Answer:
The velocity of the man from the frame of reference of a stationary observer is, V₂ = 5 m/s
Explanation:
Given,
Your velocity, V₁ = 2 m/
The velocity of the person, V₂ =?
The velocity of the person relative to you, V₂₁ = 3 m/s
According to the relative velocity of two
V₂₁ = V₂ -V₁
∴ V₂ = V₂₁ + V₁
On substitution
V₂ = 3 + 2
= 5 m/s
Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s
Answer:
Explanation:
Use the one-dimensional equation
where vf is the final velocity of the dog, v0 is the initial velocity of the dog, a is the acceleration of the dog, and t is the time it takesto reach that final velocity. For us:
0 = 2 + -.43t and
-2 = -.43t so
t = 4.7 seconds
Answer:
La velocidad de la luz en el vacío es una constante universal con el valor de 299 792 458 m/s (186 282,397 mi/s),aunque suele aproximarse a 3·108 m/s. Se simboliza con la letra c, proveniente del latín celéritās (en español, celeridad o rapidez).
¿Cuál es la consecuencia que a velocidad de la luz sea constante?
Respuesta. En modificaciones del vacío más sutiles, como espacios curvos, efecto Casimir, poblaciones térmicas o presencia de campos externos, la velocidad de la luz depende de la densidad de energía de ese vacío.
Answer:
- quality factor (Q) = 69.99
- inductor = 1.591 x 10⁻⁴ H
- capacitor = 3.248 x 10⁻¹⁰ F
Explanation:
Given;
resonance frequency (F₀) = 700 kHz
resistor, R = 10 Ohm
bandwidth (BW) = 10 kHz
bandwidth (BW) ![= \frac{R}{2\pi L}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7BR%7D%7B2%5Cpi%20L%7D)
![BW = \frac{R}{2\pi L}](https://tex.z-dn.net/?f=BW%20%3D%20%5Cfrac%7BR%7D%7B2%5Cpi%20L%7D)
make L (inductor) the subject of the formula
![L = \frac{R}{2\pi *BW} = \frac{10}{2\pi *10,000} =1.591 *10^{-4} \ H = \ 0.1591\ mH](https://tex.z-dn.net/?f=L%20%3D%20%5Cfrac%7BR%7D%7B2%5Cpi%20%2ABW%7D%20%20%3D%20%20%5Cfrac%7B10%7D%7B2%5Cpi%20%2A10%2C000%7D%20%3D1.591%20%2A10%5E%7B-4%7D%20%5C%20H%20%3D%20%5C%200.1591%5C%20mH)
![F_o =\frac{1}{2\pi\sqrt{LC} } \\\\\sqrt{LC} = \frac{1}{2\pi F_o} \\\\LC = \frac{1}{4\pi ^2F_o^2}= \frac{1}{4\pi ^2(700,000)^2} = 5.168*10^{-14}](https://tex.z-dn.net/?f=F_o%20%3D%5Cfrac%7B1%7D%7B2%5Cpi%5Csqrt%7BLC%7D%20%7D%20%5C%5C%5C%5C%5Csqrt%7BLC%7D%20%3D%20%5Cfrac%7B1%7D%7B2%5Cpi%20F_o%7D%20%5C%5C%5C%5CLC%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5E2F_o%5E2%7D%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5E2%28700%2C000%29%5E2%7D%20%3D%205.168%2A10%5E%7B-14%7D)
make C (capacitor) the subject of the formula
![C = \frac{5.168*10^{-14}}{1.591*10^{-4}} = 3.248*10^{-10} \ F = \ 3.248*10^{-4} \ \mu F](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B5.168%2A10%5E%7B-14%7D%7D%7B1.591%2A10%5E%7B-4%7D%7D%20%3D%203.248%2A10%5E%7B-10%7D%20%5C%20F%20%3D%20%5C%203.248%2A10%5E%7B-4%7D%20%5C%20%5Cmu%20F)
quality factor (Q) ![= \frac{1}{R} \sqrt{\frac{L}{C}} \ = \frac{1}{10} \sqrt{\frac{1.591*10^{-4}}{3.248*10^{-10}}}=69.99](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B1%7D%7BR%7D%20%5Csqrt%7B%5Cfrac%7BL%7D%7BC%7D%7D%20%5C%20%3D%20%5Cfrac%7B1%7D%7B10%7D%20%5Csqrt%7B%5Cfrac%7B1.591%2A10%5E%7B-4%7D%7D%7B3.248%2A10%5E%7B-10%7D%7D%7D%3D69.99)
quality factor (Q) = 69.99
Answer:
50.3N
Explanation:
Work done = force x distance
422J. = force x 8.39m
÷8.39 both side to get force
Force is 50.3N to 1 d.p.
Check:
50.3 x 8.39=422.017J
Same as 422J to 1 d.p