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3 years ago

6

If your weight is force created by gravity on your body, and therefore is the Earth pulling you down, what is the reaction force

?

1 answer:

IRISSAK [1]3 years ago

3 0

Answer:

The reaction force is the force exerted by your body on the earth pulling upwards.

Explanation:

This can be explained using Newton's third law of motion which states that if object A exerts a force on object B, object B exerts a force of equal magnitude and opposite direction on object A. One characteristics of these force are that they act o 2 different bodies. In this example, the earth exerts a force on you. The Newton's reaction force would be you exerting a force of equal magnitude on the earth.

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den301095 [7]

Less than or equal to the magnitude of the vector

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3 years ago

If the temperature is held constant during this process and the final pressure is 683 torrtorr , what is the volume of the bulb

Anna [14]

Answer:

Explanation:

Let the volume of the unknown bulb = X L

The volume of the system , after opening valve = (X + 0.72 L )

Use Boyles law gas equation,

P1V1 = P2V2 ( at temperature is constant )

Given:

P1 = 1.2 atm

P2 = 683 torr

Converting mmHg to atm,

1 atm = 760 mmHg(torr)

683 mmHg = 683/760

= 0.8987 atm

1.2X = 0.8987*(X + 0.720)

1.2X = 0.8987X + 0.6471

0.3013X = 0.6471

X = 2.15 L

5 0

3 years ago

A metal bar has a frictionless axle going through its center of mass. You notice that the bar is not level (flat), but that it i

ANTONII [103]

Answer:

yep

Explanation:

5 0

3 years ago

A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.

babunello [35]

A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.

Read more on Brainly.com - brainly.com/question/18743384#readmore

8 0

3 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