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3 years ago

You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you

Physics

1 answer:

lord [1]3 years ago

5 0

Answer:

The speed of light will be c=3x10^8m/s

Explanation:

This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c

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A cannon is fired from the edge of a cliff, which is 60m above the sea. The cannonball's initial velocity is 88.3m/s and it is f

wel

Answer:

a. 11.29 s b. 94.72 m/s at -39.8° c. 821.57 m

Explanation:

a. Using y - y₀ = ut - 1/2gt² where u = vertical component of velocity = v₀sinθ where v₀ = 88.3 m/s and θ = 34.5°, y₀ = + 60 m and y = water surface = 0 m, g = 9.8 m/s² and t = time it takes the cannon to reach the water surface.

So y - y₀ = ut - 1/2gt²

y - y₀ = (v₀sinθ)t - 1/2gt²

substituting the values of the variables into the equation, we have

0 - 60 = (88.3 m/s × sin34.5°)t - 1/2 × 9.8 m/s²× t²

- 60 = 50t - 4.9t²

So, 4.9t² - 50t - 60 = 0

Using the quadratic formula to find t,

$t = \frac{-(-50) +/- \sqrt{(-50)^{2} - 4 X 4.9 X -60} }{2 X 4.9} \\t = \frac{50 +/- \sqrt{2500 + 1176} }{9.8} \\t = \frac{50 +/- \sqrt{3676} }{9.8} \\t = \frac{50 +/- 60.63 }{9.8} \\t = \frac{50 + 60.63 }{9.8} or t = \frac{50 - 60.63 }{9.8} \\t = \frac{110.63 }{9.8} or t = \frac{-10.63 }{9.8} \\t = 11.29 sor -1.085$

Since t cannot be negative, t = 11.29 s

b. We first need to find the impact vertical velocity component. Using

v = u - gt where u = initial vertical velocity component = v₀sinθ and t = 11.29 s and g = 9.8 m/s². So,

v = v₀sinθ - gt

= 88.3 m/s × sin34.5° - 9.8 m/s² × 11.29 s

= 50.01 m/s - 110.64 m/s

= -60.63 m/s

Since the horizontal velocity is constant u' = v₀cosθ = 88.3 m/s × cos34.5° = 72.77 m/s.

The impact velocity is thus the resultant of the horizontal velocity and final impact velocity. So, V = √(v² + u'²)

= √((-60.63 m/s)² + (72.77 m/s)²)

= √((3676 m²/s² + 5295.48 m²/s²)

= √(8971.48 m²/s²

= 94.72 m/s

The angle θ = tan⁻¹(v/u') = tan⁻¹(-60.63 m/s ÷ 72.77 m/s) = tan⁻¹(-0.8332) = -39.8°

So the impact velocity is 94.72 m/s at -39.8°

c. The horizontal distance out from the base of the cliff that the ball strikes the water is the range, R = u't = 72.77 m/s × 11.29 s = 821.57 m

5 0

3 years ago

A 125 g pendulum bob hung on a string of length 35 cm has the same period as when the bob is hung from a spring and caused to os

Bingel [31]

Answer:

k = 3.5 N/m

Explanation:

It is given that the time period the bob in pendulum is the same as its time period in spring mass system:

$Time\ Period\ of\ Pendulum = Time\ Period\ of\ Spring-Mass\ System\\2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{m}{k}$

$\frac{l}{g} = \frac{m}{k}\\\\ k = g\frac{m}{l}$

where,

k = spring constant = ?

g = acceleration due to gravity = 9.81 m/s²

m = mass of bob = 125 g = 0.125 kg

l = length of pendulum = 35 cm = 0.35 m

Therefore,

$k = (9.81\ m/s^2)(\frac{0.125\ kg}{0.35\ m})\\\\$

4 0

3 years ago

Calculate the de Broglie wavelength of (a) a mass of 1.0 g traveling at 1.0 m s−1 , (b) the same, traveling at 1.00 × 105 km s−1

lesantik [10]

Answer:

a) $\lambda=6.63\times10^{-31}m$

b) $\lambda=6.63\times10^{-39}m$

c) $\lambda=9.97\times10^{-11}m$

d) $\lambda=4.03\times10^{-36}$ m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

$\lambda=\frac{h}{P}=\frac{h}{mv}$

with h the Plank's constant ( $6.63\times10^{-34}\frac{m^{2}kg}{s}$ ) and P the momentum of the object that is mass (m) times velocity (v).

a) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}$

$\lambda=6.63\times10^{-31}m$

b) $\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}$

$\lambda=6.63\times10^{-39}m$

c) $\lambda=\frac{6.63\times10^{-34}}{(6.65\times10^{-27}*1000)}$

$\lambda=9.97\times10^{-11}m$

d) $\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}$

$\lambda=4.03\times10^{-36}$ m

e) $\lambda=\frac{6.63\times10^{-34}}{(74*0)}$

λ=∞

6 0

3 years ago