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3 years ago

1. Which atmospheric layer is closest to planet earth?

Physics

1 answer:

Sphinxa [80]3 years ago

4 0

Troposphere: 0 to 12 km (0 to 7 miles)

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When you measure the boiling point of mercury, you are investigating a ___. a.chemical change b.chemical property c.physical cha

Amanda [17]

D. physical property

the bonds between molecules of mercury are breaking so it's physical and it's not changing the chemical composition of the substance

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3 years ago

A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run

pickupchik [31]

Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

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4 years ago

A bungee jumper with mass 65.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting

ololo11 [35]

Explanation:

It is given that,

Mass of a bungee jumper is 65 kg

The time period of the oscillation is 38 s, hitting a low point eight more times.It means its time period is

$T=\dfrac{38}{8}\\\\T=4.75\ s$

After many oscillations, he finally comes to rest 25.0 m below the level of the bridge.

For an oscillating object, the time period is given by :

$T=2\pi \sqrt{\dfrac{m}{k}}$

k = spring stiffness constant

So,

$k=\dfrac{4\pi ^2m}{T^2}\\\\k=\dfrac{4\pi ^2\times 65}{(4.75)^2}\\\\k=113.43\ N/m$

When the cord is in air,

mg=kx

x = the extension in the cord

$x=\dfrac{mg}{k}\\\\x=\dfrac{65\times 9.8}{113.6}\\\\x=5.6\ m$

So, the unstretched length of the bungee cord is equal to 25 m - 5.6 m = 19.4 m

5 0

3 years ago

What are some thermal insulators you can find in a normal house

adoni [48]

Body heat :)

i'm going du.mb here

6 0

3 years ago

A camera takes a properly exposed photo with a 4.0 mm diameter aperture and a shutter speed of 1/1000 s. If the photographer foc

hoa [83]

Answer:

option E

Explanation:

given,

diameter = 4 mm

shutter speed = 1/1000 s

diameter of aperture = ?

shutter speed = 1/250 s

exposure time to the shutter time

$E V = log_2(\dfrac{N^2}{t})$

N is the diameter of the aperture and t is the time of exposure

now,

$log_2(\dfrac{N^2}{t_1})=log_2(\dfrac{N^2}{t_2})$

$\dfrac{N_1^2}{t_1}=\dfrac{N_2^2}{t_2}$

inserting all the values

$\dfrac{4^2}{1000}=\dfrac{N_2^2}{250}$

N₂² = 4

N₂ = 2 mm

hence , the correct answer is option E

4 0

3 years ago