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3 years ago

Describe the energy transformations that occur from the time a skydiver jumps out of a plane until landing on the ground.

2 answers:

6 0

When the Skydiver jump out a plane, his Potential Energy is being converted or transform into Kinetic energy due to gravity. Hope this helps

omeli [17]3 years ago

4 0

Answer:

Before jumping from the plane, the skydiver has potential energy. When the skydiver jumps, the potential energy is transformed into kinetic energy, which increases until the skydiver reaches terminal velocity. Potential energy is then transformed into thermal energy.

Explanation:

Thats the answer

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Given that average speed is distance traveled divided by time, determine the values of m and n when the time it takes a beam of

emmainna [20.7K]

Answer:

$5,2$

Explanation:

From the question we are told that:

Speed of light $C=3.0×10^8 m/s.$

Generally the equation for Average Speed is mathematically given by

$V_{avg}=\frac{d}{t}$

Where

d=Distance between the Earth and the sun

$d=1.5*10^11m$

Therefore

$t=\frac{d}{V_{avg}}$

$t=\frac{1.5*10^11m}{3.0×10^8 m/s.}$

$t=5*10^2s$

Since m and n is given in the form of

$m*10^n$

Therefore

$m=5 & n=2$

$5,2$

3 0

3 years ago

In the design of wall and column forms, the two most important factors are which of the following? a. rate of placement of the c

Simora [160]

Answer:

c. length of the wall or column and the rate of placement of the concrete

Explanation:

when designing for wall and column form-works, it is of utmost important to know the length of the wall and the type of concrete placement to be used.

Concrete placement has methods and precaution to be taken when doing the form work

if the concrete placement is manually (hand or funnel) the form work height should not be more than 1 m to enable easy compaction and vibration of concrete in the form.

Also, if the form work length is too long and it is not well reinforced, it tends to burg if the force apply during concrete placement or during vibration is much.

5 0

3 years ago

Look at the image to answer the question correctly.

r-ruslan [8.4K]

Answer:

1- b: 2- a : 3- c : 4- d

Explanation:

it starts 2 move away from strting point, then no motion, then moves toward the start, the slows up.

3 0

2 years ago

An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

<u>Given:</u>

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

<u>Assume:</u>

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

3 years ago

Which would BEST describes what occurs when a ball is thrown against a wall? A) The ball will not bounce off the wall. B) The ba

oee [108]

Answer:

D) The ball exerts a force on the wall and the wall exerts a force back.

Explanation:

Newton's third law of motion states that:

"When an object A exerts a force on another object B, then object B exerts an equal and opposite force on object A"

In this problem, we can identify (for instance) object A with tha ball and object B with the wall. Therefore, if we apply Newton's third law, we get:

The ball (object A) exerts a force on the wall (object B), therefore the wall (object B) exerts an equal and opposite force on the ball (object A). So, option D is the correct one.

3 0

3 years ago

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