Answer:
Explanation:
Given
Cannon is fired with a velocity of [tex]u=72.50\ m/s[/tex]
Using Equation of motion
[tex]y=ut+\frac{1}{2}at^2[/tex]
where
[tex]y=displacement[/tex]
[tex]u=initial\ velocity[/tex]
[tex]a=acceleration[/tex]
[tex]t=time[/tex]
after time [tex]t=3.3 s[/tex]
[tex]y=72.50\times 3.3-\frac{1}{2}\times 9.8\times (3.3)^2[/tex]
[tex]y=239.25-53.36[/tex]
[tex]y=185.89\ m[/tex]
So after 3.3 s cannon ball is at a height of 185.89 m
Air temperature in a desert can reach 58.0°C (about 136°F). What is the speed of sound (in m/s) in air at that temperature?
The speed of sound in air increases with temperature. By taking the difference between the given temperature and 0°C, then multiplying by the rate of speed increase per degree Celsius, you can find the speed of sound. This comes out to approximately 365.8 m/s at 58.0°C.
Explanation:The speed of sound in air varies depending on the temperature of the air. In general, the speed of sound increases by approximately 0.6 m/s for each degree Celsius increase in temperature. Therefore, we need to find the difference between the given temperature (58.0°C) and 0°C, which is 58, and then multiply that by 0.6 to find the increase in speed due to temperature.
Step 1: Find the temperature difference = 58.0°C - 0°C = 58°C
Step 2: Multiply the temperature difference by the rate of speed increase, which is 0.6 m/s/°C. We get: (58°C) x (0.6 m/s/°C) = 34.8 m/s.
Step 3: To find the speed of sound at the higher temperature, add this increase to the speed of sound at 0°C, which is 331 m/s.
Step 4: So, the speed of sound in air at 58.0°C is 331 m/s + 34.8 m/s = 365.8 m/s (approximately).
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In a circuit with parallelresistors, the smaller resistance dominates; in a circuit with resistors inseries, the larger one dominates. Make some speci c examples of resistorsin parallel and in series to explain this rule of thumb.
Answer:
In a circuit with parallelresistors, the smaller resistance dominates; in a circuit with resistors inseries, the larger one dominates. Make some specific examples of resistorsin parallel and in series to explain this rule of thumb.
Explanation:
Lets start with Resistors in series:
Suppose a 12v battery is connected in a circuit with 3 resistors in series.
Voltage = 12 V
R1 = 1 ohm
R2 = 6 ohm
R3 = 13 ohm
Total resistance is simply sum of all resistors connected in series as following:
Rs = R1 + R2 + R3
Rs = 1 + 6 + 13
Rs = 20 ohm
Here Rs = Total Resistance in series circuit
Now total current can be found by using ohm's law V = IR
I = V/R
Here R = R total
I = 12/20 = 0.6 Ampere
In series circuit current remains same.
so I1 = I2 = I3 = I
Now we can find Power dissipation across every resistor to show the dominant Resistor as:
P1 = I² R1 = (0.6 A)² ( 1 ohm) = 0.36 watt. (a)
P2 = I² R2 = (0.6 A)² ( 6 ohm) = 2.16 watt. (b)
P3 = I² R3 = (0.6 A)² ( 13 ohm) = 4.68 watt. (c)
From equation a, b and c we can conclude that more power is dissipated across R3 which is the Larger than both other resistors R1 and R2.
Now lets take the case of Parallel Circuit with same Values of Voltage and resistors But now in parallel.
In parallel circuit Total resistance is calculated as following:
1/Rp = 1/R1 + 1/R2 + 1/R3
1/Rp = 1/1 + 1/6 + 1/13 = 1 + 0.1667 + 0.0769
1/Rp = 1.2436
Rp = 0.8041 ohm
Now total current is as following:
I = V/R
I = 12/0.8041 = 14.92 A
Now we calculate individual currents :
In parallel circuit Voltage remains same.
I1 = V/R1
I1 = 12/1 = 12 A
Similarly,
I2 = V/R2
I2 = 12/ 6 = 2 A
and
I3 = V/R3
I3 = 12/13 = 0.92 A
Now Power dissipation can be calculated as :
P1 = I1² * R1
P1 = 12² * 1 = 144 * 1
P1 = 144 Watt (1)
Similarly,
P2 = I2² * R2
P2 = 2² * 6 = 4 * 6
P2 = 24 Watt (2)
And
P3 = I3² * R3
P3 = (0.92)² * 13 = 0.8464 * 13
P3 = 11 Watt. (3)
Now from Equation 1, 2 and 3 we can conclude that More power is dissipated across Lower resistor .
Hence in a circuit with parallel resistors, the smaller resistance dominates; in a circuit with resistors in series, the larger one dominates.
Two stars that are 109 km apart are viewed by a telescope and found to be separated by an angle of 10-5 radians. The eyepiece of the telescope has a focal length of 1.5 cm and the objective has a focal length of 3 meters. How far away are the stars from the observer? Give your answer in kilometers.
Answer:
x = 2 x 10¹⁶ Km
Explanation:
distance between two star,d = 10⁹ Km
separation between them, θ = 10⁻⁵ radians
focal length of the eyepiece = 1.5 cm = 0.015 m
focal length of the objective = 3 m
observer distance from star, x = ?
we know,
[tex]tan \theta = \dfrac{d}{x}[/tex]
for small angle
[tex]\theta = \dfrac{d}{x}[/tex].......(1)
angular magnification of telescope
[tex]M = \dfrac{f_{objective}}{f_{eyepiece}}=\dfrac{3}{0.015} = 200[/tex]
Angular magnification of the telescope is also calculated by
[tex]M = \dfrac{observed\ angle}{original\ angle}[/tex]
[tex]M = \dfrac{\theta_0}{\theta}[/tex]
now,
[tex]\dfrac{\theta_0}{\theta}=200[/tex]
[tex]\dfrac{\theta_0}{200}=\theta[/tex]
from equation (1)
[tex]\dfrac{\theta_0}{200}=\dfrac{d}{x}[/tex]
[tex]x=\dfrac{200d}{\theta_0}[/tex]
[tex]x=\dfrac{200\times 10^9}{10^{-5}}[/tex]
x = 2 x 10¹⁶ Km
Distance between the observer and the star is x = 2 x 10¹⁶ Km
The number of confirmed exoplanets is (a) less than 10; (b) roughly 50; (c) more than 500; (d) more than 5000.
Answer:
(c) more than 500
Explanation:
Until 2019, more than 3000 planetary systems have been discovered that contain more than 4000 exoplanets, since some of these systems contain multiple planets. Most known extrasolar planets are gas giants equal to or more massive than the planet Jupiter, with orbits very close to its star.
Final answer:
The number of confirmed exoplanets is more than 5000. Through missions like Kepler, the catalog of exoplanets includes systems with a variety of planet arrangements and types, indicating an incredibly diverse universe with potentially billions of Earth-size planets. The correct answer is option is d) more than 5000.
Explanation:
The number of confirmed exoplanets is (d) more than 5000. As of July 2015, NASA's Kepler mission had detected a total of 4,696 possible exoplanets, and 1,030 of those candidates had been confirmed as planets. Advancing to 2018, astronomers had data on nearly 3,000 exoplanet systems.
By 2022, this knowledge expanded to include data on over 800 systems, many with multiple planets, and an understanding that one quarter of stars may have exoplanet systems. This implies the existence of at least 50 billion planets in our Galaxy alone. Recent studies have shown that planets like Earth are the most common type of planet, leading to an estimated 100 billion Earth-size planets around Sun-like stars in the Galaxy.
It's important to note that detections have been made possible using the Doppler and transit techniques, and while most of the exoplanets found are more massive than Earth, the shortage of small rocky planets detected is an observational bias, as they are more difficult to detect.
The ensemble of exoplanets discovered is incredibly diverse, suggesting a variety of planet formations and arrangements in these systems. Some systems have been observed to have rocky planets closer to their stars than in our solar system, and others have large gas giants, known as 'hot Jupiters', very close to their stars.
An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance between right and left wheels is 1.5 m. What are the rotating speeds of each driving wheel as fractions of the drive shaft speed?
Explanation:
The given data is as follows.
Inner wheel Radius = 20 m,
Distance between left and right wheel = 1.5m,
Let us assume speed of drive shaft is N rpm.
Formula to calculate angular velocity is as follows.
Angular velocity of automobile = w = [tex]\frac{V}{R}[/tex]
where, V = linear velocity of automobile m/min,
R = turning radius from automobile center in meter
In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.
Now, we assume that
u = linear velocity of inner wheel
and, u' = linear velocity of outer wheel.
Formula for angular velocity of inner wheel w = ,
Formula for angular velocity of outer wheel w =
Now, for inner wheels
w =
= [tex]\frac{u}{(R - d)}[/tex]
u = [tex]V \times \frac{(R - d)}{R}[/tex]
= [tex]V \times (1 - \frac{d}{R})[/tex]
If radius of wheel is r it will cover distance in one min.
Since, velocity of wheel is u it will cover distance u in unit time(min)
Thus, u = [tex]2\pi rn[/tex] = [tex]V \times (1 - \frac{d}{R})[/tex]
Now, rotation per minute of inner wheel is calculated as follows.
n = [tex]\frac{V}{2 \pi r \times (1 - \frac{d}{R})}[/tex]
= [tex]\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}[/tex] (since 2d = 1.5m given, d = 0.75m),
= [tex]\frac{V}{r} \times 0.1532[/tex]
So, rotation per minute of outer wheel; n' =
= [tex]\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}[/tex]
= [tex]\frac{V}{r} \times 0.1651[/tex]
In a sharp left turn, the car's right wheel rotates 1.0375 times the speed of the drive shaft, while the left wheel rotates at the speed of the drive shaft. This difference is due to the right wheel covering a larger distance than the left wheel.
The principle in operation here is that in a sharp turn, the outer wheel (in this case, the right wheel) has to cover a larger distance than the inner wheel (left wheel). Therefore, the right wheel will rotate faster than the left one.
Now, calculating the difference in rotation speeds of the wheels, we consider the radii of the paths of the two wheels. For the left wheel, the radius is 20 m and for the right wheel, the radius is greater by half the wheelbase distance, that is, 20.75 m.
The ratio of the speeds is equal to the ratio of the radii of the two paths. Therefore, the speed of the right wheel as a fraction of the drive shaft speed is 20.75/20 = 1.0375 and the speed of the left wheel as a fraction of the drive shaft speed is 20/20 = 1.
So, in conclusion, during a sharp turn to the left, the right wheel rotates 1.0375 times the speed of the drive shaft, while the left wheel rotates at the same speed as the drive shaft.
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A ship sets sail from Rotterdam, The Netherlands, heading due north at 7.00 m/s relative to the water. The local ocean current is 1.53 m/s in a direction 40° north of east. What is the velocity of the ship in meters per second relative to the Earth in degrees north of east?
Answer:
8.07 m/s, 81.7º NE.
Explanation:
The ship, due to the local ocean current, will be deviated from its original due north bearing.In order to find the magnitude of the velocity of the ship, we need to convert a vector equation, in an algebraic one.If we choose two axes coincident with the N-S and W-E directions, we can find the components of the velocity along these directions.Clearly, the velocity of the ship, relative to water, is only due north, so it has no component along the W-E axis.The local ocean current, as it is directed at an angle between both axes, has components along these axes.These components can be found from the projections of the velocity vector along these axes, as follows:[tex]vocx = voc* cos 40 = 1.53 m/s * 0.766 = 1.17 m/s\\vocy = voc* sin 40 = 1.53 m/s * 0.643 = 0.98 m/s[/tex]
The component along the N-S axis (y-axis) of the velocity of the ship will be the sum of the velocity relative to water, plus the component of the ocean current along this same axis:[tex]vshy = vsw + vwy = 7.00 m/s + 0.98 m/s = 7.98 m/s[/tex]
The component along the W-E axis, is just the component of the local ocean current in this direction:vshx = 1.17 m/s
We can find the magnitude of the velocity vector, applying the Pythagorean theorem, as follows:[tex]v = \sqrt{vshx^{2} + vshy^{2} } =\sqrt{(7.98m/s)^{2} +(1.17m/s)^{2} } =8.07 m/s[/tex]
The direction of the vector relative to the W-E axis (measured in counterclockwise direction) is given by the relative magnitude of the x and y components, as follows:[tex]tg \theta = \frac{vshy}{vshx} = \frac{7.98}{1.17} = 6.82 \\ \theta = tg^{-1} (6.82)\\ \theta= 81.7\deg[/tex]
The velocity of the ship, relative to Earth, is 8.07 m/s, 81.7º North of East.How much heat must be absorbed by 125 g of ethanol to change its temperature from 21.5 oC to 34.8 oC?
Q: How much heat must be absorbed by 125 g of ethanol to change its temperature from 21.5 oC to 34.8 oC? The specific heat of ethanol is 2.44 J/(gC).
Answer:
4056.5 J
Explanation:
The formula for the specific heat capacity of ethanol is given as
Q = cm(t₂-t₁)..................... Equation 1
Where q = quantity of heat, c = specific heat capacity of ethanol, m = mass of ethanol, t₁ = initial temperature of ethanol, t₂ = final temperature of ethanol.
Given: m = 125 g, t₁ = 25.5 °C, t₂ = 34.8 °C
Constant; c = 2.44 J/g.°C
Substitute into equation 1
Q = 125(2.44)(34.8-21.5)
Q = 125(2.44)(13.3)
Q = 4056.5 J.
Hence the amount of heat absorbed = 4056.5 J
Give the name and symbol of the prefixes used with SI units to indicate multiplication by the following exact quantities. (a) 103 (b) 10−2 (c) 0.1 (d) 10−3 (e) 1,000,000 (f) 0.000001
Answer:
Please see below as the answer is self - explanatory
Explanation:
a) 10³ = kilo (kilogram = 10³ grams, kilometer= 10³ meters)
Symbol : k.
b) 10⁻² = centi (it is the 100th part of a unit, like centimeter, or centigram) Symbol: c.
c) .1 = deci (it is the tenth part (deci comes from the word that means "ten"in latin) of a unit: decimeter, decigram)
Symbol: d.
d) 10⁻3 = mili (it is a 1000th part of a unit: milimeter, miligram), the name comes from the word used in latin to mean "one thousand".
Symbol: m
e) 1,000,000 = 10⁶ = mega (megawatt)
Symbol: M
f= 0.000001 = 10⁻6 = micro (micrometer, microsecond),
Symbol: μ.
If a jumping frog can give itself the same initial speed regardless of the direction in which it jumps (forward or straight up), how is the maximum vertical height to which it can jump related to its maximum horizontal range Rmax = v20/g?
Answer:
Explanation:
If u is the initial velocity at an angle \theta with horizontal then
Horizontal range of Frog can be given by
[tex]R=ut+\frac{1}{2}at^2[/tex]
where u=initial velocity
a=acceleration
t=time
Here initial horizontal velocity
[tex]u_x=u\cos \theta [/tex]
and there is no acceleration in the horizontal motion
Therefore
[tex]R=u\cos \theta \times t+0[/tex]
Considering vertical motion
[tex]Y=ut+\frac{1}{2}at^2[/tex]
here Initial vertical velocity [tex]u_y=u\sin \theta [/tex]
acceleration [tex]a=g[/tex]
for complete motion Y=0 i.e.displacement is zero
[tex]0=u\sin \theta \times t-\frac{1}{2}gt^2[/tex]
[tex]t=\frac{2u\sin \theta }{g}[/tex]
Therefore Range is
[tex]R=\frac{u^2\sin 2\theta }{g}[/tex]
Range will be maximum when [tex]\theta =45[/tex]
[tex]R=\frac{u^2}{g}----1[/tex]
and Maximum height [tex]h_{max}=\frac{u^2\sin ^2 \theta }{2g}[/tex]
for [tex]\theta =45[/tex]
[tex]h_{max}=\frac{u^2}{4g}----2[/tex]
Divide 1 and 2
[tex]\frac{R_{max}}{h_{max}}=\frac{\frac{u^2}{g}}{\frac{u^2}{4g}}[/tex]
[tex]\frac{R_{max}}{h_{max}}=4[/tex]
Say for a particular population body temperature measured in Celsius has a mean of 37 and the standard deviation is equal to 0.5. What is the mean and standard deviation of the population when temperature is measured in Fahrenheit units?
Answer
given,
Mean of temperature in Celsius = 37 Standard deviation in Celsius = 0.5
relation between Fahrenheit and Celsius
[tex]F = \dfrac{9}{5}C + 32[/tex]
Mean of temperature in Fahrenheit
[tex]Mean_F =\dfrac{9}{5}\times Mean_C + 32[/tex]
[tex]Mean_F =\dfrac{9}{5}\times 37 + 32[/tex]
[tex]Mean_F = 98.6\ F[/tex]
Standard deviation of Fahrenheit.
[tex]SD_F = \dfrac{9}{5}\times SD_C [/tex]
addition of 32 will not change the Standard deviation.
[tex]SD_F = \dfrac{9}{5}\times 0.5[/tex]
[tex]SD_F = 0.9[/tex]
Two timpani (tunable drums) are played at the same time. One is correctly tuned so that when it is struck, sound is produced that has a wavelength of 2.20 m. The second produces sound with a wavelength of 2.10 m. If the speed of sound is 343 m/s, what beat frequency is heard?
Answer:
Explanation:
Given
Wavelength of first timpani [tex]\lambda _1=2.2\ m[/tex]
Frequency corresponding to this Wavelength
[tex]f_1=\frac{v}{\lambda _1}[/tex]
where [tex]v=velocity\ of\ sound (343 m/s)[/tex]
[tex]f_1=\frac{343}{2.2}[/tex]
[tex]f_1=155.9\ Hz[/tex]
Wavelength of Second timpani [tex]\lambda _2=2.1\ m[/tex]
Frequency corresponding to this Wavelength
[tex]f_2=\frac{v}{\lambda _2}[/tex]
[tex]f_2=\frac{343}{2.1}=163.33\ Hz[/tex]
Beat frequency [tex]=f_2-f_1[/tex]
Beat frequency [tex]=163.33-155.9=7.43[/tex]
so approximately 7 beats per second
Final answer:
The beat frequency heard when two timpani are played together, with wavelengths of 2.20 m and 2.10 m and the speed of sound being 343 m/s, is 7.42 Hz.
Explanation:
When two timpani (tunable drums) are played at the same time with wavelengths of 2.20 m and 2.10 m, respectively, and the speed of sound is 343 m/s, we can calculate the frequencies of these sounds and subsequently determine the beat frequency heard.
The frequency of a wave is given by the formula f = v / λ (where f is the frequency, v is the speed of sound, and λ is the wavelength).
For the first timpani, the frequency is 343 m/s / 2.20 m = 155.91 Hz; for the second, it is 343 m/s / 2.10 m = 163.33 Hz.
The beat frequency, which is the absolute difference between these two frequencies, will be -
= 163.33 Hz - 155.91 Hz
= 7.42 Hz.
If the humidity in a room of volume 410 m3 at 25 ∘C is 70%, what mass of water can still evaporate from an open pan? Express your answer using two significant figures.
Answer:
[tex]m=1864.68\ g[/tex]
Explanation:
Given:
volume of air in the room, [tex]V=410\ m^3[/tex]
temperature of the room, [tex]T=25+273=298\ K[/tex]
Saturation water vapor pressure at any temperature T K is given as:
[tex]p_{sw}=\frac{e^{(77.3450 + 0.0057\times T - \frac{ 7235}{T} )}}{T^{8.2}}[/tex]
putting T=298 K we have
[tex]p_{sw}=3130\ Pa[/tex]
The no. of moles of water molecules that this volume of air can hold is:
Using Ideal gas law,
[tex]P.V=n.R.T[/tex]
[tex]n=\frac{P_{sw}.V}{R.T}[/tex]
[tex]n=\frac{3130\times 410}{8.314\times 298}[/tex]
[tex]n=518\ moles[/tex] is the maximum capacity of the given volume of air to hold the moisture.
Currently we have 80% of n, so the mass of 20% of n:
[tex]m=(20\%\ of\ n)\times M}[/tex]
where;
M= molecular mass of water
[tex]m=0.2\times 518\times 18[/tex]
[tex]m=1864.68\ g[/tex] is the mass of water that can vaporize further.
How much energy is required to raise the temperature of 11.4 grams of gaseous helium from 24.2 °C to 38.3 °C ?
Answer:
Q= 835J
Explanation:
Specific heat capacity of Helium=5.193J/gk
Q=msΔT
ΔT=T2-T1
Q=change in energy
m=mass of substance
S= Specific heat capacity
Q= Change in energy
T2= 38.3 degrees Celsius =(38.3+273)k= 311.3k
T1=24.2 degree Celsius =24.2+273k
T1=297.2k
ΔT=T2-T1=(311.3-297.2)k=14.1K
Q=(11.4g)*(5.193J/gk)*(14.1k)
Q=834.72282J
Approximately, Q= 835J
After a great many contacts with the charged ball, how is the charge on the rod arranged (when the charged ball is far away)?
a. There is positive charge on end B and negative charge on end A.
b. There is negative charge spread evenly on both ends.
c. There is negative charge on end A with end B remaining neutral.
d. There is positive charge on end A with end B remaining neutral.
Answer: Option (b) is the correct answer.
Explanation:
Since, there is a negative charge present on the ball and a positive charge present on the rod. So, when the negatively charged metal ball will come in contact with the rod then positive charges from rod get conducted towards the metal ball.
Hence, the rod gets neutralized. But towards the metal ball there is a continuous supply of negative charges. Therefore, after the neutralization of positive charge from the rod there will be flow of negative charges from the metal ball towards the rod.
Thus, we can conclude that negative charge spread evenly on both ends.
Option (b) is the correct answer.
b. There is negative charge spread evenly on both ends.
The following information should be considered:
Since, there should be the negative charge present on the ball and a positive charge present on the rod. Due to this, at the time when the negatively charged metal ball will come in contact with the rod so positive charges from rod get conducted towards the metal ball. Therefore, the rod gets neutralized. However towards the metal ball there is a continuous supply of negative charges.Learn more: https://brainly.com/question/10024737?referrer=searchResults
Use your observations to determine qualitatively how the strength of the electric interaction between charged objects depends on the distance between them. Explain your reasoning (this is trickier than it might seem at first).
Answer:
Explanation:
The experiment confirm the inverse square law which relates the force between two charged particles to the product of the charges and the distance between the charges.
From the general equation, we notice the force is inversely related to the distance between the charges, when the distance is halved, the force increase by a factor of 4, hence a decrease in distance leads to a corresponding increase in the force value.
Also the electric field intensity a charge exert on another charge within the region of its field is independent on the distance because distance has no effect on electric field strength.
Collapse question part Part 4 (d) What is the unit vector in the direction of the spacecraft's velocity? (Express your answer in vector form.)
Answer:
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
Explanation:
Given:
v = (-23.2, -104.4, 46.4) m/s
Above expression describes spacecraft's velocity vector v.
Find:
Find unit vector in the direction of spacecraft velocity v.
Solution:
Step 1: Compute magnitude of velocity vector.
mag (v) = sqrt ( 23.2^2 + 104.4^2 + 46.4^2)
mag (v) = 116.58 m/s
Step 2: Compute unit vector unit (v)
unit (v) = vec (v) / mag (v)
unit (v) = [ -23.2 i -104.4 j + 46.4 k ] / 116.58
unit (v) = [ -0.199 i - 0.8955 j + 0.39801 k ]
The unit vector in the direction of the spacecraft’s velocity is found by dividing the velocity vector by its magnitude. Accordingly, if one knows the component values of the spacecraft's velocity, one can calculate the components for the unit vector by dividing each component by the magnitude of the velocity.
Explanation:The unit vector in the direction of the spacecraft's velocity can be determined from its velocity vector. The unit vector is a vector of length 1 that points in the same direction as the given vector. It can be found by dividing the velocity vector of the spacecraft by its magnitude. If V represents the velocity vector, then the unit vector U is calculated as U = V / |V| where |V| is the magnitude of the vector V. Thus, if you know the component values of the spacecraft's velocity, you can calculate the respective components for the unit vector by dividing each component of the velocity vector by the velocity's magnitude.
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A rocket carrying a satellite is accelerating straight up from the earth’s surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.
Answer:
197.263157895 m/s
169.491525424 m/s
Explanation:
x Denotes position
t Denotes time
Average velocity is given by
[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s[/tex]
The average velocity is 197.263157895 m/s
[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s[/tex]
The average velocity is 169.491525424 m/s
The magnitude of the average velocity of the rocket during the 4.75-second part of the flight is 197.3 m/s, while for the first 5.90 seconds of the flight, the average velocity is 169.5 m/s.
Explanation:To find the magnitude of the average velocity of the rocket, we use the formula average velocity = displacement / time.
(a) The displacement during the 4.75-second part of the flight is 1.00 km - 63 m = 937 m (we converted km to m to keep units consistent). Hence, the average velocity in this part of the flight is 937 m / 4.75 s = 197.3 m/s.
(b) For the first 5.90 seconds of flight, the displacement is 1.00 km = 1000 m (the height above the ground) while time is 5.90 s. Therefore, for this duration, the average velocity is 1000 m / 5.90 s = 169.5 m/s.
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A parallel-plate capacitor is made of two conducting plates of area A separated by a distance d. The capacitor carries a charge Q and is initially connected to a battery that maintains a constant potential difference between the plates. The battery is then disconnected from the plates and the separation between the plates is doubled. 1) Which of the following remains constant? Oa. Voltage across the capacitor b. Capacitance of the capacitor c. Charge on the capacitor
Answer:
C. Charge on the capacitor
Explanation:
Read further: Capacitors consist of two parallel conductive
plates (usually a metal) which are prevented
from touching each other (separated) by an
insulating material called the “dielectric”. When
a voltage is applied to these plates an
electrical current flows charging up one plate
with a positive charge with respect to the
supply voltage and the other plate with an
equal and opposite negative charge.
Then, a capacitor has the ability of being able
to store an electrical charge Q (units in
Coulombs ) of electrons. When a capacitor is
fully charged there is a potential difference,
p.d. between its plates, and the larger the area
of the plates and/or the smaller the distance
between them (known as separation) the
greater will be the charge that the capacitor
can hold and the greater will be its
Capacitance.
The capacitors ability to store this electrical
charge ( Q ) between its plates is proportional
to the applied voltage, V for a capacitor of
known capacitance in Farads.
The correct option is c which is charge on the capacitance. When a parallel-plate capacitor is disconnected from the battery, and the separation between its plates is doubled, the charge on the capacitor remains constant. Changes in plate separation affect capacitance and voltage, but not the existing charge on the plates.
The question asks which of the following remains constant when the battery is disconnected from a parallel-plate capacitor and the separation between the plates is doubled: voltage across the capacitor, capacitance of the capacitor, or the charge on the capacitor. The key to answering this question lies in understanding how capacitors work and the relationship between charge (Q), capacitance (C), and voltage (V).
Capacitance is given by C = εA/d, where A is the area of the plates, d is the separation between the plates, and ε is the permittivity of the material between the plates. When the battery is disconnected, the external voltage source is removed, but the charge on the plates does not have a path to dissipate. Therefore, the charge on the capacitor remains constant, even when the plate separation is changed. Doubling the separation would affect the capacitance and the voltage across the capacitor but not the charge.
A man drops a baseball from the top of a building. If the ball is held at a height of 1m before it is dropped, and takes 6.8 seconds to hit the ground, how high is the building in meters? (Neglect air resistance)
To solve this problem we will apply the linear motion kinematic equations.
The equation that describes the position as a function of the initial velocity, acceleration and time is given by the relation
[tex]s = v_0 t +\frac{1}{2} at^2[/tex]
Here,
[tex]v_0 =[/tex] Initial velocity
t = Time
a = Acceleration, at this case due to gravity
There is not initial velocity then we have that the equation to the given time is
[tex]s = \frac{1}{2} (9.8)(6.8)^2[/tex]
[tex]s = 226.8m[/tex]
If the ball is held at a height of 1m before it is dropped, we have that the Building height is
[tex]h = 226.8-1[/tex]
[tex]h = 225.8m[/tex]
Two identical loudspeakers are driven in phase by the same amplifier. The speakers are positioned a distance of 3.2 m apart. A person stands 4.1 m away from one speaker and 4.8 m away from the other. Calculate the second lowest frequency that results in destructive interference at the point where the person is standing. Assume the speed of sound to be 343 m/s.
Answer:
f = 735 Hz
Explanation:
given,
Person distance from speakers
r₁ = 4.1 m r₂ = 4.8 m
Path difference
d = r₂ - r₁ = 4.8 - 4.1 = 0.7 m
For destructive interference
[tex]d = \dfrac{n\lambda}{2}[/tex]
where, n = 1, 3,5..
we know, λ = v/f
[tex]d = \dfrac{n v}{2f}[/tex]
v is the speed of the sound = 343 m/s
f is the frequency
[tex]f = \dfrac{n v}{2d}[/tex]
for n = 1
[tex]f = \dfrac{343}{2\times 0.7}[/tex]
f = 245 Hz
for n = 3
[tex]f = \dfrac{3\times 343}{2\times 0.7}[/tex]
f = 735 Hz
Hence,the second lowest frequency of the destructive interference is 735 Hz.
A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has built a 2.0 kg robot "mouse" that runs up and down the rope. What minimum magnitude of the acceleration should the robot have for the rope to fail? Express your answer with the appropriate units.
Answer:
6.8 m/s2
Explanation:
Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is
W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N
For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of
a = F/m = 13.6 / 2 = 6.8 m/s2
So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail
The robot must have a minimum acceleration of 16.6 m/s^2 for the rope to fail.
Explanation:To determine the minimum magnitude of acceleration the robot should have for the rope to fail, we need to consider the forces acting on the rope. The weight of the robot is 2.0 kg multiplied by the acceleration, which we need to find. The tension in the rope is equal to the weight of the rope plus the weight of the robot. Since the rope breaks if the tension exceeds 43 N, we can set up the equation:
Tension = Weight of rope + Weight of robot
43 N = (1.0 kg)(9.8 m/s^2) + (2.0 kg)(acceleration)
Solving for the acceleration, we get:
acceleration = (43 N - 9.8 N) / 2.0 kg = 16.6 m/s^2
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An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 2.00 mm. If a 23.0-V potential difference is applied to these plates, calculate the following.
(a) Calculate the electric field between the plates.
(b) Calculate the surface charge density.
(c) Calculate the capacitance.
(d) Calculate the charge on each plate.
Answer:
a. 11.5kv/m
b.102nC/m^2
c.3.363pF
d. 77.3pC
Explanation:
Data given
[tex]area=7.60cm^{2}\\ distance,d=2mm\\voltage,v=23v[/tex]
to calculate the electric field, we use the equation below
V=Ed
where v=voltage, d= distance and E=electric field.
Hence we have
[tex]E=v/d\\E=\frac{23}{2*10^{-3}} \\E=11.5*10^{3} v/m\\E=11.5Kv/m[/tex]
b.the expression for the charge density is expressed as
σ=ξE
where ξ is the permitivity of air with a value of 8.85*10^-12C^2/N.m^2
If we insert the values we have
[tex]8.85*10^{-12} *11500\\1.02*10^{-7}C/m^{2} \\102nC/m^{2}[/tex]
c.
from the expression for the capacitance
[tex]C=eA/d[/tex]
if we substitute values we arrive at
[tex]C=\frac{8.85*10^{-12}*7.6*10^{-4}}{2*10^{-3} } \\C=\frac{6.726*10^{-15} }{2*10^{-3} } \\C=3.363*10^{-12}F\\C=3.363pF[/tex]
d. To calculate the charge on each plate, we use the formula below
[tex]Q=CV\\Q=23*3.363*10^{-12}\\ Q=7.73*10^{-12}\\ Q=77.3pC[/tex]
For the air-filled capacitor:
(a) The electric field is 11.5 kV/m
(b) The surface charge density is 1.02×10⁻⁷C/m²
(c) The capacitance is 3.36pF
(d) Charge on the plate: 76pC
Capacitor:Given that an air-filled capacitor consists of two parallel plates such that the:
Area of the plates, A = 7.6 cm² = 7.6×10⁻⁴ m²
distance between the plates, d = 2mm = 2×10⁻³m
voltage applied, V = 23V
(a) Electric field is given by:
E = V\d
E = 23/2×10⁻³
E = 11.5 kV/m
(b) The electric field for a parallel plate capacitor in terms of the surface charge density is given by:
E = σ/ε₀
where σ is the surface charge density:
σ = ε₀E
σ = 8.85×10⁻¹²×11.5×10³
σ = 1.02×10⁻⁷C/m²
(c) the capacitance is given by:
C = ε₀A/d
C = (8.85×10⁻¹²)×(7.6×10⁻⁴)/2×10⁻³
C = 3.36×10⁻¹²F
C = 3.36 pF
(d) The charge (Q) on the plate is given by:
Q = σA
Q = 1.02×10⁻⁷× 7.6×10⁻⁴
Q = 76 pC
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On a dry road, a car with good tires may be able to brake withconstant deceleration of 4.92 m/s^2. (a) how long does such a car,initially traveling at 24.6 m/s, take to stop?(b) How far does ittravel in this time? (c) Graph x vs t and v vs t for thedeceleration.
Answer:
a) [tex]t=5\ s[/tex]
b) [tex]s=61.5\ m[/tex]
Explanation:
Given:
acceleration of the car, [tex]a=-4.92\ m.s^{-1}[/tex]
initial velocity of the car, [tex]u=24.6\ m.s^{-1}[/tex]
final velocity of the car, [tex]v=0\ m.s^{-1}[/tex]
a)
Using eq. of motion:
[tex]v=u+a.t[/tex]
[tex]0=24.6-4.92\times t[/tex]
[tex]t=5\ s[/tex]
b)
Distance travelled before stopping:
[tex]s=u.t+\frac{1}{2} a.t^2[/tex]
[tex]s=24.6\times 5-0.5\times 4.92\times 5^2[/tex]
[tex]s=61.5\ m[/tex]
c)
The car takes deceleration in 5 seconds to stop and travels a distance of 61.5 meters.
a) To find the time it takes for the car to stop, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the deceleration, and t is the time elapsed. Rearranging the equation to solve for t, we have t = (v - u) / a. Substituting the given values, we get t = (0 - 24.6) / -4.92 = 5 seconds.
b) To find the distance traveled during this time, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the deceleration, and t is the time elapsed. Substituting the given values, we have s = 24.6(5) + (1/2)(-4.92)(5)^2 = 61.5 meters.
c) The graph of x vs t would be a straight line with a negative slope, representing the car's distance decreasing over time. The graph of v vs t would also be a straight line with a negative slope, representing the car's velocity decreasing over time.
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A metallic sheet has a large number of slits, 5.0 mm wide and 16 cm apart, and is used as a diffraction grating for microwaves. A wide parallel beam of microwaves is incident normally on the sheet. What is the smallest microwave frequency for which only the central maximum occurs? (The speed of these EM waves is c = 3.00 × 10 8 m/s.)
Answer:
[tex]6\times 10^{10}\ Hz[/tex]
Explanation:
d = Slit gap = 5 mm
Slit distance = 16 cm
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]\lambda[/tex] = Wavelength
We have the relation
[tex]dsin\theta=\lambda[/tex]
Here, [tex]\theta=90[/tex]
So
[tex]d=\lambda\\\Rightarrow \lambda=5\ mm[/tex]
Frequency is given by
[tex]f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{5\times 10^{-3}}\\\Rightarrow f=6\times 10^{10}\ Hz[/tex]
The frequency is [tex]6\times 10^{10}\ Hz[/tex]
What is relative velocity? Suppose you want to design an airbag system that can protect the driver in a head on collision at a speed of 100 km/hr or 60 mph. Estimate how fast the airbag must inflate to effectively protect the driver? The car crumples within a distance of 1m.
Answer:
0.072 seconds
Explanation:
t = Time taken
u = Initial velocity = 100 km/h
v = Final velocity
s = Displacement = 1 m
a = Acceleration
[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(\dfrac{100}{3.6})^2}{2\times 1}\\\Rightarrow a=-385.802469136\ m/s^2[/tex]
[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-\dfrac{100}{3.6}}{-385.802469136}\\\Rightarrow t=0.072\ s[/tex]
The time taken is 0.072 seconds
A metal cylinder is measured to have a length of 5.0 cm and a diameter of 1.26 cm. Compute the volume of the cylinder. Write your final answer with the correct number of significant figures.
Answer:
V = 6.23 cm³
Explanation:
given,
Length of the cylinder, h = 5 cm
Diameter of the cylinder, d = 1.26 cm
r = 0.63 cm
Volume of the cylinder = ?
We know,
[tex]V = \pi r^2 h[/tex]
[tex]V = \pi \times 0.63^2\times 5[/tex]
V = 6.23 cm³
Volume of the cylinder is equal to V = 6.23 cm³
The volume of a metal cylinder with a length of 5.0 cm and a diameter of 1.26 cm (-approximately 4.9 cm^3- when calculated using the formula: Volume = π * (d/2)^2 * h.
Explanation:To calculate the volume of a cylinder, you use the formula: Volume = π * (d/2)^2 * h. Here, d represents the diameter of the cylinder, h represents the height, and π (pi) is a constant approximately equal to 3.14159. Plugging the values into the formula, we get:
Volume = π * (1.26 cm/2)^2 * 5.0 cm
After performing the above computations, the volume of the metal cylinder is approximately 4.9 cm^3 when rounded to two significant figures.
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The total electric flux from a cubical box 26.0 cm on a side is 1840 N m2/C. What charge is enclosed by the box?
Final answer:
The charge enclosed by the cubical box with a total electric flux of 1840 N·m2/C is calculated using Gauss's law and is found to be 16.29 nC.
Explanation:
The question deals with the concept of electric flux and its relation to the enclosed charge using Gauss's law, which is a fundamental principle in electromagnetism. According to Gauss's law, the total electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (ε0).
The formula for Gauss's law in integral form is Φ = Q / ε0, where Φ is the electric flux, Q is the charge enclosed, and ε0 is the electric constant (approximately 8.854 x 10-12 C2/N·m2). Given that the total electric flux from a cubical box is 1840 N·m2/C, and using the value of ε0, the enclosed charge (Q) can be calculated.
To find the charge, we rearrange the equation as Q = Φ·ε0 and substitute the given values to get Q = 1840 N·m2/C × 8.854 x 10-12 C2/N·m2, resulting in Q = 1.629 x 10-8 C or 16.29 nC (nanocoulombs).
The charge enclosed by the cubical box,is approximately 1.63 × 10⁻⁸ C .
To determine the charge enclosed by a cubical box, we can use Gauss's Law. According to Gauss's Law, the electric flux (Φ) through a closed surface is given by:
Φ = Q / ε₀
where:
Φ is the total electric flux (1840 N·m²/C)Q is the charge enclosed by the surfaceε₀ is the permittivity of free space (8.854 × 10⁻¹² C²/N·m²)Rearranging this formula to solve for Q, we get:
Q = Φ × ε₀
Substitute the given values:
Q = 1840 N·m²/C × 8.854 × 10⁻¹² C²/N·m²
Q ≈ 1.63 × 10⁻⁸ C
Therefore, the charge enclosed by the cubical box is approximately 1.63 × 10⁻⁸ C.
A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward with a force FS whose direction makes an angle of 30.0° with the ramp (Fig. E4.4). (a) How large a force FS is necessary for the component Fx parallel to the ramp to be 90.0 N?(b) How large will the component Fy perpendicular to the ramp be then?
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°).
Explanation:To find the force FS necessary for the component Fx parallel to the ramp to be 90.0 N, we can use the equation Fx = FS * cos(30°) = 90.0 N. Rearranging the equation, FS = 90.0 N / cos(30°). Therefore, FS ≈ 103.9 N.
To find the component Fy perpendicular to the ramp, we can use the equation Fy = FS * sin(30°). Substituting the value of FS, Fy ≈ 103.9 N * sin(30°). Therefore, Fy ≈ 51.9 N.
A common practice in cooking is the addition of salt to boiling water (Kb = 0.52 °C kg/mole). One of the reasons for this might be to raise the temperature of the boiling water. If 2.85 kg of water is boiling at 100 °C, how much NaCl (MW = 58.44 g/mole) would need to be added to the water to increase the boiling point by 2 °C? Must show your work.
Final answer:
To raise the boiling point of 2.85 kg of water by 2°C, one needs to add approximately 320.41 grams of sodium chloride (NaCl), calculated based on the boiling point elevation formula and considering NaCl's dissociation into ions.
Explanation:
To calculate how much NaCl is needed to increase the boiling point of 2.85 kg of water by 2°C, we use the boiling point elevation formula: ΔT = i*Kb*m, where ΔT is the change in boiling point, i is the van 't Hoff factor (which is 2 for NaCl because it dissociates into Na+ and Cl- ions), Kb is the ebullioscopic constant of water (0.52 °C kg/mole), and m is the molality of the solution. First, we solve for m knowing that we want to increase the boiling point by 2°C. With Kb = 0.52 °C kg/mole and i=2, we have 2°C = (2)*(0.52 °C kg/mol)*m. From here, m = 1.923 mol/kg.
Next, to find the mass of NaCl needed, we convert molality to moles of solute needed using the mass of the solvent (water) in kg, then multiply by the molar mass of NaCl. Since molality = moles of solute / kg of solvent, moles of NaCl = molality * kg of solvent = 1.923 mol/kg * 2.85 kg = 5.48 moles. The mass of NaCl required = moles * molar mass = 5.48 mol * 58.44 g/mol = 320.41 g.
Therefore, to increase the temperature of the boiling water by 2°C, we need to add approximately 320.41 grams of NaCl.
On a dry day, the temperature in Boulder (altitude: 5330’) is 40F. What is the temperature (in F) on nearby Bear Peak (altitude: 8460’)?
Answer: °C = 4.44°C
Explanation: The centigrade scale and Fahrenheit scale are related by the formulae below
9 *°C = 5(°F - 32)
Where °C = measurement of temperature in centigrade scale.
°F = measurement of temperature in Fahrenheit scale =40°F
By substituting the parameters, we have that
9 * °C = 5 (40 - 32)
9* °C = 5(8)
9 * °C = 40
°C = 40/9
°C = 4.44°C
Temperature decreases by approximately 3.5°F for every 1000 feet increase in altitude. Given the altitude difference of 3130 feet from Boulder to Bear Peak and Boulder's temperature of 40°F, we can calculate that the temperature at Bear Peak is expected to be approximately 29°F.
Explanation:The question asks for the temperature at a higher altitude given the temperature at a lower one. This involves understanding how temperature changes with altitude. It is stated that, on average, temperature decreases by about 3.5°F for every 1000 feet you climb in altitude, a phenomenon known as the lapse rate.
Given that, if the starting temperature in Boulder (altitude: 5330’) is 40F, and we need to find the temperature on Bear Peak (altitude: 8460’), we need to calculate the difference in altitude and the corresponding temperature decrease.
The altitude difference is 8460 - 5330 = 3130 feet. From this height difference, we can determine the corresponding temperature change by multiplying 3130 feet by 3.5°F per 1000 feet, yielding a temperature decrease of about 11°F.
To find the temperature at the higher altitude (Bear Peak), we then subtract this temperature change from the starting temperature. Thus, 40°F - 11°F gives 29°F as the expected temperature on Bear Peak.
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