Answer:
Question not completed, so I analysed the question first
Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?
Explanation:
Let use variables to solve the problems
Let the first trip to be mountain take x hours
Let the trip back home take y hours
Let the speed to while going to the mountain be a miles/hour
Then, while going home it was b miles/hour faster than while going to the mountain.
Then, speed going home is (a+b)miles / hour
The formula for speed is given as
Speed=distance/time
The constant through out the journey is distance, the two journey has the same distance.
Then,
Distance =speed×time
For first journey going to the mountain
Distance = a×x=ax miles
For the second journey going home
Distance =y×(a+b)
Distance Mountain= distance home
ax=y(a+b)
Make a subject of the formula
ax=ya+yb
ax-ya=yb
a(x-y)=yb
a=yb/(x-y)
Therefore, distance from mountain is
Distance=speed ×time
Distance= a×x=ax
Now, applying the questions
So from the questions
x=6hours, y=4hours
Also, b=22miles/hour
Then,
a=yb/(x-y)
a=4×22/(6-4)
a=88/2
a=44miles/hour
Then, the house distance from the mountain is
Distance=ax
Distance =44×6
Distance =264miles
Answer:
480 miles.
Explanation:
Let, S = rate on his way to the mountains.
Assume, Sgoing x time going = Sreturning x time returning
= S × 12 hours = (S + 20mph) ×8 hours
= 12 × S = 8 × S + 160.
4 × S = 160
S = 40 miles/hour
The trip took 12 hours at 40 miles per hour, so distance was:
= 12 hours × 40 mph
= 480 miles.
Explain Rutherford's experiment?
Answer:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
Explanation:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.
The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.
Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.
A bungee jumper attains a speed of 30 m/s just as the bungee cord begins to stretch. If the period of stretch is 2 s while coming to a halt, the jumper's average deceleration in g’s is about ______?
Answer:[tex]1.53g[/tex]
Explanation:
average deceleration= ?
inial velocity: [tex]u=0[/tex]
final velocity: [tex]v=30m/s[/tex]
time: [tex]t=2seconds[/tex]
The first law of kinematics :
[tex]v=u+at[/tex]
find a the subject of the formula
[tex]a=v-u/ t[/tex]
[tex]a=\frac{0-30} 2[/tex]
[tex]a=-30/2[/tex]
[tex]a=-15m/s^{2}[/tex]
The deceleration about g(acceleration due to gravity) will be:
[tex]15/9.8[/tex]
[tex]1.53g[/tex]
The bungee jumper's average deceleration in g's is approximately -1.53 g's.
Explanation:To find the average deceleration, we need to calculate the change in velocity and divide it by the time it took to come to a halt. The change in velocity is the final velocity minus the initial velocity, which is 0 m/s minus 30 m/s, giving us -30 m/s. The time is given as 2 s. Plugging these values into the formula for average deceleration, we get:
Average Deceleration = (Change in Velocity) / (Time)
Average Deceleration = (-30 m/s) / (2 s) = -15 m/s²
To convert this to g's, we need to divide by the acceleration due to gravity (9.8 m/s²).
Average Deceleration in g's = (-15 m/s²) / (9.8 m/s²) ≈ -1.53 g's
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The total power consumption by all humans on earth is approximately 1013 W. Let’s compare this to the power of incoming solar radiation. The intensity of radiation from the sun at the top of the atmosphere is 1380 W/m2. The earth's radius is 6.37×106 m.
Answer:
Power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.
Explanation:
Intensity of sunlight = I = 1380 w/m^2
Area of earth = A = 4*pi*r^2 = 4*pi*(6.37*10^6)^2 = 5.09*10^14 m^2
he intensity is defined as the total power spread over the area of earth (Area of Sphere with radius equal to distance between earth and sun) and given by the following formula:
Intenity of sunlight = Power/Area of earth
I = P/A
P = IA
P = (1380)(5.09*10^14)
P = 7.036*10^17 W
if we take ratio:
7.036*10^17/1013 = 6.94 * 10^14
Hence, power coming from solar radiations is 6.94 * 10^14 times higher that the power consumption of all humans.
You go to the doctor and he gives you 13 milligrams of radioactive dye. After 12 minutes, 8 milligrams of dye remain in your system. To leave the doctor's office, you must pass through a radiation detector without sounding the alarm. If the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived
Answer:
Explanation:
Applying the exponential function of decay
M=Cexp(-kt)
At t =0 the mass is 13mmg
Therefore
13=Cexp(0)
C=13
M=13exp(-kt)
After 12mins, M=8mmg
8=13exp(-K×12)
8/13=exp(-12k)
0.615=exp(-12k)
Take In of both sides
In(0.615)=-12k
-0.4855=-12k
Then, k=0.0405
Then the equations become
M=13exp(-0.0405t)
We need to find t at M=2mmg
M=13exp(-0.0405t)
2=13exp(-0.0405t)
2/13=exp(-0.0405t)
0.1538=exp(-0.0405t)
Take In of both sides
In(0.1538)=-0.0405t
-1.872=-0.0405t
Then t=-1.82/-0.0405
t=46.22mintes
Solar energy heats the surface of the earth including the ground rocks and even roadways as the temperatures of the surfaces increase heat energy is released back
Answer: into the atmosphere
Explanation:
As this energy is released into the atmosphere, bubbles of warm air is formed which is released and it is replaced by cooler air. This process is responsible for many of the weather patterns in our atmosphere, and is known as....
convection
While convection is a form of heat transfer from one place to the other involving the movement of fluids. So in the case narrated above the fluid which serves as a medium is Air.
You push a 45 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.25. Now you double the force on the box. How long would it take for the velocity of the crate to double to 2.0 m/s
Final answer:
To double the velocity of the wooden box, you need to double the force applied to it. Use the equation Force = coefficient of kinetic friction x normal force to calculate the force required to push the crate at a constant speed. Then, double this force to find the force required to double the velocity.
Explanation:
To double the velocity of the crate to 2.0 m/s, you need to double the force applied to the crate. The force required to push the crate at a constant speed can be calculated using the equation:
Force = coefficient of kinetic friction x normal force
The normal force is equal to the weight of the crate, which is given by:
Normal force = mass x gravity
With the given values of the mass of the crate (45 kg), the coefficient of kinetic friction (0.25), and the acceleration due to gravity (9.8 m/s²), you can calculate the force required to push the crate at a constant speed. Then, you can double this force to find the force required to double the velocity.
Consider two uniform solid spheres where one has twice the mass and twice the diameter of the other. The ratio of the larger moment of inertia to that of the smaller moment of inertia is:_________.a) 2b) 8c) 4d) 10e) 6
Answer:
b) 8.
Explanation:
Below is an attachment containing the solution.
How long can it light a flashlight bulb that draws 0.60 a?
Final answer:
To calculate the flashlight's power usage, we use power and current formulas based on given charge and voltage values. Similarly, to estimate battery life for a flashlight bulb, we divide the battery's Amp-Hour capacity by the bulb's current draw.
Explanation:
The question is about determining the average current used by a flashlight bulb over a time period and how long the battery would last. To calculate the power usage, we apply the formula P = IV, where P is the power in watts, I is the current in amperes, and V is the voltage in volts. Given that 600 C (coulombs) of charge passes through the flashlight in 0.500 hours (which is 1800 seconds) and the voltage is 3.00 V, we can find the average current using the formula I = Q/t, where Q is the charge in coulombs and t is the time in seconds.
To find out how long a battery will keep a flashlight bulb burning, we divide the battery capacity in ampere-hours (Ah) by the current drawn by the bulb. If a 1.00-W bulb is used and the battery is rated at 1.00 Ah and 1.58 V, we first need to use the power formula P = IV to find the current drawn by the bulb.
The flashlight bulb that draws 0.60 A can be lit for approximately 1.67 hours using a 1.00 Ah alkaline battery. This is calculated by dividing the battery capacity by the current drawn by the bulb.
To determine how long a flashlight bulb that draws 0.60 A can be lit, we must consider the battery's capacity and voltage. Here is a step-by-step explanation:
First, identify the battery capacity. Assume we have an alkaline battery rated at 1.00 Ah.
Next, understand that 1.00 Ah means the battery can supply 1.00 ampere for 1 hour.
Since the flashlight draws 0.60 A, we calculate the time the battery can light the bulb using the formula:
Time (hours) = Battery Capacity (Ah) / Current (A)
Substitute the values:
Time = 1.00 Ah / 0.60 A = 1.67 hours
Thus, the flashlight bulb can be lit for approximately 1.67 hours or 1 hour and 40 minutes before the battery is depleted.
\ describes the size and distance relationship of our sun and the nearest star?
Answer: Two marbles separated by 300 kilometers
Explanation: Hope i helped have a great day and please mark brainliest i would appreciate it!
The physical model of the sun’s interior has been confirmed by observations of
The physical model of the sun's interior has been confirmed by observations of neutrino and seismic vibrations.
Explanation:
Sun's interior is composed of very high temperature and solar flares. So it is very difficult to understand the interior of the sun. But by using the vibrations of neutrino and seismic waves emitted by the solar waves, the physical model can be assumed.
As the interior of the sun performs continuous chain of hydrogen cycle. So the continuous emission of energy from the chain reaction releases neutrino. So these vibrations in neutrino and seismic vibrations, the physical model can be assumed easily.
A block of mass 2 kg is traveling in the positive direction at 3 m/s. Another block of mass 1.5 kg, traveling in the same direction at 4 m/s, collides elastically with the first block. Find the final velocities of the blocks. How much kinetic energy did the system lose
Answer:
a. The final velocity of the block of mass 2 kg is 3 m/s or 3.86 m/s. The final velocity of the block of mass 1.5 kg is 4 m/s or 2.86 m/s b. The kinetic energy change is 0 J or -12.235 J. Since the collision is elastic, we choose ΔK = 0
Explanation:
From principle of conservation of momentum,
momentum before impact = momentum after impact
Let m₁ = 2 kg, m₂ = 1.5 kg and v₁ = 3 m/s, v₂ = 4 m/s represent the masses and initial velocities of the first and second blocks of mass respectively. Let v₃ and v₄ be the final velocities of the blocks. So,
m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄
(2 × 3 + 1.5 × 4) = 2v₃ + 1.5v₄
6 + 6 = 2v₃ + 1.5v₄
12 = 2v₃ + 1.5v₄
2v₃ + 1.5v₄ = 12 (1)
Since the collision is elastic, kinetic energy is conserved. So
1/2m₁v₁² + 1/2m₂v₂² = 1/2m₁v₃² + 1/2m₂v₄²
1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 1/2 ×2v₃² + 1/2 × 1.5v₄²
9 + 12 = v₃² + 0.75v₄²
21 = v₃² + 0.75v₄²
v₃² + 0.75v₄² = 21 (2)
From (1) v₃ = 6 - 0.75v₄ (3) . Substituting v₃ into (2)
(6 - 0.75v₄)² + 0.75v₄² = 21
36 - 9v₄ + 0.5625v₄² + 0.75v₄² = 21
36 - 9v₄ + 1.3125v₄² - 21 = 0
1.3125v₄² - 9v₄ + 15 = 0
Using the quadratic formula,
v₄ = [-(-9) ± √[(-9)² - 4 × 1.3125 × 15]]/(2 × 1.3125)
= [9 ± √[81 - 78.75]]/2.625
= [9 ± √2.25]/2.625
= [9 ± 1.5]/2.625
= [9 + 1.5]/2.625 or [9 - 1.5]/2.625
= 10.5/2.625 or 7.5/2.625
= 4 m/s or 2.86 m/s
Substititing v₄ into (3)
v₃ = 6 - 0.75v₄ = 6 - 0.75 × 4 = 6 - 3 = 3 m/s
or
v₃ = 6 - 0.75v₄ = 6 - 0.75 × 2.86 = 6 - 2.145 = 3.855 m/s ≅ 3.86 m/s
b. The kinetic energy change ΔK = K₂ - K₁
K₁ = initial kinetic energy of the two blocks = 1/2m₁v₁² + 1/2m₂v₂²
= 1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 9 + 12 = 21 J
K₂ = final kinetic energy of the two blocks = 1/2m₁v₃² + 1/2m₂v₄². Using v = 3 m/s and v = 4 m/s
= 1/2 × 2 × 3² + 1/2 × 1.5 × 4² = 9 + 12 = 21 J.
ΔK = K₂ - K₁ = 21 - 21 = 0
Using v = 3.86 m/s and v = 2.86 m/s
K₂ = 1/2 × 2 × 3.86² + 1/2 × 1.5 × 2.86² = 14.8996 - 6.1347 = 8.7649 J ≅ 8.765 J
ΔK = K₂ - K₁ = 8.765 - 21 = -12.235 J
Since the collision is elastic, we choose ΔK = 0
Which conditions are usually the effect of a low air pressure system?
The given question is incomplete as the options are missing. The options related to this question are as follows-
(A) clear dry weather
(B) hot dry weather
(C) cloudy wet weather
(D) cold dry weather
Answer:
Option (C)
Explanation:
The surface temperature often increases because of increased absorption of solar radiation, the air present at the surface gets heated up more readily, as a result of which the air becomes less dense, and eventually rises up. This gives rise to the creation of a low air pressure system. It often forms clouds comprising of increase relative humidity, and generates wind and thereby causes precipitation. It also causes heavy storms when the atmospheric conditions are too intense.
Thus, the type of weather associated with this is wet and cloudy weather.
Hence, the correct answer is option (C).
Answer:
D
Explanation:
Cloudy wet weather
The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 hz. Which frequencies will not resonate in the same pipe?
Answer:
The frequency 400 hz is not possible .
Explanation:
Given that,
Frequency = 200 hz
Length = l
Suppose, The given frequencies are,
600 Hz, 1000 Hz, 1400 Hz, 1800 Hz and 400 hz.
The possible resonance frequencies are
We need to calculate the fundamental frequency
Using formula of fundamental frequency for pipe
[tex]F=\dfrac{nv}{4L}[/tex]
Where, n = odd number
Put the value of frequency
[tex]200= \dfrac{nv}{4L}[/tex]
We need to calculate the first over tone
Using formula of fundamental frequency
n = 3,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{2}=3\times\dfrac{v}{4l}[/tex]
[tex]F_{2}=3\times200[/tex]
[tex]F_{2}=600\ Hz[/tex]
We need to calculate the second over tone
Using formula of fundamental frequency
n = 5,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{3}=5\times200[/tex]
[tex]F_{3}=1000\ Hz[/tex]
We need to calculate the third over tone
Using formula of fundamental frequency
n = 7,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{4}=7\times200[/tex]
[tex]F_{4}=1400\ Hz[/tex]
We need to calculate the fourth over tone
Using formula of fundamental frequency
n = 9,
[tex]F=\dfrac{nv}{4L}[/tex]
Put the value into the formula
[tex]F_{5}=9\times200[/tex]
[tex]F_{5}=1800\ Hz[/tex]
Hence, The frequency 400 hz is not possible .
Final answer:
In a pipe that is closed at one end and open at the other, only odd multiples of the fundamental frequency will resonate. Frequencies that are even multiples of the fundamental 200 Hz, such as 400 Hz, 600 Hz, 800 Hz, will not resonate.
Explanation:
The lowest-pitch tone to resonate in a pipe of length l that is closed at one end and open at the other end is 200 Hz. This pipe supports harmonic frequencies following the sequence fn = n(v/4L), where n = 1, 3, 5..., v is the speed of sound, and L is the length of the pipe. Therefore, frequencies that will not resonate in the same pipe are those that are even multiples of the fundamental frequency, such as 400 Hz, 600 Hz, 800 Hz, and so on.
Two small, irregularly-shaped moons, Phobos and Deimos, orbit Mars. They are believed to be captured asteroids. What are the approximate orbital periods of Phobos and Deimos respectivelyA. 7 days, 12 hours; 1 day, 2 hours
B. 7 hours 35 minutes; 1 day, 6 hours
C. 14 days, 10 minutes; 2 days, 12 hours
D. 15 hours; 2 days, 12 hours
Answer:
Option B
Explanation:
The orbital periods of Phobos and Deimos can be calculated using the Newton's form of Kepler's third law:
[tex] T^{2} = \frac {4 \pi^{2}}{G*M_{m}} \cdot a^{3} [/tex]
where T: is the period, G: is the gravitational constant = 6.67x10⁻¹¹ m³kg⁻¹s⁻², Mm: is the mass of Mars = 6.42x10²³ kg, [tex]a_{P}[/tex]: is the average radius of orbit for the satellite Phobos = 9376 km, and [tex]a_{D}[/tex]: is the average radius of orbit for the satellite Deimos = 23463 km.
The orbital period of Phobos is:
[tex] T = \sqrt {\frac {4 \pi^{2}}{6.67 \cdot 10^{-11} m^{3} kg^{-1} s^{-2}*6.42 \cdot 10^{23} kg} \cdot (9.376 \cdot 10^{6} m)^{3}} = 2.75 \cdot 10^{4} s = 7 hours 36 min [/tex]
The orbital period of Deimos is:
[tex] T = \sqrt {\frac {4 \pi^{2}}{6.67 \cdot 10^{-11} m^{3} kg^{-1} s^{-2}*6.42 \cdot 10^{23} kg} \cdot (2.35 \cdot 10^{7} m)^{3}} = 1.09 \cdot 10^{5} s = 1 day 6 hours [/tex]
Therefore, the approximate orbital periods of Phobos and Deimos are 7 hours 35 minutes and 1 day 6 hours, respectively, so the correct answer is option B.
I hope it helps you!
Calculate the ratio of the drag force on a jet flying at 1190 km/h at an altitude of 7.5 km to the drag force on a prop-driven transport flying at half that speed and altitude. The density of air is 0.53 kg/m3 at 7.5 km and 0.74 kg/m3 at 3.8 km. Assume that the airplanes have the same effective cross-sectional area and drag coefficient C.
Answer:
[tex]\frac{D_{jet}}{D_{prop}}=2.865[/tex]
Explanation:
Given data
Speed of jet Vjet=1190 km/h
Speed of prop driven Vprop=595 km/h
Height of jet 7.5 km
Height of prop driven transport 3.8 km
Density of Air at height 10 km p7.8=0.53 kg/m³
Density of air at height 3.8 km p3.8=0.74 kg/m³
The drag force is given by:
[tex]D=\frac{1}{2}CpAv^2\\[/tex]
The ratio between the drag force on the jet to the drag force on prop-driven transport is then given by:
[tex]\frac{D_{jet}}{D_{prop}}=\frac{(1/2)Cp_{7.5}Av_{jet}^2}{1/2)Cp_{3.8}Av_{prop}^2} \\\frac{D_{jet}}{D_{prop}}=\frac{p_{7.5}v_{jet}^2}{p_{3.8}v_{prop}}\\\frac{D_{jet}}{D_{prop}}=\frac{(0.53)(1190)^2}{(0.74)(595)^2}\\ \frac{D_{jet}}{D_{prop}}=2.865[/tex]
A photovoltaic array of (solar cells) is 10.0% efficient in gathering solar energy and converting it to electricity. show answer Incorrect Answer 50% Part (a) If the average intensity of sunlight on one day is 710 W/m2, what area, in square meters, should your array have to gather energy at the rate of 75 W
Answer:
[tex]A = 1.056\,m^{2}[/tex]
Explanation:
The portion of solar energy converted into electric energy is given by the following equation:
[tex]\dot E = \eta \cdot I\cdot A[/tex]
The area needed to produce energy is derived by clearing the corresponding variable:
[tex]A = \frac{\dot E}{\eta \cdot I}[/tex]
[tex]A = \frac{75\,W }{(0.1)\cdot (710\,\frac{W}{m^{2}} )}[/tex]
[tex]A = 1.056\,m^{2}[/tex]
In a parallel portion of a series-parallel circuit, the voltage across the branches can be found by multiplying the sum of the branch currents by the equivalent resistance of the resistors in the parallel portion.True / False.
Answer:
It's true.
Explanation:
It's true. When we connect two resistors in parallel the current is divided between the two in such a way that the sum of the currents on each resistor should be equal to the current on that branch. By finding the equivalent resistance we can use Ohm's law to determine the voltage drop across the resistors. This voltage drop is the same for both, since they're connected in parallel.
This statement is true. In a series-parallel circuit, the voltage across the branches in the parallel portion can be found by multiplying the sum of the branch currents by the equivalent resistance of the resistors in the parallel portion.
Explanation:In a series-parallel circuit, the voltage across the branches in the parallel portion can be found by multiplying the sum of the branch currents by the equivalent resistance of the resistors in the parallel portion. This statement is true.
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How does the water table change around a pumping water well?
The ball was kicked in the air and it iss about to hit the . if horizonta componenet of its final velocity is 10m/s and vertical component of its final velocity is -10m/s .what is the magnotide of th final vellocity of the ball?
Answer:
The total velocity of the ball will be 14.14 m/s.
Explanation:
Horizontal Velocity component = 10 m/s
Vertical Velocity component = -10 m/s
Total velocity of the ball will be found from the following equation:
(Total velocity) ^2 = (Horizontal Velocity) ^2 + (Vertical Velocity) ^2
Total Velocity ^2 = 10^2 + (-10)^2
Total Velocity^2 = 100 + 100
Total Velocity = [tex]\sqrt{200}[/tex]
Total Velocity = 14.14 m/s
What is the coefficient of static friction between the coin and the turntable?
Answer:The coefficient of static friction between the turntable and the coin is 0.1
Explanation:
The coefficient of static friction is the friction force between two objects when neither of the objects is moving. ... A value of 1 means the frictional force is equal to the normal force. It is a misconception that the coefficient of friction is limited to values between zero and one.
In a Millikan oil-drop experiment, a uniform electric field of 5.71 x 10^5 N/C is maintained in the region between two plates separated by 6.49 cm. Find the potential difference (in V) between the plates.
Answer:
37057.9V
Explanation:
Electric potential is defined as the work done in moving a unit positive charge from infinity to a point.
Electric potential (E) = Potential Difference (V)/distance between plates(d)
Given; electric field of 5.71 x 10^5 N/C; distance between plates =6.49cm = 0.0649m
Since E = V/d
V = Ed
V = 5.71×10^5×0.0649
V = 37057.9Volts
The potential difference (in V) between the plates is 37057.9V
Answer:
V = 3.71×10⁴ V
Explanation:
Potential difference: This can be defined as the work done in moving a positive charge from infinity to any point in an electric field.
The S.I unit of potential difference is Volt (V).
The expression for potential difference is
V = E×d.............................. Equation 1
Where V = potential difference between the plates, E = Electric field , d = distance of separation between the plates
Given: E = 5.71×10⁵ N/C, d = 6.49 cm = 0.0649 m.
Substitute into equation 1
V = 5.71×10⁵×0.0649
V = 3.71×10⁴ V
Two identical loudspeakers 2.00 m apart are emitting sound waves into a room where the speed of sound is 340 m/s. Abby is standing 5.50 m in front of one of the speakers perpendicular to the line joining the speakers, and hears a maximum in the intensity of the sound. What is the lowest possible frequency of sound for which this is possible? Express your answer with the appropriate units.
Answer:
The lowest possible frequency of sound is 971.4 Hz.
Explanation:
Given that,
Distance between loudspeakers = 2.00 m
Height = 5.50 m
Sound speed = 340 m/s
We need to calculate the distance
Using Pythagorean theorem
[tex]AC^2=AB^2+BC^2[/tex]
[tex]AC^2=2.00^2+5.50^2[/tex]
[tex]AC=\sqrt{(2.00^2+5.50^2)}[/tex]
[tex]AC=5.85\ m[/tex]
We need to calculate the path difference
Using formula of path difference
[tex]\Delta x=AC-BC[/tex]
Put the value into the formula
[tex]\Delta x=5.85-5.50[/tex]
[tex]\Delta x=0.35\ m[/tex]
We need to calculate the lowest possible frequency of sound
Using formula of frequency
[tex]f=\dfrac{nv}{\Delta x}[/tex]
Put the value into the formula
[tex]f=\dfrac{1\times340}{0.35}[/tex]
[tex]f=971.4\ Hz[/tex]
Hence, The lowest possible frequency of sound is 971.4 Hz.
A 49-year-old female was referred for mammography, but she is very apprehensive after reading about the risk of ionizing radiation. How should the radiographer handle this situation?
Answer:
Mammography is the process in which low energy radiations are used to diagnose and screening. The purpose of this process is the early detection of the breast cancer. These low energy radiations may have some risks like damaging and burning of cells.
In the current scenario, woman is apprehensive because she has read about the risks of using ionizing radiations. The radiographer should tell her the benefits of the mammography will outweigh its potential consequences. Screening, for instance, will let her know if she is suffering from breast cancer. Cancer is very dangerous disease as compare to very small burning.
In this way radiographer should handle the situation.
Star A and star B appear equally bright in the sky. Star A is twice as far away from Earth as star B. How do the luminosities of stars A and B compare?
Answer:
The answer to the question is
The luminosity of stars A is four times that of star B
Explanation:
Flux (F) produced by a source of light is directly proportional to the brightness or Luminosity (L), and varies inversely to its distance d, that is [tex]F \alpha \frac{L}{d^2}[/tex]
Therefore if the two stars present the same flux then we have
[tex]\frac{L_1}{d_1^2} = \frac{L_2}{(2d_1)^2}[/tex] then crossing out like terms gives [tex]\frac{L_1}{1} = \frac{L_2}{2^2}[/tex] or 4·L₁ = L₂
The luminosity of star A is 4 times the that of star B
Final answer:
Star A, being twice as far away from Earth as Star B but appearing equally bright, must have a luminosity that is four times greater than Star B's luminosity, due to the inverse square law of light.
Explanation:
When comparing the luminosities of two stars that appear equally bright from Earth, but one is twice as far away as the other, we must account for the inverse square law of light. This law dictates that the intensity (brightness) of light from a source (in this case, a star) decreases proportionally to the square of the distance from the source.
Therefore, if Star A is twice as far away from Earth as Star B, yet they appear to have the same brightness, Star A must have a luminosity four times greater than that of Star B. This is because, to compensate for the increased distance, Star A must emit more light to be perceived as equally bright as Star B from Earth.
(a) Calculate the magnitude of the gravitational force exerted by the Moon on a 75 kg human standing on the surface of the Moon. (The mass of the Moon is 7.41022 kg and its radius is 1.7106 m.)
Answer:
128 N
Explanation:
Using
F = Gm'm/r²....................... Equation 1
Where F = Force, G = Universal constant, m = mass of the human, m' = mass of the moon, r = radius of the moon
Given: m = 75 kg, m' = 7.4×10²² kg, r = 1.7×10⁶ m
Constant: G = 6.67×10⁻¹¹ Nm²/kg²
Substitute into equation 1
F = (6.67×10⁻¹¹ )(75)(7.4×10²²)/(1.7×10⁶)²
F = (3.7×10¹⁴)/(2.89×10¹²)
F = 1.28×10²
F = 128 N
Explanation:
Below is an attachment containing the solution.
Two points are located on a rigid wheel that is rotating with decreasing angular velocity about a fixed axis. Point A is located in the rim of the wheel and pint B is halfway between the rim and the axis. Which one of the following statements concerning this situation is true?
1. The angular velocity at point A is greater than that of point B
2. Both points have the same centripetal acceleration
3. Both points have the same tangential acceleration4. Both points have the same instantaneous angular velocity
5. Each second, point A turns through a greater angle than point B
Answer:
4. Both points have the same instantaneous angular velocity
Explanation:
Angular velocity is a measure of the the number of rotations per unit time. This does not depend on the radius of the wheel. Hence, all points on the wheel have the same angular velocity. This invalidates option 1.
The centripetal acceleration is given by the product to the square of the angular velocity and the radius or distance from the centre. A and B are located at different distances from the centre. Hence, they have different centripetal acceleration. This invalidates option 2.
The tangential acceleration depends on the linear velocity which itself is a product of the angular velocity and the distance from the centre. Hence, it is different for both points because they are at different distances from the centre.
Since both A and B are fixed points on the wheel, they move through equal angles in the same time. In fact, for any other fixed point, they all move through the same angle in the same time. This invalidates option 5.
Final answer:
All points on a rotating wheel share the same instantaneous angular velocity; however, points farther from the axis will experience greater centripetal acceleration. The correct statement is that both points have the same instantaneous angular velocity.
Explanation:
The question concerns the properties of points located at different radii of a rotating wheel, specifically relating to angular velocity, centripetal acceleration, and tangential acceleration. We can address the situation by considering the principles of circular motion. When a rigid wheel is rotating about a fixed axis, all points on the wheel have the same instantaneous angular velocity since every point on the wheel rotates through the same angle in the same amount of time.
Centripetal acceleration is proportional to the radius and the square of the angular velocity. Since point A, being at the rim, is farther from the axis than point B, point A experiences a greater centripetal acceleration. On the other hand, tangential acceleration is related to the angular acceleration and the radius. If the wheel is rotating with decreasing angular velocity, both point A and B experience the same tangential acceleration because it is a property of the wheel's rotation, not the points' individual locations.
The correct statement in this scenario is that both points have the same instantaneous angular velocity, which makes option 4 the true statement.
A spectrophotometer measures the transmittance or the absorbance. True or False
Answer: FALSE
Explanation: Could you help me with a question?
Which is true about the spacing of the streamlines in a wire?
The statement " Close spacing represents greater current densities" is true about the spacing of the streamlines in a wire (option F)
Why is this correct?
In a scenario where a conductor is wider on the left and narrower on the right, the electric field lines and current density are depicted by streamlines. With current flowing from the wider end to the narrower end, the total charge and current remain constant. However, the current density fluctuates, being higher at the narrower end.
Hence, when observing streamlines, closer spacing within the wire signifies a higher current density, specifically in the narrower sections compared to the wider ones.
Complete question:
Which is true about the spacing of the streamlines in a wire?
A. Wide spacing represents faster random-motion velocities.
B. Wide spacing represents greater electric field vectors.
C. Wide spacing represents greater current densities.
D. Close spacing represents faster random-motion velocities.
E. Close spacing represents greater electric field vectors.
F. Close spacing represents greater current densities.
A box weighing 460 N is pushed along a horizontal floor at constant velocity by a force of 270 N parallel to the floor. What is the coefficient of kinetic friction between the box and the floor
Answer:
μ= F÷N
μ= 270/460= 0.587
Explanation:
The friction force always acts in the opposite direction of the intended or actual motion.
How many times did thomas edison fail before inventing the lightbulb
Answer:
he failed thousands of times
Explanation:
There is no known number for his failings. Edison may have failed in many of his experiments and in his schooling, but he had something better working in his favor. He had great determination and persistence.
He failed thousands of times in an attempt to develop an electric light, the great Edison simply viewed each unsuccessful experiment as the elimination of a solution that wouldn’t work, thereby moving him that much closer to a successful solution.