The drawing shows a laser beam shining on a plane mirror that is perpendicular to the floor. The angle of incidence is 33.0°. The beam emerges from the laser at a point that is 1.10 m from the mirror and 1.80 m above the floor. After reflection, how far from the base of the mirror does the beam strike the floor?
Answers
Answer:
x= 2.7734 m
Explanation:
Given the angle of incidence θ= 33.0°
distance from laser point and mirror = 1.10 m
laser point is 1.80 m above the floor
let the distance between mirror base and laser beam striking the floor after reflection = x
figure has been attached and from the from the figure we can write
[tex]tan\theta= \frac{1.80}{x}[/tex]
[tex]tan33\degree= \frac{1.80}{x}[/tex]
[tex]tan33\degree= 0.649[/tex]
putting value and solving we get
x= 2.7734 m
Related Questions
Suppose you are inside a smooth-flying [no bumps] jet plane moving at constant speed in a straight line across the Atlantic Ocean. Can you detect this motion [i.e., can you tell that you are moving] without directly or indirectly making use of anything outside of your own reference frame? If so, how could you detect this; and if not, why not?
Answers
Answer:
No we cannot perform any experiment that tells us weather we are moving or not at a constant speed.
Explanation:
Motion of any body is relative to other reference bodies. We can perceive motion only if our position with respect to a fixed object changes. This is the fundamental concept of reference frame in classical physics or Newtonian Physics that motion is always from a reference.
We choose a fixed body as reference and ,measure the distance we cover from this fixed point and also our speed with respect to this fixed point. The choice of frame of reference is completely dependent on the observer.
Since in the given case a reference cannot be established outside the plane thus we cannot detect our motion.
Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is the fourth lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 343 m/s.
Answers
Answer:
f = 632 Hz
Explanation:
As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength
here we know that
[tex]\Delta x = (2n + 1)\frac{\lambda}{2}[/tex]
given that path difference from two loud speakers is given as
[tex]\Delta x = 5.80 m - 3.90 m[/tex]
[tex]\Delta x = 1.90 m[/tex]
now we know that it will have fourth lowest frequency at which destructive interference will occurs
so here we have
[tex]\Delta x = 1.90 = \frac{7\lambda}{2}[/tex]
[tex]\lambda = \frac{2 \times 1.90}{7}[/tex]
[tex]\lambda = 0.54 m[/tex]
now for frequency we know that
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{343}{0.54} = 632 Hz[/tex]
A meteorite has a speed of 95.0 m/s when 750 km above the Earth. It is falling vertically (ignore air resistance) and strikes a bed of sand in which it is brought to rest in 3.35 m . Part A What is its speed just before striking the sand?
Answers
The meteorite's speed just before striking the sand is approximately 3836.82 m/s, calculated using conservation of energy principles from its initial potential energy at 750 km above Earth.
Let's break down the solution step by step:
Step 1: Calculate the initial potential energy of the meteorite when it is 750 km above the Earth's surface.
[tex]\[PE_{initial} = mgh\][/tex]
Where:
- [tex]\(m\)[/tex] is the mass of the meteorite,
- [tex]\(g\)[/tex] is the acceleration due to gravity, and
- [tex]\(h\)[/tex] is the height above the Earth's surface.
Given:
- [tex]\(h = 750 \, \text{km} = 750 \times 10^3 \, \text{m}\)[/tex] (converted to meters)
- [tex]\(g = 9.8 \, \text{m/s}^2\)[/tex] (acceleration due to gravity)
[tex]\[PE_{initial} = mg \times h\][/tex]
[tex]\[PE_{initial} = (m)(9.8 \, \text{m/s}^2)(750 \times 10^3 \, \text{m})\][/tex]
Step 2: Calculate the final kinetic energy of the meteorite just before it strikes the sand.
[tex]\[KE_{final} = \frac{1}{2}mv^2\][/tex]
Where:
- [tex]\(v\)[/tex] is the final velocity of the meteorite just before it strikes the sand.
Given:
- [tex]\(d = 3.35 \, \text{m}\)[/tex] (distance traveled while coming to rest)
Step 3: Apply the principle of conservation of energy.
Since energy is conserved, the initial potential energy of the meteorite is converted into kinetic energy just before it strikes the sand.
[tex]\[PE_{initial} = KE_{final}\][/tex]
[tex]\[mg \times h = \frac{1}{2}mv^2\][/tex]
[tex]\[9.8 \times 750 \times 10^3 = \frac{1}{2} \times v^2\][/tex]
[tex]\[v^2 = \frac{2 \times 9.8 \times 750 \times 10^3}{1}\][/tex]
[tex]\[v^2 = 2 \times 9.8 \times 750 \times 10^3\][/tex]
[tex]\[v^2 = 2 \times 7.35 \times 10^6\][/tex]
[tex]\[v^2 = 14.7 \times 10^6\][/tex]
[tex]\[v^2 = 1.47 \times 10^7\][/tex]
[tex]\[v = \sqrt{1.47 \times 10^7}\][/tex]
[tex]\[v \approx 3836.82 \, \text{m/s}\][/tex]
So, the speed of the meteorite just before striking the sand is approximately [tex]\(3836.82 \, \text{m/s}\)[/tex].
Suppose you are an astronaut on a spacewalk, far from any source of gravity. You find yourself floating alongside your spacecraft but 10 m away, with no propulsion system to get back to it. In your tool belt you have a hammer, a wrench, and a roll of duct tape. How can you get back to your spacecraft?
a. Move like you are flying to the spaceship
b. Move like you are swimming to the spaceship
c. Throw the items away from the spaceship
d. Throw the items to the spaceship.
Answers
Answer:
c. Throw the items away from the spaceship.
Explanation:
By the Principle of action and reaction yu can get back to your spacecraft throwing the items away from the spaceship.
Answer:
c. Throw the items away from the spaceship
Explanation:
The propulsion that throwing items away from the spaceship will propel you to the opposite direction in which you are throwing the objects, this means that if you throw the items away from the spaceship you will be getting force of rpopulsion towards the spaceship.
A GPS tracking device is placed in a police dog to monitor its whereabouts relative to the police station. At time t1 = 23 min, the dog's displacement from the station is 1.2 km, 33° north of east. At time t2 = 57 min, the dog's displacement from the station is 2.0 km, 75° north of east. Find the magnitude and direction of the dog's average velocity between these two times.
Answers
Answer:
Explanation:
Let east be towards X-axis ant north be Y-axis. Let initial position of Dog be at
A . O be the police station (centre). Vector OA can be written as follows
OA = 1.2 Cos 33 + 1.2 Sin 33
O B =2 Cos 75 + 2 Sin 75.
Displacement A → B = OB - OA =2Cos75i +2 Sin 75j -1.2 Cos 33i -1.2 Sin 33j
= 2 x .2588 i+ 2x .966j - 1.2 x .8387i - 1.2 x .5446j = .5176i + 1.932 j- 1.0064i - .65356j
= -.4888 i + 1.27844j
Magnitude of displacement =√( .4888)² + ( 1.27844)²
= 1.405 km
Average velocity =1.405 / 57-23 km / min = 1.405 /34 x 60 =2.48 km/h
angle with x-axis ( east towards north ) ∅
Tan∅ =- 1.27844/.4888 = - 2.615
∅ = -69° or 111° towards north from east or 21° towards west from north.
Which of the following represents a fully plastic collision? 1. Two billiard balls hit each other then move in opposite directions. 2. Two train cars hit then detach. 3. A ball hits a wall with no loss of energy. 4. A bullet hits a block and becomes embedded. 5. None of these.
Answers
Answer:
Option (4)
Explanation:
There are two types of collision.
Perfectly elastic collision: the collision in which the momentum and kinetic energy is conserved. There is no loss of energy in other forms of energy.
Perfectly plastic collision: The collision in which the momentum is conserved and kinetic energy is not conserved. The two bodies stick after the collision.
Here, the bullet hits the block and then embedded in the block, it is the example of plastic collision.
Two 1.50-V batteries—with their positive terminals in the same direction—are inserted in series into the barrel of a flashlight. One battery has an internal resistance of 0.240 Ω, the other an internal resistance of 0.180 Ω. When the switch is closed, a current of 600 mA occurs in the lamp. (a) What is the bulb's resistance? Ω (b) What fraction of the chemical energy transformed appears as internal energy in the batteries? %
Answers
a. The resistance of the bulb is equal to 4.58 Ohms.
b. The fraction of the chemical energy transformed which appears as internal energy in the batteries is 8.4%.
Given the following data:
Voltage of battery = 1.5 VoltsInternal resistance A = 0.240 OhmsInternal resistance B = 0.180 OhmsCurrent = 600 mA = 0.6 Ampsa. To find the resistance of the bulb:
First of all, we would determine the total electromotive force (E) of the electric circuit:
[tex]E = 2 \times 1.50[/tex]
E = 3.0 Volts
Total internal resistance = [tex]0.240 + 0.180 = 0.42 \;Ohms[/tex]
Mathematically, electromotive force (E) is given by the formula:
[tex]E = V + Ir[/tex] ....equation 1.
Where:
E is the electromotive force (E).V is the the voltage or potential difference.I is the current.r is the internal resistance.According to Ohm's law, voltage is given by:
[tex]V = IR[/tex] ....equation 2
Substituting eqn 2 into eqn 1, we have:
[tex]E = IR + Ir\\\\E = I(R + r)\\\\R + r = \frac{E}{I} \\\\R = \frac{E}{I} - r\\\\R = \frac{3}{0.6} - 0.42\\\\R = 5 - 0.42[/tex]
Resistance, R = 4.58 Ohms
b. To determine what fraction of the chemical energy transformed appears as internal energy in the batteries:
First of all, we would determine the electromotive force (E) in the batteries.
[tex]E_B = IR\\\\E_B = 0.6 \times 0.42[/tex]
[tex]E_B[/tex] = 0.252 Volt
[tex]Percent = \frac{E_B}{E} \times 100\\\\Percent = \frac{0.252}{3} \times 100\\\\Percent = \frac{25.2}{3}[/tex]
Percent = 8.4 %
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The resistance of the bulb and the fraction of the chemical energy transformed into internal energy in the batteries can be determined using Ohm's law and the power dissipation formula. The total resistance in the series circuit is the sum of the individual resistances, and the total voltage is the sum of the individual voltages.
Explanation:The total voltage supplied by the batteries is the sum of their voltages, so Vtot = 1.5V + 1.5V = 3.0V. The total internal resistance of the batteries is the sum of their internal resistances, Rt = 0.240Ω + 0.180Ω = 0.420Ω. For the bulb's resistance, we can rearrange Ohm's law to R = V/I. Knowing the total voltage (3.0V) and the current (600 mA or 0.600 A), we can find the total resistance in the circuit.
Subtracting the total internal resistance gives us the resistance of the bulb. As for the second part of the question, the power dissipated in the internal resistances can be found using P = I²r for each battery, and then add these together. The total power supplied by the batteries is P = IV, using the total current and total voltage. The fraction of the chemical energy that appears as internal energy in the batteries is then Pinternal / Ptotal.
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A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than the other. He fires a gun and, after a while, hears three echoes. The second echo arrives 1.76 s after the first, and the third echo arrives 1.38 s after the second. Assuming that the speed of sound is 343 m/s and that there are no reflections of sound from the ground, find the distance (in m) between the cliffs.
Answers
Answer:
Distance=538.51m
Explanation:
The echo is heard after covering double distance .
Therefore 2d=(t1+t2)×speed.
2d={(1.76+1.38)×343}
2d=743.02
d=1077.02÷2
=538.51m
An engineering firm is designing a ski lift. The wire rope needs to travel with a linear velocity of 2.0 meters per second, and the angular velocity of the bullwheel will be 10 revolutions per minute. What diameter bullwheel should be used to drive the wire rope?
Answers
Answer:
The diameter of the bull-wheel is 3.82
Explanation:
Given that,
Velocity = 2.0 m/s
Angular velocity = 10 rev/m
[tex]\omega=10\times\dfrac{2\pi}{60}[/tex]
[tex]\omega=1.0472\ rad/s[/tex]
We need to calculate the diameter of bull-wheel
Using formula of angular velocity
[tex]v= r\omega[/tex]
[tex]r=\dfrac{v}{\omega}[/tex]
Put the value into the formula
[tex]r=\dfrac{2.0}{1.0472}[/tex]
[tex]r=1.91\ m[/tex]
The diameter of the bull-wheel
[tex]D=2r[/tex]
[tex]D=2\times1.91[/tex]
[tex]D=3.82\ m[/tex]
Hence, The diameter of the bull-wheel is 3.82 m.
A diffraction grating with 140 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles in the third-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm? (angles in radians) The 434 nm line:
Answers
Answer:
0.003181 radians
0.003005 radians
Explanation:
Number of slits = 140 /cm
λ = Wavelength = 434 nm = 434×10⁻⁹ m
m = 3 Third order spectrum
Space between slits
[tex]d=\frac{0.01}{140} =7.14\times 10^{-5}\ m[/tex]
Now,
[tex]dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 434\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.018228\\\Rightarrow \theta=0.01823^{\circ}=0.01823\times \frac{\pi}{180}=0.003181 radians[/tex]
0.003181 radians
When λ = 410 nm = 410×10⁻⁹ m
[tex]dsin\theta = m\lambda\\\Rightarrow \theta=sin^{-1}\left(\frac{m\lambda}{d}\right)\\\Rightarrow \theta=sin^{-1}\left(\frac{3\times 410\times 10^{-9}}{7.14\times 10^{-5}}\right)\\\Rightarrow \theta=sin^{-1}0.01722\\\Rightarrow \theta=0.01722^{\circ}=0.01722\times \frac{\pi}{180}=0.003005 radians[/tex]
0.003005 radians
A uniform thin rod is hung vertically from one end and set into small amplitude oscillation. If the rod has a length of 2.6 m, this rod will have the same period as a simple pendulum of length ____ cm. Round your answer to the nearest whole number.
Answers
Answer:
Length of pendulum, l = 1.74 meters
Explanation:
The time period of simple pendulum is, [tex]T=2\pi\sqrt{\dfrac{l}{g}}[/tex]
Where
l is the length of simple pendulum
The time period of uniform thin rod is hung vertically from one end is, [tex]T=2\pi\sqrt{\dfrac{2l'}{3g}}[/tex]
l' is the length of uniform rod, l' = 2.6 m
It is given that the rod and pendulum have same time period. So,
[tex]2\pi\sqrt{\dfrac{l}{g}}=2\pi\sqrt{\dfrac{2l'}{3g}}[/tex]
After solving above expression, the value of length of the pendulum is, l = 1.74 meters. Hence, this is the required solution.
A father racing his son has 1/4 the kinetic energy of the son, who has 1/3 the mass of the father. The father speeds up by 1.2 m/s and then has the same kinetic energy as the son. What are the original speeds of (a) the father and (b) the son?
Answers
Answer:
Explanation:
KE_s: Kinetic Energy Son
KE_f: Kinetic Energy Father.
Relationship
KE_f: = (1/4) KE_s
m_s: = (1/3) m_f
v_f: = velocity of father
v_s: = velocity of the son
Relationship
1/2 mf (v_f + 1.2)^2 = 1/2 m_s (v_s)^2 Multiply both sides by 2.
mf (v_f + 1.2)^2 = m_s * (v_s)^2 Substitute for the mass of the m_s
mf (v_f + 1.2)^2 = (m_f/3) * (v_s)^2 Divide both sides by father's mass
(v_f + 1.2)^2 = 1/3 * (v_s)^2 multiply both sides by 3
3*(v_f + 1.2)^2 = (v_s) ^2 Take the square root both sides
√3 * (v_f + 1.2) = v_s
Note
You should work your way through all the cancellations to find the last equation shown aboutWe have another step to go. We have to use the first relationship to get the final answer.KE_f = (1/4) KE_s Multiply by 4
4* KE_f = KE_s Substitute (again)
4*(1/2) m_f (v_f + 1.2)^2 = 1/2* (1/3)m_f *v_s^2 Divide by m_f
2* (v_f + 1.2)^2 = 1/6 * (v_s)^2 multiply by 6
12*(vf + 1.2)^2 = (v_s)^2 Take the square root
2*√(3* (v_f + 1.2)^2) = √(v_s^2)
2*√3 * (vf + 1.2) = v_s
Use the second relationship to substitute for v_s so you can solve for v_f
2*√3 * ( v_f + 1.2) = √3 * (v_f + 1.2) Divide by sqrt(3)
2(v_f + 1.2) = vf + 1.2
Edit
2vf + 2.4 = vf + 1.2
2vf - vf + 2.4 = 1.2
vf = 1.2 - 2.4
vf = - 1.2
This answer is not possible, but 2 of us are getting the same answer. The other person is someone whose math I would never question. She rarely makes an error. And I do mean rarely. Could you check to see that you have copied this correctly?
A baseball approaches home plate at a speed of 44.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 53.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.20 ms. What is the average vector force the ball exerts on the bat during their interaction?
Answers
Explanation:
It is given that,
Speed of the baseball, u = 44 m/s
Speed of the baseball, v = 53 m/s
Mass of the ball, m = 145 g = 0.145 kg
Time of contact between the ball and the bat, t = 2.2 ms = 0.0022 s
[tex]F=ma[/tex]
[tex]F=\dfrac{mv}{t}[/tex]
[tex]F_1=\dfrac{0.145\ kg\times 44\ m/s}{0.0022\ s}[/tex]
F₁ = 2900 N...........(1)
[tex]F=ma[/tex]
[tex]F=\dfrac{mv}{t}[/tex]
[tex]F_2=\dfrac{0.145\ kg\times 53\ m/s}{0.0022\ s}[/tex]
F₂ = 3493.18 N.........(2)
In average vector form force is given by :
[tex]F=F_1+F_2[/tex]
[tex]F=(2900i+(-3493.18)\ N[/tex]
[tex]F=(2900i-3493.18j)\ N[/tex]
Hence, this is the required solution.
A 400-turn circular coil (radius = 1.0 cm) is oriented with its plane perpendicular to a uniform magnetic field which has a magnitude that varies sinusoidally with a frequency of 90 Hz. If the maximum value of the induced emf in the coil is observed to be 4.2 V, what is the maximum value of the magnitude of the varying magne?
Answers
Answer:
The Magnetic field is 59.13 mT.
Explanation:
Given that,
Number of turns = 400
Radius = 1.0 cm
Frequency = 90 Hz
Emf = 4.2 V
We need to calculate the angular velocity
Using formula of angular velocity
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\times3.14\times90[/tex]
[tex]\omega=565.2\ rad/s[/tex]
We need to calculate the magnetic flux
Relation between magnetic flux and induced emf
[tex]\epsilon=NA\omega B[/tex]
[tex]B=\dfrac{\epsilon}{NA\omega}[/tex]
Put the value into the formula
[tex]B=\dfrac{4.2}{400\times\pi\times(1.0\times10^{-2})^2\times 565.2}[/tex]
[tex]B=0.05913\ T[/tex]
[tex]B=59.13\ mT[/tex]
Hence, The Magnetic field is 59.13 mT.
A playground tire swing has a period of 2.0 s on Earth. What is the length of its chain?
Answers
Answer:
Length of the chain, l = 0.99 m
Explanation:
Given that,
A playground tire swing has a period of 2.0 s on Earth i.e.
T = 2 s
We need to find the length of this chain. The relationship between the length and the time period is given by :
[tex]T=2\pi \sqrt{\dfrac{l}{g}}[/tex]
Where
l = length of chain
g = acceleration due to gravity
[tex]l=\dfrac{T^2g}{4\pi^2}[/tex]
[tex]l=\dfrac{(2)^2\times 9.8}{4\pi^2}[/tex]
l = 0.99 meters
So, the length of the chain is 0.99 meters. Hence, this is the required solution.
The 630 nm light from a helium neon laser irradiates a grating. The light then falls on a screen where the first bright spot is separated from the central maxim by 0.51m. Light of another wavelength produces its first bright spot at 0.39 m from its central maximum. Determine the second wavelength.
Answers
Answer:
The second wavelength is 482 nm.
Explanation:
Given that,
Wavelength = 630 nm
Distance from central maxim = 0.51 m
Distance from central maxim of another wavelength = 0.39 m
We need to calculate the second wavelength
Using formula of width of fringe
[tex]\beta=\dfrac{\lambda d}{D}[/tex]
Here, d and D will be same for both wavelengths
[tex]\lambda[/tex] = wavelength
[tex]\beta [/tex] = width of fringe
The width of fringe for first wavelength
[tex]\beta_{1}=\dfrac{\lambda_{1} d}{D}[/tex]....(I)
The width of fringe for second wavelength
[tex]\beta_{2}=\dfrac{\lambda_{2} d}{D}[/tex]....(II)
Divided equation (I) by equation (II)
[tex]\dfrac{\beta_{1}}{\beta_{2}}=\dfrac{\lambda_{1}}{\lambda_{2}}[/tex]
[tex]\lambda_{2}=\dfrac{630\times10^{-9}\times0.39}{0.51}[/tex]
[tex]\lambda_{2}=4.82\times10^{-7}[/tex]
[tex]\lambda=482\ nm[/tex]
Hence, The second wavelength is 482 nm.
To find the second wavelength, we can use the formula for the wavelength of light from a diffraction grating. In this case, we know the first wavelength is 630 nm, the first bright spot is separated from the central maximum by 0.51 m, and we need to find the second wavelength.
Explanation:To find the second wavelength, we can use the formula for the wavelength of light from a diffraction grating: wavelength = (m * d * sin(theta)) / n. In this case, we know the first wavelength is 630 nm, the first bright spot is separated from the central maximum by 0.51 m, and we need to find the second wavelength. The m and n values are the same for both cases, so we can set up the equation:
630 nm = (m * 0.51 m * sin(theta)) / n
wavelength = (m * 0.51 m * sin(theta)) / n
Next, we can solve for the second wavelength by rearranging the equation:
wavelength = (630 nm * n) / (m * 0.51 m * sin(theta))
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A 1.50-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction between the block and the wall is 0.54. What is the minimum compression in the spring to prevent the block from slipping down?
Answers
Answer:
0.032 m
Explanation:
Consider the forces acting on the block
m = mass of the block = 1.50 kg
[tex]f_{s}[/tex] = Static frictional force
[tex]F_{n}[/tex] = Normal force on the block from the wall
[tex]F_{s}[/tex] = Spring force due to compression of spring
[tex]F_{g}[/tex] = Force of gravity on the block = mg = 1.50 x 9.8 = 14.7 N
k = spring constant = 860 N/m
μ = Coefficient of static friction between the block and wall = 0.54
x = compression of the spring
Spring force is given as
[tex]F_{s}[/tex] = kx
From the force diagram of the block, Using equilibrium of force along the horizontal direction, we get the force equation as
[tex]F_{n}[/tex] = [tex]F_{s}[/tex]
[tex]F_{n}[/tex] = kx eq-1
Static frictional force is given as
[tex]f_{s}[/tex] = μ [tex]F_{n}[/tex]
Using eq-1
[tex]f_{s}[/tex] = μ k x eq-2
From the force diagram of the block, Using equilibrium of force along the vertical direction, we get the force equation as
[tex]f_{s}[/tex] = [tex]F_{g}[/tex]
Using eq-2
μ k x = 14.7
(0.54) (860) x = 14.7
x = 0.032 m
If one gram of matter could be completely converted into energy, the yield would be
A. 0.51 MeV.
B. 931 MeV.
C. 3 × 1013 J.
D. 9 × 1013 J.
E. 9 × 1016 J.
Answers
Answer:
Energy, [tex]E=9\times 10^{13}\ J[/tex]
Explanation:
It is given that,
Mass of matter, m = 1 g = 0.001 Kg
If this matter is completely converted into energy, we need to find the yield. The mass of an object can be converted to energy as :
[tex]E=mc^2[/tex]
c = speed of light
[tex]E=0.001\ kg\times (3\times 10^8\ m/s)^2[/tex]
[tex]E=9\times 10^{13}\ J[/tex]
So, the energy yield will be [tex]9\times 10^{13}\ J[/tex]. Hence, this is the required solution.
Radio receivers are usually tuned by adjusting the capacitor of an LC circuit. If C = C1 for a frequency of 600 kHz, then for a frequency of 1200 kHz C must be adjusted to what value?
Answers
Answer:
[tex]C_2=\frac{C_1}{4}[/tex]
Explanation:
Given:
Initial capacitance, C = C₁
Initial Frequency, F₁ = 600kHz
Final Frequency, F₂ = 1200kHz
let the adjusted capacitance be, C₂
Now
the frequency (f) is given as:
[tex]f={\frac{1}{2\pi \sqrt{LC}}}[/tex]
where, L = inductance (same for the same material)
thus substituting the values we have
[tex]600={\frac{1}{2\pi \sqrt{LC_1}}}[/tex] ...............(1)
and
[tex]1200={\frac{1}{2\pi \sqrt{LC_2}}}[/tex] ..............(2)
dividing the equation 1 with 2, we get
[tex]\frac{600}{1200}=\frac{{\frac{1}{2\pi \sqrt{LC_1}}}}{{\frac{1}{2\pi \sqrt{LC_2}}}}[/tex]
or
[tex]\frac{1}{2}=\sqrt{\frac{C_2}{C_1}}[/tex]
or
[tex]\frac{C_2}{C_1}=\frac{1}{4}[/tex]
or
[tex]C_2=\frac{C_1}{4}[/tex]
hence, the new capacitance C, must be one-fourth times the initial capacitance
The capacitance for the frequency of 1200 kHz must be one-fourth times the capacitance for the frequency of 800 kHz.
What is capacitance?Capacitance is the ability of a circuit to store energy in the form of an electrical charge.it is an energy-storing device. generally, it is defined by the ratio of electric charge stored to the potential difference.
Given:
Initial capacitance is C
Capacitance for the frequency of 1200 kHz is [tex]\rm{C=C_1}[/tex]
The capacitance for the frequency of 800 kHz is [tex]\rm{C=C_2}[/tex]
The frequency (f) is given by
[tex]f=\frac{1}{2\pi \sqrt{LC} }[/tex]
The capacitance for the frequency of 1200 kHz finds by
[tex]f_1=\frac{1}{2\pi \sqrt{LC_1} }[/tex]
[tex]1200=\frac{1}{2\pi \sqrt{LC_1} }[/tex] ------------------1
The capacitance for the frequency of 800 kHz
[tex]f_2=\frac{1}{2\pi \sqrt{LC_2} }[/tex]
[tex]600=\frac{1}{2\pi \sqrt{LC_2} }[/tex] ------------------2
On dividing the equation 2 by 1
[tex]\frac{1}{2} =\sqrt{\frac{C_2}{C_1} }[/tex]
[tex]{\frac{C_2}{C_1} }=\frac{1}{4}[/tex]
[tex]C_2=\frac{C_1}{4}[/tex]
Hence the capacitance for the frequency of 1200 kHz must be one-fourth times the capacitance for the frequency of 800 kHz.
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A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
Answers
Answer:
374 N
Explanation:
N = normal force acting on the skier
m = mass of the skier = 82.5
From the force diagram, force equation perpendicular to the slope is given as
N = mg Cos18.7
μ = Coefficient of friction = 0.150
frictional force is given as
f = μN
f = μmg Cos18.7
F = force applied by the rope
Force equation parallel to the slope is given as
F - f - mg Sin18.7 = 0
F - μmg Cos18.7 - mg Sin18.7 = 0
F = μmg Cos18.7 + mg Sin18.7
F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7
F = 374 N
A 4.0 × 102-nm thick film of kerosene (n = 1.2) is floating on water. White light is normally incident on the film. What is the visible wavelength in air that has a maximum intensity after the light is reflected? Note: the visible wavelength range is 380 nm to 750 nm.
Answers
Answer:
the visible wavelength is 480 nm
Explanation:
Given data
thick film = 4.0 × 10² nm
n = 1.2
wavelength range = 380 nm to 750 nm
to find out
the visible wavelength in air
solution
we know that index of water is 1 and kerosene is 1.2
we can say that when light travel reflected path difference is = 2 n t
and for maximum intensity it will be k × wavelength
so it will be 2 n t = k × wavelength
2 × 1.2 × 4.0 × 10² = k × wavelength
wavelength = 2 × 1.2 × 4.0 × 10² / k
here k is 2 for visible
so wavelength = 2 × 1.2 × 4.0 × 10² / 2
wavelength = 480 nm
the visible wavelength is 480 nm
Newton’s Second Law establishes the relationship between mass, net applied force, and acceleration given by F=ma. Consider a 4 kg box of holiday candy on a horizontal surface such as a table. There is a 10N applied force to the right and a 2N force to the left. The box accelerates with a magnitude of _________?
Answers
Answer:
a= 2 m/s^2
Explanation:
take to the right as positive
let Ftot be the total forces acting on the box , m be the mass of the box and a be the acceleration of the box.
Ftot = 10 - 2 = 8 N
and,
Ftot = ma
a = Ftot/m
= 8/4
= 2 m/s^2
therefore, the acceleration of the box is of magnitude of 2 m/s^2.
A parallel-plate capacitor is formed from two 6.0-cm-diameter electrodes spaced 2.0 mm apart. The electric field strength inside the capacitor is 1.0×106 N/C1.0×106 N/C. What is the charge (in nC) on each electrode?
Answers
Answer:
2.5 x 10^-8 C
Explanation:
Diameter = 6 cm, radius = 3 cm = 0.03 m, d = 2 mm = 2 x 10^-3 m
E = 1 x 10^6 N/C, q = ?
q = C V
As we know that, V = E x d and C = ∈0 A / d
q = ∈0 x A x E x d / d
q = ∈0 x A x E
q = 8.854 x 10^-12 x 3.14 x 0.03 x 0.03 x 1 x 10^6
q = 2.5 x 10^-8 C
Electric field exerts a force on all charged particles. The charge on the electrode is 2.505 x 10⁻⁸ C.
What is an electric field?An electric field can be thought to be a physical field that surrounds all the charged particles and exerts a force on all of them.
Given to us
Plate dimensions diameter, d = 6 cm
Area of the plate, A = πr² = π(0.03)² = 0.00283 m²
Distance between the two plates, d = 2 mm = 0.002 m
Electric field strength, E = 1.0 x 10⁶ N/C
We know that electric field inside a parallel plate capacitor is given as,
[tex]E = \dfrac{Q}{A\epsilon_0}[/tex]
Substitute the value,
[tex]1 \times 10^6 = \dfrac{Q}{0.00283 \times 8.854 \times 10^{-12}}[/tex]
Q = 2.505 x 10⁻⁸ C
Hence, the charge on the electrode is 2.505 x 10⁻⁸ C.
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A certain gasoline engine has an efficiency of 35.3%. What would the hot reservoir temperature be for a Carnot engine having that efficiency, if it operates with a cold reservoir temperature of 160°C? (°C)
Answers
Answer:
The temperature of hot reservoir is 669.24 K.
Explanation:
It is given that,
Efficiency of gasoline engine, [tex]\eta=35.3\%=0.353[/tex]
Temperature of cold reservoir, [tex]T_C=160^{\circ}C=433\ K[/tex]
We need to find the temperature of hot reservoir. The efficiency of Carnot engine is given by :
[tex]\eta=1-\dfrac{T_C}{T_H}[/tex]
[tex]T_H=\dfrac{T_C}{1-\eta}[/tex]
[tex]T_H=\dfrac{433}{1-0.353}[/tex]
[tex]T_H=669.24\ K[/tex]
So, the temperature of hot reservoir is 669.24 K. Hence, this is the required solution.
Problem One: A beam of red light (656 nm) enters from air into the side of a glass and then into water. wavelength, c. and speed in both the glass and the water. Find the a. the frequencies, b. the
Answers
Answer:
Part a)
[tex]f_w = f_g = 4.57 \times 10^{14} Hz[/tex]
Part b)
[tex]\lambda_w = 492 nm[/tex]
[tex]\lambda_g = 437.3 nm[/tex]
Part c)
[tex]v_w = 2.25 \times 10^8 m/s[/tex]
[tex]v_g = 2.0 \times 10^8 m/s[/tex]
Explanation:
Part a)
frequency of light will not change with change in medium but it will depend on the source only
so here frequency of light will remain same in both water and glass and it will be same as that in air
[tex]f = \frac{v}{\lambda}[/tex]
[tex]f = \frac{3 \times 10^8}{656 \times 10^{-9}}[/tex]
[tex]f = 4.57 \times 10^{14} Hz[/tex]
Part b)
As we know that the refractive index of water is given as
[tex]\mu_w = 4/3[/tex]
so the wavelength in the water medium is given as
[tex]\lambda_w = \frac{\lambda}{\mu_w}[/tex]
[tex]\lambda_w = \frac{656 nm}{4/3}[/tex]
[tex]\lambda_w = 492 nm[/tex]
Similarly the refractive index of glass is given as
[tex]\mu_w = 3/2[/tex]
so the wavelength in the glass medium is given as
[tex]\lambda_g = \frac{\lambda}{\mu_g}[/tex]
[tex]\lambda_g = \frac{656 nm}{3/2}[/tex]
[tex]\lambda_g = 437.3 nm[/tex]
Part c)
Speed of the wave in water is given as
[tex]v_w = \frac{c}{\mu_w}[/tex]
[tex]v_w = \frac{3 \times 10^8}{4/3}[/tex]
[tex]v_w = 2.25 \times 10^8 m/s[/tex]
Speed of the wave in glass is given as
[tex]v_g = \frac{c}{\mu_g}[/tex]
[tex]v_g = \frac{3 \times 10^8}{3/2}[/tex]
[tex]v_g = 2 \times 10^8 m/s[/tex]
When 0.511 g of sodium metal is added to an excess of hydrochloric acid, 5310 J of heat are produced. What is the enthalpy of the reaction as written?
Answers
Answer:
The the enthalpy of the reaction is 241.36 KJ.
Explanation:
Given that,
Weight of sodium = 0.511 g
We need to calculate the number of moles
Using formula of moles
[tex]moles=\dfrac{given\ mass}{molar\ mass}[/tex]
[tex]moles=\dfrac{0.511}{23}[/tex]
[tex]moles= 0.022[/tex]
We need to calculate the energy in 1 mole
In 0.022 moles of sodium metal = 5310 J
In 1 moles of sodium metal =[tex]\dfrac{5310}{0.022}[/tex]
The Energy in 1 moles of sodium is 241363.63 J.
Hence, The the enthalpy of the reaction is 241.36 KJ.
Final answer:
To find the reaction enthalpy per mole of sodium reacting with hydrochloric acid, divide the heat produced (5310 J) by the number of moles of sodium (0.0222 mol). This calculation results in an enthalpy change of -239 kJ/mol for sodium.
Explanation:
To calculate the enthalpy of the reaction per mole of sodium metal reacting with hydrochloric acid, we can use the information provided about the heat produced during the reaction. Given that 0.511 g of sodium metal yields 5310 J of heat, we first need to determine the amount in moles of sodium. We know that the molar mass of sodium (Na) is approximately 23.0 g/mol. Thus, 0.511 g Na amounts to 0.511 g / 23.0 g/mol = 0.0222 mol Na.
Given the heat released, we can now calculate the enthalpy change per mole of sodium reacted. The enthalpy change (H) for the reaction can be calculated by dividing the heat by the number of moles of sodium:
H = Heat Produced / Moles of Sodium
H = 5310 J / 0.0222 mol Na
H = -239,189 J/mol Na (note the negative sign because the reaction is exothermic)
Converting this result to kilojoules, the enthalpy change per mole of sodium is approximately -239.189 kJ/mol Na, which we often express as -239 kJ/mol to match conventional significant figures for such values.
In a circus performance, a large 3.0 kg hoop with a radius of 1.3 m rolls without slipping. If the hoop is given an angular speed of 6.8 rad/s while rolling on the horizontal and is allowed to roll up a ramp inclined at 24◦ with the horizontal, how far (measured along the incline) does the hoop roll? The acceleration of gravity is 9.81 m/s2 .
Answers
Answer:
The distance is 19.58 m.
Explanation:
Given that,
Mass = 3.0 kg
Radius = 1.3 m
Angular speed = 6.8 rad/s
Angle = 24°
Acceleration of gravity = 9.81 m/s²
We need to calculate the distance
Using formula of kinetic energy
Total Initial kinetic energy,
[tex]K.E = \dfrac{1}{2}(mv^2+I\omega^2)[/tex]
[tex]K.E = \dfrac{1}{2}(mv^2+mr^2\omega^2)[/tex]
[tex]K.E =\dfrac{(mr^2+mr^2)\omega^2}{2}[/tex]
[tex]K.E =mr^2\omega^2[/tex]....(I)
Now, Total potential energy
[tex]P.E=mgs\sin\theta[/tex]
[tex]P.E=mgs\sin24^{\circ}[/tex]....(II)
Equating equation (I) and (II)
[tex]mr^2\omega^2==mgs\sin24^{\circ}[/tex]
[tex](1.3)^2\times6.8^2=9.81\times s\times\sin24^{\circ}[/tex]
[tex]s = \dfrac{1.3^2\times6.8^2}{9.81\times\sin24^{\circ}}[/tex]
[tex]s=19.58\ m[/tex]
Hence, The distance is 19.58 m.
Alpha particles of charge q = +2e and mass m = 6.6 × 10-27 kg are emitted from a radioactive source at a speed of 1.6 × 107 m/s. What magnetic field strength would be required to bend them into a circular path of radius r = 0.18 m?
Answers
Answer:
Magnetic field, B = 1.84 T
Explanation:
It is given that,
Charge on alpha particle, q = +2e = [tex]3.2\times 10^{-19}\ C[/tex]
Mass of alpha particle, [tex]m=6.6\times 10^{-27}\ kg[/tex]
Speed of alpha particles, [tex]v=1.6\times 10^7\ m/s[/tex]
We need to find the magnetic field strength required to bend them into a circular path of radius, r = 0.18 m
So, [tex]F_m=F_c[/tex]
[tex]F_m\ and\ F_c[/tex] are magnetic force and centripetal force respectively
[tex]qvB=\dfrac{mv^2}{r}[/tex]
[tex]B=\dfrac{mv}{qr}[/tex]
[tex]B=\dfrac{6.6\times 10^{-27}\ kg\times 1.6\times 10^7\ m/s}{3.2\times 10^{-19}\ C\times 0.18\ m}[/tex]
B = 1.84 T
So, the value of magnetic field is 1.84 T. Hence, this is the required solution.
The magnetic field strength required to bend them into a circular path is 1.83 T.
Force of the emitted chargesThe magnetic force on the emitted charge is given as;
F = qvB
The centripetal force of the emitted charge is given as;
F = mv²/r
Magnetic field strengthThe magnetic field strength required to bend them into a circular path is calculated as follows;
qvB = mv²/r
[tex]B = \frac{mv}{rq}[/tex]
[tex]B = \frac{6.6 \times 10^{-27} \times 1.6 \times 10^7}{(2\times 1.6 \times 10^{-19} ) \times 0.18} \\\\B = 1.83 \ T[/tex]
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A boat sails south with the help of a wind blowing in the direction S36°E with magnitude 300 lb. Find the work done by the wind as the boat moves 130 ft. (Round your answer to the nearest whole number.) ft-lb
Answers
To find the work done by the wind on the boat, we first find the southward component of the wind's force, since the boat is sailing south. We find this to be about 243 lb. Multiplying this force by the distance the boat travels, we find that the wind does approximately 31590 ft-lb of work on the boat.
Explanation:In order to find the work done by the wind on the boat, we need to find the component of the wind force that acts in the same direction as the displacement of the boat. The wind is blowing in the direction of S36°E, meaning it has a southward component and an eastward component. However, since the boat is moving south, only the southward component of the wind's force will do work on the boat.
We use the equation F_s = F * cos(θ) to find the southward component of the force, where F is the magnitude of the total wind force and θ is the angle between the force of the wind and the direction of displacement. Plugging in the given values, we get F_s = 300 lb * cos(36°) = 243 lb.
To find work, we use the equation W = F * d * cos(θ), where F is the force, d is the distance traveled, and θ is the angle between the force and the displacement. Since the force and the displacement are in the same direction, the angle θ is 0, so cos(θ) is 1. Plugging in the appropriate values, we get W = 243 lb * 130 ft * 1 = 31590 ft-lb.
Rounded to the nearest whole number, the wind does approximately 31590 ft-lb of work on the boat.
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Which of the following is a property of an object not a result of something happening to the object? Select one: weight a. b. inertia C. momentum d. force
Answers
Answer:
option (b)
Explanation:
Th weight of the body is equal to the product of mass of the body and acceleration due to gravity. So, it depends on the acceleration due to gravity.
Inertia is the inherent property of the body which always resists any change in the body. It does not depend on any factor. mass is the measure of inertia.
Momentum is the product of mass of body and the velocity of the body.
Force is the product of mass and the acceleration of the body.