greater than for the normal eye.
2. This distance is shorter than for the normal eye.
Answer:
a) The distance from the cornea vertex to the retina is 2.37 cm
b) This distance is shorter than for the normal eye.
Explanation:
a) Let refractive index of air,
n(air) = x = 1
Let refractive index of lens,
n(lens) = y = 1.4
Object distance, s = 36 cm
Radius of curvature, R = 0.65 cm
The distance from the cornea vertex to the retina is the image distance because image is formed in the retina.
Image distance, s' = ?
(x/s) + (y/s') = (y-x)/R
(1/36) + (1.4/s') = (1.4 - 1)/0.65
1.4/s' = 0.62 - 0.028
1.4/s' = 0.592
s' = 1.4/0.592
s' = 2.37 cm
Distance from the cornea vertex to the retina is 2.37 cm
(b) For a normal eye, the distance between the cornea vertex and the retina is 2.60 cm. Since 2.37 < 2.60, this distance is shorter than for normal eye.
A box on a ramp is connected by a rope to a winch. The winch is turned so that the box moves down the ramp at a constant speed. The box experiences kinetic friction with the ramp. Which forces on the box do zero work as the box moves down the ramp?
a. Weight (gravitational force)
b. Normal force
c. Kinetic friction force
d. Tension force
e. None
Answer:
Option B:
The normal force
Explanation:
The normal force does no work as the box slides down the ramp.
Work can only be done when the force succeeds in moving the object in the direction of the force.
All the other forces involved have a component that is moving the box in their direction.
However, the normal force does not, as it points downwards into the ramp. Since the normal force is pointing into the ramp, and the box is sliding down the ramp, we can say that no work is being done by the normal force because the box is not moving in its direction (which would have been the box moving into the ramp)
A 25 kg box is 220 N pulled at constant speed up a frictionless inclined plane by a force that is parallel to the incline. If the plane is inclined at an angle of 25o above the horizontal, the magnitude of the applied force is
Answer:
F = 103.54N
Explanation:
In order to calculate the magnitude of the applied force, you take into account that the forces on the box are the applied force F and the weight of the box W.
The box moves with a constant velocity. By the Newton second law you have that the sum of forces must be equal to zero.
Furthermore, you have that the sum of forces are given by:
[tex]F-Wsin\theta=0[/tex] (1)
F: applied force = ?
W: weight of the box = Mg = (25kg)(9.8m/s^2) = 245N
θ: degree of the incline = 25°
You solve the equation (1) for F:
[tex]F=Wsin\theta=(245N)sin(25\°)=103.54N[/tex] (2)
The applied force on the box is 103.54N
what is the orbital speed for a satellite 3.5 x 10^8m from the center of mars? Mars mass is 6.4 x 10^23 kg
Answer:
v = 349.23 m/s
Explanation:
It is required to find the orbital speed for a satellite [tex]3.5\times 10^8\ m[/tex] from the center of mass.
Mass of Mars, [tex]M=6.4\times 10^{23}\ kg[/tex]
The orbital speed for a satellite is given by the formula as follows :
[tex]v=\sqrt{\dfrac{GM}{r}} \\\\v=\sqrt{\dfrac{6.67\times 10^{-11}\times 6.4\times 10^{23}}{3.5\times 10^8}} \\\\v=349.23\ m/s[/tex]
So, the orbital speed for a satellite is 349.23 m/s.
A heavy, 6 m long uniform plank has a mass of 30 kg. It is positioned so that 4 m is supported on the deck of a ship and 2 m sticks out over the water. It is held in place only by its own weight. You have a mass of 70 kg and walk the plank past the edge of the ship. How far past the edge do you get before the plank starts to tip, in m
Answer:
about 1 meter
Explanation:
The distance past the edge that the man will get before the plank starts to tip is; 0.4285 m
We are given;
Mass of plank; m = 30 kg
Length of plank; L = 6m
Mass of man; M = 70 kg
Since the plank has 2 supports which are the deck of the ship, then it means that, we can take moments about the right support before the 2m stick out of the plank.
Thus;
Moment of weight of plank about the right support;
τ_p = mg((L/2) - 2)
τ_p = 30 × 9.8((6/2) - 2)
τ_p = 294 N.m
Moment of weight of man about the right support;
τ_m = Mgx
where x is the distance past the edge the man will get before the plank starts to tip.
τ_m = 70 × 9.8x
τ_m = 686x
Now, moment of the board is counterclockwise while that of the man is clockwise. Thus;
τ_m = τ_p
686x = 294
x = 294/686
x = 0.4285 m
Read more at; https://brainly.com/question/22150651
A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point circled A, at a potential of 1.50 103 V, to point circled B, at 4.00 103 V. What is its speed at point circled B?
Answer:
[tex]v_B=3.78\times 10^5\ m/s[/tex]
Explanation:
It is given that,
Charge on helium nucleus is 2e and its mass is [tex]6.63\times 10^{-27}\ kg[/tex]
Speed of nucleus at A is [tex]v_A=6.2\times 10^5\ m/s[/tex]
Potential at point A, [tex]V_A=1.5\times 10^3\ V[/tex]
Potential at point B, [tex]V_B=4\times 10^3\ V[/tex]
We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :
increase in kinetic energy = increase in potential×charge
[tex]\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s[/tex]
So, the speed at point B is [tex]3.78\times 10^5\ m/s[/tex].
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. When the flask and mercury are warmed together to a temperature of 52.0°C , a volume of 8.50 cm^3 of mercury overflows the flask.Required:If the coefficient of volume expansion of mercury is βHg = 1.80×10^−4 /K , compute βglass, the coefficient of volume expansion of the glass. Express your answer in inverse kelvins.
Answer:
the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Explanation:
Given that:
Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³
temperature of the glass flask and mercury= 1.00° C
After heat is applied ; the final temperature = 52.00° C
Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C
Volume of the mercury overflow = 8.50 cm^3 = 8.50 × 10⁻⁶ m³
the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K
The increase in the volume of the mercury = 10⁻³ m³ × 51.00 × 1.80 × 10⁻⁴
The increase in the volume of the mercury = [tex]9.18*10^{-6} \ m^3[/tex]
Increase in volume of the glass = 10⁻³ × 51.00 × [tex]\beta _{glass}[/tex]
Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask
the mercury overflow = [tex](9.18*10^{-6} - 51.00* \beta_{glass}*10^{-3})\ m^3[/tex]
[tex]8.50*10^{-6} = (9.18*10^{-6} -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]-6.8*10^{-7} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]6.8*10^{-7} = ( 51.00* \beta_{glass}* 10^{-3} )\ m^3[/tex]
[tex]\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}= ( \beta_{glass} )[/tex]
[tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
Thus; the coefficient of volume expansion of the glass is [tex]\mathbf{ ( \beta_{glass} )= 1.333 *10^{-5} / K}[/tex]
High voltage power is often carried in wire bundles made up of individual strands. In your initial post to the discussion, discuss the forces on the strands of wire due to the current flowing through them. What would happen if the force acted opposite of the known behavior
Answer:
More current will be loss through the metal wire strands if the force on them was repulsive, and more stress will be induced on the wire strands due to internal and external flexing.
Explanation:
A wire bundle is made up of wire strands bunched together to increase flexibility that is not always possible in a single solid metal wire conductor. In the strands of wire carrying a high voltage power, each strand carries a certain amount of current, and the current through the strands all travel in the same direction. It is know that for two conductors or wire, separated by a certain distance, that carries current flowing through them in the same direction, an attractive force is produced on these wires, one on the other. This effect is due to the magnetic induction of a current carrying conductor. The forces between these strands of the high voltage wire bundle, pulls the wire strands closer, creating more bond between these wire strands and reducing internal flex induced stresses.
If the case was the opposite, and the wires opposed themselves, the effect would be that a lot of cost will be expended in holding these wire strands together. Also, stress within the strands due to the repulsion, will couple with external stress from the flexing of the wire, resulting in the weakening of the material.
The biggest problem will be that more current will be lost in the wire due to increased surface area caused by the repulsive forces opening spaces between the strand. This loss is a s a result of the 'skin effect' in wire transmission, in which current tends to flow close to the surface of the metal wire. The skin effect generates power loss as heat through the exposed surface area.
Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If the person generates a new pulse like the first but more quickly, the pulse would be a) same size b) wider c) narrower If the person generates another pulse like the first but he moves his hand further, the pulse would be a) same size b) taller c) shorter If the person generates another pulse like the first but the rope is tightened, the pulse will move a) at the same rate b) faster c) slower Now the person moves his hand back and forth several times to produce several waves. You freeze the movie and get this snapshot. Underline your answer for each situation: If you advance the movie one frame, the knot at point A would be a) in the same place b) higher c) lower d) to the right e) to the left If you advance the movie one frame, the pattern of the waves will be _________relative to the hand. a) in the same place b) shifted right c) shifted left d) shifted up e) shifted down If the person starts over and moves his hand more quickly, the peaks of the waves will be a) the same distance apart b) further apart c) closer together If you lower the frequency of a wave on a string you will lower its speed. b) increase its wavelength. c). lower its amplitude. d) shorten its period.
Answer:
a) correct answer is b higher , b) correct answer is b higher , c) correct answer is b faster , d) traveling wave , e)
Explanation:
A traveling wave is described by the expression
y = A sin (kx - wt)
where k is the wave vector and w is the angular velocity
let's examine every situation presented
a) a new faster pulse is generated
A faster pulse should have a higher angular velocity
equal speed is related to the period and frequency
w = 2π f = 2π / T
therefore in this case the period must decrease so that the angular velocity increases
the correct answer is c narrower
b) Generate a pulse, but move your hand more.
Moving the hand increases the amplitude (A) of the pulse
the correct answer is b higher
c) generates a pulse but the force is tightened
Set means that more tension force is applied to the string, so the velicate changes
v = √ (T /μ)
the correct answer is b faster
d) move your hand back and forth
in this case you would see a pulse series whose sum corresponds to a traveling wave
e) Advance a frame the movie
in this case the wave will be displaced a whole period to the right
the correct answer is b
f) move your hand faster
the waves will have a maximum fast, so they are closer
answer C
g) decrease wave frequency
Since the speed of the wave is a constant m ak, decreasing the frequency must increase the wavelength to keep the velocity constant.
the correct answer is b increases its wavelength
The robot HooRU is lost in space, floating around aimlessly, and radiates heat into the depths of the cosmos at the rate of 13.1 W. HooRU's surface area is 1.55 m2 and the emissivity of its surface is 0.287. Ignoring the radiation that HooRU absorbs from the cold universe, what is HooRU's temperature T?
Answer:
The temperature is [tex]T = 168.44 \ K[/tex]
Explanation:
From the question ewe are told that
The rate of heat transferred is [tex]P = 13.1 \ W[/tex]
The surface area is [tex]A = 1.55 \ m^2[/tex]
The emissivity of its surface is [tex]e = 0.287[/tex]
Generally, the rate of heat transfer is mathematically represented as
[tex]H = A e \sigma T^{4}[/tex]
=> [tex]T = \sqrt[4]{\frac{P}{e* \sigma } }[/tex]
where [tex]\sigma[/tex] is the Boltzmann constant with value [tex]\sigma = 5.67*10^{-8} \ W\cdot m^{-2} \cdot K^{-4}.[/tex]
substituting value
[tex]T = \sqrt[4]{\frac{13.1}{ 0.287* 5.67 *10^{-8} } }[/tex]
[tex]T = 168.44 \ K[/tex]
a What CE describes the way energy is stored in a sandwich
A nonuniform electric field is given by the expression = ay î + bz ĵ + cx , where a, b, and c are constants. Determine the electric flux (in the +z direction) through a rectangular surface in the xy plane, extending from x = 0 to x = w and from y = 0 to y = h. (Use any variable or symbol stated above as necessary.)
How much force is needed to cause a 15 kilogram bicycle to accelerate at a rate of 10
meters per second per second?
O A. 15 newtons
OB. 1.5 newtons
C. 150 newtons
OD. 10 newtons
A small, rigid object carries positive and negative 3.00 nC charges. It is oriented so that the positive charge has coordinates (−1.20 mm, 1.20 mm) and the negative charge is at the point (1.70 mm, −1.30 mm).
Required:
a. Find the electric dipole moment of the object.
b. The object is placed in an electric field E = (7.80 103 î − 4.90 103 ĵ). Find the torque acting on the object.
c. Find the potential energy of the object–field system when the object is in this orientation.
d. Assuming the orientation of the object can change, find the difference between the maximum and the minimum potential energies of the system,
Answer:
Umax = 105.8nJ
Umin =-105.8nJ
Umax-Umin = 211.6nJ
Explanation:
Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivere
Complete question:
Two conductors made of the same material are connected across the same potential difference. Conductor A has seven times the diameter and seven times the length of conductor B. What is the ratio of the power delivered to A to power delivered to B.
Answer:
The ratio of the power delivered to A to power delivered to B is 7 : 1
Explanation:
Cross sectional area of a wire is calculated as;
[tex]A = \frac{\pi d^2}{4}[/tex]
Resistance of a wire is calculated as;
[tex]R = \frac{\rho L}{A} \\\\R = \frac{4\rho L}{\pi d^2} \\\\[/tex]
Resistance in wire A;
[tex]R = \frac{4\rho _AL_A}{\pi d_A^2}[/tex]
Resistance in wire B;
[tex]R = \frac{4\rho _BL_B}{\pi d_B^2}[/tex]
Power delivered in wire;
[tex]P = \frac{V^2}{R}[/tex]
Power delivered in wire A;
[tex]P = \frac{V^2_A}{R_A}[/tex]
Power delivered in wire B;
[tex]P = \frac{V^2_B}{R_B}[/tex]
Substitute in the value of R in Power delivered in wire A;
[tex]P_A = \frac{V^2_A}{R_A} = \frac{V^2_A \pi d^2_A}{4 \rho_A L_A}[/tex]
Substitute in the value of R in Power delivered in wire B;
[tex]P_B = \frac{V^2_B}{R_B} = \frac{V^2_B \pi d^2_B}{4 \rho_B L_B}[/tex]
Take the ratio of power delivered to A to power delivered to B;
[tex]\frac{P_A}{P_B} = (\frac{V^2_A \pi d^2_A}{4\rho_AL_A} ) *(\frac{4\rho_BL_B}{V^2_B \pi d^2_B})\\\\ \frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{\rho_AL_A} )*(\frac{\rho_BL_B}{V^2_B d^2_B})\\\\[/tex]
The wires are made of the same material, [tex]\rho _A = \rho_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{V^2_A d^2_A}{L_A} )*(\frac{L_B}{V^2_B d^2_B})\\\\[/tex]
The wires are connected across the same potential; [tex]V_A = V_B[/tex]
[tex]\frac{P_A}{P_B} = (\frac{ d^2_A}{L_A} )* (\frac{L_B}{d^2_B} )[/tex]
wire A has seven times the diameter and seven times the length of wire B;
[tex]\frac{P_A}{P_B} = (\frac{ (7d_B)^2}{7L_B} )* (\frac{L_B}{d^2_B} )\\\\\frac{P_A}{P_B} = \frac{49d_B^2}{7L_B} *\frac{L_B}{d^2_B} \\\\\frac{P_A}{P_B} =\frac{49}{7} \\\\\frac{P_A}{P_B} = 7\\\\P_A : P_B = 7:1[/tex]
Therefore, the ratio of the power delivered to A to power delivered to B is
7 : 1
Potential difference of a battery is 2.2 V when it is connected
across a resistance of 5 ohm, if suddenly the potential difference
falls to 1.8V, its internal resistance will be
Answer:
1.1ohms
Explanation:
According to ohms law E = IR
If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5
I = 0.36A (This will be the load current).
Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.
Voltage drop = 2.2V - 1.8V = 0.4V
Then we calculate the internal resistance using ohms law.
According to the law, V = Ir
V= voltage drop
I is the load current
r = internal resistance
0.4 = 0.36r
r = 0.4/0.36
r = 1.1 ohms
Two red blood cells each have a mass of 9.0 x 10-14 kg and carry a negative charge spread uniformly over their surfaces. The repulsion from the excess charge prevents the cells from clumping together. One cell carries -2.5pC and the other -3.30 pC, and each cell can be modeled as a sphere 3.75 × 10-6 m in radius. If the red blood cells start very far apart and move directly toward each other with the same speed.
1. What initial speed would each need so that they get close enough to just barely touch?
2. What is the maximum acceleration of the cells as they move toward each other and just barely touch?
Answer:
Explanation:
Given that:
The mass of the cell is 9.0 x 10^-14 kg
The charges of the cell is -2.5pC and the other -3.30 pC
[tex]q_1=-2.5\times10^{-12}C \ \ and \ \ q_2=-3.75\times10^{-12}C[/tex]
Radius is 3.75 × 10-6 m
The final distance is twice the radius
i.e [tex]2*(3.75 \times 10^{-6}) = 7.5*10^{-6}m[/tex]
The formula for the velocity of the cell is
[tex]mv^2=\frac{q_1q_2}{4\pi \epsilon 2 r} \\[/tex]
[tex]v=\sqrt{\frac{q_1q_2}{4\pi \epsilon 2 r} }[/tex]
[tex]=\sqrt{\frac{(-2.5\times10^{-12})(-3.3\times10^{-12}}{4(3.14)(8.85\times10^{-112}(2\times3.75\times10^{-6})(9\times10^{-14})} } \\\\=\sqrt{\frac{(-8.25\times10^{-24})}{(7503.03\times10^{-32})} } \\\\=\sqrt{109955.5779} \\\\=331.60m/s[/tex]
The maximum acceleration of the cells as they move toward each other and just barely touch is
[tex]ma= \frac{q_1q_2}{4\pi \epsilon (2r)^2} \\\\a= \frac{q_1q_2}{4\pi \epsilon (2r)^2(m)}[/tex]
[tex]=\frac{(-2.5\times10^{-12})(-3.3\times10^{-12})}{4(3.14)(8.85\times10^{-12})(2\times3.75\times10^{-6})^2(9\times10^{-14})}[/tex]
[tex]=\frac{(-8.25\times10^{-24})}{(56272.725\times10^{-38})} \\\\=1.47\times10^{10}m/s^2[/tex]
The answers obtained are;
1. The initial speed of each of the red blood cells is [tex]v= 331.66\,m/s[/tex].
2. The maximum acceleration of the cells is [tex]a=1.47\times 10^{10}\,m/s^2[/tex].
The answer is explained as shown below.
We have, the mass of the red blood cell;
[tex]m=9\times 10^{-14}\,kg[/tex]Also, the charges of the cells are;
[tex]q_1=-2.5\times 10^{-12}\,C[/tex] and[tex]q_2=-3.30\times 10^{-12}\,C[/tex]The distance between the charges when they barely touch will be two times the radius of each charge.
[tex]r=2\times r\,'=2\times3.75\times10^{-6}\,m=7.5\times10^{-6}\,m[/tex]Kinetic Energy of moving charges1. As both the cells are negatively charged they will repel each other.
So, for the cells to come nearly close, their kinetic energies must be equal to the electric potential between them.[tex]\frac{1}{2}mv^2+ \frac{1}{2}mv^2=k\frac{q_1 q_2}{r^2}[/tex]Where, [tex]k=9\times10^9\,Nm^2/C^2[/tex] is the Coulomb's constant.Now, substituting all the known values in the equation, we get;
[tex](9\times 10^{-14}\,kg)\times v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m}[/tex][tex]v^2=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{7.5\times10^{-6}\,m\times(9\times 10^{-14}\,kg)} =110000\,m^2/s^2[/tex]
[tex]\implies v=\sqrt{110000\,m^2/s^2}=331.66\,m/s[/tex]Electrostatic force between two charges2. Also as the force between them is repulsive, there must be an acceleration to make them barely touch each other.
[tex]ma=k\frac{q_1 q_2}{r^2}[/tex]Substituting the known values, we get;
[tex](9\times 10^{-14}\,kg)\times a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2}[/tex]
[tex]\implies a=9\times 10^9Nm^2/C^2\times\frac{(-2.5\times 10^{-12}\,C)\times(-3.30\times 10^{-12}\,C)}{(7.5\times10^{-6}\,m)^2\times(9\times 10^{-14}\,kg) }[/tex]
[tex]a=1.47\times 10^{10}\,m/s^2[/tex]Find out more information about moving charges here:
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A charged particle q moves at constant velocity through a crossed electric and magnetic fields (E and B, which are both constant in magnitude and direction). Write the magnitude of the electric force on the particle in terms of the variables given. Do the same for the magnetic force
Answer:
The magnitude of the electric force on the particle in terms of the variables given is, F = qE
The magnitude of the magnetic force on the particle in terms of the variables given is, F = q (v x B)
Explanation:
Given;
a charged particle, q
magnitude of electric field, E
magnitude of magnetic field, B
The magnitude of the electric force on the particle in terms of the variables given;
F = qE
The magnitude of the magnetic force on the particle in terms of the variables given;
F = q (v x B)
where;
v is the constant velocity of the charged particle
Answer:
The magnitude of the electric force acting on a charged particle moving through an electric field = |qE|
The magnitude of the magnetic force of a charged particle moving at a particular velocity through a magnetic field = |qv × B|
Explanation:
The electric force acting on a charged particle, q, moving through an electric field, E, is given as a product of the charge on the particle (a scalar quantity) and the electric field (a vector quantity).
Electric force = qE
The magnitude of the electric force = |qE|
That is, magnitude of the product of the charge and the electric field vector.
The magnetic force acting on a charged particle, q, moving with a velocity, v, through a magnetic field, B is a vector product of qv [a product of the charge of the particle (a scalar quantity) and the velocity of the particle (a vector quantity)] and B (a vector quantity).
It is given mathematically as (qv × B)
The magnitude of the magnetic force is the magnitude of the vector product obtained.
Magnitude of the magnetic force = |qv × B|
Hope this Helps!!!
In 1949, an automobile manufacturing company introduced a sports car (the "Model A") which could accelerate from 0 to speed v in a time interval of Δt. In order to boost sales, a year later they introduced a more powerful engine (the "Model B") which could accelerate the car from 0 to speed 2.92v in the same time interval. Introducing the new engine did not change the mass of the car. Compare the power of the two cars, if we assume all the energy coming from the engine appears as kinetic energy of the car.
Answer: [tex]\frac{P_B}{P_A}[/tex] = 8.5264
Explanation: Power is the rate of energy transferred per unit of time: P = [tex]\frac{E}{t}[/tex]
The energy from the engine is converted into kinetic energy, which is calculated as: [tex]KE = \frac{1}{2}.m.v^{2}[/tex]
To compare the power of the two cars, first find the Kinetic Energy each one has:
K.E. for Model A
[tex]KE_A = \frac{1}{2}.m.v^{2}[/tex]
K.E. for model B
[tex]KE_B = \frac{1}{2}.m.(2.92v)^{2}[/tex]
[tex]KE_B = \frac{1}{2}.m.8.5264v^{2}[/tex]
Now, determine Power for each model:
Power for model A
[tex]P_{A}[/tex] = [tex]\frac{m.v^{2} }{2.t}[/tex]
Power for model B
[tex]P_B = \frac{m.8.5264.v^{2} }{2.t}[/tex]
Comparing power of model B to power of model A:
[tex]\frac{P_B}{P_A} = \frac{m.8.5264.v^{2} }{2.t}.\frac{2.t}{m.v^{2} }[/tex]
[tex]\frac{P_B}{P_A} =[/tex] 8.5264
Comparing power for each model, power for model B is 8.5264 better than model A.
That 85 kg paratrooper from the 50's was moving at constant speed of 56 m/s because the air was applying a frictional drag force to him that matched his weight. If he fell this way for 40 m, how much heat was generated by this frictional drag force in J
Answer:
46648 J
Explanation:
mass m= 85 Kg
velocity v = 56 m/s
distance covered s =40 m
According to Question,
frictional drag force to him that matched his weight
[tex]\Rightarrow F_d =mg\\=85\times9.81=833 N[/tex]
Therefore, work done by practometer against the drag force = heat was generated by this frictional drag force in J
W=Q= F_d×s
=833×56 = 46648 J
A 4.5 kg ball swings from a string in a vertical circle such that it has constant sum of kinetic and gravitational potential energy. Ignore any friction forces from the air or in the string. What is the difference in the tension between the lowest and highest points on the circle
Answer:
88.29 N
Explanation:
mass of the ball = 4.5 kg
weight of the ball will be = mass x acceleration due to gravity(9.81 m/s^2)
weight W = 4.5 x 9.81 = 44.145 N
centrifugal forces Tc act on the ball as it swings.
At the top point of the vertical swing,
Tension on the rope = Tc - W.
At the bottom point of the vertical swing,
Tension on the rope = Tc + W
therefore,
difference in tension between these two points will be;
Net tension = tension at bottom minus tension at the top
= Tc + W - (Tc - W) = Tc + W -Tc + W
= 2W
imputing the value of the weight W, we have
2W = 2 x 44.145 = 88.29 N
A fox locates rodents under the snow by the slight sounds they make. The fox then leaps straight into the air and burrows its nose into the snow to catch its meal. If a fox jumps up to a height of 85 cm , calculate the speed at which the fox leaves the snow and the amount of time the fox is in the air. Ignore air resistance.
Answer:
v = 4.08m/s₂
Explanation:
In 1898, the world land speed record was set by Gaston Chasseloup-Laubat driving a car named Jeantaud. His speed was 39.24 mph (63.15 km/h), much lower than the limit on our interstate highways today. Repeat the calculations of Example 2.7 (assume the car accelerates for 6 miles to get up to speed, is then timed for a one-mile distance, and accelerates for another 6 miles to come to a stop) for the Jeantaud car. (Assume the car moves in the +x direction.)
Find the acceleration for the first 6 miles.
Answer:
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
Explanation:
Given that:
the initial speed v₁ = 0 m/s i.e starting from rest ; since the car accelerates at a distance Δx = 6 miles in order to teach that final speed v₂ of 63.15 km/h.
So; the acceleration for the first 6 miles can be calculated by using the formula:
v₂² = v₁² + 2a (Δx)
Making acceleration a the subject of the formula in the above expression ; we have:
v₂² - v₁² = 2a (Δx)
[tex]a = \dfrac{v_2^2 - v_1^2 }{2 \Delta x}[/tex]
[tex]a = \dfrac{(63.15 \ km/s)^2 - (0 \ m/s)^2 }{2 (6 \ miles)}[/tex]
[tex]a = \dfrac{(17.54 \ m/s)^2 - (0 \ m/s)^2 }{2 (9.65*10^3 \ m)}[/tex]
[tex]a =0.0159 \ m/s^2[/tex]
Thus;
Assume the car moves in the +x direction;
the acceleration [tex]a^{\to} = (0.0159 \ \ m/s^2 )i[/tex]
Two guitarists attempt to play the same note of wavelength 6.50 cm at the same time, but one of the instruments is slightly out of tune. Consequently, a 17.0-Hz beat frequency is heard between the two instruments. What were the possible wavelengths of the out-of-tune guitar’s note? Express your answers, separated by commas, in centimeters to three significant figures IN cm.
Answer:
The two value of the wavelength for the out of tune guitar is
[tex]\lambda _2 = (6.48,6.52) \ cm[/tex]
Explanation:
From the question we are told that
The wavelength of the note is [tex]\lambda = 6.50 \ cm = 0.065 \ m[/tex]
The difference in beat frequency is [tex]\Delta f = 17.0 \ Hz[/tex]
Generally the frequency of the note played by the guitar that is in tune is
[tex]f_1 = \frac{v_s}{\lambda}[/tex]
Where [tex]v_s[/tex] is the speed of sound with a constant value [tex]v_s = 343 \ m/s[/tex]
[tex]f_1 = \frac{343}{0.0065}[/tex]
[tex]f_1 = 5276.9 \ Hz[/tex]
The difference in beat is mathematically represented as
[tex]\Delta f = |f_1 - f_2|[/tex]
Where [tex]f_2[/tex] is the frequency of the sound from the out of tune guitar
[tex]f_2 =f_1 \pm \Delta f[/tex]
substituting values
[tex]f_2 =f_1 + \Delta f[/tex]
[tex]f_2 = 5276.9 + 17.0[/tex]
[tex]f_2 = 5293.9 \ Hz[/tex]
The wavelength for this frequency is
[tex]\lambda_2 = \frac{343 }{5293.9}[/tex]
[tex]\lambda_2 = 0.0648 \ m[/tex]
[tex]\lambda_2 = 6.48 \ cm[/tex]
For the second value of the second frequency
[tex]f_2 = f_1 - \Delta f[/tex]
[tex]f_2 = 5276.9 -17[/tex]
[tex]f_2 = 5259.9 Hz[/tex]
The wavelength for this frequency is
[tex]\lambda _2 = \frac{343}{5259.9}[/tex]
[tex]\lambda _2 = 0.0652 \ m[/tex]
[tex]\lambda _2 = 6.52 \ cm[/tex]
This question involves the concepts of beat frequency and wavelength.
The possible wavelengths of the out-of-tune guitar are "6.48 cm" and "6.52 cm".
The beat frequency is given by the following formula:
[tex]f_b=|f_1-f_2|\\\\[/tex]
f₂ = [tex]f_b[/tex] ± f₁
where,
f₂ = frequency of the out-of-tune guitar = ?
[tex]f_b[/tex] = beat frequency = 17 Hz
f₁ = frequency of in-tune guitar = [tex]\frac{speed\ of\ sound\ in\ air}{\lambda_1}=\frac{343\ m/s}{0.065\ m}=5276.9\ Hz[/tex]
Therefore,
f₂ = 5276.9 Hz ± 17 HZ
f₂ = 5293.9 Hz (OR) 5259.9 Hz
Now, calculating the possible wavelengths:
[tex]\lambda_2=\frac{speed\ of\ sound}{f_2}\\\\\lambda_2 = \frac{343\ m/s}{5293.9\ Hz}\ (OR)\ \frac{343\ m/s}{5259.9\ Hz}\\\\[/tex]
λ₂ = 6.48 cm (OR) 6.52 cm
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a wall, a 55.6 kg painter is standing on a 3.15 m long homogeneous board that is resting on two saw horses. The board’s mass is 14.5 kg. The saw horse on the right is 1.00 m from the right. How far away can the painter walk from the saw horse on the right until the board begins to tip?
Answer:
0.15 m
Explanation:
First calculating the center of mass from the saw horse
[tex]\frac{3.15}{2} -1=0.575 m[/tex]
from the free body diagram we can write
Taking moment about the saw horse
55.9×9.81×y=14.5×0.575×9.81
y= 0.15 m
So, the painter walk from the saw horse on the right until the board begins to tip is 0.15 m far.
In a contest, two tractors pull two identical blocks of stone thesame distance over identical surfaces. However, block A is moving twice as fast as block B when it crosses the finish line. Which statement is correct?a) Block A has twiceas much kinetic energy as block B.b) Block B has losttwice as much kinetic energy to friction as block A.c) Block B has losttwice as much kinetic energy as block A.d) Both blocks havehad equal losses of energy to friction.e) No energy is lostto friction because the ground has no displacement.
Answer:
d) Both blocks have had equal losses of energy to friction
Explanation:
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces
Moreover, the block A is twice as fast than block B at the time of crossing the finish line
So based on the above information, it contains the losses of identical friction
And we also know that
Friction energy loss is
[tex]= \mu \times m \times g \times D[/tex]
It would be the same for both the blocks
hence, the option d is correct
The correct answer will be both blocks have had equal losses of energy to friction.
What is friction?Friction is defined as when any object is slides on a surface by means of any external force then the force in the opposite direction generated between the surface and the body restrict the motion of the body this force is called as the friction.
As it is mentioned in the question that two tractors pull two same stone blocks having the identical distance over the same surfaces.
Moreover, the block A is twice as fast as block B at the time of crossing the finish line.
So based on the above information, it contains the losses of identical friction.
And we also know that
Friction energy loss is
[tex]E_f=\mu m g D[/tex]
It would be the same for both the blocks
Hence both blocks have had equal losses of energy to friction.
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A brass ring of diameter 10.00 cm at 19.0°C is heated and slipped over an aluminum rod with a diameter of 10.01 cm at 19.0°C. Assuming the average coefficients of linear expansion are constant. What if the aluminum rod were 10.02 cm in diameter?
Answer:
the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
Explanation:
The change in length of a bar can be expressed with the relation;
[tex]\Delta L = L_f - L_i[/tex] ---- (1)
Also ; the relative or fractional increase in length is proportional to the change in temperature.
Mathematically;
ΔL/L_i ∝ kΔT
where;
k is replaced with ∝ (the proportionality constant )
[tex]\dfrac{ \Delta L}{L_i}=\alpha \Delta T[/tex] ---- (2)
From (1) ;
[tex]L_f = \Delta L + L_i[/tex] --- (3)
from (2)
[tex]{ \Delta L}=\alpha \Delta T*{L_i}[/tex] ---- (4)
replacing (4) into (3);we have;
[tex]L_f =(\alpha \Delta T*{L_i} ) + L_i[/tex]
On re-arrangement; we have
[tex]L_f = L_i + \alpha L_i (\Delta T )[/tex]
from the given question; we can say that :
[tex](L_f)_{brass}}} = (L_f)_{Al}[/tex]
So;
[tex]L_{brass} + \alpha _{brass} L_{brass}(\Delta T) = L_{Al} + \alpha _{Al} L_{Al}(\Delta T)[/tex]
Making the change in temperature the subject of the formula; we have:
[tex]\Delta T = \dfrac{L_{Al}-L_{brass}}{\alpha _ {brass} L_{brass}-\alpha _{Al}L_{Al}}[/tex]
where;
[tex]L_{Al}[/tex] = 10.02 cm
[tex]L_{brass}[/tex] = 10.00 cm
[tex]\alpha _{brass}[/tex] = 19 × 10⁻⁶ °C ⁻¹
[tex]\alpha_{Al}[/tex] = 24 × 10⁻⁶ °C ⁻¹
[tex]\Delta T = \dfrac{10.02-10.00}{19*10^{-6} \ \ {^0}C^{-1} *10.00 -24*10^{-6} \ \ {^0}C^{-1} *10.02}[/tex]
[tex]\Delta T[/tex] = −396.1965135 ° C
[tex]\Delta T[/tex] ≅ −396.20 °C
Given that the initial temperature [tex]T_i = 19^0 C[/tex]
Then ;
[tex]\Delta T = T_f - T_i[/tex]
[tex]T_f = \Delta T + T_I[/tex]
Thus;
[tex]T_f =(-396.20 + 19.0)^0 C[/tex]
[tex]\mathbf{T_f = -377.2^0 C}[/tex]
Thus; the final temperature is [tex]\mathbf{T_f = -377.2^0 C}[/tex]
An electron of mass 9.11 x 10^-31 kg has an initial speed of 4.00 x 10^5 m/s. It travels in a straight line, and its speed increases to 6.60 x10^5 m/s in a distance of 5.40 cm. Assume its acceleration is constant.
Required:
a. Determine the magnitude of the force exerted on the electron.
b. Compare this force (F) with the weight of the electron (Fg), which we ignored.
Answer:
a. F = 2.32*10^-18 N
b. The force F is 2.59*10^11 times the weight of the electron
Explanation:
a. In order to calculate the magnitude of the force exerted on the electron you first calculate the acceleration of the electron, by using the following formula:
[tex]v^2=v_o^2+2ax[/tex] (1)
v: final speed of the electron = 6.60*10^5 m/s
vo: initial speed of the electron = 4.00*10^5 m/s
a: acceleration of the electron = ?
x: distance traveled by the electron = 5.40cm = 0.054m
you solve the equation (2) for a and replace the values of the parameters:
[tex]a=\frac{v^2-v_o^2}{2x}=\frac{(6.60*10^5m/s)^2-(4.00*10^5m/s)^2}{2(0.054m)}\\\\a=2.55*10^{12}\frac{m}{s^2}[/tex]
Next, you use the second Newton law to calculate the force:
[tex]F=ma[/tex]
m: mass of the electron = 9.11*10^-31kg
[tex]F=(9.11*10^{-31}kg)(2.55*10^{12}m/s^2)=2.32*10^{-18}N[/tex]
The magnitude of the force exerted on the electron is 2.32*10^-18 N
b. The weight of the electron is given by:
[tex]F_g=mg=(9.11*10^{-31}kg)(9.8m/s^2)=8.92*10^{-30}N[/tex]
The quotient between the weight of the electron and the force F is:
[tex]\frac{F}{F_g}=\frac{2.32*10^{-18}N}{8.92*10^{-30}N}=2.59*10^{11}[/tex]
The force F is 2.59*10^11 times the weight of the electron
At the local playground, a 21-kg child sits on the right end of a horizontal teeter-totter, 1.8 m from the pivot point. On the left side of the pivot an adult pushes straight down on the teeter-totter with a force of 151 N. Part A In which direction does the teeter-totter rotate if the adult applies the force at a distance of 3.0 m from the pivot?Part B
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.5 m from the pivot?
(clockwise/counterclockwise)
Part C
In which direction does the teeter-totter rotate if the adult applies the force at a distance of 2.0 m from the pivot?
(clockwise/counterclockwise)
Answer:
By convention a negative torque leads to clockwise rotation and a positive torque leads to counterclockwise rotation.
here weight of the child =21kgx9.8m/s2 = 205.8N
the torque exerted by the child Tc = - (1.8)(205.8) = -370.44N-m ,negative sign is inserted because this torque is clockwise and is therefore negative by convention.
torque exerted by adult Ta = 3(151) = 453N , counterclockwise torque.
net torque Tnet = -370.44+453 =82.56N , which is positive means counterclockwise rotation.
b) Ta = 2.5x151 = 377.5N-m
Tnet = -370.44+377.5 = 7.06N-m , positive ,counterclockwise rotation.
c)Ta = 2x151 = 302N-m
Tnet = -370.44+302 = -68.44N-m, negative,clockwise rotation.