All categories

3 years ago

What is the height of the Golden Bridge?

Physics

2 answers:

Mademuasel [1]3 years ago

7 0

Answer:

hi your answer to this question is 220'

Explanation:

hope this helped you

professor190 [17]3 years ago

3 0

If your talking about the Golden Gate Bridge then the height is 746 ft (227.4m)

You might be interested in

Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball

Likurg_2 [28]

Answer:

k = 4422.35 KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k

k=Spring constant

So we can say that

( 2 π f)² = m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³ k

33167.69 = 7.5 x 10⁻³ k

k=4422.35 x 10³ N/m

k = 4422.35 KN/m

Therefore spring constant will be 4422.35 KN/m

4 0

3 years ago

A pendulum

RSB [31]

Explanation:

The time period rotation of the bob is t and the tension in the thread is T. ... A simple pendulum of length. ... Balancing the forces in Horizontal and vertical direction: ... Circular motion 2.

1 answer

8 0

3 years ago

What does popular sovereignty

Lana71 [14]

I'm a little confused by your question.

If you mean what is popular sovereignty, it is the belief that the governments is made and sustained by the consent of the people that it governs.

6 0

3 years ago

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

3 years ago

Read 2 more answers

Science in the study of the natural world through. And.

Lyrx [107]

You need to finish the question

6 0

3 years ago