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Digiron [165]
3 years ago
7

This is for science

Physics
1 answer:
andriy [413]3 years ago
6 0

Answer:

Look at this picture hopefully it helps

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The theory of plate tectonics describes _______.
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I believe the answer is C
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Kepler's third law is founded on a mathematical formula that is based on
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Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum
Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

m is the mass of the man

v is the speed after jumping

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0
3 years ago
A helicopter descends from a height of 600 m with uniform negative acceleration, reaching the ground at rest in 5.00 minutes. de
uranmaximum [27]
Given:
h = 600 m, the height of descent
t = 5 min = 5*60 = 300 s, the time of descent.

Let a =  the acceleration of descent., m/s².
Let u =  initial velocity of descent, m/s.
Let t = time of descent, s.
The final velocity is v = 0 m/s because the helicopter comes to rest on the ground.
Note that u,  v, and a are measured as positive upward.

Then
 u + at = v
(u m/s) + (a m/s²)*(t s) = 0
u = - at
u = - 300a                  (1)

Also,
u*t + (1/2)at² = -h
(um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600
ut + (1/2)at² = 600       (2)
From (1), obtain
-300a +(1/2)(a)(90000) = -600
44700a = -600
a = - 1.3423 x 10⁻² m/s²

From (1), obtain
u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s

Answer:
The acceleration is 0.0134 m/s² downward.
The initial velocity is 4.0 m/s upward.

3 0
3 years ago
Vygotsky believed that scaffolding limited communication between the teacher and the learner.
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Answer: The given statement is False.

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He described scaffolding as the support given to child by more knowledgeable others ( like parents or teachers). This means that scaffolding allowed communication between the teacher and the learner.

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