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3 years ago

11

Oxygen and nutrients are moved through the body by the: a. circulatory system c. lymphatic system b. digestive system d. reprodu

ctive system

1 answer:

vagabundo [1.1K]3 years ago

3 0

A. The Circulatory System.

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An airplane flies 700 miles with a tail wind in 2.0 hrs. If it takes 2.5 hrs to cover the same distance against the headwind, th

nikklg [1K]

Answer:

B).315mph

Explanation:

Let the speed of the plane = p

Let the speed of the wind = w

Set up the system equation as;

Relative V: Time: Distance:

in wind direction: p + w 2 700

against wind: p - w 2.5 700

2(p + w) = 700

2.5(p - w) = 700

2p + 2w = 700

2.5p - 2.5w = 700

2.5 x: 5p + 5w = 1750

2 x: 5p - 5w = 1400

10p = 3150

p = 315 mph

Therefore, he speed of the plane in still air is 315 mph

6 0

4 years ago

Please answer ASAP .

Arisa [49]

Answer:

Explanation:

a = gsinθ

a = 9.8sin30

a = 4.9 m/s²

v² = u² + 2as

v² = 0² + 2(4.9)(10)

v² = 98

v = √98 = 9.8994949...

v = 9.9 m/s

8 0

3 years ago

An object moving due to gravity can be described by the motion equation y=y0+v0t−12gt2, where t is time, y is the height at tha

LenaWriter [7]

Answer:

$t=6.4534 s$

Explanation:

This is an exercise where you need to use the concepts of <em>free fall objects</em>

Our <u>knowable variables</u> are initial high, initial velocity and the acceleration due to gravity:

$y_{0}=75m$

$v_{oy} =20m/s$

$g=9.8 m/s^{2}$

At the end of the motion, the <u><em>rock hits the ground</em></u> making the final high y=0m

$y=y_{o}+v_{oy}*t-\frac{1}{2}gt^{2}$

If we <em>evaluate the equation</em>:

$0=75m+(20m/s)t-\frac{1}{2}(9.8m/s^{2})t^{2}$

This is a classic form of <u><em>Quadratic Formula</em></u>, we can solve it using:

$t=\frac{-b ± \sqrt{b^{2}-4ac } }{2a}$

$a=-4.9\\b=20\\c=75$

$t=\frac{-(20) + \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=-2.37s$

$t=\frac{-(20) - \sqrt{(20)^{2}-4(-4.9)(75) } }{2(-4.9)}=6.4534s$

Since the <u><em>time can not be negative</em></u>, the <em>reasonable answer</em> is

$t=6.4534s$

8 0

3 years ago

Consider three identical metal spheres, A, B, and C. Sphere A carries a charge of 9q. Sphere B carries a charge of -q. Sphere C

Leona [35]

Answer:

Explanation:

A and B are touched .

charge on each of them after touching = (9q - q) / 2 = 4q

when C is touched with A

charge on A and C each after touching

= 4q + 0 / 2 = 2q

When C is touched with B

charge on each of them

(2q + 4q ) / 2 = 3 q

Final charge on C = 3q

6 0

4 years ago

What is the escape speed for an electron initially at rest on the surface of a sphere with a radius of 1.3 cm and a uniformly di

Vlad [161]

Answer:

2.37 * 10^4 m/s

Explanation:

Constants :

Mass of electron = 9.11 * 10^(-31) kg

Electric charge of an electron = 1.602 * 10^(-19) C

Parameters given:

Radius of sphere = 1.3cm = 0.013m

Charge of sphere = 2.3 * 10^(−15) C

Using the law of conservation of energy, we have that:

K. E.(initial) + P. E.(initial) = K. E.(final) + P. E.(final)

K. E.(final) = 0, since final velocity is zero and P. E.(final) = 0 since the electron reaches a final distance of infinity.

Hence,

K. E.(initial) = P. E.(initial)

0.5mv^2 = (kqQ)/r

Where k = Coulumbs constant

Q = charge of the sphere.

r = radius of the sphere.

=> 0.5*m*v^2 = (kqQ)/r

0.5 * 9.11 * 10^(-31) * v^2 = (9 * 10^9 * 1.602 * 10^(-19) * 2.3 * 10^(-15))/0.013

4.555 * 10^(-31) * v^2 = 2550.88 * 10^(-25)

=> v^2 = 2550.88 * 10^(-25) / 4.555 * 10^(-31)

v^2 = 560 * 10^6 = 5.60 * 10^8

=> v = 2.37 * 10^4 m/s

4 0

3 years ago