Answer:
a) The acceleration is 2.14 m/s^{2}
b) The distance traveled by the car is 65.61 m
c) The average velocity is 18.75 m/s
Explanation:
Using the equations that describe an uniformly accelerated motion:
a) [tex]a=\frac{v_f - v_o}{t} =\frac{22.5m/s - 15.0 m/s}{3.50s}[/tex]
b) [tex]d= d_0 + v_0 t + \frac{1}{2} a t^{2} = 0 +15.0 x 3.5 + \frac{2.14x3.50^{2} }{2} = 65.61 m[/tex]
c) [tex]v_m =\frac{d}{t}=\frac{65.61}{3.5} =18.75 m/s[/tex]
The index of refraction of a thin lens is 1.5. Its one surface is convex (radius of curvature 20 cm) and the other planar. Calculate the focal length of the lens. What kind of an image does it form when the object is real and at 40 cm before the lens? What is the focal length if the lens is turned around? How does this influence image formation?
Answer:
f = 40 cm
image formed at a distance of 40 cm from lens is magnified and virtual.
when lens is turned around focal length is f = 40 cm
Explanation:
given data:
[tex]R_1 = 20 cm[/tex]
[tex]R_2 = \infty[/tex]
Refraction index of lens = 1.5
focal length of lens is given as
[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1} - \frac{1}{R_1})[/tex]
Putting all value to get focal length value
[tex]\frac{1}{f} = (1.5 -1) \frac{1}{20}[/tex]
[tex]\frac{1}{f} =\frac{0.5}{20}[/tex]
f = 40 cm
image formed at a distance of 40 cm from lens is magnified and virtual.
when lens is turned around
[tex]R_1 = \infty [/tex]
[tex]R_2 = -20cm[/tex]
Refraction index of lens = 1.5
focal length of lens is given as
[tex]\frac{1}{f} = (n-1) (\frac{1}{R_1} - \frac{1}{R_1})[/tex]
Putting all value to get focal length value
[tex]\frac{1}{f} = (1.5 -1) \frac{1}{20}[/tex]
[tex]\frac{1}{f} =\frac{0.5}{20}[/tex]
f = 40 cm
there is no change can be seen between two condition. image will form at 40 cm from lens
If two automobiles collide, they usually do not stick together.Does
this mean the collision is elastic? Explain why a head-oncollision
is likely to be more dangerous than other types ofcollisions.
Explanation:
We know that if
e = 1 then collision is called perfectly elastic collision.
0<e <1 then collision is called partial elastic collision.
e =0 then collision is called inelastic collision.
So when two automobile collide then they usually do not stick together then this collision is called as elastic collision.
When object collide head to head it become more dangerous than other type of collision because when object come toward each other and due to suddenly velocity of object become zero due to this it produce large amount of force.Usually this force produce two time more as compare to when object moving in same direction.
A woman on a bridge 90.0 m high sees a raft floating at
aconstant speed on the river below. She drops a stone fromrest in
an attempt to hit the raft. The stone is releasedwehn the raft has
6.00 m more to travel before passing under thebridge. The stone
hits the water 2.00 m in front of theraft. Find the speed of the
raft.
Answer:
0.93 m/s
Explanation:
t = Time taken
u = Initial velocity = 0
v = Final velocity
s = Displacement = 90 m
a = Acceleration = 9.81 m/s²
[tex]s=ut+\frac{1}{2}at^2\\\Rightarrow 90=0\times t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{90\times 2}{9.81}}\\\Rightarrow t=4.3\ s[/tex]
So, the raft covered 6-2 = 4 m in 4.3 seconds
Speed = Distance / Time
[tex]\text{Speed}=\frac{4}{4.3}=0.93\ m/s[/tex]
Speed of the raft is 0.93 m/s
To find the speed of the raft, we can use the principle of conservation of energy. When the woman drops the stone, it starts with potential energy due to its height and then converts to kinetic energy as it falls.
Explanation:To find the speed of the raft, we can use the principle of conservation of energy. When the woman drops the stone, it starts with potential energy due to its height and then converts to kinetic energy as it falls. The kinetic energy of the stone when it hits the water is equal to the potential energy it had initially. We can use the equation:
mgh = 0.5mv^2
Where m is the mass of the stone, g is the acceleration due to gravity, h is the height of the bridge, and v is the speed of the stone when it hits the water. Rearranging the equation, we can solve for v:
v = √(2gh)
Substituting the given values h = 90.0 m and g = 9.8 m/s^2, we can calculate the speed of the stone when it hits the water. This speed is equal to the speed of the raft.
When 9.72 g of an unknown non-electrolyte is dissolved in 50.0 g of cyclohexane, the boiling point increased to 84.93 degrees C from 80.7 degrees C. If the Kbp of the solvent is 2.79 K/m, calculate the molar mass of the unknown solute.
Answer : The molar mass of unknown compound is 128.22 g/mole
Explanation :
Mass of unknown compound = 9.72 g
Mass of solvent = 50.0 g
Formula used :
[tex]\Delta T_b=i\times K_b\times m\\\\T_2-T_1=i\times K_b\times\frac{\text{Mass of unknown compound}}{\text{Molar mass of unknown compound}\times \text{Mass of solvent in Kg}}[/tex]
where,
[tex]\Delta T_b[/tex] = elevation in boiling point
[tex]T_1[/tex] = temperature of solvent = [tex]80.7^oC=273+80.7=353.7K[/tex]
[tex]T_2[/tex] = temperature of solution = [tex]84.93^oC=273+84.93=357.93K[/tex]
i = Van't Hoff factor = 1 (for non-electrolyte)
[tex]K_f[/tex] = boiling point constant for solvent = 2.79 K/m
m = molality
Now put all the given values in this formula, we get:
[tex](357.93-353.7)K=1\times (2.79K/m)\times \frac{9.72g\times 1000}{\text{Molar mass of unknown compound}\times 50.0g}[/tex]
[tex]\text{Molar mass of unknown compound}=128.22g/mole[/tex]
Therefore, the molar mass of unknown compound is 128.22 g/mole
An electron is released from rest at a distance d = 100 m from an infinite conducting plane. The electron will begin to move towards the plane due to charge induction in the plane. How long will take for the electron to strike the plane?
Answer:
t=89.44 sec
Explanation:
Given that
d= 100 m
Mass of electron
[tex]m=9.11\times 10^{-31}kg[/tex]
Force acting on electron
[tex]F=\dfrac{Kq^2}{d^2}[/tex]
Now by putting the values of K and charge on electron
[tex]F=\dfrac{Kq^2}{d^2}[/tex]
[tex]F=\dfrac{9\times 10^9(1.6\times 10^{-19})^2}{100^2}[/tex]
[tex]F=2.3\times 10^{-32}[/tex]
As we know that
F= m a
So acceleration of electron
a=F/m
[tex]a=\dfrac{2.3\times 10^{-32}}{9.11\times 10^{-31}}\ m/s^2[/tex]
[tex]a=0.025\ m/s^2[/tex]
We know that
[tex]S=ut+\dfrac{1}{2}at^2[/tex]
here electron move from rest so u= 0
[tex]100=\dfrac{1}{2}\times 0.025^2\times t^2[/tex]
t=89.44 sec
So time taken by electron is 89.44 sec.
Force on electron, released from rest at a distance from conducting plane indirectly proportional to this distance coulombs law. Time required for electron to strike the plane is 89.44 seconds.
What is coulombs law?Coulombs law states the the force of attraction or repulsion between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of distance between them.
It can be given as,
[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]
Here, [tex]k[/tex] is coulombs constant.
Given information-
The electron is released from rest at a distance 100 m from an infinite conducting plane.
The acceleration of a object is the ratio of force applied on it to the mass of the object. Thus the acceleration of the electron is,
[tex]a=\dfrac{F}{m}[/tex]
Force is the ratio of charge on particles and the square of distance between them, multiplied by coulomb's constant. Thus,
[tex]a=\dfrac{k\dfrac{q_1q_2}{r^2}}{m}\\a=\dfrac{k{q_1q_2}}{m{r^2}}[/tex]
As the mass of the electron is [tex]9.11\times10^{-31}kg[/tex] and the charge on a electron is [tex]1.6\times10^{-19} C[/tex]. Thus put the value in above expression as,
[tex]a=\dfrac{(9\times10^9)(1.6\times10^{-19})(1.6\times10^{-19})}{9.11\times10^{-31}\times100^2}\\a=0.025\rm m/s^2[/tex]
As the value of acceleration is 0.025 meter per second squared and initial velocity is zero. Thus by the distance formula of equation of motion,
[tex]100=0+\dfrac{1}{2}\times0.025\times t^2\\t=89.44 \rm sec[/tex]
Therefore, the time required for the electron to strike the plane is 89.44 seconds.
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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.
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 concert loudspeaker suspended high off the ground emits 32.0 W of sound power. A small microphone with a 1.00 cm^2 area is 52.0 m from the speaker. Part complete What is the sound intensity at the position of the microphone?How much sound energy impinges on the microphone each second?
Answer:
Sound Intensity at microphone's position is [tex]9.417\times 10^{- 4} W/m^{2}[/tex]
The amount of energy impinging on the microphone is [tex]9.417\times 10^{- 8} W/m^{2}[/tex]
Solution:
As per the question:
Emitted Sound Power, [tex]P_{E} = 32.0 W[/tex]
Area of the microphone, [tex]A_{m} = 1.00 cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex]
Distance of microphone from the speaker, d = 52.0 m
Now, the intensity of sound, [tex]I_{s}[/tex] at a distance away from the souce of sound follows law of inverse square and is given as:
[tex]I_{s} = \frac{P_{E}}{Area} = \frac{P_{E}}{4\pi d^{2}}[/tex]
[tex]I_{s} = \frac{32.0}{4\pi (52.0)^{2}} = 9.417\times 10^{- 4} W/m^{2}[/tex]
Now, the amount of sound energy impinging on the microphone is calculated as:
If [tex]I_{s}[/tex] be the Incident Energy/[tex]m^{2}/s[/tex]
Then
The amount of energy incident per 1.00 [tex]cm^{2} = 1.00\times 10^{- 4} m^{2}[/tex] is:
[tex]I_{s}(1.00\times 10^{- 4}) = 9.417\times 10^{- 4}\times 1.00\times 10^{- 4} = 9.417\times 10^{- 8} J[/tex]
Final answer:
The sound intensity at the position of the microphone is calculated using the formula Intensity = Power/Area. The amount of sound energy impinging on the microphone each second is found using the formula Energy = Power * Time.
Explanation:
To find the sound intensity at the position of the microphone, we can use the formula:
Intensity = Power/Area
Given that the sound power is 32.0W and the microphone has an area of 1.00cm^2 (converted to m^2 by dividing by 10000), we can calculate:
Intensity = 32.0W / (1.00cm^2 / 10000)
Next, to find the amount of sound energy impinging on the microphone each second, we can use the formula:
Energy = Power * Time
Since the time is 1 second, we have:
Energy = 32.0W * 1s
Therefore, the sound intensity at the position of the microphone is the calculated value, and the amount of sound energy impinging on the microphone each second is 32.0 joules.
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
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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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]
The following sound waves have what velocity? (a) f = 36.3 Hz, λ = 11.0 m m/s (b) f = 363.0 Hz, λ = 4.80 m m/s (c) f = 3,630.0 Hz, λ = 11.0 cm m/s (d) f = 36,300.0 Hz, λ = 4.80 cm m/s
Answer:
(a) v = 399.3 m/s, (b) v = 1742.4 m/s, (c) v = 399.3 m/s, (d) v = 1742.4 m/s
Explanation:
The velocity of a wave can be defined as:
[tex]v = \lambda f[/tex] (1)
Where [tex]\lambda[/tex] and f are the wavelength and frequency of the sound wave.
The values for each case will be replaced in equation (1).
(a) f = 36.3 Hz, λ = 11.0 m
[tex]v = (11.0 m)(36.3 Hz)[/tex]
But 1 Hz = s⁻¹, therefore:
[tex]v = (11.0 m)(36.3 s^{-1})[/tex]
[tex]v = 399.3 m.s^{-1}[/tex]
[tex]v = 399.3 m/s[/tex]
So the sound wave has a velocity of 399.3 m/s.
(b) f = 363.0 Hz, λ = 4.80 m
[tex]v = (4.80 m)(363.0 Hz)[/tex]
[tex]v = (4.80 m)(363.0 s^{-1})[/tex]
[tex]v = 1742.4 m.s^{-1}[/tex]
[tex]v = 1742.4 m/s[/tex]
So the sound wave has a velocity of 1742.4 m/s.
(c) f = 3,630.0 Hz, λ = 11.0 cm
Before using equation (1) it is necessary to express [tex]\lambda[/tex] in meters.
[tex]11.0 cm . \frac{1 m}{100 cm}[/tex] ⇒ [tex]0.11 m[/tex]
[tex]v = (0.11 m)(3630.0 Hz)[/tex]
[tex]v = (0.11 m)(3630.0 s^{-1})[/tex]
[tex]v = 399.3 m.s^{-1}[/tex]
[tex]v = 399.3 m/s[/tex]
So the sound wave has a velocity of 399.3 m/s.
(d) f = 36,300.0 Hz, λ = 4.80 cm
[tex]4.80 cm . \frac{1 m}{100 cm}[/tex] ⇒ [tex]0.048 m[/tex]
[tex]v = (0.048 m)(36300.0 Hz)[/tex]
[tex]v = (0.048 m)(36300.0 s^{-1})[/tex]
[tex]v = 1742.4 m.s^{-1}[/tex]
[tex]v = 1742.4 m/s[/tex]
So the sound wave has a velocity of 1742.4 m/s.
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
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 pot contains 257 g of water at 99°C. If this water is heated and all evaporates to form steam at 100°C, what is the change in the entropy of the H2O?
Answer:
1556.906J/K=ΔS
Explanation:
Entropy is a thermodynamic property that measures the level of molecular disorder in a substance.
This property is already calculated for all pressure and temperature values.
Therefore, to solve this problem we must use thermodynamic tables for water and calculate the specific entropy in the two states, finally multiply by the mass to find the entropy change.
the entropy change is given by the following equation
ΔS=m(s2-s1)
where
ΔS=change in the entropy
s= especific entropy
m=mass=257g=0.257Kg
for the state 1: entropy for liquid water at 99°C
s1=1296J/kgK
for the state 2: entropy for steam at 100°C
s2=7354J/kgK
solving
ΔS=0.257(7354-1296)=1556.906J/K
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 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.
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 sprinter runs 50m. You are given a stopwatch. Describe how you would measure: 1. the average speed of the sprinter over the entire 50m
2. the instantaneous speed of the sprinter when he crosses the finish line
Answer:
Explanation:
Distance = 50 m
1. To find the average speed, first start the stop watch as the sprinter starts running and then stop it when he reaches the finish line.
Now note the time taken by the sprinter to run for 50 m.
The average speed of the sprinter is defined as the ratio of total distance covered to the total time taken.
Average speed = total distance / total time
2. To find the instantaneous speed, check the seed of the sprinter as he is at the finish line.
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.
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
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]
One way to heat a gas is to compress it. A gas at 3.0 atm at 27°C is compressed to three tenths of its original volume, and it reaches 35 atm pressure. What is its new temperature?
Answer:
T =3227 °C
Explanation:
Given data:
P1 = 3.0 atm
T1 = 27 degree celcius
P2 = 35 atm
from ideal gas equation
PV/T=const
[tex]\frac{P_1 V_1}{T_1} =\frac{P_2 V_2}{T_2}[/tex]
[tex]\frac{3*V}{(27+273)}=\frac{35*V*10}{T}[/tex]
solving for T WE GET
[/tex]0.01V = \frac{350V}{T}[/tex]
T=3500K
T=3500-273=3227 °C
T =3227 °C
An auditorium measures 35.0 m x 30.0 m x 5.0 m. The density of air is 1.20 kg/m^3. (a) What is the volume of the room in cubic feet? (b) What is the weight of air in the room in pounds?
Answer:
(a) 1852259 [tex]ft^3[/tex] (b) 489085.47 pound
Explanation:
We have given auditorium measures 35 m×30 m×5 m
We know that 1 meter = 3.28 feet
So the measure of auditorium = 35×3.28 feet ×30×3.28 feet× 5×3.28 feet
(a) So the volume of the auditorium [tex]=35\times 3.28\times 30\times 3.28\times 5\times 3.28=185259.648ft^3[/tex]
Density is given as [tex]d=1.20kg/m^3[/tex]
(b) weight of air = volume × density [tex]=185259.648\times 1.2=222311.577kg[/tex]
We know that 1 kg = 2.20 pound
So 222311.577 kg =222311.577×2.20=489085.47 pound
The volume of the room is 185,197 cubic feet and the weight of air in the room is 13,889 pounds.
Explanation:To convert the volume of the auditorium from cubic meters to cubic feet, we can use the conversion factor 1 cubic meter = 35.3147 cubic feet. With dimensions of 35.0 m x 30.0 m x 5.0 m, the volume of the auditorium is 5250 cubic meters. Multiplying this by the conversion factor, we find that the volume of the room is approximately 185,197 cubic feet.
To calculate the weight of air in the room, we can multiply the volume of air by its density. The density of air is given as 1.20 kg/m³. Using the volume of the room in cubic meters (5250 m³), we can multiply it by the density to find the mass of air, which is 6300 kg. To convert this to pounds, we can multiply by the conversion factor 1 kg = 2.20462 pounds. The weight of air in the room is therefore approximately 13,889 pounds.
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.
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 electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is = 2.1 × 10^-7 C/m^2, and the plates are separated by a distance of 1.2 × 10^-2 m. How fast is the electron moving just before it reaches the positive plate?
Explanation:
An electron is released from rest, u = 0
We know that charge per unit area is called the surface charge density i.e. [tex]\sigma=\dfrac{q}{A}=2.1\times 10^{-7}\ C/m^2[/tex]
Distance between the plates, [tex]d=1.2\times 10^{-2}\ m[/tex]
Let E is the electric field,
[tex]E=\dfrac{\sigma}{\epsilon_o}[/tex]
[tex]E=\dfrac{2.1\times 10^{-7}}{8.85\times 10^{-12}}[/tex]
E = 23728.81 N/C
Now, [tex]ma=qE[/tex]
[tex]a=\dfrac{qE}{m}[/tex]
[tex]a=\dfrac{1.6\times 10^{-19}\times 23728.81}{9.1\times 10^{-31}}[/tex]
[tex]a=4.17\times 10^{15}\ m/s^2[/tex]
Let v is the speed of the electron just before it reaches the positive plate. So, third equation of motion becomes :
[tex]v^2=2ad[/tex]
[tex]v^2=2\times 4.17\times 10^{15}\times 1.2\times 10^{-2}[/tex]
[tex]v=10.003\times 10^6\ m/s[/tex]
Hence, this is the required solution.
If the length of a wire is increased by 20% keeping its volume constant. what will be the % change in heat produced when connected across same potential difference. please explain properly!!
Answer:decreases by 30.55%
Explanation:
Given
length of wire is increased by 20 % keeping volume constant
Let the length of wire be L and its area of cross section be A
Thus new length=1.2 L
Volume is constant
[tex]AL=1.2 L\times A'[/tex]
A'=0.833 A
and resistance is given by
[tex]R=\frac{\rho L}{A}[/tex]
where [tex]\rho [/tex]=resistivity
New resistance [tex]R'=\frac{\rho\times 1.2L}{0.833A}[/tex]
R'=1.44 R
heat produced for same potential
[tex]H_1=\frac{V^2t}{R}[/tex]
[tex]H_2=\frac{V^2t}{1.44R}=0.694H_1[/tex]
% change in heat
[tex]\frac{H_2-H_1}{H_1}\times 100[/tex]
[tex]=\frac{0.694-1}{1}[/tex]
=30.55 decreases
Answer:
30.55 %
Explanation:
Assumptions:
l = initial length of the wireL = final length of the wirev = initial volume of the wireV = final volume of the wirea = initial cross sectional area of the wireA = final cross sectional area of the wireh = initial heat of generated by the wireH = final heat generated by the wireP = potential difference across the wiret = time for which the potential difference is created across the wirer = initial resistance of the wireR = final resistance of the wire[tex]\Delta H[/tex] = change in heat producedAccording to the question, we have
[tex]L = l + 20\ \% l = \dfrac{120l}{100}\\V=v\\\Rightarrow LA=la\\\Rightarrow A= \dfrac{la}{L}\\\Rightarrow A= \dfrac{la}{\dfrac{120l}{100}}\\\Rightarrow A= \dfrac{100a}{120}[/tex]
Using the formula of resistance of a wire in terms of its length, cross sectional area and the resistivity of the material, we have
[tex]r = \dfrac{\rho l}{a}\\R=\dfrac{\rho L}{A}=\dfrac{\rho\times \dfrac{120l}{100} }{\dfrac{100a}{120}}=(\dfrac{120}{100})^2\dfrac{\rho l}{a}= 1.44r\\[/tex]
Using the formula of heat generated by the wire for potential diofference created across its end for time t, we have
[tex]h = \dfrac{P^2}{r}t\\H = \dfrac{P^2}{R}t= \dfrac{P^2}{1.44r}t\\\therefore \Delta H = h-H\\\Rightarrow \Delta H = \dfrac{P^2}{r}t-\dfrac{P^2}{1.44r}t\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(-\dfrac{1}{1.44})\\\Rightarrow \Delta H = \dfrac{P^2t}{r}(\dfrac{0.44}{1.44})\\\therefore \textrm{Percentage change in the heat produced}= \dfrac{\Delta H}{h}\times 100\ \%= \left (\dfrac{\dfrac{P^2t}{r}(\dfrac{0.44}{1.44})}{\dfrac{P^2}{r}t} \right )\times 100\ \% = 30.55\ \%[/tex]
Hence, the percentage change in the heat produced in the wire is 30.55 %.
An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at that time, y0 is the initial height (at t=0), v0 is the initial velocity, and g=9.8m/s2 (the acceleration due to gravity). If you stand at the edge of a cliff that is 75 m high and throw a rock directly up into the air with a velocity of 20 m/s, at what time will the rock hit the ground? (Note: The Quadratic Formula will give two answers, but only one of them is reasonable.) View Available Hint(s)
Answer: 6.45 s
Explanation:
We have the following equation:
[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex] (1)
Where:
[tex]y=0[/tex] is the height when the rock hits the ground
[tex]y_{o}=75 m[/tex] the height at the edge of the cilff
[tex]V_{o}=20 m/s[/tex] the initial velocity
[tex]g=9.8 m/s^{2}[/tex] acceleration due gravity
[tex]t[/tex] time
[tex]0=75 m+(20 m/s)t-(4.9 m/s^{2})t^{2}[/tex] (2)
Rearranging the equation:
[tex]-(4.9 m/s^{2})t^{2} + (20 m/s)t + 75 m=0[/tex] (3)
At this point we have a quadratic equation of the form [tex]at^{2}+bt+c=0[/tex], and we have to use the quadratic formula if we want to find [tex]t[/tex]:
[tex]t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}[/tex] (4)
Where [tex]a=-4.9[/tex], [tex]b=20[/tex], [tex]c=75[/tex]
Substituting the known values and choosing the positive result of the equation:
[tex]t=\frac{-20\pm\sqrt{20^{2}-4(-4.9)(75)}}{2(-4.9)}[/tex] (5)
[tex]t=6.453 s[/tex] This is the time it takes to the rock to hit the ground
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