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3 years ago

About how far does the suns light have to travel to reach Earth?

Physics

1 answer:

Ksju [112]3 years ago

7 0

It takes 499.0 seconds for light to travel from the Sun to the Earth, a distance called 1 Astronomical Unit.

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A rock hanging from a string has a net force of zero acting on it ? true or false ?

sattari [20]

Answer:

true cuase it is true its not false

8 0

3 years ago

2000, the Millennium Bridge, a new footbridge over the River Thames in London, England, was opened to the public. However, after

sweet [91]

Answer:

n = 1810

A = 25 mm

Explanation:

Given:

Lateral force amplitude, F = 25 N

Frequency, f = 1 Hz

mass of the bridge, m = 2000 kg/m

Span, L = 144 m

Amplitude of the oscillation, A = 75 mm = 0.075 m

time, t = 6T

now,

Amplitude as a function of time is given as:

$A(t)=A_oe^{\frac{-bt}{2m}}$

or amplitude for unforce oscillation

$\frac{A_o}{e}=A_oe^{\frac{-b(6T)}{2m}}$

$\frac{6bt}{2m}=1$

$b=\frac{m}{3T}$

Now, provided in the question Amplitude of the driven oscillation

$A=\frac{F_{max}}{\sqrt{(k-m\omega_d^2)+(b\omega_d^2)}}$

the value of the maximum amplitude is obtained $(k=m\omega_d^2)$

thus,

$A=\frac{F_{max}}{(b\omega_d}$

Now, for n people on the bridge

Fmax = nF

thus,

max amplitude

$0.075=\frac{nF}{((\frac{m}{3T})2\pi}$

n = 1810

hence, there were 1810 people on the bridge

b) $A=\frac{F_{max}}{(b\omega_d}$

since the effect of damping in the millenium bridge is 3 times

thus,

b=3b

therefore,

$A=\frac{F_{max}}{(3b\omega_d}$

$A=\frac{1}{(3}A_o$

$A=\frac{1}{(3}0.075$

A = 0.025 m = 25 mm

6 0

3 years ago

You have a resistor of resistance 200 ? and a 6.00-?F capacitor. Suppose you take the resistor and capacitor and make a series c

Ne4ueva [31]

Answer:

eazy

Explanation:

hi bro it is very hard so I can't help and I don't the answer

4 0

4 years ago

What is the velocity of sound in air and vacuum

guapka [62]

Answer:

approximately 2.99 × 10⁸ m/s

8 0

3 years ago

Read 2 more answers

The electrons that produce the picture in a

vredina [299]

Given:
u = 10⁵ m/s, the entrance velocity
v = 2.5 x 10⁶ m/s, the exit velocity
s = 1.6 cm = 0.016 m, distance traveled

Let a = the acceleration.
Then
u² + 2as = v²
(10⁵ m/s)² + 2*(a m/s²)*(0.016 m) = (2.5 x 10⁶ m/s)²
0.032a = 6.25 x 10¹² - 10¹⁰ = 6.24 x 10¹²
a = 1.95 x 10¹⁴ m/s²

Answer: 1.95 x 10¹⁴ m/s²

4 0

3 years ago