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NARA [144]
2 years ago
13

The Earth's biosphere is consists of

Physics
1 answer:
Firdavs [7]2 years ago
8 0

Answer:

The part of the earth and its atmosphere in which living organisms exist or that is capable of supporting life.

Explanation:

It’s in the definition of biosphere. “Bio” means “life,” so “biosphere” is that portion of the Earth capable of supporting life.

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At what speed would a 3.00 x 10^4 kg airplane have to fly and with a momentum of 1.60 x 10^9 kg.m/s
Ymorist [56]

Answer:

5.3×10⁴ m/s

Explanation:

From the question,

Momentum = mass× velocity

M = mV................ Equation 1

Where M = momentum of the airplane, m = mass of the airplane, V = Velocity of the airplane

make V the subject of the equation

V = M/m.................. Equation 2

Given: M = 1.6×10⁹ Kg.m/s, m = 3.0×10⁴ kg

Substitute these values into equation 2

V = 1.6×10⁹/3.0×10⁴

V = 5.3×10⁴ m/s

3 0
3 years ago
3) If an airplane is in flight cruising at constant velocity, what can be said about the net work
Crazy boy [7]

Explanation:

<h2>Yes!</h2>

<h3>In physics, constant velocity occurs when there is no net force acting on the object causing it to accelerate. In terms of airplane flight, the two main forces influencing its velocity forward are drag and thrust. At a constant altitude, when the force of thrust equals the opposing force of drag, then the airplane will experience uniform motion in one direction. This can be further explained by Newton’s First Law. </h3>
6 0
3 years ago
Why does air blows from one place to another​
Maslowich

Air blows from one place to another because gases move from high-pressure areas to low-pressure areas

In simple words

it happens because of pressure differences.

3 0
2 years ago
Read 2 more answers
.A hard rubber ball, released at chest height, falls to the pavement and bounces back to nearly the same height. When it is in c
ohaa [14]

Answer:

 a = 1.1 10⁵ m / s²

Explanation:

This is a momentum exercise, where we use the relationship between momentum and momentum

          I = ∫ F dt = Δp

= p_f - p₀

as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction

        F t = 2 (m v)

therefore the average force is

         F = 2 m v / t

where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy

starting point. Done the ball is released with zero initial velocity

        Em₀ = U = mgh

final point. Upon reaching the ground, just before the deformation begins

        Em_f = K = ½ m v²

energy is conserved in this system

        Em₀ = Em_f

        m g h = ½ m v²

        v = √ (2gh)

This is the velocity of the body when it reaches the ground, so the force remains

        F = 2m √(2gh)   /t

where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds

knowing this force let's use Newton's second law

          F = m a

          a = F / m

 

          a = 2 √(2gh) / t

We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s

         a = 2 √ (2 9.8 1.5) / 10⁻³

         a = 1.1 10⁵ m / s²

6 0
2 years ago
At a distance of 11 cm from a presumably isotropic, radioactive source, a pair of students measure 65 cps (cps = counts per seco
Alborosie

To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.

The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,

A_1*m=M*A_2

A_i =Area

M,m = Counts per second

Our radios are given by

r_1 = 11cm

R_2 = 20cm

m = 65cps

Therefore replacing we have that,

A_1*m=M*A_2

4\pi r_1^2*m = M * 4\pi R_2^2 M

r^2*m=MR^2

M = \frac{m*r^2}{R^2}

M = \frac{65*11^2}{20^2}

M = 19.6625cps

Therefore the number of counts expect at a distance of 20 cm is 19.66cps

7 0
3 years ago
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