Answer:
2.73×10¯³⁴ m.
Explanation:
The following data were obtained from the question:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Wavelength (λ) =?
Next, we shall determine the energy of the ball. This can be obtained as follow:
Mass (m) = 0.113 Kg
Velocity (v) = 43 m/s
Energy (E) =?
E = ½m²
E = ½ × 0.113 × 43²
E = 0.0565 × 1849
E = 104.4685 J
Next, we shall determine the frequency. This can be obtained as follow:
Energy (E) = 104.4685 J
Planck's constant (h) = 6.63×10¯³⁴ Js
Frequency (f) =?
E = hf
104.4685 = 6.63×10¯³⁴ × f
Divide both side by 6.63×10¯³⁴
f = 104.4685 / 6.63×10¯³⁴
f = 15.76×10³⁴ Hz
Finally, we shall determine the wavelength of the ball. This can be obtained as follow:
Velocity (v) = 43 m/s
Frequency (f) = 15.76×10³⁴ Hz
Wavelength (λ) =?
v = λf
43 = λ × 15.76×10³⁴
Divide both side by 15.76×10³⁴
λ = 43 / 15.76×10³⁴
λ = 2.73×10¯³⁴ m
Therefore, the wavelength of the ball is 2.73×10¯³⁴ m.
Answer:
The force when θ = 33° is 1.7625 times of the force when θ = 18°
Explanation:
The force on a moving charge through a magnetic field is given by
F = qvB sin θ
q = charge of the moving particle
v = Velocity of the moving charge
B = Magnetic field strength
θ = angle between the magnetic field and the velocity (direction of the motion) of the moving charge
Because qvB are all constant, we can call the expression K.
F = K sinθ
when θ = 18°,
F = K sin 18° = 0.309K
when θ = 33°, let the force be F₁
F₁ = K sin 33° = 0.5446K
(F₁/F) = (0.5446K/0.309K) = 1.7625
F₁ = 1.7625 F
Hope this Helps!!!
1. Safety equipment is available
2. Person attempting the task has some general knowledge about wiring
3. Not All Cable is Color-Coded
Cable-sheath color coding started in 2001 and is still voluntary. If you have older wiring, don’t assume it complies with the current color coding. However, most manufacturers now follow the standard color code.
4. Stranded wire is more flexible than solid. If you’re pulling wire through conduit, stranded wire makes it easier to get around corners and bends in the conduit. However, if the situation requires pushing wires through conduit, you’ll want to use solid wire.
The planar simple harmonic wave travels in the positive direction of x axis with wave velocity u=2m/s, and the vibration curve of the particle at the origin in cosinusoidal form is shown in the figure.
Try to find (1) the vibration function of the particle at the origin, (2) the wave function of the planar simple harmonic wave according to the origin.
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
![\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%28t%29%7D%7Bdt%7D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Bate%5E%7B-6t%7D%5D%3Da%5B%281%29e%5E%7B-6t%7D%2Bt%28e%5E%7B-6t%7D%28-6%29%29%5D)
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):
hence, the maximum speed is v_max = ((1/6)e^-1)a
