Answer:
Horizontal launch
[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]
Vertical launch
[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]
Explanation:
The launch speed is calculated by means of the Principle of Energy Conservation:
[tex]U_{k} = K[/tex]
[tex]\frac{1}{2}\cdot k \cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}[/tex]
[tex]v = x \cdot \sqrt{\frac{k}{m} }[/tex]
[tex]v = (0.15\,m)\cdot \sqrt{\frac{100\,\frac{N}{m} }{0.3\,kg} }[/tex]
[tex]v \approx 2.739\,\frac{m}{s}[/tex]
The velocities for each scenario are presented herein:
Horizontal launch
[tex]\vec v = 2.739\cdot i \,\left[\frac{m}{s} \right][/tex]
Vertical launch
[tex]\vec v = 2.739\cdot j \,\left[\frac{m}{s} \right][/tex]
Final answer:
The launch velocity of the T-shirt, when the spring with a spring constant of 100 N/m is compressed by 0.15 m, is calculated using conservation of energy. It turns out to be approximately 2.74 m/s for both the horizontal and vertical launches.
Explanation:
To calculate the launch velocity of the T-shirt from the cannon in both horizontal and vertical scenarios, we'll use the conservation of energy principle. This principle states that the potential energy stored in the compressed spring is converted into kinetic energy of the moving T-shirt.
Calculation of Launch Velocity (Horizontal and Vertical)
The energy stored in the compressed spring can be given by the formula:
[tex]E = 1/2 k x^2[/tex]
where E is the energy in joules, k is the spring constant, and x is the compression distance.
For the given spring with k = 100 N/m compressed by 0.15 m, the energy stored is:
[tex]E = 1/2 (100) (0.15)^2 = 1.125 joules[/tex]
Now we set this equal to the kinetic energy of the T-shirt upon release:
[tex]KE = 1/2 m v^2[/tex]
where KE is the kinetic energy, m is the mass of the T-shirt, and v is the velocity.
To find the launch velocity, we solve for v given the mass m = 0.3 kg and energy E = KE:
[tex]v = sqrt((2 * E) / m) = sqrt((2 * 1.125) / 0.3) = sqrt(7.5) = approximately 2.74 m/s[/tex]
This velocity will be the same for both the horizontal and the vertical launch as friction is ignored and the energy is purely transferred into the kinetic energy of the T-shirt.
A 47 g particle undergoes SHM with an amplitude of 7.1 mm, a maximum acceleration of magnitude 5.4 x 103 m/s2, and an unknown phase constant φ. What are (a) the period of the motion, (b) the maximum speed of the particle, and (c) the total mechanical energy of the oscillator? What is the magnitude of the force on the particle when the particle is at (d) its maximum displacement and (e) half its maximum displacement?
Answer:
Explanation:
check the picture attached for the explanation. I hope it helps . Thank you
Answer:
Explanation:
a) the period of the motion:
In order to calculate, we have the following formula.
f = 1/T whereas, T = 1/f. So, we need frequency first to calculate period of the motion.
In the question, we have been given amplitude and maximum acceleration. So, we can use the formula of maximum acceleration to calculate the frequency. Here's how:
Maximum acceleration = (2πf)² x A
where, A is amplitude.
a(max) = maximum acceleration
By rearranging the equation and making the frequency our subject and plugging in the values of given quantities then we have:
f = 138.80 hertz
Now, we can calculate, period of the motion by plugging in the value of frequency in the equation of period mentioned above.
Period of the motion = T = 1/f
T= 1/138.80
T = 7.204 x [tex]10^{-3}[/tex] seconds.
b) Maximum Speed of the particle
In order to calculate max. speed of the particle, we have to use following formula:
Max. Speed = 2πf x A
Max. Speed = ( 2 x 3.14 x 138.80 x 7.1x[tex]10^{-3}[/tex])
Max. Speed = 6.19 m/s
c) Total mechanical energy of the oscillator
Total mechanical energy of the oscillator is the sum of kinetic energy and potential energy and for which we have formula to calculate total mechanical energy of the oscillator:
Total Mechanical Energy = T.E = K.E + P.E
T.E = 2m[tex]\pi ^{2}[/tex][tex]f^{2}[/tex][tex]A^{2}[/tex]
we do have values of all the quantities and now by just plugging in the values we will get the total mechanical energy of the oscillator.
T.E = 2 x 0.047 x [tex]3.14^{2}[/tex] x [tex]138.80^{2}[/tex] x [tex]0.0071^{2}[/tex]
T.E = 0.90 J
d) Magnitude of the force on the particle when the particle is at its maximum displacement.
F = -KA here, amplitude is used because it is the maximum displacement from the mean position.
where K = -m ω² x A
ω = 2πf = 2 x 3.14 x 138.80
ω = 871.66
F = -(0.047 x 871.66² x 0.0071²)
F = - 1.8 N
e) when the displacement is at half
F = -Kx A/2
A/2 = 0.0071 / 2
A/2 = 0.0035
F = - m ω² x A/2
F = - (0.047 x 871.66² x 0.0035)
F = - 125 N
A solenoid having an inductance of 5.41 μH is connected in series with a 0.949 kΩ resistor. (a) If a 16.0 V battery is connected across the pair, how long will it take in seconds for the current through the resistor to reach 77.9% of its final value? (b) What is the current through the resistor at a time t = 1.00τL?
Answer:
(A) [tex]9.14\times 10^{-9}sec[/tex]
(B) [tex]6.20\times 10^{-3}A[/tex]
Explanation:
We have given inductance [tex]L=5.41\mu H=5.41\times 10^{-6}H[/tex]
Resistance [tex]R=0.949kohm=0.949\times 10^3ohm[/tex]
Time constant of RL circuit is equal to [tex]\tau =\frac{L}{R}[/tex]
[tex]\tau =\frac{5.41\times 10^{-6}}{0.949\times 10^3}=5.70\times 10^{-9}sec[/tex]
Battery voltage e = 16 volt
(a) It is given current becomes 79.9% of its final value
Current in RL circuit is given by
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
According to question
[tex]0.799i_0=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]e^{\frac{-t}{\tau }}=0.201[/tex]
[tex]{\frac{-t}{\tau }}=ln0.201[/tex]
[tex]{\frac{-t}{5.7\times 10^{-9} }}=-1.6044[/tex]
[tex]t=9.14\times 10^{-9}sec[/tex]
(b) Current at [tex]t=\tau sec[/tex]
[tex]i=i_0(1-e^{\frac{-t}{\tau }})[/tex]
[tex]i=\frac{16}{0.949\times 10^3}(1-e^{\frac{-\tau }{\tau }})[/tex]
[tex]i=6.20\times 10^{-3}A[/tex]