Answer : The new or final volume of gas will be, 11.4 L
Explanation :
Charles' Law : It is defined as the volume of gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
[tex]V\propto T[/tex]
or,
[tex]\frac{V_1}{V_2}=\frac{T_1}{T_2}[/tex]
where,
[tex]V_1[/tex] = initial volume of gas = 12.1 L
[tex]V_2[/tex] = final volume of gas = ?
[tex]T_1[/tex] = initial temperature of gas = [tex]9^oC=273+9=282K[/tex]
[tex]T_2[/tex] = final temperature of gas = [tex]-7^oC=273+(-7)=266K[/tex]
Now put all the given values in the above formula, we get the final volume of the gas.
[tex]\frac{12.1L}{V_2}=\frac{282K}{266K}[/tex]
[tex]V_2=11.4L[/tex]
Therefore, the new or final volume of gas will be, 11.4 L
Final answer:
Using Charles's Law, the new volume of the helium gas in the balloon when taken from an inside temperature of 9°C to an outside temperature of -7°C, at constant pressure, is calculated to be 11.4 L.
Explanation:
To calculate the new volume of the helium gas in the balloon when it is taken outside to a colder temperature, we can use Charles's Law, which states that for a given mass of gas at constant pressure, the volume is directly proportional to its temperature in kelvins (V/T = k). We need to convert the temperatures from Celsius to Kelvin (K = °C + 273.15) and then apply Charles's Law (V1/T1 = V2/T2).
First, convert the temperature from Celsius to Kelvin:
Inside temperature: T1 = 9 °C + 273.15 = 282.15 K
Outside temperature: T2 = -7 °C + 273.15 = 266.15 K
Next, apply Charles's Law to find the new volume (V2):
V1/T1 = V2/T2
Plugging in the known values:
12.1 L / 282.15 K = V2 / 266.15 K
Solving for V2, we get:
V2 = (12.1 L × 266.15 K) / 282.15 K
V2 = 11.4 L (rounded to three significant digits)
Therefore, the new volume of the balloon when taken outside will be 11.4 L.
A janitor standing on the top floor of a building wishes to determine the depth of the elevator shaft. They drop a rock from rest and hear it hit bottom after 2.42 s. (a) How far (in m) is it from where they drop the rock to the bottom of the shaft? The speed of sound at the temperature of air in the shaft is 336 m/s. (Round your answer to at least three significant figures. Use g = 9.80 m/s2 as needed.) m (b) If the travel time for the sound is ignored, what percent error is introduced in the determination of depth of the shaft? %
Answer:
Part a)
H = 26.8 m
Part b)
error = 7.18 %
Explanation:
Part a)
As the stone is dropped from height H then time taken by it to hit the floor is given as
[tex]t_1 = \sqrt{\frac{2H}{g}}[/tex]
now the sound will come back to the observer in the time
[tex]t_2 = \frac{H}{v}[/tex]
so we will have
[tex]t_1 + t_2 = 2.42[/tex]
[tex]\sqrt{\frac{2H}{g}} + \frac{H}{v} = 2.42[/tex]
so we have
[tex]\sqrt{\frac{2H}{9.81}} + \frac{H}{336} = 2.42[/tex]
solve above equation for H
[tex]H = 26.8 m[/tex]
Part b)
If sound reflection part is ignored then in that case
[tex]H = \frac{1}{2}gt^2[/tex]
[tex]H = \frac{1}{2}(9.81)(2.42^2)[/tex]
[tex]H = 28.7 m[/tex]
so here percentage error in height calculation is given as
[tex]percentage = \frac{28.7 - 26.8}{26.8} \times 100[/tex]
[tex]percentage = 7.18 [/tex]
The janitor can determine the depth of the elevator shaft by timing the drop of the rock and using the displacement for free-falling objects formula, considering the time for the sound to travel up the shaft. The percent error if the sound travel time is ignored can be calculated by comparing the depth calculated with and without the sound travel time.
Explanation:The janitor wants to determine how far it is from the point they dropped the rock to the bottom of the elevator shaft by measuring the time it takes for the rock to drop and hit the bottom. Here we can use the formula for the displacement of items in free-fall: d = 0.5gt², where d is the displacement (depth of the shaft), g is the acceleration due to gravity (9.8 m/s²), and t is the time it takes for the rock to fall (2.42s).
So, the depth of the shaft is d = 0.5 x 9.80 m/s² x (2.42 s)² = 28.63 m. However, we need to account for the time it takes for the sound to travel back up the shaft. The speed of sound is given as 336 m/s, and the time it takes to travel a certain distance is the distance divided by the speed, so in this case, it’s 28.63 m / 336 m/s = 0.085 s. Subtracting this from the total time gives us the true fall time of the rock, which we can plug back into the displacement formula to get the corrected depth of the shaft.
The percent error introduced if we ignore the travel time for the sound is then the depth obtained with sound travel time subtracted from the depth obtained without it, divided by the depth obtained with sound travel time, times 100.
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Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphere does the volume of a shell equal that of the interior sphere? Assume the shell thickness to be t = 1 nm.
Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. On log-log axes, plot the ratio of the shell volume to the interior sphere volume for an interior sphere radius of 10 microns to 1 nm. Assume the shell thickness to be t = 1 nm.
Answer:
3.85 nm
Explanation:
The volume of a sphere is:
V = 4/3 * π * r^3
The volume of a shell is the volume of the big sphere minus the volume of the small sphere
Vs = 4/3 * π * r2^3 - 4/3 * π * r1^3
Vs = 4/3 * π * (r2^3 - r1^3)
If the difference between the radii is 1 nm
r2 = r1 + 1 nm
Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)
Vs = 4/3 * π * ((r1 + 1)^3 - r1^3)
Vs = 4/3 * π * (r1^3 + 3*r1^2 + 3*r1 + 1 - r1^3)
Vs = 4/3 * π * (3*r1^2 + 3*r1 + 1)
The volme of the shell is equall to the volume of the inner shell:
4/3 * π * (r1^3) = 4/3 * π * (3*r1^2 + 3*r1 + 1)
r1^3 = 3*r1^2 + 3*r1 + 1
0 = -r1^3 + 3*r1^2 + 3*r1 + 1
Solving this equation electronically:
r1 = 3.85 nm
A ball is tossed with enough speed straight up so that it is in the air serveral seconds. Assume upwards direction is positive and downward is negative. What is the change in its velocity during this 1-s interval?
Answer:
- 32.17 fts/s in the imperial system, -9.8 m/s in the SI
Explanation:
We know that acceleration its the derivative of velocity with respect to time, this is (in 1D):
[tex]a = \frac{dv}{dt}[/tex]
So, if we wanna know the change in velocity, we can take the integral:
[tex]v(t_f) - v(t_i) = \int\limits^{t_f}_{t_i} {\frac{dv}{dt} \, dt = \int\limits^{t_f}_{t_i} a \, dt[/tex]
Luckily for us, the acceleration in this problem is constant
[tex]a \ = \ - g[/tex]
the minus sign its necessary, as downward direction is negative. Now, for a interval of 1 second, we got:
[tex]t_f = t_i + 1 s[/tex]
[tex]v(t_i + 1 s) - v(t_i) = \int\limits^{t_i + 1 s}_{t_i} a \, dt[/tex]
[tex]v(t_i + 1 s) - v(t_i) = \int\limits^{t_i + 1 s}_{t_i} (-g ) \, dt[/tex]
[tex]v(t_i + 1 s) - v(t_i) = [-g t]^{t_i + 1 s}_{t_i} [/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g (t_i+1s) + g t_i [/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g t_i + -g 1s + g t_i[/tex]
[tex]v(t_i + 1 s) - v(t_i) = -g 1s [/tex]
taking g in the SI
[tex]g=9.8 \frac{m}{s^2}[/tex]
this is:
[tex]v(t_i + 1 s) - v(t_i) = - 9.8 \frac{m}{s} [/tex]
or, in imperial units:
[tex]g=32.17 \frac{fts}{s^2}[/tex]
this is:
[tex]v(t_i + 1 s) - v(t_i) = - 32.17 \frac{fts}{s} [/tex]
A satellite of mass 5000 kg orbits the Earth (mass = 6.0 x 1024 kg) and has a period of 6000 s. In the above problem the altitude of the satellite above the Earth's surface is
Answer:
altitude of satellite from earth surface is 7140 km
Explanation:
given data:
mass of satellite = 5000kg
altitude of satellite from earth surface can be calculated by using given relation
[tex]p^2 = \frac{(4* \pi^2)(a^3)}{(G*M)}[/tex]
where,
p - satellite's period in seconds,
a - satellite's altitude in meters,
G - gravitational constant (6.67 x 10^-11 m³/kgs²) and
M - mass of Earth 5.98 x 10^24 kg
solving for altitude we have
[tex]a^3 =\fracp{(G*M*p^2)}{(4*\pi^2)}[/tex]
[tex]a^3 = 3.637 * 10^{20}[/tex]
a = 7,138,000 meters = 7,140 km
altitude of satellite from earth surface is 7140 km
What is the wavelength of a monochromatic light beam, where the photon energy is 2.70 × 10^−19 J? (h = 6.63 ×10^−34 J⋅s, c = 3.00 × 10^8 m/s, and 1 nm = 10^−9 m)
Answer:
Wavelength = 736.67 nm
Explanation:
Given
Energy of the photon = 2.70 × 10⁻¹⁹ J
Considering:
[tex]Energy=h\times frequency[/tex]
where, h is Plank's constant having value as 6.63 x 10⁻³⁴ J.s
The relation between frequency and wavelength is shown below as:
c = frequency × Wavelength
Where, c is the speed of light having value = 3×10⁸ m/s
So, Frequency is:
Frequency = c / Wavelength
So, Formula for energy:
[tex]Energy=h\times \frac {c}{\lambda}[/tex]
Energy = 2.70 × 10⁻¹⁹ J
c = 3×10⁸ m/s
h = 6.63 x 10⁻³⁴ J.s
Thus, applying in the formula:
[tex]2.70\times 10^{-19}=6.63\times 10^{-34}\times \frac {3\times 10^8}{\lambda}[/tex]
Wavelength = 736.67 × 10⁻⁹ m
1 nm = 10⁻⁹ m
So,
Wavelength = 736.67 nm
If the atmospheric pressure in a tank is 23 atmospheres at an altitude of 1,000 feet, the air temperature in the tank is 700F, and the volume of the tank is 800 f3, determine the weight of the air in the tank.
Answer:
W = 289.70 kg
Explanation:
Given data:
Pressure in tank = 23 atm
Altitude 1000 ft
Air temperature in tank T = 700 F
Volume of tank = 800 ft^3 = 22.654 m^3
from ideal gas equation we have
PV =n RT
Therefore number of mole inside the tank is
[tex]\frac{1}{n} = \frac{RT}{PV}[/tex]
[tex] = \frac{8.206\times 10^{-5}} 644.261}{23\times 22.654}[/tex]
[tex]= 1.02\times 10^{-4}[/tex]
[tex]n = 10^4 mole[/tex]
we know that 1 mole of air weight is 28.97 g
therefore, tank air weight is [tex]W = 10^4\times 28.91 g = 289700 g[/tex]
W = 289.70 kg
An observer is approaching at stationary source at 17.0 m/s. Assuming the speed of sound is 343 m/s, what is the frequency heard by the observer if the source emits a 2550-Hz sound?
Answer:
the frequency heard by the observer is equal to 2677 Hz
Explanation:
given,
velocity of the observer = 17 m/s
speed of the sound = 343 m/s
velocity of the source = 0 m/s
frequency emitted from the source = 2550 Hz
[tex]f = f_0(\dfrac{v-v_0}{v-v_s})[/tex]
[tex]f = 2550\times (\dfrac{343+17}{343-0})[/tex]
velocity of observer is negative as it is approaching the source. f = 2676.38 Hz ≈ 2677 Hz
hence, the frequency heard by the observer is equal to 2677 Hz
A car traveling 56.0 km/h is 25.0 m from a haystack when the driver slams on the brakes. The car hits the haystack 2.11 s later. How fast is the car traveling at impact?
Answer:
The car is traveling at [tex]8.1366\frac{m}{s}[/tex]
Explanation:
The known variables are the following:
[tex]V_{0} = 56 \frac{km}{h} = 15.56 \frac{m}{s}\\ D=25m\\ t=2.11s\\ V_{f}=?[/tex]
First, from the equation of motion we find the deceleration:
[tex]D=V_{0}*t+\frac{1}{2} a*t^{2} \\ a=\frac{2(D-V_{0})}{t^{2} } \\ a=3.5182\frac{m}{s^2}[/tex]
Then, with the equation for the speed:
[tex]V_{f}=V_{o}+a*t\\ V_{f}=8.1366\frac{m}{s}[/tex]
The speed of the car at impact is approximately 25.72 m/s.
Explanation:To find the speed of the car at impact, we first need to calculate the deceleration. We can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Given that the initial velocity (u) is 56.0 km/h, the distance (s) is 25.0 m, and the time taken (t) is 2.11 s, we can rearrange the equation to solve for a: a = (v - u) / t.
Plugging in the values, we have a = (0 - 56.0 km/h) / 2.11 s = -26.5 m/s². Since the car is decelerating, the acceleration is negative.
Now, to find the final velocity (v) at impact, we can use the equation v² = u² + 2as. Rearranging the equation, we have v² = u² + 2(-26.5 m/s²)(25.0 m) = u² - 2(26.5 m/s²)(25.0 m).
Plugging in the values, we have v² = (56.0 km/h)² - 2(26.5 m/s²)(25.0 m).
Converting the initial velocity (u) to m/s, we get u = 56.0 km/h * (1000 m/1 km) * (1 h/3600 s) = 15.6 m/s.
Substituting the values, we have v² = (15.6 m/s)² - 2(26.5 m/s²)(25.0 m) = 0.16 m²/s² - 26.5 m²/s² * 25.0 m = 0.16 m²/s² - 662.5 m²/s² = -662.34 m²/s².
Taking the square root of both sides, we get v ≈ √(-662.34 m²/s²) ≈ -25.72 m/s.
Since speed is a positive quantity, the speed of the car at impact is approximately 25.72 m/s.
Part A) A small turbo prop commuter airplane, starting from rest on a New York airport runway, accelerated for 19.0s before taking off. It's speed at takeoff is 50.0 m/s. Calculate the acceleration of the plane , assuming it remains constant. Part B) In this problem, how far did the plane move while accelerating for 19.0s?
Answer:
a)a=2.63m/s^2
b)x=475.25m
Explanation:
To solve the first part of this problem we use the following equation
Vf=final speed=50m/s
Vo=initial speed=0
t=time
a=aceleration
a=(Vf-Vo)/t
a=(50-0)/19=2.63m/s^2
B )
For the second part of this problem we use the following equation
X=(Vf^2-Vo^2)/2a
X=(50^2-0^2)/(2*2.63)=475.25m
A special standard weight, known to weight 10 N to high accuracy, is used to check the accuracy of two spring scales. On the first scale the weight gives a reading of 8.4 N. On the second scale the weight gives a reading of 9.1 N. The percent difference between the measurements of the two scales is %. (Never negative.)
Answer:
16 % and 9 %
Explanation:
Accurate reading = 10 N
reading of first balance = 8.4 N
reading of second balance = 9.1 N
Percentage difference in first reading
= [tex]\frac{True reading - reading of first balance}{True reading}\times 100 %[/tex]
= [tex]\frac{10-8.4}{10}\times 100 %[/tex] = 16 %
Percentage difference in second reading
= [tex]\frac{True reading - reading of second balance}{True reading}\times 100 %[/tex]
= [tex]\frac{10-9.1}{10}\times 100 %[/tex] = 9 %
The expression "1in. = 2.54cm" is properly called: A. an equation B. a conversion factor C. a rule of thumb D. an equivalence statement
Answer:
B. a conversion factor
Explanation:
The expression "1 in. = 2.54 cm" is called a conversion factor.
With this expression inches can be converted to centimeters.
Inversely the expression can also be used to convert centimeters to inches
By rearranging the equation we get
[tex]\frac{1}{2.54}\ inches=1\ cm\\\Rightarrow 1\ cm=0.394\ inches[/tex]
The statement which shows the equal nature of two different expressions is called an equation
Rule of thumb is a general approximation of a result of a test or experiment.
Hi your answer is B on edge
For two different inertial frames of reference would acceleration and change in velocity be different? How would energy be affected?
Answer and Explanation:
The inertial reference frame is one with constant velocity or non-accelerated frame of reference.
The value of acceleration and velocity change will vary in the two frames and will not be same.
As in case, we observe the acceleration and velocity of a moving train from the platform and the one observed in the train itself will be different.
In case of energy, it is dependent on the frame of reference but the energy change is independent of the frame of reference.
What is the net charge of a metal ball if there are 21,749 extra electrons in it?
Answer:
[tex]Q=3.47\times 10^{-15}\ C[/tex]
Explanation:
Given that,
Number of extra electrons, n = 21749
We need to find the net charge on the metal ball. Let Q is the net charge.
We know that the charge on an electron is [tex]q=1.6\times 10^{-19}\ C[/tex]
To find the net charge if there are n number of extra electrons is :
Q = n × q
[tex]Q=21749\times 1.6\times 10^{-19}\ C[/tex]
[tex]Q=3.47\times 10^{-15}\ C[/tex]
So, the net charge on the metal ball is [tex]3.47\times 10^{-15}\ C[/tex]. Hence, this is the required solution.
A 0.5 kg block is attached to a spring (k = 12.5 N/m). The damped frequency is 0.2% lower than the natural frequency, (a) What is the damping constant? (b) How does the amplitude vary with time? (c) Determine the critical damping constant?
For a mass-spring system with specified parameters, the damping constant is found to be 4.999 Ns/m. The amplitude reduces exponentially over time. The critical damping constant, defining the transition from underdamped to overdamped behavior, is 5 Ns/m.
Explanation:This problem pertains to a system of mass springs undergoing damped oscillations. The damping constant (b) can be calculated using the formula b = 2*m*ωd. Here ωd signifies the damped frequency and is given by ωd = ω * (1 - γ/2), where ω is the natural frequency and γ is the damping ratio of 0.2% = 0.002. The natural frequency ω = sqrt(k/m) = sqrt(12.5/0.5) = 5 rad/s. Substituting these values in the equation gives b = 2*0.5*4.999 = 4.999 Ns/m.
The amplitude (A) of a damped oscillation decreases exponentially with time due to energy loss, described by A(t) = A0 * e^(-γωt/2), wherein A0 is the initial amplitude.
The critical damping constant (bc), describing the boundary between underdamped and overdamped systems, is equal to 2*sqrt(m*k) = 2*sqrt(0.5*12.5) = 5 Ns/m.
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When a particular wave is vibrating with a frequency of 4.00 Hz, a transverse wave of wavelength 60 cm is produced. Determine the speed of waves along the wire.
Answer:
Speed of the wave, v = 2.4 m/s
Explanation:
Given that,
Frequency of vibrations, f = 4 Hz
Wavelength of the transverse wave, [tex]\lambda=60\ cm=0.6\ m[/tex]
We need to find the speed of the waves along the wire. Let it is equal to v. So, the relationship is given by :
[tex]v=f\times \lambda[/tex]
[tex]v=4\times 0.6[/tex]
v = 2.4 m/s
So, the speed of waves along the wire is 2.4 m/s. Hence, this is the required solution.
Final answer:
The speed of the transverse wave on the wire, given a frequency of 4.00 Hz and a wavelength of 60 cm, is calculated to be 2.40 meters per second (m/s) using the formula Speed = Frequency × Wavelength.
Explanation:
To determine the speed of a wave, you can use the formula:
Speed = Frequency × Wavelength
In the given scenario, the frequency is 4.00 Hz, and the wavelength is 60 cm, which is equivalent to 0.60 meters (since 1 meter = 100 cm). Thus, the speed of the wave on the wire can be calculated as follows:
Speed = 4.00 Hz × 0.60 m
Speed = 2.40 m/s
Therefore, the speed of the transverse wave along the wire is 2.40 meters per second (m/s).
A 280 g object is attached to a spring and executes simple harmonic motion with a period of 0.270 s. If the total energy of the system is 4.75 J. (a) Find the maximum speed of the object _______ m/s (b) Find the force constant of the spring_________ N/m (c) Find the amplitude of the motion_________m
Answer:
Explanation:
given,
mass of the object = 280 g = 0.28 kg
time period = 0.270 s
total energy of the system = 4.75 J
[tex]\dfrac{1}{2}\ m\ V^2 = 4.75 J[/tex]
maximum speed of the object V = [tex]\sqrt{ \dfrac{2 \times 4.75}{0.28} }[/tex]
V= 5.82 m / s
(b) force constant of the spring K = m ω²
where ω = angular frequency = 2π / T
T= time period = 0.25 s
ω = 25.13 rad / s
K = 0.28 × 25.13²
K = 176.824 N / m
(c). Amplitude of motion A = [tex]\dfrac{V}{\omega}[/tex]
= [tex]\dfrac{5.82}{25.13}[/tex]
A = 0.232 m
What uniform magnetic field, applied perpendicular to a beam
ofelectrons moving at 1.30 x 106 m/s, is required to
makethe elctrons travel in a ciruclar arc of radius 0.350 m?
Answer:
The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]
Explanation:
Given that,
Speed [tex]v=1.30\times10^{6}\ m/s[/tex]
Radius = 0.350 m
We need to calculate the magnetic field
Using formula of magnetic field
[tex]B =\dfrac{mv}{qr}[/tex]
Where, m = mass of electron
v = speed of electron
q = charge of electron
r = radius
Put the value into the formula
[tex]B=\dfrac{9.1\times10^{-31}\times1.30\times10^{6}}{1.6\times10^{-19}\times0.350}[/tex]
[tex]B=2.11\times10^{-5}\ T[/tex]
Hence, The magnetic field is [tex]2.11\times10^{-5}\ T[/tex]
Cheetahs can accelerate to a speed of 21.8 m/s in 2.55 s and can continue to accelerate to reach a top speed of 28.1 m/s . Assume the acceleration is constant until the top speed is reached and is zero thereafter. Let the +x direction point in the direction the cheetah runs. Express the cheetah's top speed vtop in miles per hour (mi/h) .
Starting from a crouched position, how much time totall does it take a cheetah to reach its top speed and what distance d does it travel in that time?
If a cheetah sees a rabbit 122.0 m away, how much time ttotal will it take to reach the rabbit, assuming the rabbit does not move and the cheetah starts from rest?
Answer:
Explanation:
Given
Cheetah speed=21.8 m/s in 2.55 sec
i.e. its [tex]a=\frac{21.8}{2.55}=8.55 m/s^2[/tex]
Cheetah top speed=28.1 m/s
And 1 m/s is equal to 2.23694 mph
therefore 28.1 m/s is [tex]2.236\times 28.1=62.858 mph[/tex]
Time taken to reach top speed
v=u+at
[tex]28.1=0+8.55\times t[/tex]
[tex]t=\frac{28.1}{8.55}=3.286 s [/tex]
Distance traveled during this time
[tex]v^2-u^2=2as[/tex]
[tex]28.1^2=2\times 8.55\times s[/tex]
[tex]s=\frac{789.61}{2\times 8.55}=46.176 m[/tex]
If cheetah sees a rabbit 122 m away
time taken to reach rabbit
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]122=0+\frac{1}{2}\times 8.55\times t^2[/tex]
[tex]t^2=\frac{244}{8.55}[/tex]
[tex]t=\sqrt{28.53}=5.34 s[/tex]
A small rock is thrown vertically upward with a speed of 27.0 m/s from the edge of the roof of a 21.0-m-tall building. The rock doesn't hit the building on its way back down and lands in the street below. Ignore air resistance. Part A: What is the speed of the rock just before it hits the street?
Part B: How much time elapses from when the rock is thrown until it hits the street?
Answer:
A. 33.77 m/s
B. 6.20 s
Explanation:
Frame of reference:
Gravity g=-9.8 m/s^2; Initial position (roof) y=0; Final Position street y= -21 m
Initial velocity upwards v= 27 m/s
Part A. Using kinematics expression for velocities and distance:
[tex]V_{final}^{2}=V_{initial}^{2}+2g(y_{final}-y_{initial})\\V_{f}^{2}=27^{2}-2*9.8(-21-0)=33.77 m/s[/tex]
Part B. Using Kinematics expression for distance, time and initial velocity
[tex]y_{final}=y_{initial}+V_{initial}t+\frac{1}{2} g*t^{2}\\==> -0.5*9.8t^{2}+27t+21=0\\==> t_1 =-0.69 s\\t_2=6.20 s[/tex]
Since it is a second order equation for time, we solved it with a calculator. We pick the positive solution.
The speed of the rock just before it hits the street is approximately 20.3 m/s, and the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.
Explanation:Part A: To determine the speed of the rock just before it hits the street, we can use the concept of conservation of energy. When the rock is at the top of its trajectory, its potential energy is at its maximum and its kinetic energy is at its minimum. When the rock is just before hitting the street, its potential energy is at its minimum (zero) and its kinetic energy is at its maximum. Using the equations for potential energy and kinetic energy, we can calculate the speed of the rock:
Initial potential energy = final potential energy + final kinetic energy
mgh = 1/2m*v^2
where m is the mass of the rock, g is the acceleration due to gravity, h is the height of the building, and v is the final velocity of the rock.
Since the mass of the rock cancels out, we have:
gh = 1/2v^2
Plugging in the values, g = 9.8 m/s^2 and h = 21.0 m, we can solve for v:
v = sqrt(2gh)
v = sqrt(2*9.8*21.0)
v = sqrt(411.6)
v = 20.3 m/s
Therefore, the speed of the rock just before it hits the street is approximately 20.3 m/s.
Part B: To calculate the time elapsed from when the rock is thrown until it hits the street, we can use the equation for time of flight of a vertically thrown object:
t = 2v/g
where t is the time of flight, v is the initial upward velocity of the rock, and g is the acceleration due to gravity.
Plugging in the values, v = 27.0 m/s and g = 9.8 m/s^2, we can solve for t:
t = 2*27.0/9.8
t = 5.5 s
Therefore, the time elapsed from when the rock is thrown until it hits the street is approximately 5.5 seconds.
An electric field of intensity 3.0kN is applied along the x axis. Calculate the electric flux through a rectangular plane 0.350m wide and 0.700 m long if the following conditions are true: a) the plane is parallel to the yz plane b) The plane is parallel to thte xy plane c) The plane contains the y axis and its normal makes an angle of 30 degrees with the x axis.
Answer:
Explanation:
Area of plane = .35 x .7 = 0.245 m²
a) When plane is perpendicular to field ( plane is parallel to yz plane. )
Flux = field x area
3000 x .245 = 735 weber
b ) When plane is parallel to xy plane , the plane also becomes parallel to electric field so no flux will pass though the given plane.
Flux through the plane = 0
c ) Since normal to the plane makes 30 degree with x axis, it will also make 30 degree with direction of the field.
Flux through the plane
= electric field x area x cos 30
3000 x .245 x .866
636.51 weber.
Aaron is flying a helicopter with an air-relative velocity of 12.5 m/s to the West. A constant wind is also blowing the helicopter towards the South at 4.55 m/s relative to the ground. What is the velocity vector of the helicopter relative to the ground? What is the speed and direction relative to the East? How long does it take to travel 1 km?
Answer:
Explanation:
Given
Velocity of helicopter relative to air 12.5 m/s to west
In vector form
[tex]V_{ha}=-12.5\hat{i}[/tex]
where [tex]V_h[/tex]= velocity of helicopter relative to ground
Also the velocity of air
[tex]V_a=4.55\hat{j}[/tex]
[tex]V_{ha}=V_h-V_a[/tex]
[tex]V_h=V_{ha}+V_a[/tex]
[tex]V_h=-12.5\hat{i}+4.55\hat{j}[/tex]
Speed relative to east[tex]=\sqrt{12.5^2+4.55^2}[/tex]
=13.302
For Direction
[tex]tan\theta =\frac{4.55}{-12.5}[/tex]
[tex]180-\theta =20[/tex]
[tex]\theta =160^{\circ}[/tex] relative to east
For 1 km travel it takes
[tex]t=\frac{1000}{13.302}=75.17 s[/tex]
supply 5 N∙m of torque to a bolt located at the origin using a wrench that lies along the positive yy-axis. Assume the bolt head is at the origin and the threaded portion of the bolt lies along the negative xx-axis. If you apply a 25 N force at the end of the wrench in the direction 2???????? − 3????????, what length of wrench would be required to supply the desired torque? Under this force are we tightening or loosening the bolt?
Answer:
Explanation:
We need to supply 5 Nm to the bolt.
If we have a force of 25 N in the direction (2*i - 3*j), the torque is related to the length of the wrench.
T = F * d
d = T / F
d = 5 / 25 = 0.2 m
Assuming the wrench is perpendicular to the force.
The vector of the force is:
[tex]F = 25 N \frac{2i - 3j}{\sqrt{2^2 + (-3)^2}} = 13.9i - 20.8[/tex]
This is a countr-clockwise direction. If the bolt is right handed (most are) we are tightening it.
A student is driving to school on a road with a speed limit of 40mph (17.9 m/s). A stoplight ahead of her turns yellow, so she begins to slow down to come to a stop. On a dry day, she is able to make her car slow down with an acceleration of magnitude of 8.8 m/s^2. However, if the road is wet, she can only manage an acceleration with a magnitude 3.9 m/s^2. Part A:
On a dry day, how far (in meters) from the intersection would the student need to start braking in order to stop in time?
Part B:
If it were raining instead, how far (in meters) from the intersection would the student need to start braking?
Answer:
18.21 m
41.1 m
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration
Part A
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -8.8}\\\Rightarrow s=18.21\ m[/tex]
On a dry day she would have to start braking 18.21 m away from the intersection
Part B
[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-17.9^2}{2\times -3.9}\\\Rightarrow s=41.1\ m[/tex]
On a wet day she would have to start braking 41.1 m away from the intersection
Which properties should an object, system, or process have in order for it to be a useful standard of measurement for a physical quantity such as length or time? (Choose all that apply) -precise-restricted availability-defined in SI units-small-reproducible
Answer:
Precise and Reproducible
Explanation:
For a system or a process to be useful a measurement standard for any physical quantity, it must be:
Precise, Precision is important in any measurement.The values of the measurement obtained should be reproducibleThe measurement may be large or smallIt may or may not be in SI unitsIt should be available easily.Answer:
precise
reproducible
Explanation:
Give a critical discussion on why hardness test needs to perform in engineering practice.
b) Discuss 2 possible source of errors (other than parallax error) and how it's can effect the result.
Answer:
Explanation:
a) Hardness is a measure of the resistance of a material to permanent deformation (plastic) on its surface,
Hardness tests play an important role in material testing, quality control and component acceptance.
Hardness test are needed to be perform as a quality assurance procedure, to validate materials are according to the specific hardness required,
We depend on the data to verify the quality of the components to determine if a material has the necessary properties for its intended use.
Through the years, the establishment of increasingly productive and effective means of testing, has given way to new cutting-edge methods that perform and interpret hardness tests more effectively than ever. The result is a greater capacity and dependence on "letting the instrument do the work", contributing to substantial increases in performance and consistency and continuing to make hardness tests very useful in industrial and R&D applications.
b)
Instrumental errors: Instrument calibration is extremely important. An instrument with expired calibration may be generating erroneous data systematically.Enviromental error: An example is when surface preparation of the sample to be tested is poor, then the error can be presented when measuring the indentation on the sample to determine the hardness value."The hardness test is a critical aspect of engineering practice for several reasons:
1. Material Performance: Hardness is a measure of a material's resistance to deformation, which is directly related to its wear resistance, strength, and durability. By performing hardness tests, engineers can predict how a material will perform under service conditions.
2. Quality Control: Hardness testing is used to ensure that materials meet the required specifications. It is a non-destructive (in some cases) or semi-destructive method to verify the quality of the material without compromising its integrity.
3. Material Selection: The results of hardness tests help engineers to select appropriate materials for different applications. A material that is too soft or too hard for a particular application can lead to premature failure or unnecessary costs.
4. Heat Treatment Verification: Heat treatment processes such as quenching, annealing, and tempering are used to alter the mechanical properties of materials. Hardness testing is used to confirm that the desired properties have been achieved after such treatments.
5. Failure Analysis: In the event of a material failure, hardness testing can provide valuable information about the cause of the failure. It can indicate if the material was too brittle or too soft for the intended application.
6. Comparative Analysis: Hardness tests provide a simple way to compare the properties of different materials or different batches of the same material.
7. Research and Development: In the development of new materials, hardness testing is an essential tool for characterizing materials and understanding their properties.
B) Two possible sources of error in hardness testing, other than parallax error, and their effects on the results are:
1. Indenter Geometry Errors: The geometry of the indenter (e.g., ball, cone, or pyramid) must be precise according to the test standard being used. Wear, damage, or incorrect shape of the indenter can lead to inaccurate measurements. For instance, a worn or blunt indenter will give lower hardness values than the actual hardness of the material.
2. Loading Rate and Duration Errors: The rate at which the load is applied and the duration for which it is held can significantly affect the hardness measurement. Some materials exhibit time-dependent plasticity (creep), which can affect the size of the indentation. If the loading rate or duration is not consistent with the standards or previous tests, the results may not be comparable or may not accurately represent the material's hardness. For example, applying the load too quickly may result in an underestimation of hardness due to the material's inability to fully resist the indenter.
To minimize these errors, it is essential to follow standardized testing procedures, maintain and calibrate equipment regularly, and ensure that the test conditions are consistent. Additionally, operators should be well-trained and aware of the potential sources of error to ensure the accuracy and reliability of the hardness test results."
What is the forward velocity of a 250 lb stone projectile that leaves a medieval catapult 55 degrees from the ground at 50 m/s?
Answer:
28.5 m/s
Explanation:
The speed has two orthogonal components, horizontal and vertical. To decompose the speed into these components we can use these trigonometric equations:
Vh = V * cos(a)
Vv = V * sin(a)
The forwards velocity is the horizontal component, so we use
Vh = V * cos(a)
Vh = 50 * cos(55)
Vh = 50 * 0.57
Vh = 28.5 m/s
The acceleration of a particle is given by a = 3t – 4, where a is in meters per second squared and t is in seconds. Determine the velocity and displacement for the time t = 3.6 sec. The initial displacement at t = 0 is s0 = – 8 m, and the initial velocity is v0 = – 5 m/sec.
Answer:
[tex]v=0.04m/s\\[/tex]
[tex]s=-28.592m\\[/tex]
Explanation:
[tex]a = 3t-4[/tex]
[tex]v(t)=\int\limits^t_0 {a(t)} \, dt =3/2*t^{2}-4t+v_0\\[/tex]
if t=3.6s and initial velocity, v0, is -5m/s
[tex]v=0.04m/s\\[/tex]
[tex]s(t)=\int\limits^t_0 {v(t)} \, dt =1/2*t^{3}-2t^{2}+v_0*t+s_0\\[/tex]
if t=3.6s and the initial displacement, s0, is -8m:
[tex]s=-28.592m\\[/tex]
An engine extracts 441.3kJ of heat from the burning of fuel each cycle, but rejects 259.8 kJ of heat (exhaust, friction,etc) during each cycle. What is the thermal efficiency of the engine?
Answer:
[tex]\eta = 0.411[/tex]
Explanation:
As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine
So here we know that the
input energy = 441.3 kJ
energy rejected = 259.8 kJ
so we have
[tex]Q_1 - Q_2 = W[/tex]
[tex]W = 441.3 kJ - 259.8 kJ = 181.5 kJ[/tex]
now efficiency is defined as
[tex]\eta = \frac{W}{Q_1}[/tex]
[tex]\eta = \frac{181.5}{441.3}[/tex]
[tex]\eta = 0.411[/tex]
What is the amount of heat, in Sl units, necessary to melt 1 lb of ice?
Answer:
Heat required to melt 1 lb of ice is 151.469 KJ
Explanation:
We have given mass of ice = 1 lb
We know that 1 lb = 0.4535 kg
Latent heat of fusion for ice =334 KJ/kg
Amount if heat for fusion of ice is given by
[tex]Q=mL[/tex], here m is mass of ice and L is latent heat of fusion
So heat [tex]Q=mL=0.4535\times 334=151.469kj[/tex]
So heat required to melt 1 lb of ice is equal to 151.469 KJ
An aircraft is in a steady level turn at a flight speed of 200 ft/s and a turn rate about the local vertical of 5 deg/s. Thrust frm the engine is along the flight direction. Given the weight of the aircraft as 50,000 lb and L/D of 10, determine the lift, drag and thrust required for the given equilibrium flight. Assume g = 32.2 ft/s
Answer:
L =4166.66 N and D = 46296.29 N and T = 96047.34 N
Explanation:
given,
flight speed = 200 ft/s
turn rate = 5 deg/s
weight of aircraft = 50000 lb
L/D = 10
for equilibrium in the horizontal position
w - L cos ∅ - D sin ∅ = 0 .............(1)
D cos ∅ - L sin ∅ = 0................(2)
L/D = tan ∅ = 10
∅ = 84.29°
50000 - L cos 84.29°- D sin 84.29°= 0
D cos 84.29° - L sin 84.29° = 0
on solving the above equation we get
L =4166.66 N and D = 46296.29 N
thrust force calculation:
T = W sin ∅ + D
= 50000×sin 84.29 + 46296.29
T = 96047.34 N