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3 years ago

How many seconds will it take for a the International Space Station to travel 450 km at a rate of 100 m/s?

Physics

1 answer:

SVEN [57.7K]3 years ago

8 0

Time = (distance) / (speed)

Time = (450 km) / (100 m/s)

Time = (450,000 m) / (100 m/s)

Time = <em>4500 seconds </em>(that's 75 minutes)

Note:

This is about HALF the speed of the passenger jet you fly in when you go to visit Grandma for Christmas.

If the International Space Station flew at this speed, it would immediately go ker-PLUNK into the ocean.

The speed of the International Space Station in its orbit is more like 3,100 m/s, not 100 m/s.

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A car is traveling around a horizontal circular track with radius r = 210 m at a constant speed v = 23 m/s as shown. The angle θ

Fantom [35]

Centripetal acceleration of car is given by formula

$a_c = \frac{v^2}{R}$

now plug in the values in this

$a_c = \frac{23^2}{210}$

$a_c = 2.52 m/s^2$

Part b)

At position A we have

x component of acceleration is given as

$a_x = -a_c cos23$

$a_x = -2.32 m/s^2$

Part c)

At position A we have

y component of acceleration is given as

$a_y = -a_c sin23$

$a_y = -0.98 m/s^2$

Part d)

At position B we have

x component of acceleration is given as

$a_x = -a_c cos53$

$a_x = -1.52 m/s^2$

Part e)

At position B we have

Y component of acceleration is given as

$a_y = a_c sin53$

$a_y = 2.01 m/s^2$

6 0

3 years ago

Most towns use a water tower to store water and provide pressure in the pipes that deliver water to customers. The figure below

MatroZZZ [7]

The effective height of the water for Smith's house will be 24.61m.

<h3>How to calculate the height?</h3>

Based on the information given, the volume of the water in sphere will be:

= 4/3πr³ = (5.80 × 10^5)/1000

= 4.18r³ = 580

r³ = 138.7

r = 5.18m

The effective height of the water will be:

= 18.0 + 2(5.18)

= 28.36

The gauge pressure at Faucet of Jones house will be:

= pgh

= 1000(9.8)(28.36)

= 277.9kPa

The effective height of the water for Smith's house will be:

= 18.0 + 2(5.18) - 3.75

= 24.61m

The gauge pressure at Faucet of Jones house will be:

= 1000 × 9.8 × 24.61

= 241.2kPa

Learn more about height on:

brainly.com/question/983412

#SPJ1

8 0

1 year ago

A runner traveled 7.7 miles an hour and a half what was their average speed

Alenkasestr [34]

Answer:5.13333333...

Explanation: 7.7 is the distance and an hour and a half is the average speed. You would have to divide the total distance by the total time. So it would be 7.7 divided by 1.5 which would equal 5.13333333333...

6 0

3 years ago

A particular light source gives off light waves with a measured wavelength of

Umnica [9.8K]

The frequency of the light source is 1.5 x 10¹⁵ Hz.

<h3>Frequency of the light source</h3>

The frequency of the light source is determined using the following equations;

c = fλ

where;

c is speed of light

f is the frequency

λ is the wavelength

f = (3 x 10⁸) / (2 x 10⁻⁷)

f = 1.5 x 10¹⁵ Hz

Thus, the frequency of the light source is 1.5 x 10¹⁵ Hz.

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3 0

2 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

3 years ago