Answer:
E = 0.18 J
Explanation:
given,
Potential of the battery,V = 9 V
Charge on the circuit, Q = 20 m C
= 20 x 10⁻³ C
energy delivered in the circuit
E = Q V
E = 20 x 10⁻³ x 9
E = 180 x 10⁻³
E = 0.18 J
Energy delivered in the circuit is equal to E = 0.18 J
Answer:
(a) v = 5.42m/s
(b) vo = 4.64m/s
(c) a = 2874.28m/s^2
(d) Δy = 5.11*10^-3m
Explanation:
(a) The velocity of the ball before it hits the floor is given by:
(1)
g: gravitational acceleration = 9.8m/s^2
h: height where the ball falls down = 1.50m
![v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B2%289.8m%2Fs%5E2%29%281.50m%29%7D%3D5.42%5Cfrac%7Bm%7D%7Bs%7D)
The speed of the ball is 5.42m/s
(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.
You use the following formula:
(2)
vo: velocity of the ball where it starts its motion upward
You solve for vo and replace the values of the parameters:
![v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}](https://tex.z-dn.net/?f=v_o%3D%5Csqrt%7B2gh_%7Bmax%7D%7D%3D%5Csqrt%7B2%289.8m%2Fs%5E2%29%281.10m%29%7D%3D4.64%5Cfrac%7Bm%7D%7Bs%7D)
The velocity of the ball is 4.64m/s
(c) The acceleration is given by:
![a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7Bv_o-v%7D%7B3.50%2A10%5E%7B-3%7Ds%7D%3D%5Cfrac%7B4.64m%2Fs-%28-5.42m%29%2Fs%7D%7B3.50%2A10%5E%7B-3%7Ds%7D%3D2874.285%5Cfrac%7Bm%7D%7Bs%5E2%7D)
![a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B%5CDelta%20v%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7Bv_o-v%7D%7B3.50%2A10%5E%7B-3%7Ds%7D%3D%5Cfrac%7B4.64m%2Fs-5.42m%2Fs%7D%7B3.50%2A10%5E%7B-3%7Ds%7D%3D-222.85%5Cfrac%7Bm%7D%7Bs%5E2%7D)
The acceleration of the ball is 2874.28/s^2
(d) The compression of the ball is:
![\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m](https://tex.z-dn.net/?f=%5CDeta%20y%3D%5Cfrac%7Bv%5E2%7D%7B2%28a%29%7D%3D%5Cfrac%7B%285.42m%2Fs%29%5E2%7D%7B2%282874.28m%2Fs%5E2%29%7D%3D5.11%2A10%5E%7B-3%7Dm)
THe compression of the ball when it strikes the floor is 5.11*10^-3m
Answer with Explanation:
We are given that
Diameter of pipe,![d_1=0.573 m](https://tex.z-dn.net/?f=d_1%3D0.573%20m)
![v_1=13.5 m/s](https://tex.z-dn.net/?f=v_1%3D13.5%20m%2Fs)
![v_2=5.83 m/s](https://tex.z-dn.net/?f=v_2%3D5.83%20m%2Fs)
Volume flow rate of the petroleum along the pipe=![Q_{refinery}=A_1v_1=v_(\frac{\pi d^2_1}{4})](https://tex.z-dn.net/?f=Q_%7Brefinery%7D%3DA_1v_1%3Dv_%28%5Cfrac%7B%5Cpi%20d%5E2_1%7D%7B4%7D%29)
![Q_{refinery}=13.5\times (\pi\times \frac{0.573)^2}{4})=3.48 m^3/s](https://tex.z-dn.net/?f=Q_%7Brefinery%7D%3D13.5%5Ctimes%20%28%5Cpi%5Ctimes%20%5Cfrac%7B0.573%29%5E2%7D%7B4%7D%29%3D3.48%20m%5E3%2Fs)
By equation of continuity
![A_1v_1=A_2v_2](https://tex.z-dn.net/?f=A_1v_1%3DA_2v_2)
![\frac{\pi d^2_1}{4}v_1=\frac{\pi d^2_2}{4}v_2](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%20d%5E2_1%7D%7B4%7Dv_1%3D%5Cfrac%7B%5Cpi%20d%5E2_2%7D%7B4%7Dv_2)
![d^2_2=\frac{v_1}{v_2}d^2_1](https://tex.z-dn.net/?f=d%5E2_2%3D%5Cfrac%7Bv_1%7D%7Bv_2%7Dd%5E2_1)
![d_2=\sqrt{\frac{v_1}{v_2}}d_1](https://tex.z-dn.net/?f=d_2%3D%5Csqrt%7B%5Cfrac%7Bv_1%7D%7Bv_2%7D%7Dd_1)
![d_2=0.573\sqrt{\frac{13.5}{5.83}}](https://tex.z-dn.net/?f=d_2%3D0.573%5Csqrt%7B%5Cfrac%7B13.5%7D%7B5.83%7D%7D)
![d_2=0.87 m](https://tex.z-dn.net/?f=d_2%3D0.87%20m)
![d_2=0.87\time 100=87 cm](https://tex.z-dn.net/?f=d_2%3D0.87%5Ctime%20100%3D87%20cm)
1 m=100 cm
The temperature rises until the water reaches the next change of state — boiling. As the particles move faster and faster, they begin to break the attractive forces between each otherand move freely as steam — a gas. The process by which a substance moves from the liquid state to the gaseousstate is called boiling.
Answer:
a♦1 E_average = n E₀ / 2
, b) E_average= infinity
Explanation:
The energy values form an arithmetic series, whose sum is
S = n (a₁ + aₙ) / 2 = n (2a₁ + (n-1) r)/ 2
Where n is the number of terms, a₁ is the first term, aₙ the last term and r is the difference between two consecutive numbers in the series
r = 2E₀ - 0 = 2E₀
Therefore the sum is
S = n (0 + n E₀) / 2
S = n² E₀ / 2
The average value is
E_average = S / n
E_average = n E₀ / 2
b) the case of harmonic oscillation
We have two possibilities.
- if we take a finite number and terms gives the same previous value
- If we take an infinite number of fears the series gives infinity and the average is also infinite
E_average= infinity