Answer:
[tex]t=6.19s[/tex]
Explanation:
With a initial velocity is zero, kinematics equation is:
[tex]y=1/2*g*t^2[/tex]
[tex]t=\sqrt{2y/g}=\sqrt{2*188/9.81}=6.19s[/tex]
A sodium atom will absorb light with a wavelength near 589 nm if the light is within 10 MHz of the resonant frequency. The atomic mass of sodium is 23. (i) Calculate the number of "yellow" photons of wavelength 2 = 589 nm that must be absorbed to stop a sodium atom initially at room temperature (V-600 m/s). [7 marks] (ii) What is the minimum time needed to cool a sodium atom?
Answer:
i)20369 photons
ii) 40 ps
Explanation:
Momentum of one Sodium atom:
[tex]P=m*v =600m/s*23amu*\frac{1 kg}{6.02*10^{23}amu}\\P=2.29*10^{-23}kgm/s[/tex]
In other to stop it, it must absorb the same momentum in photons:
[tex]P=2.29*10^{-23}kgm/s=n_{photons}*\frac{h_{planck}}{\lambda}\\=n*\frac{6.63*10^{-34}}{589*10^{-9}} \\==>n=20369 photons[/tex]
Now, for the minimun time, we use the speed of light and the wavelength. For the n photons:
[tex]t=n*T=n*\frac{\lambda}{c} =20369*\frac{589nm}{3*10^{8}m/s}=4*10^{-11} second=40 ps[/tex]
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
Your school science club has devised a special event for homecoming. You’ve attached a rocket to the rear of a small car that has been decorated in the green and gold school colors. The rocket provides a constant acceleration for 9.0 s. As the rocket shuts off, a parachute opens and slows the car at a rate of 5.0 m/s2 . The car passes the judges’ box in the center of the grandstand, 990 m from the starting line, exactly 12 s after you fire the rocket. What is the car’s speed as it passes the judges
Answer:
V=120m/s
Explanation:
A body that moves with constant acceleration means that it moves in "a uniformly accelerated movement", which means that if the velocity is plotted with respect to time we will find a line and its slope will be the value of the acceleration, it determines how much it changes the speed with respect to time.
When performing a mathematical demonstration, it is found that the equations that define this movement are as follows.
Vf=Vo+a.t (1)
{Vf^{2}-Vo^2}/{2.a} =X (2)
X=Xo+ VoT+0.5at^{2} (3)
X=(Vf+Vo)T/2 (4)
Where
Vf = final speed
Vo = Initial speed
T = time
A = acceleration
X = displacement
In conclusion to solve any problem related to a body that moves with constant acceleration we use the 3 above equations and use algebra to solve
for this problem you must divide the problem into two parts 1 and 2, when the rocket accelerates (1), and when the rocket decelerates (2).
Then you raise equations 3 and 1 in both parts.
finally you use algebraic methods to find the value of speed
I attach the complete procedure
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 %
If the distance between the Earth and the Sun were increased by a factor of 2.51, by what factor would the strength of the force between them change? [Hint: Use Newton's Law of Universal Gravitation, and give your answer to 2 decimal places only]
Answer:
Force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times
Explanation:
According to law of gravitation we know that force between two object is given by
[tex]F=\frac{GM_1M_2}{R^2}[/tex], here G is universal gravitational constant [tex]M_1[/tex] is mass of one object and [tex]M_2[/tex] is mass of other object , and R is the distance between them
From the relation we can see that force is inversely proportional to square of distance between them
So force between sun and earth will decrease by [tex]2.51^2=6.3[/tex] times
Final answer:
The strength of the gravitational force between the Earth and the Sun would decrease by a factor of approximately 6.30 if the distance between them were increased by a factor of 2.51, according to the inverse square law of universal gravitation.
Explanation:
If the distance between the Earth and the Sun were increased by a factor of 2.51, the strength of the force between them would change according to Newton's law of universal gravitation. This law states that the force is inversely proportional to the square of the distance between two masses. Therefore, if the distance increases by a factor of 2.51, the force of attraction would decrease by a factor of (2.51)2.
To calculate the specific factor by which the force decreases, we square 2.51 which gives us 6.3001. The gravitational force would thus be reduced by this factor. Since the law involves an inverse square relationship, as the distance increases by a factor of 2.51, the strength of the gravitational force decreases by a factor of approximately 6.30.
You plan a trip on which you want to average 91 km/h. You cover the first half of the distance at an average speed of only 49 km/h. What must your average speed be in the second half of the trip to meet your goal? Note that the velocities are based on half the distance, not half the time. km/hIs this reasonable? Yes No
Answer:
637 km/h, not reasonable
Explanation:
Total average speed = 91 km/h
Average speed of first half, v1 = 49 km/h
Let the average speed for the next half is v2.
Let the total distance is 2s.
Time taken for first half of the journey, t1 = s / v1 = s / 49
Time take for the next half of the journey, t2 = s / v2
The average speed is defined as the total distance traveled to the total time taken
[tex]91=\frac{2s}{\frac{s}{49}+\frac{s}{v_{2}}}[/tex]
[tex]\frac{s}{49}+\frac{s}{v_{2}}=\frac{2s}{91}[/tex]
[tex]\frac{1}{v_{2}}=\frac{2}{91}-\frac{1}{49}[/tex]
[tex]\frac{1}{v_{2}}=\frac{7}{4459}
v2 = 637 km/h
This is very high speed which is not reasonable.
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providing a steady force of 161 N that is parallel to the motion of the sled. The combined mass of the girl and the sled is 87.7 kg, and the coefficient of kinetic friction between the snow and the runners of the sled is 0.151. How much time is required for the sled to travel down a 271-m slope, starting from rest?
Answer:
The sled required 9.96 s to travel down the slope.
Explanation:
Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.
Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).
The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.
The magnitude of the friction force is calculated as follows:
Ff = μ · Fn
where :
μ = coefficient of kinetic friction
Fn = normal force
The normal force has the same magnitude as the y-component of the gravity force:
Fgy = Fg · cos 30º = m · g · cos 30º
Where
m = mass
g = acceleration due to gravity
Then:
Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º
Fgy = 744 N
Then, the magnitude of Fn is also 744 N and the friction force will be:
Ff = μ · Fn = 0.151 · 744 N = 112 N
The x-component of Fg, Fgx, is calculated as follows:
Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N
The resulting force, Fr, will be the sum of all these forces:
Fw + Fgx - Ff = Fr
(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).
Fr = 161 N + 430 N - 112 N = 479 N
With this resulting force, we can calculate the acceleration of the sled:
F = m·a
where:
F = force
m = mass of the object
a = acceleration
Then:
F/m = a
a = 479N/87.7 kg = 5.46 m/s²
The equation for the position of an accelerated object moving in a straight line is as follows:
x = x0 + v0 · t + 1/2 · a · t²
where:
x = position at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.
x = 1/2· a ·t²
Let´s find the time at which the position of the sled is 271 m:
271 m = 1/2 · 5.46 m/s² · t²
2 · 271 m / 5.46 m/s² = t²
t = 9.96 s
The sled required almost 10 s to travel down the slope.
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.
Learn more about Electric Field Magnitude here:https://brainly.com/question/28561944
#SPJ3
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.
What is the repulsive force between two pith balls that are 9.00 cm apart and have equal charges of -28.0 nC?
Answer:
Force, [tex]F=8.71\times 10^{-4}\ N[/tex]
Explanation:
Given that,
Charges on pith balls, [tex]q_1=q_2=-28\ nC=-28\times 10^{-9}\ C[/tex]
Distance between balls, d = 9 cm = 0.09 m
Let F is the repulsive force between two pith balls. We know that the repulsive force between two charges is given by :
[tex]F=k\dfrac{q_1^2}{d^2}[/tex]
[tex]F=9\times 10^9\times \dfrac{(-28\times 10^{-9})^2}{(0.09)^2}[/tex]
F = 0.000871 N
or
[tex]F=8.71\times 10^{-4}\ N[/tex]
So, the repulsive force between the pith balls is [tex]8.71\times 10^{-4}\ N[/tex]. Hence, this is the required solution.
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.
How many seconds will elapse between seeing lightning and hearing the thunder if the lightning strikes 4.5 mi (23,760 ft) away and the air temperature is 74.0°F?
Answer:
t = 20.96 seconds
Explanation:
given,
lightning strikes at = 4.5 mi( 23,760 ft)
air temperature =74.0°F
74.0°F = 23.3 °C
speed of the sound at 23.3 °C
V = 331.5 + 0.6 × T
V = 331.5 + 0.6 × 23.3
= 345.48 m/s
distance given = 4.5 mile
1 mile = 1609.4
4.5 mile = 4.5 × 1609.4 = 7242.3 m
time taken =
[tex]t =\dfrac{d}{v}[/tex]
[tex]t =\dfrac{7242.3}{345.48}[/tex]
t = 20.96 seconds
hence, time taken by the sound to reach by the observer is 20.96 sec.
Final answer:
To estimate the time elapsed between seeing lightning and hearing thunder, one must know the speed of sound, which varies with temperature, and the distance to the lightning strike. At a distance of 4.5 miles and an air temperature of 74.0°F, approximately 21 seconds will elapse.
Explanation:
Calculating the Time Difference Between Lightning and Thunder
The question involves the relationship between the speed of light and sound to determine how long it will take for the sound of thunder to reach an observer after the lightning is seen. Since light travels at approximately 300,000 kilometers per second (3 × 108 meters per second), lightning is seen almost instantaneously. However, the sound of thunder, which is caused by the rapid expansion of superheated air due to a lightning strike, travels much slower.
At a temperature of 74.0°F, we need to convert this to Celsius since the speed of sound in air is given by the formula v = 331.4 m/s + 0.6T, where T is in Celsius. The temperature in Celsius can be found through T(°C) = (T(°F) - 32) × 5/9, giving us T(°C) = (74 - 32) × 5/9 = 23.33°C.
The speed of sound at this temperature is v = 331.4 m/s + 0.6 × 23.33 = 345.4 m/s. To find the time it takes for the sound to travel 4.5 miles (or 23,760 feet), we need to convert the distance to meters since the speed of sound is given in meters per second. We have 1 mile = 1,609.34 meters, so 4.5 miles = 7,242 meters approximately. Finally, we calculate the time as time = distance/speed = 7,242 m / 345.4 m/s ≈ 20.97 s.
Therefore, approximately 21 seconds will elapse between seeing the lightning and hearing the thunder if the lightning strikes 4.5 miles away and the air temperature is 74.0°F.
A ladder leans against a non vertical wall with friction that make a angle of 50 degree with ground. how do show the free body diagram. The wall is not vertical
Answer:
As from the given question,
The free body diagram of a ladder leans against a non vertical wall with friction that make a angle of 50 degree with ground is show below.
Friction force can be calculated by formula:
Friction force = μ N
As we know that
At equilibrium,
The sum of the total horizontal force is equal to '0'.
The sum of the total vertical force is equal to '0'.
A light source of wavelength λ illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 2.00 eV. A second light source with half the wavelength of the first ejects photoelectrons with a maximum kinetic energy of 6.00 eV. What is the work function of the metal?
Answer:
[tex]W=2eV[/tex]
Explanation:
Energy from a light source (photons):
E=h*c/λ
Photoelectric effect and Work function (W):
E=W+Ek
Ek: maximum kinetic energy from photoelectrons
then:
h*c/λ=W+Ek
Case 1:
[tex]h*c/lambda_{1}=W+Ek_{1}[/tex] (1)
Case 2:
[tex]h*c/lambda_{2}=W+Ek_{2}[/tex]
but [tex]lambda_{2}=lambda_{1}/2[/tex]
[tex]2h*c/lambda_{1}=W+Ek_{2}[/tex] (2)
If we divide (2) by (1):
[tex]2=\frac{W+Ek_{2}}{W+Ek_{1}}[/tex]
[tex]W=Ek_{2}-2Ek{1}=2eV[/tex]
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.
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
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
Convert the following binary numbers to a decimal: 1001, 10101, 1010001, and 1010.1010.
Answer:
1) 1001
=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 1 = (9)₁₀
2) 10101
=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰
= 16 + 4 + 1 = (21)₁₀
3) 1010001
=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 64 + 16 + 1 = (81)₁₀
4) 1010.1010
= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴
= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀
Answer:
1) 1001
=1 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 8 + 1 = (9)₁₀
2) 10101
=1 × 2⁴ + 0 × 2³ + 1 × 2² + 0 × 2¹ + 1 × 2⁰
= 16 + 4 + 1 = (21)₁₀
3) 1010001
=1 × 2⁶ + 0 × 2⁵ + 1 × 2⁴ + 0 × 2³ + 0 × 2² + 0 × 2¹ + 1 × 2⁰
= 64 + 16 + 1 = (81)₁₀
4) 1010.1010
= 1 × 2³ + 0 × 2² + 1 × 2¹ + 0 × 2⁰ + 1 × 2⁻¹ + 0 × 2⁻² + 1 × 2⁻³ + 0 × 2⁻⁴
= 8 + 2 + 0.5 + 0.125 = (10.625)₁₀
Explanation:
A research team developed a robot named Ellie. Ellie ran 1,000 meters for 200 seconds from the research building, rested for 100 seconds, and walked back to the research building for 1000 seconds. To find out Ellie’s average velocity for each case while running, resting, and walking, begin by plotting a graph between position and time. 1.List the velocity from greatest to least among running, resting, and walking
2.List the speed from greatest to least among running, resting, and walking.
Answer:
1. Running velocity (5 m/s)
2. Resting velocity (0 m/s)
3. Walking velocity (-1 m/s)
1. Running speed (5 m/s)
2. Walking speed (1 m/s)
3. Resting speed (0 m/s)
Explanation:
Attached you will find the plot of position vs time of Ellie´s movement.
The velocity is the displacement of the object over time relative to the system of reference. The speed, in change, is the traveled distance over time in disregard of the system of reference.
So, the velocity is calculated as follows:
v = Δx / Δt
where
Δx = final position - initial position
Δt = elapsed time
1) The average velocity of Ellie while running is:
v = 1000 m - 0 m / 200 s = 5 m/s
While resting:
v = 0 m - 0 m / 100 s = 0 m/s
And while walking back:
v = 0 m - 1000 m / 1000 s = - 1 m/s
Note that in this last case, the initial position is 1000 m because Ellie is 1000 m from the origin of the system of reference when she walks back. The final position will be the origin of the system of reference, 0 m.
Comparing with the graphic, the velocity is the slope of the function position(t).
Then:
1. Running velocity (5 m/s)
2. Resting velocity (0 m/s)
3. Walking velocity (-1 m/s)
2) The speed is the distance traveled over time:
Running speed = 1000 m / 200 s = 5m /s
Resting speed = 0 m / 100 s = 0 m/s
Walking speed = 1000 m/ 1000 s = 1 m/s
Then:
1. Running speed (5 m/s)
2. Walking speed (1 m/s)
3. Resting speed (0 m/s)
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]
Vector A is 3 m long and vector B is 4 m long. The length
ofthe sum of the vectors must be:
A. 5 m
B. 7 m
C. 12 m
D. some value from 1 m to 7 m.
Answer:d
Explanation:
Given
Magnitude of Vector A is 3 m
and Magnitude of vector B is 4 m
So the maximum value of resultant can be 7 m when both are at an angle of [tex]0^{\circ}[/tex]
and its minimum value can be 1 m when both are at angle of [tex]180 ^{\circ}[/tex]
So the resultant must lie between 1 m to 7 m
The length of the sum of the vectors A and B with no direction given must be some value from 1 m to 7 m. Thus, the correct option is D.
What is Vector sum?Vector addition is the property or operation of addition of two or more vectors together into a vector sum. The sign used with the vectors depends upon the direction in which they move. If the vectors are moving in the same direction then they will be added and if they are moving in opposite directions then they are substracted.
Given, Magnitude of Vector A is 3 m and Magnitude of vector B is 4 m.
So, in this case the maximum value of resultant vector can be 7 m when both are moving in the same direction with an angle of 0° and it has minimum value which can be 1 m when both are at angle of 180° in opposite direction.
So, the resultant vector must lie between 1 m to 7 m.
Therefore, the correct option is D.
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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 comet is cruising through the solar system at a speed of 50,000 kilometers per hour foe 4 hours time. What is the total distance traveled by the comet during that time
Answer:
The distance covered by comet is [tex]200,000 km[/tex]
Explanation:
Speed is defined as the rate of change of distance with time. It is given by the equation speed= [tex]\bold{\frac{distance}{time}}[/tex]
Thus distance= [tex]speed*time[/tex]
In this problem it is given that speed of comet= [tex]\frac{50,000km}{hr}[/tex]
time travelled by the comet= 4 hours
Thus distance= [tex]speed*time[/tex]
= [tex]500000*4[/tex]
= [tex]\bold{200,000km}[/tex]
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
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.
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.
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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.
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 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