What respiratory structure controls breathing?

The sunâ€™s energy hits earthâ€™s surface at the most direct angle at which locations?

3241004551 [841]

Answer:

at the Equator

Explanation:

The four seasons are determined by four main positions in the Earth's orbit in its turn around the Sun (ecliptic plane), which are called solstices and equinoxes: winter solstice (Capricorn point, December 22), spring equinox (Aries point, around March 21-22), summer solstice (Cancer point, June 21) and autumn equinox (Libra point, around September 22-23).

In the equinoxes, the axis of rotation of the Earth is perpendicular to the sun's rays, which fall vertically over the equator. In solstices, the axis is inclined 23.5º, so that the sun's rays fall vertically on the Tropic of Cancer (summer in the northern hemisphere) or Capricorn (summer in the southern hemisphere).

When falling vertically on Ecuador, it generates a greater impact on the surface of the Tierre reaching a greater amount of energy and therefore UV rays.

7 0

3 years ago

Your name is Galileo Galilei and you toss a weight upward at 20 feet per second from the top of the Leaning Tower of Pisa (heigh

Veseljchak [2.6K]

Answer:

a) v(t) = -32.2 ft/s² · t + 20 ft/s

b) h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²

c) The weight will reach its maximum height after 0.62 s. The maximum height will be 190 feet.

Explanation:

Hi there!

a) Since the only force that acts on the weight is the gravity force, the object is under a constant downward acceleration g = -32.2 ft/s² (it is negative because we consider the upward direction as positive). The acceleration is the variation of the velocity over time (dv/dt). Then:

dv/dt = g

Separating variables:

dv = g dt

Integrating from the initial velocity, v0, to v and from t = 0 to t, we obtain:

v - v0 = g t

v = g t + v0

Then:

v(t) = -32.2 ft/s² · t + 20 ft/s

b) The velocity of the weight is the variation of the height over time:

dh/dt = v(t)

dh/dt = g t + v0

Separating varibles:

dh = g t dt + v0 dt

Integrating from initial height, h0, to h and from t = 0 to t:

h - h0 = 1/2 · g · t² + v0 · t

h = h0 + v0 · t + 1/2 · g · t²

Then:

h(t) = 184 ft + 20 ft/s · t - 1/2 · 32.2 ft/s² · t²

h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²

c) When the weight reaches its maximum height, its velocity will be zero. Then, using the equation of velocity we can obtain the time at which the weight is at the maximum height:

v(t) = -32.2 ft/s² · t + 20 ft/s

0 = -32.2 ft/s² · t + 20 ft/s

-20 ft/s/ -32.2 ft/s² = t

t = 0.62 s

The weight will reach its maximum height after 0.62 s.

The maximum height will be h(0.62 s):

h(t) = 184 ft + 20 ft/s · t - 16.1 ft/s² · t²

h(0.62 s) = 184 ft + 20 ft/s · (0.62 s) - 16.1 ft/s² · (0.62 s)²

h(0.62 s) = 190 ft

The maximum height will be 190 feet.

3 0

3 years ago

Read 2 more answers

One model for a certain planet has a core of radius R and mass M surrounded by an outer shell of inner radius R, outer radius 2R

Drupady [299]

(a) 120.8 m/s^2

The gravitational acceleration at a generic distance r from the centre of the planet is

$g=\frac{GM'}{r^2}$

where

G is the gravitational constant

M' is the mass enclosed by the spherical surface of radius r

r is the distance from the centre

For this part of the problem,

$r=R=1.17\cdot 10^6 m$

so the mass enclosed is just the mass of the core:

$M'=M=2.48\cdot 10^{24}kg$

So the gravitational acceleration is

$g=\frac{(6.67\cdot 10^{-11})(2.48\cdot 10^{24}kg)}{(1.17\cdot 10^6 m)^2}=120.8 m/s^2$

(b) 67.1 m/s^2

In this part of the problem,

$r=3R=3(1.17\cdot 10^6 m)=3.51\cdot 10^6 m$

and the mass enclosed here is the sum of the mass of the core and the mass of the shell, so

$M'=M+4M=5M=5(2.48\cdot 10^{24}kg)=1.24\cdot 10^{25}kg$

so the gravitational acceleration is

$g=\frac{(6.67\cdot 10^{-11})(1.24\cdot 10^{25}kg)}{(3.51\cdot 10^6 m)^2}=67.1 m/s^2$

8 0

4 years ago

A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from

Harman [31]

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

$F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s$

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

5 0

4 years ago

A wagon is rolling forward on level ground. Friction is negligible. The person sitting in the wagon is holding a rock. The total

Harlamova29_29 [7]

Answer:

a)0.48 m/s

b) 0.583 m/s

Explanation:

As the wagon rolls,

momentum'p'= m x v => 95.8 x 0.530 = 50.774 Kgm/s

(a)Rock is thrown forward,

momentum of rock = 0.325 x 15.1 = 4.9075 Kgm/s

Conservation of momentum says momentum of wagon is given by

50.774 - 4.9075 = 45.8665

Therefore, Speed of wagon = 45.8665 / (95.8-0.325) = 0.48 m/s

(b) Rock is thrown backward,

momentum of wagon = 50.774 + 4.9075 = 55.68 Kgm/s

Therefore, speed of wagon = 55.68 / (95.8-0.325) = 0.583 m/s

4 0

4 years ago