A narrow region between two air masses of different densities is a

Explain what the ionsphere is and how it interacts with some radio waves

mrs_skeptik [129]

Answer:

not sure

Explanation:

8 0

3 years ago

Identify the phyla of the organisms on the basis of following distinct characteristics:

MrRissso [65]

Answer:

Phylum Annelida commonly referred as segmented worms possess long , cylindrical and segmented body .

Phylum Aschelminthes commonly referred as round worms possess long , cylindrical , unsegmented body and show sexual dimorphism .

Phylum Echinodermata which includes star fish have tube feet as locomotory organ .

Phylum Porifera commonly referred as pore bearing animals and are diploplastic which includes euspongia etc.

3 0

3 years ago

A person jumps out a fourth-story window 14 m above a firefighter safety net. The survivor stretches the net 1.8 m before coming

Monica [59]

Answer:

The deceleration is $a = - 76.27 m/s^2$

Explanation:

From the question we are told that

The height above firefighter safety net is $H = 14 \ m$

The length by which the net is stretched is $s = 1.8 \ m$

From the law of energy conservation

$KE_T + PE_T = KE_B + PE_B$

Where $KE_T$ is the kinetic energy of the person before jumping which equal to zero(because to kinetic energy at maximum height )

and $PE_T$ is the potential energy of the before jumping which is mathematically represented at

$PE_T = mg H$

and $KE_B$ is the kinetic energy of the person just before landing on the safety net which is mathematically represented at

$KE_B = \frac{1}{2} m v^2$

and $PE_B$ is the potential energy of the person as he lands on the safety net which has a value of zero (because it is converted to kinetic energy )

So the above equation becomes

$mgH = \frac{1}{2} m v^2$

=> $v = \sqrt{2 gH }$

substituting values

$v = 16.57 m/s$

Applying the equation o motion

$v_f = v + 2 a s$

Now the final velocity is zero because the person comes to rest

So

$0 = 16.57 + 2 * a * 1.8$

$a = - \frac{16.57^2 }{2 * 1.8}$

$a = - 76.27 m/s^2$

4 0

3 years ago

When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le

77julia77 [94]

To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

$x+2.93$

Both cover a magnitude of 8.32 ft, therefore

$x +(x+2.97) = 8.32$

Now solving for x we have,

$x + (x + 2.93) = 8.32$

$2x + 2.93 = 8.32$

$2x = 8.32 - 2.93$

$x = \frac{ 8.32 - 2.93}{2}$

$x = 2.695 ft$

Therefore the shorter stick is 2.695ft long.

7 0

3 years ago

A mass is attached to a spring with an unknown spring constant. The spring gains 10 J of elastic potential energy if stretched b

Firlakuza [10]

From the given information in the question, the correct option is Option 1: 14 cm.

A non-stretched elastic spring has a conserved potential energy which gives it the ability to perform work. The elastic potential energy can be expressed as:

PE = $\frac{1}{2}$ k $x^{2}$

Where PE is the energy, k is the spring constant and x is extension.

i. Given that: PE = 10 J and x = 10 cm, then;

PE = $\frac{1}{2}$ k $x^{2}$

10 = $\frac{1}{2}$ k $10^{2}$

20 = 100k

k = 0.2 J/cm

ii. To determine how far the spring is needed to be stretched, given that PE = 20 J.

PE = $\frac{1}{2}$ k $x^{2}$

20 = $\frac{1}{2}$ (0.2) $x^{2}$

40 = 0.2 $x^{2}$

$x^{2}$ = 200

x = $\sqrt{200}$

= 14.1421

x = 14.14 cm

So that;

x is approximately 14.00 cm.

Thus, the spring need to be stretched to 14.00 cm to give the spring 20 J of elastic potential energy.

For more information, check at: brainly.com/question/1352053.

8 0

3 years ago