Answer:
diameter = 9.951 × [tex]10^{-6}[/tex] m
Explanation:
given data
NA = 0.1
refractive index = 1.465
wavelength = 1.3 μm
to find out
What is the largest core diameter for which the fiber remains single-mode
solution
we know that for single mode v number is
V ≤ 2.405
and v = [tex]\frac{2*\pi *r}{ wavelength} NA[/tex]
here r is radius
so we can say
[tex]\frac{2*\pi *r}{ wavelength} NA[/tex] = 2.405
put here value
[tex]\frac{2*\pi *r}{1.3*10^{-6}} 0.1[/tex] = 2.405
solve it we get r
r = 4.975979 × [tex]x^{-6}[/tex] m
so diameter is = 2 × 4.975979 × [tex]10^{-6}[/tex] m
diameter = 9.951 × [tex]10^{-6}[/tex] m
A convex thin lens with refractive index of 1.50 has a focal length of 30cm in air. When immersed in a certain transparent liquid, it becomes a negative lens of focal length of 188 cm. Determine the refractive index of the liquid.
Answer:
[tex]n_l = 1.97[/tex]
Explanation:
given data:
refractive index of lens 1.50
focal length in air is 30 cm
focal length in water is -188 cm
Focal length of lens is given as
[tex]\frac{1}{f} =\frac{n_2 -n_1}{n_1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
[tex]\frac{1}{f} =\frac{n_{g} -n_{air}}{n_{air}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
[tex]\frac{1}{f} =\frac{n_{g} -1}{1} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
focal length of lens in liquid is
[tex]\frac{1}{f} =\frac{n_{g} -n_{l}}{n_{l}} * \left [\frac{1}{r1} -\frac{1}{r2} \right ][/tex]
[tex]=\frac{n_{g} -n_{l}}{n_{l}} [\frac{1}{(n_{g} - 1) f}[/tex]
rearrange fro[tex] n_l[/tex]
[tex]n_l = \frac{n_g f_l}{f_l+f(n_g-1)}[/tex]
[tex]n_l = \frac{1.50*(-188)}{-188 + 30(1.50 -1)}[/tex]
[tex]n_l = 1.97[/tex]
Final answer:
To calculate the refractive index of a liquid in which a convex lens acts as a diverging lens, the lensmaker's formula and given focal lengths are used. The formula is manipulated to solve for the refractive index of the liquid, providing an approximation of 0.2394 for the unknown index.
Explanation:
To find the refractive index of the liquid in which a convex lens becomes a diverging lens, we need to apply the lensmaker's formula, considering refractive indices of the lens material and the surrounding medium. The lens has a focal length of 30 cm in air, and its refractive index is 1.50. When the lens is immersed in the liquid, its focal length changes to -188 cm, indicating that it now diverges light rays.
We use the formula for the focal length of a lens in a medium:
1 / f = (n_lens / n_medium - 1) * (1 / R1 - 1 / R2)
which is derived from the lensmaker's formula, where f is the focal length, n_lens is the refractive index of the lens, n_medium is the refractive index of the surrounding medium, and R1 and R2 are the radii of curvature for the lens surfaces. For a thin lens, this can be simplified to:
f in medium = (n_medium / n_lens) * f in air.
Rearranging for n_medium, we get:
n_medium = (f in air / f in medium) * n_lens.
Plugging in the values, we calculate:
n_medium = (30 cm / -188 cm) * 1.50 = -0.2394 (approximately).
The negative sign indicates that we need to take the absolute value, so the refractive index of the liquid is approximately 0.2394.
A coin is tossed upward from an initial height of 7 m above the ground, with an initial speed of 7.4 m/s. The magnitude of the gravitational acceleration g = 9.8 m/s^2
Take the point of release to be y0 = 0. Choose UPWARD as positive y direction. Pay attention to the signs of position, velocity and acceleration.
Keep 2 decimal places in all answers. a. Find the coin’s maximum height in meters above the ground?
b. How long in seconds is the coin in the air? You need to set up a quadratic equation with time t. Solve it for time t. Only take the positive solution.
c. What is its speed in m/s when it hits the ground? Note: speed has no sign, or always positive.
Answer:
a) The coin´s maximum height is 9.79 m above the ground.
b) The coin is 2.17 s in the air.
c) The speed is 13.82 m/s when the coin hits the ground
Explanation:
The equations for the position and velocity of the coin are the following:
y = y0 + v0 · t + 1/2 · g · t²
v = v0 + g · t
Where
y = height at time t
y0 = initial height
v0 = initial velocity
t = time
g = acceleration due to gravity
v = velocity at time t
a) At its max-height, the velocity of the coin is 0. Using the equation of velocity, we can obtain the time at which the velocity is 0.
v = v0 + g · t
0 = 7.4 m/s - 9.8 m/s² · t
- 7.4 m/s / - 9.8 m/s² = t
t = 0.76 s
Now calculating the height of the coin at t = 0.76 s, we will obtain the maximum height:
y = y0 + v0 · t + 1/2 · g · t²
y = 0 m + 7.4 m/s · 0.76 s - 1/2 · 9.8 m/s² · (0.76 s)²
y = 2.79 m
The coin´s maximum height above the ground is 7 m + 2.79 m = 9.79 m
b) After the coin reaches its maximum height, it falls to the ground. The initial position will be the max-height (2.8 m) and the final position is the ground (-7 m). The initial velocity, v0, will be 0, because the coin is at the max-height. Then, using the equation of position we can calculate the time the coin is falling:
y = y0 + v0 · t + 1/2 · g · t²
-7 m = 2.79 m - 1/2 · 9.8 m/s² · t²
-2 ·(-7 m - 2.79 m)/ 9.8 m/s² = t²
t = 1.41 s
The coin is (1.41 s + 0.76 s) 2.17 s in the air
c) Using the equation of velocity, we can calculate the speed at time 1.41 s, when the coin hits the ground.
v = v0 + g · t
v = 0 m/s - 9.8 m/s² · (1.41 s)
v = -13.82 m/s
The speed is 13.82 m/s when the coin hits the ground.
What does it mean when we say that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side?
Answer:
It means at a diverging boundary when magma spreads to both sides it is almost identical in both the side.
Explanation:
At a diverging boundary when magma spreads to both sides it is almost identical in both the side.
If you take a picture of one side of the boundry and the spreading of sea floor and place it before a mirror you can see the image is identical to the picture of other side.
Therefore, the meaning of saying, each side of the sea floor away from the mid ocean ridge is a mirror image of the other side is explained above.
The statement that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side refers to the symmetric nature of seafloor spreading. Molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust, creating symmetric magnetic striping patterns. This is akin to the bilaterally symmetric structure observed in certain organisms.
Explanation:When we say that each side of the seafloor away from the mid ocean ridge is a mirror image of the other side, we are referencing the symmetric nature of seafloor spreading. Just as a mirror image reflects an object exactly, so does the seafloor on one side of the mid-ocean ridge reflect the seafloor on the opposite side. This is because molten material rises from the Earth's mantle at the mid-ocean ridge, cools, and forms new oceanic crust. This new crust then moves away from the ridge due to tectonic forces, and the process repeats, creating a pattern of symmetrical magnetic striping on the seafloor.
This 'mirroring' effect is similar to the bilaterally symmetric structure seen in certain organisms, where a plane cut from the front to back of the organism produces distinct left and right sides that are mirror images of each other. You can see this symmetry in the images of the Moon provided – two different sides, yet mirroring similar physical traits.
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What from the following list of statements about vectors is definitely true? (section 3.3) The magnitude of a vector can be smaller than length of one of its components. Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction. Magnitude of a vector can be zero if only one of components is zero. If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B. Magnitude of a vector can be zero only if all components of a vector are zero.
Answer:
"Magnitude of a vector can be zero only if all components of a vector are zero."
Explanation:
"The magnitude of a vector can be smaller than length of one of its components."
Wrong, the magnitude of a vector is at least equal to the length of a component. This is because of the Pythagoras theorem. It can never be smaller.
"Magnitude of a vector is positive if it is directed in +x and negative if is is directed in -X direction."
False. Magnitude of a vector is always positive.
"Magnitude of a vector can be zero if only one of components is zero."
Wrong. For the magnitude of a vector to be zero, all components must be zero.
"If vector A has bigger component along x direction than vector B, it immediately means, the vector A has bigger magnitude than vector B."
Wrong. The magnitude of a vector depends on all components, not only the X component.
"Magnitude of a vector can be zero only if all components of a vector are zero."
True.
The correct statements are A and E.
A) True. The magnitude of a vector can indeed be smaller than the length of one of its components, especially when the vector has components in multiple directions.
B) False. The magnitude of a vector is always positive or zero, regardless of its direction. It is not negative if directed in the -x direction.
C) False. The magnitude of a vector can be zero only if all of its components are zero. Having only one component as zero does not guarantee a zero magnitude.
D) False. The magnitude of a vector depends on the vector's components in all directions, not just the component along the x direction. It cannot be immediately concluded that vector A has a bigger magnitude than vector B solely based on the x-component.
E) True. The magnitude of a vector can be zero only if all components of the vector are zero. This is a fundamental property of vectors; they have zero magnitude only if they have no components in any direction.
Complete question:
Which of the following statements about vectors is definitely true? (Section 3.3)
A) The magnitude of a vector can be smaller than the length of one of its components.
B) The magnitude of a vector is positive if it is directed in the +x direction and negative if it is directed in the -x direction.
C) The magnitude of a vector can be zero if only one of its components is zero.
D) If vector A has a bigger component along the x direction than vector B, it immediately means that vector A has a bigger magnitude than vector B.
E) The magnitude of a vector can be zero only if all components of the vector are zero.
A m=7.2 gram object is accelerated at a rate of a=2.9 m/s^2. What force (in millinewtons) does the object experience? No need to add the unit (already given).
Answer:
20.88 mN
Explanation:
given data:
mass of object = 7.2 gm
acceleration of object = 2.9 m/s2
we know that force is given as
F = ma
where m is mass of object and a is acceleration of moving object.
putting all value to get required force
= 7.2*10^{-3}\ kg *2.9 m/s2
= 20.88*10^{-3} N
force in milli newton is
= 20.88*10^{-3} * 1000 = 20.88 mN
A golfer hits a shot to a green that is elevated 3 m above the point where the ball is struck. The ball leaves the club at a speed of 16.6 m/s at an angle of 40.9° above the horizontal. It rises to its maximum height and then falls down to the green. Ignoring air resistance, find the speed of the ball just before it lands.
Answer:14.72 m/s
Explanation:
Given
Initial velocity (u)=16.6 m/s
[tex]\theta =40.9^{\circ}[/tex]
Horizontal velocity component ([tex]u_x[/tex])=16.6cos40.9=12.54 m/s
As the ball comes down so its vertical displacement is zero except 3 m elevation
Thus [tex]v_y=\sqrt{\left ( 16.6sin40.9\right )^2+2\left ( -9.81\right )\left ( 3\right )}[/tex]
[tex]v_y=\sqrt{10.868^2-58.86}[/tex]
[tex]v_y=\sqrt{59.253}[/tex]
[tex]v_y=7.69 m/s[/tex]
there will be no change is horizontal velocity as there is no acceleration
Therefore Final Velocity
[tex]v=\sqrt{u_x^2+v_y^2}[/tex]
[tex]v=\sqrt{12.54^2+7.69^2}[/tex]
v=14.72 m/s
A 55-liter tank is full and contains 40kg of fuel. Find using Sl units: • Density p. • Specific Weight y • Specific Gravity Answer tolerance = 1%. Be sure to include units. The sign of the answers will not be graded, use a positive value for your answer. Your answers: p= (Enter a positive value) y = (Enter a positive value) SG = (Enter a positive value)
Answer:
p = 727.273 kg/m3, y = 7134.545 N/m^3, SG = 0.7273
Explanation:
Density is simply the amount of mass of a substance per unit of volume. It can be found by dividing the mass in kg by the volume im m^3:
[tex]p = \frac{m}{v} = \frac{40kg}{55lt*\frac{1 m^3}{1000 lt}} = 727.273 kg/m^3[/tex]
Specific weight is the weight of the substance per unit of volume. The weight is the mass of the material times the gravity, and it represents the force that the earth exerts on an object. Another way of calculate this value, its multiplying the density of the fuel times the gravity. Then:
[tex]y = p*g = 727.273 kg/m^3 * 9.81 m/s^2 = 7134.545 N/m^3[/tex]
Specific Gravity is the ratio of the density of the substance to the density of a reference substance. For liquids, the reference substance is water at 4°C, which has a density of about 1000 kg/m^3.
[tex]SG =\frac{ p_{fuel}}{p_{water}} = \frac{727.273 kg/m^3}{1000 kg/m^3} = 0.7273[/tex]
Which of the following lead-acid battery voltages has the highest value? A. Floating voltage B. Discharge voltage c. Open-circuit voltagels. D. Equalize voltage
Answer:
D. Equalize voltage
Explanation:
Of the following lead-acid battery the battery with voltage value is Equalize voltage. EQUALIZING lead acid battery is process of de-sulphating the battery by carrying out a controlled overcharge. When battery plates acquire sulphate coating over time, their efficiency reduces, by overcharging this sulpahte coating is blown-off and the battery is rejuvenated.
What is the difference in energy between a beta particle at rest and one traveling 0.35c?
Answer:
They have a difference in energy of 35 eV.
Explanation:
The energy at rest of a particle is given by:
[tex]E_{R} = m_{0}c^2[/tex] (1)
Where [tex]m_{0}[/tex] is the mass of the particle at rest and c is the speed of light.
Beta particles are high energy and high velocity electrons or positrons ejected from the nucleus of an atom as a consequence of a radioactive decay. Either if the beta particle is an electron¹ or a positron² it will have the same mass.
Hence, the mass of the beta particle at rest in equation (1) will be equal to the mass of an electron:
[tex]m_{e} = 9.1095x10^{-31} Kg[/tex]
Replacing the values of [tex]m_{e}[/tex] and c in equation (1) it is gotten:
[tex]E_{R} = (9.1095x10^{-31} Kg)(3.00x10^{8} m/s)^{2}[/tex]
[tex]E_{R} = 8.19x10^{-14} Kg.m^{2}/s^{2}[/tex]
But [tex]1 J = Kg.m^{2}/s^{2}[/tex], therefore:
[tex]E_{R} = 8.19x10^{-14} J[/tex]
It is better to express the rest energy in electronvolts (eV):
[tex]1eV = 1.60x10^{-19} J[/tex]
[tex]8.19x10^{-14} J . \frac{1 eV}{1.60x10^{-19} J}[/tex] ⇒ [tex]511.875 eV[/tex]
[tex]E_{R} = 511.875 eV[/tex]
So the energy of the beta particle at rest is 511.875 eV.
Case for the one traveling at 0.35c:
Since it is traveling at 35% of the speed of light it is necessary to express equation (1) in a relativistic way, that can be done adding the Lorentz factor to it:
[tex]E = \frac{m_{0}c^{2}}{sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex] (2)
Where v is the velocity of the particle (for this case 0.35c).
[tex]E = \frac{511.875 eV}{sqrt{1-\frac{(0.35c)^{2}}{c^{2}}}}[/tex]
[tex]E = \frac{511.875 eV}{sqrt{1-\frac{0.1225c^{2}}{c^{2}}}}[/tex]
[tex]E = \frac{511.875 eV}{sqrt{1-0.1225}}[/tex]
[tex]E = \over{511.875 eV}{sqrt{0.8775}}[/tex]
[tex]E = \over{511.875 eV}{0.936}[/tex]
[tex]E = 546.875 eV[/tex]
The difference in energy between the two particles can be determined using the relativistic form of the kinetic energy:
[tex]K = E – E_{R}[/tex] (3)
Where E is the energy of the particle traveling at 0.35c and [tex]E_{R}[/tex] is the energy of the beta particle at rest.
[tex]K = 546.875 eV – 511.875 eV[/tex]
[tex]K = 35 eV[/tex]
They have a difference in energy of 35 eV.
Key terms:
¹Electron: Fundamental particle of negative electric charge.
²Positron: Is an electron with positive electric charge (similar to an electron in all its properties except in electric charge and magnetic moment).
A uniform plank of length 4.4m and weight 228N
restshorizontally on two supports, with 1.1m of the plank hanging
overthe right support. To what distance x can a person who
weighs449N walk on the overhanging part of the plank before it
begins totip.
Answer:0.558 m
Explanation:
Given
weight of Plank=228 N
weight of man=449 N
Length of plank=4.4 m
let [tex]R_a[/tex] and [tex]R_b[/tex] be the reactions at A & B
and Reaction at A becomes zero when plank is about to rotate at B
Taking moment about B at that instant so that plank is just about to rotate.
[tex]228\times 1.1=449\times x[/tex]
Where x is the maximum distance that man can walk
x=0.558 m
In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm. (a) Find the magnitude of the electrostatic force of attraction, Fe between the electron and the proton. (b) Find the magnitude of the gravitational force of attraction Fg , between the electron and the proton, and find the ratio, Fe /Fg . me = 9.11 x 10-31kg, e = 1.602 x 10-19C mp = 1.67 x 10-27kg k = 9 x 109 Nm2 /C2 G = 6.67 x 10-11 Nm2 /kg2
Answer:
(a): [tex]F_e = 8.202\times 10^{-8}\ \rm N.[/tex]
(b): [tex]F_g = 3.6125\times 10^{-47}\ \rm N.[/tex]
(c): [tex]\dfrac{F_e}{F_g}=2.27\times 10^{39}.[/tex]
Explanation:
Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053[tex]\times 10^{-9}[/tex] m.
Part (a):
According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges [tex]q_1[/tex] and [tex]q_2[/tex] respectively is given by
[tex]F_e = \dfrac{k|q_1||q_2|}{r^2}[/tex]
where,
[tex]k[/tex] = Coulomb's constant = [tex]9\times 10^9\ \rm Nm^2/C^2.[/tex][tex]r[/tex] = distance of separation between the charges.For the given system,
The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, [tex]q_1 = +1.6\times 10^{-19}\ C.[/tex]
The charge on the electron, [tex]q_2 = -1.6\times 10^{-19}\ C.[/tex]
These two are separated by the distance, [tex]r = 0.053\times 10^{-9}\ m.[/tex]
Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by
[tex]F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.[/tex]
Part (b):
The gravitational force of attraction between two objects of masses [tex]m_1[/tex] and [tex]m_1[/tex] respectively is given by
[tex]F_g = \dfrac{Gm_1m_2}{r^2}.[/tex]
where,
[tex]G[/tex] = Universal Gravitational constant = [tex]6.67\times 10^{-11}\ \rm Nm^2/kg^2.[/tex][tex]r[/tex] = distance of separation between the masses.For the given system,
The mass of proton, [tex]m_1 = 1.67\times 10^{-27}\ kg.[/tex]
The mass of the electron, [tex]m_2 = 9.11\times 10^{-31}\ kg.[/tex]
Distance between the two, [tex]r = 0.053\times 10^{-9}\ m.[/tex]
Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by
[tex]F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.[/tex]
The ratio [tex]\dfrac{F_e}{F_g}[/tex]:
[tex]\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.[/tex]
It takes 15 min to drive 6.0 mi in a straight line to the local hospital. It takes 10 min to go the last 3.0 mi, 2.0 min to go the last mile, and only 30 s (0.50 min) to go the last 0.50 mi. What is the average speed of the trip? Take the positive xx ‑direction to be from the starting point toward the hospital.
Final answer:
The average speed for the trip to the hospital, which covers a distance of 6.0 miles in 15 minutes, is 24 miles per hour (mph).
Explanation:
To calculate the average speed of the trip to the hospital, you take the total distance traveled and divide it by the total time taken. The student has specified that it takes 15 minutes to drive 6.0 miles in total. Therefore, the average speed can be calculated as follows:
Speed = Total Distance / Total Time
Speed = 6.0 miles / 15 minutes
Speed = 0.4 miles per minute
To convert this to miles per hour (mph), multiply by 60 (since there are 60 minutes in an hour):
Speed = 0.4 miles/minute × 60 minutes/hour
Speed = 24 mph.
Therefore, the average speed for the entire trip is 24 mph.
You attach a meter stick to an oak tree, such that the top of the meter stick is 1.87 meters above the ground. Later, an acorn falls from somewhere higher up in the tree. If the acorn takes 0.166 seconds to pass the length of the meter stick, how high above the ground was the acorn before it fell (assuming that the acorn didn\'t run into any branches or leaves on the way down)?
Answer:
3.25 m
Explanation:
t = Time taken = 0.166 seconds
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration = 9.81 m/s²
s = 1 because meter stick is 1 meter in length
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{1-\frac{1}{2}\times 9.81\times 0.166^2}{0.166}\\\Rightarrow u=5.21\ m/s[/tex]
Here, the initial velocity of point B is calculated from the time which is given. This velocity will be the final velocity of the acorn which falling from point A.
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{5.21^2-0^2}{2\times 9.81}\\\Rightarrow s=1.38\ m[/tex]
The distance of the acorn from the ground is 1.87+1.38 = 3.25 m
How is the position vs. time graph with the cart moving towards the sensor with decreasing speed different than the position vs. time graph with the cart moving away from the sensor at increasing speed? How are the graphs similar?
Answer:
if Y is the position and X the time: in the first one you will see a crescent function that starts sharp and starts to curve down as the time pases. as the cart is slowing down, you will need more time to move the same as before.
Y (position)
I
sensor-------------------------------------------------------------------
I o
I o
I o
I o
I o
I o
I o
I------------------------------------------------------------------------------------- X (time)
in the second case the cart starts close to the sensor and starts getting sharper and sharper as the time pases. This is because the velocity is increasing, so for each second that pases, you will travel more distance that the second before it.
Y (position)
I
sensor ----------------------
I o
I o
I o
I o
I o
I o
I o
I------------------------------------------------------------------------------------- X (time)
i hope you can understand it, kinda hard to do graphs here.
A 5.0 c charge is 10 m from a small test charge. what is the magnitude of the electric field at the location of the test charge ?
Answer:
[tex]4.50*10^8\frac{N}{C}[/tex]
Explanation:
The electric field is generated by a charge and represents the force exerted on a test charge, that is, the electric force per unit of charge. Therefore the equation for the electric field can be obtained from Coulomb's law.
[tex]E=\frac{F}{q}\\E=\frac{kq}{r^2}\\E=\frac{5C*8.99*10^9\frac{Nm^2}{C^2}}{(10 m)^2}=4.50*10^8\frac{N}{C}[/tex]
The magnitude of the electric field created by a larger charge Q at the location of a test charge is calculated by the formula E = k*Q/r² where k is Coulomb's constant, Q is the charge creating the field, and r is the distance between the charges. Substituting the given values into the formula, we find the electric field magnitude to be 4.495 x 10^8 N/C.
Explanation:The magnitude of the electric field created by a point charge Q can be calculated using Coulomb's law as follows:
The equation for calculating the electric field E in relation to the force F imparted on a small test charge q is defined as E = F/q.
However, the Coulomb’s law gives the force F between two charges as F = k*Q*q/r², where, k is the Coulomb's constant (8.99 × 10^9 N.m²/C²), Q is the charge creating the field, q is the test charge, and r is the distance between them.
In the case where we want to find the electric field created by a larger charge Q at the location of a small charge q, we consider the force on the small charge exerted by Q, and therefore, we rewrite the equation as E = k*Q/r².
Therefore for your question, the magnitude of the electric field (E) at the location of the test charge is calculated as E = (8.99 x 10^9 N.m²/C² * 5.0 C) / (10m)² = 4.495 x 10^8 N/C.
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The heaviest flying bird is the trumpeter swan, which weighsin
at about 38 pounds at its heaviest. What is the weight
innnewtons?
Answer:169.1 N
Explanation:
Given
bird weigh 38 pounds
and we know
1 Pound is equal to 0.453592 kg
Thus 38 pounds is equal to 17.236 kg
Thus the weight of bird is [tex]17.236\times 9.8=169.09 N \approx 169.1 N[/tex]
Answer:
The weight in newtons is 169.1.
Explanation:
This question can be solved as a simple rule of three problem.
We have that each pound has 4.45 newtons. So, how many newtons are there in 38 pounds?
So
1 pound - 4.45N
38 pounds - xN
[tex]x = 38*4.45[/tex]
[tex]x = 169.1N[/tex]
The weight in newtons is 169.1.
Derive the formula for the electric field E to accelerate the charged particle to a fraction f of the speed of light c. Express E in terms of M, Q, D, f, c and v0. – (a) Using the Coulomb force and kinematic equations. (8 points) – (b) Using the work-kinetic energy theorem. ( 8 points) – (c) Using the formula above, evaluate the strength of the electric field E to accelerate an electron from 0.1% of the speed of light to 90% of the speed of light. You need to look up the relevant constants, such as mass of the electron, charge of the electron and the speed of light. (5 points)
Answer:
[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]
Part c)
[tex]E = \frac{2.07 \times 10^5}{D}[/tex]
Explanation:
Part a)
As per Coulomb's law we know that force on a charge placed in electrostatic field is given as
[tex]F = QE[/tex]
now acceleration of charge is given as
[tex]a = \frac{QE}{M}[/tex]
now if charge moved through the distance D in electric field and its speed changes from vo to fraction f of speed of light c
then we will have
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
[tex](fc)^2 - v_o^2 = 2(\frac{QE}{M})D[/tex]
so we have
[tex]E = \frac{(f^2c^2 - v_o^2)M}{2QD}[/tex]
Part b)
Now using work energy theorem we can say that total work done by electric force on moving charge will convert into kinetic energy
So we will have
[tex]QED = \frac{1}{2}M(cf)^2 - \frac{1}{2}Mv_o^2[/tex]
so we have
[tex]E = \frac{M(c^2f^2 - v_o^2)}{2QD}[/tex]
Part c)
Now if an electron is accelerated using this field
then we have
[tex]M = 9.11 \times 10^{-31} kg[/tex]
[tex]Q = 1.6 \times 10^{-19} C[/tex]
[tex]c = 3\times 10^8 m/s[/tex]
so we have
[tex]E = \frac{(9.1 \times 10^{-31})(0.9^2 - 0.001^2)\times 9 \times 10^{16}}{2(1.6 \times 10^{-19})D}[/tex]
[tex]E = \frac{2.07 \times 10^5}{D}[/tex]
A speed trap is set up with two pressure-activated
stripsplaced across a highway, 110 m apart. A car is speeding
alongat 33m/s, while the speed limit is 21 m/s. At the instant
thecar activates the first strip, the driver begins slowingdown.
What minimum deceleration is needed in order that theaverage speed
is within the limit by the time the car crosses thesecond
marker?
The question is asking to calculate the required minimum deceleration of a speeding car to fit the speed limit when it crosses the end of a measured stretch of the road. The deceleration required can be calculated using the motion equations of physics by first calculating the time it takes to travel the distance between the two strips and then using this time to calculate the required deceleration.
Explanation:The subject area of this question is kinematics, which is a branch of physics that deals with the concepts of distance, displacement, speed, velocity, and acceleration. The problem is asking for the minimum deceleration the car needs to have in order to not exceed the speed limit when it reaches the second strip. To solve this, you would use the equation v = v0 + at, where v is the final velocity, v0 is the initial velocity, a is acceleration (in this case deceleration, so it will be negative), and t is time. But first, we need to figure out the time. The time it takes to travel between the two strips can be calculated by t = d/v. The distance d is given as 110m, and the average speed v we want is 21 m/s (the speed limit).
Once we have the time, we can substitute the values into the first equation and solve for a. The values are v = 21 m/s (the speed we want to have in the end), v0 = 33 m/s (the initial speed), and we've calculated t from the previous step. For deceleration, as the speed is decreasing, a will be a negative number.
Learn more about Deceleration here:https://brainly.com/question/37877673
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An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H? An object is dropped from a height H. During the final second of its fall, it traverses a distance of 53.2 m. What was H?
Answer:
H = 171.90 m
Explanation:
given data
distance = 53.2 m
height = H
to find out
height H
solution
we know height is here H = [tex]\frac{1}{2} gt^2[/tex] ......................1
here t is time and a is acceleration
so
we find t first
we know during time (t -1) s , it fall distance (H - 53.2) m
so equation of distance
[tex]H - 53.2 = \frac{1}{2} g (t-1)^2[/tex]
[tex]H - 53.2 = \frac{1}{2} g (t^2-2t+1)[/tex]
[tex]H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g[/tex] ................2
now subtract equation 2 from equation 1 so we get
[tex]H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)[/tex]
53.2 = gt - [tex]\frac{1}{2} g [/tex]
53.2 = 9.81 t - [tex]\frac{1}{2} 9.8 [/tex]
t = 5.92 s
so from equation 1
H = [tex]\frac{1}{2} (9.81)5.92^2[/tex]
H = 171.90 m
A) Is it possible to say that a wave pulse( waves packet) has one particular wavelength? B) How are wave packets built (mathematically speaking)?
Answer: A) we can write a wave packet like Y = ∑aₙ*cos(w*n*t)
So you have lots of different wavelengths here, and you think that a wave packet has a unified frequency, but not wavelength.
B) mathematically speaking you will write a wave packet like i did up there, has a sum of many waves, where for each one you can have an intensity aₙ and a different frequency, the only thing you must see is that all the waves you are suming are moving in the same direction.
Light of wavelength 597 nm falls on a double slit, and the first bright fringe of the interference pattern is seen at an angle of 15.8° with the horizontal. Find the separation between the slits. µm
Answer:
2.2 µm
Explanation:
For constructive interference, the expression is:
[tex]d\times sin\theta=m\times \lambda[/tex]
Where, m = 1, 2, .....
d is the distance between the slits.
Given wavelength = 597 nm
Angle, [tex]\theta[/tex] = 15.8°
First bright fringe means , m = 1
So,
[tex]d\times sin\ 15.8^0=1\times \597\ nm[/tex]
[tex]d\times 0.2723=1\times \597\ nm[/tex]
[tex]d=2192.43481\ nm[/tex]
Also,
1 nm = 10⁻⁹ m
1 µm = 10⁻⁶ m
So,
1 nm = 10⁻³ nm
Thus,
Distance between slits ≅ 2.2 µm
Answer:
The separation between the slit is [tex]2.19\mu m[/tex]
Solution:
As per the question:
Wavelength of light, [tex]\lambda = 597 nm = 597\times 10^{-9} m[/tex]
[tex]\theta = 15.8^{\circ}[/tex]
Now, by Young's double slit experiment:
[tex]xsin\theta = n\lambda[/tex]
here,
n = 1
x = slit width
Therefore,
[tex]x = \frac{597\times 10^{-9}}{sin15.8^{\circ}} = 2.19\times 10^{- 6} m[/tex]
[tex]x = 2.19\mu m[/tex]
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 11.8 m/s. The magnitude of the gravitational acceleration g = 9.8 m/s2 Take the water surface to be y0 = 0. Choose UPWARD as positive y direction. Keep 2 decimal places in all answers. (a) How high (what maximum height) in meters does his body rise above the water? To solve this part, first note that the body's final velocity at the maximum height is known (implicitly) and identify its value. Then identify the unknown, and chose the appropriate equation (based on the knowns and unknown) to solve for it. After choosing the equation, solving for the unknown, checking units, think about whether the answer is reasonable.(b) How long in seconds is the dolphin in the air? Neglect any effects due to his size or orientation. Note: this is the total time of jumping up from water to hightest point and falling down to water.
Answer:
a) [tex]y_{max}=7.10m[/tex]
b) [tex]t=2.40 s[/tex]
Explanation:
From the exercise, we know the initial velocity, gravitational acceleration and initial position of the dolphin.
[tex]v_{oy}=11.8m/s[/tex]
[tex]y_{o}=0m\\ g=9.8m/s^{2}[/tex]
a) To find maximum height, we know that at that point the dolphin's velocity is 0 and it becomes coming down later.
Knowing that, we need to know how much time does it take the dolphin to reach maximum height.
[tex]v_{y}=v_{oy}+gt[/tex]
[tex]0=11.8m/s-(9.8m/s^{2} )t[/tex]
Solving for t
[tex]t=1.20 s[/tex]
So, the dolphin reach maximum point at 1.20 seconds
Now, using the equation of position we can calculate maximum height.
[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]
[tex]y=0+11.8m/s(1.20s)-\frac{1}{2}(9.8m/s^{2} )(1.20s)=7.10m[/tex]
b) To find how long is the dolphin in the air we need to analyze it's hole motion
At the end of the jump the dolphin return to the water at y=0. So, from the equation of position we have that
[tex]y=y_{o}+v_{oy}t +\frac{1}{2}gt^{2}[/tex]
[tex]0=0+11.8t-\frac{1}{2}(9.8)t^{2}[/tex]
What we have here, is a quadratic equation that could be solve using:
[tex]t=\frac{-b±\sqrt{b^{2}-4ac } }{2a}[/tex]
[tex]a=-\frac{1}{2} (9.8)[/tex]
[tex]b=11.8\\c=0[/tex]
[tex]t=0s[/tex] or [tex]t=2.40 s[/tex]
Since the answer can not be 0, the dolphin is 2.40 seconds in the air.
A cannon tilted up at a 29.0° angle fires a cannon ball at 81.0 m/s from atop a 22.0 m -high fortress wall. What is the ball's impact speed on the ground below?
Answer:
The speed of the ball when it hits the ground is 83.4 m/s
Explanation:
Please see the attached figure for a description of the problem.
The vector velocity at time "t" can be written as follows:
[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]
where:
v0 : module of the initial velocity vector.
α: launching angle.
g: acceleration due to gravity.
t: time
To calculate the impact speed, we can use this equation once we know the time at which the ball hits the ground. For this, we can use the equation for position:
[tex]r = ( x0 + v0*t*cos\alpha ; y0 + v0*t*sin\alpha + 1/2 g*t^{2})[/tex]
where:
r= vector position
x0 = horizontal initial position
y0 = vertical initial position
The problem gives us information about the vertical displacement. If we take the base of the wall as the center of the reference system, at the time at which the ball hits the ground, the module of the vertical component of the vector position, r, is 0 m (see figure). The initial vertical position is 22 m.
if ry is the vertical component of the vector r at final time:
[tex]ry = (0; y0 + v0*t*sin\alpha +1/2 g*t^{2})[/tex]
then, the module of ry is (see figure):
module ry =[tex]0 = 22m + v0*t*sin\alpha + 1/2*g*t^{2}[/tex]
Let´s replace with the given data:
[tex]0 = 22m + 39.3 m/s*t -4.9 m/s^{2}*t^{2}[/tex]
Solving the quadratic equation:
t = -0.5 and t = 8.5 s
At 8.5 s after firing, the ball hits the ground.
Now, we can find the module of the velocity vector when the ball hits the ground:
[tex]v = (v0 * cos\alpha ; v0 * sin\alpha + g*t)[/tex]
at time t = 8.5s
[tex]v = (81.0m/s * cos\alpha ; 81.0m/s * sin\alpha - 9.8 m/s^{2} *8.5s)[/tex]
v = (70.8 m/s; -44.0 m/s)
module of v = [tex]\sqrt{(70.8m/s)^{2} + (-44.0m/s)^{2}}[/tex] = 83.4 m/s
You are trying to determine the area of the floor of a rectangular closet since you will be replacing the carpet. You measure the length to be (5.0±0.1)m and the width to be (3.8±0.1)m.What is the maximum possible area based on the measurements and associated errors?What is the minimum possible area based on the measurements and associated errors?
Answer:
Explanation:
length L = 5±.1
percentage error
= 0.1/5 x 100 = 2%
breadth b = 3.8 ±.1
percentage error
= .1 / 3.8 x 100
= 2.6 %
Maximum reading of length = 5.1 m
maximum reading of breadth = 3.9m
maximum area possible = 5.1 x 3.9
= 19.9 m².
maximum possible error in the measurement of area in percentage terms
= 2+2.6 = 4.6 %
error in maximum area
19.9 x 4.6 /100
= .9 m²
Minimum possible measurement of area
4.9 x 3.7 = 18.1 m²
possible error
18.1 x 4.6 / 100
= .8 m²
A pharmacist attempts to weigh 100 milligrams of codeine sulfate on a balance with a sensitivity requirement of 4 milligrams. Calculate the maximum potential error (nearest hundredth) in terms of percentage (%).
2) A 10-milliliter graduate weighs 42.745 grams. When 6 milliliters of distilled water are measured in it, the combined weight of graduate and water is 48.540 grams. By definition, 6 milliliters of water should weigh 6 grams. Calculate the weight of the measured water and express any deviation from 6 grams as percentage of error (%, to the nearest hundredth).
Answer:
4 %
2 ) 3.42 %
Explanation:
Sensitivity requirement of 4 milligram means it is not sensitive below 4 milligram or can not measure below 4 milligram .
Given , 4 milligram is the maximum error possible .
Measured weight = 100 milligram
So percentage maximum potential error
= (4 / 100) x 100
4 %
2 )
As per measurement
weight of 6 milliliters of water
= 48.540 - 42.745 = 5.795 gram
6 milliliters of water should measure 6 grams
Deviation = 6 - 5.795 = - 0.205 gram.
Percentage of error =(.205 / 6 )x 100
= 3.42 %
An important news announcement is transmitted by radio waves to people sitting next to their radios 42 km from the station and by sound waves to people sitting across the newsroom 9.26 m from the newscaster. Taking the speed of sound in air to be 343 m/s, who receives the news first?
Answer:
news will reach to the listener who are 42 km apart
Explanation:
given,
distance of the radios from the station = 42 Km
speed of the sound in the air = 343 m/s
distance of the people = 9.26 m
[tex]time =\dfrac{distance}{speed}[/tex]
time taken by the signal to reach to the radio
speed of the electromagnetic wave to reach to the people
speed of electromagnetic wave = 3 × 10⁸ m/s
[tex]t =\dfrac{42000}{3 \times 10^8}[/tex]
t = 1.4 μs
time taken to reach to the people
[tex]time =\dfrac{distance}{speed}[/tex]
[tex]t =\dfrac{9.26}{343}[/tex]
t = 27 ms
time taken by the station to radio is less.
hence, news will reach to the listener who are 42 km apart
Calculate the rotational inertia of a wheel that has akinetic
energy of 24,400 J when rotating at 566 rev/min.
Answer:
J = 14.4 kg*m^2
Explanation:
Assuming that the wheel is not moving anywhere, and the kinetic energy is only due to rotation:
Ek = 1/2 * J * w^2
J = 2 * Ek / (w^2)
We need the angular speed in rad / s
566 rev/min * (1 min/ 60 s) * (2π rad / rev) = 58.22 rad/s
Then:
J = 2 * 24400 / (58.22^2) = 14.4 kg*m^2
What plate area is required if an air-filled, parallel-plate capacitor with a plate separation of 3.1 mm is to have a capacitance of 23 pF ?
Explanation:
Given that,
Separation between the plates, d = 3.1 mm = 0.0031 m
Capacitance of the capacitor, C = 23 pF
We need to find the area of the plate. The capacitance of a parallel plate capacitor is given by :
[tex]C=\dfrac{A\epsilon_0}{d}[/tex]
[tex]A=\dfrac{Cd}{\epsilon_0}[/tex]
[tex]A=\dfrac{23\times 10^{-12}\times 0.0031}{8.85\times 10^{-12}}[/tex]
[tex]A=0.0080\ m^2[/tex]
or
[tex]A=80\ cm^2[/tex]
So, the area of the plate is 80 square centimetres. Hence, this is the required solution.
an egg is dropped from the top of the library roof. then 2.2 seconds after that a second egg is thrown straight down with an initial speed of 58m/s. it is then observed that the two eggs splat on the ground at the same instant. how high in meters is the library
Answer:
39.8 m
Explanation:
Let h be the required height and t be the time to reach the egg which was thrown later on. The egg thrown earlier took ( t +2.2) s , with initial velocity zero to cover distance of h.,
So
h = 1/2 g( t +2.2)²
For the egg thrown with velocity of 58 m/s
h = 58 t + 1/2 g t²
Equating these two equations
1/2 g( t +2.2)² = 58 t + 1/2 g t²
1/2 g ( t² + 4.4 t + 4.84 ) = 58 t + 1/2 g t²
58 t = 21.56 t + 23.716
t = .65 s
h = 1/2 x 9.8 x (2.2+.65)²
h = 39.8 m
In 1271, Marco Polo departed Venice and traveled to Kublai Khan's court near Beijing, approximately 7900 km away in a direction we will call positive. Assume that the Earth is flat (as some did at the time) and that the trip took him 4.0 years, with 365 days in a year. (a) What was his average velocity for the trip, in meters per second? (b) A 767 could make the same trip in about 9.0 hours. What is the average velocity of the plane in meters per second?
Answer:
a) 0.063 m/s
b) 243.83 m/s
Explanation:
given,
distance = 7900 km
time = 4 years
a) average velocity for the trip = [tex]\dfrac{7900}{4\times 365}[/tex]
= 5.41 km/day
= [tex]\dfrac{5.41\times 1000}{24\times 60\times 60}[/tex]
average velocity for the trip = 0.063 m/s
b) average velocity = [tex]\dfrac{7900\times 1000}{9\times 60\times 60}[/tex]
average velocity for the trip = 243.83 m/s