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3 years ago

What are two functions of the respiratory system

Physics

2 answers:

babymother [125]3 years ago

8 0

The lungs high function to take in oxygen and expel carbon dioxide

sp2606 [1]3 years ago

7 0

The primary organs of the respiratory system are lungs, which function to take in oxygen and expel carbon dioxide as we breathe. The human respiratory system is a series of organs responsible for taking in oxygen and expelling carbon dioxide.Mar 11, 2016

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A bomb at rest at the origin of an xy-coordinate system explodes into three pieces. Just after the explosion, one piece, of mass

ValentinkaMS [17]

Answer:

Explanation:

a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.

b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.

If v be the velocity of the third part along a direction making angle θ

with x axis ,

x component of v = vcosθ

So momentum along x axis after explosion of third part = mv cosθ

= 10 v cosθ

Momentum along x of first part = - 5 x 42 m/s

momentum of second part along x direction =0

total momentum along x direction before explosion = total momentum along x direction after explosion

0 = - 5 x 42 + 10 v cosθ

v cosθ = 21

Similarly

total momentum along y direction before explosion = total momentum along y direction after explosion

0 = - 5 x 38 + 10 v sinθ

v sinθ= 21

squaring and and then adding the above equation

v² cos²θ +v² sin²θ = 21² +19²

v² = 441 + 361

v = 28.31 m/s

Tanθ = 21 / 19

θ = 48°

6 0

3 years ago

The speed of a car cruising along a highway was 90.0km h^-1. What is this value in meters per second?

faust18 [17]

Answer:

25 m/s

Explanation:

1 km/h = .277778 m/s, so 90 x .277778 = 25

5 0

3 years ago

What is the current in a wire of radius R = 2.02 mm if the magnitude of the current density is given by (a) Ja = J0r/R and (b) J

Sloan [31]

Explanation:

For this problem we have to take into account the expression

J = I/area = I/(π*r^(2))

By taking I we have

I = π*r^(2)*J

(a)

For Ja = J0r/R the current is not constant in the wire. Hence

$I(r) = \pi r^{2} J(r) = \pi r^{2} J_{0}r/R = \pi r^{3} (3.74*10^{4}A/m^{2})/(2.02*10^{-3}m)$

and on the surface the current is

$I(R) = \pi r^{2} J(R) = \pi r^{2} J_{0}R/R = \pi(2.02*10^{-3})^{2} (3.74*10^{4}) = 0.47 A$

(b)

For Jb = J0(1 - r/R)

$I(r)=\pi r^{2}J(r) =\pi r^{2} J_{0}(1 - r/R)=\pi r^{2}J_{0}(1-\frac{r}{2.02*10^{-3}} )$

and on the surface

$I(R)=\pi r^{2}J_{0}(1-R/R)=\pi r^{2}J_{0}(1-1)= 0$

(c)

Ja maximizes the current density near the wire's surface

Additional point

The total current in the wire is obtained by integrating

$I_{T}=\pi\int\limits^R_0 {r^{2}Ja(r)} \, dr = \pi \frac{J_{0}}{R}\int\limits^R_0 {r^{3}} \ dr =\pi \frac{J_{0}R^{4}}{4R}=\frac{1}{4}\pi J_{0}R^{3}=2.42*10^{-4} A$

and in a simmilar way for Jb

$I_{T}=\pi J_{0} \int\limits^R_0 {r^{2}(1-r/R)} \, dr = \pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2R}]=\pi J_{0}[\frac{R^{3}}{3}-\frac{R^{2}}{2}]$

And it is only necessary to replace J0 and R.

I hope this is useful for you

regards

7 0

3 years ago

Pleaseeeee help me with b, c, and d. There are no angles.

taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, $F_{NET}$ = F - $F_{Friction}$ = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

$F_{Friction}$ = The force of friction

∴ $F_{NET}$ = m × a = m × 0 m/s² = 0 N

Therefore;

$F_{NET}$ = F - $F_{Friction}$ = 0 N

From which we have;

F = $F_{Friction}$ = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = $F_{Friction}$ × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, $d_p$ = 0 m

∴ The work done by the normal direction = N × $d_p$ = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, $d_{wieght}$ = 0 m

Therefore, the work done by the weight = W × $d_{wieght}$ = W × 0 m = 0 J

The work done by the weight = 0 J

7 0

3 years ago

The radius RH of a black hole, also known as the event horizon, marks the location where not even light itself can escape from t

olga nikolaevna [1]

Solution:

a) We know acceleration due to gravity, g = GM/r²

Differential change, dg/dr = -2GM/r³

Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²

My height, h=dr = 1.7 m

Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²

or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³

or, 5.9*100³*G²*M² = 2c⁶

or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms

Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.

b) This limit an upper limit ,we can tolerate smaller masses only.

4 0

3 years ago