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valina [46]
2 years ago
8

Summary about down hill on a bicycle story​

Physics
1 answer:
Ne4ueva [31]2 years ago
6 0

Answer:

The poem, Going Down the Hill on a Bicycle, written by Henry Charles Beeching describes the thrilling ride of a boy going downhill. ... The poet mentions how he lifts his feet from the pedals and keeps his hands still so that he would not lose his balance and fall off the bicycle, while it is dashing down the hill.

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Which Chinese man was leader of the National People's Party, called the Guomindang, in 1925?
Len [333]
<span>C. Mao Zedong
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5 0
3 years ago
Read 2 more answers
A beam of white light enters a prism of crown glass (n = 1.5) from air (n = 1.00). Once inside, the colors in the light disperse
eduard

Answer:

= 2.33

Explanation:

.According to snell's law:

n1sin i = n2sin r ,

where n1 is refractive index of the medium in which incident ray is travelling, n2 is the refractive index of the medium in which refracted ray is travelling,

i is angle of incidence,

r is angle of refraction.

Given that,

n1 = 1,

i = 51 degrees,

r = 19.5 degrees. ,

n2= ?

So,

1*sin 51 = n2 sin 19.5  

=> n2 = sin51 / sin19.5

 = 2.33

7 0
3 years ago
The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m
Simora [160]

Answer:

  • The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

F = k \frac{q_1 q_2}{d^2}

where k is Coulomb's constant, q_1 and q_2 are the charges and d is the distance between the charges.

Working a little the equation, we can take:

d^2 = k \frac{q_1 q_2}{F}

d = \sqrt{ k \frac{q_1 q_2}{F}}

And this equation will give us the distance between the charges. Taking the values of the problem

k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00

(the force has a minus sign, as its attractive)

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}

d = \sqrt{ 28,462,500 \ m^2}}

d = 5,335.026 m

And this is the distance between the charges.

3 0
3 years ago
A car traveling at 26 m/s starts to decelerate steadily. It comes to a complete stop in 13 seconds. What is its acceleration?
Alenkasestr [34]

accn = - 26m/s divided by 13 secs

accn = 2m/s/s

8 0
3 years ago
1. which of the following should be classified as a suspension?
Mumz [18]
Where are the answers to this question?
3 0
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