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Natasha_Volkova [10]
3 years ago
11

Three oxygen isotopes

Physics
1 answer:
igomit [66]3 years ago
6 0
Stable isotopes, radioisotopes, oxygen

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A technique in which people use machines to learn how to control their bodies is known as __________.
Debora [2.8K]

Answer:

The technique in which people use machines to learn how to control their bodies is known as D, Biofeedback.

Explanation:

Biofeedback is a variety of different machines that help people learn how to control their bodies depending on their specific needs, varying from things like scalp sensors, electrocardiographs, electromyographs and more.

7 0
3 years ago
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Fynjy0 [20]
C real,inverted and smaller than the object
6 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i
evablogger [386]

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 \frac{m}{s}

To find:

the total time it is in the air = ?

Formula used:

t = \frac{v-u}{a}

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = \frac{v-u}{a}

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = \frac{0-8}{-9.8}

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

4 0
3 years ago
As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, afte
Klio2033 [76]

Answer:

 r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

Explanation:

For this exercise we must use the principle of conservation of energy

starting point. The proton very far from the nucleus

          Em₀ = K = ½ m v²

final point. The point where the proton is stopped (v = 0)

          Em_f = U = q V

where the potential is

          V = k Ze / r²

Let us consider that all the charge of the nucleus is in the center, therefore r is the distance from this point to the proton that is approaching

Energy is conserved

          Em₀ = Em_f

           ½ m v² = e (k \frac{Ze}{r^2})

          r^2 = \frac{ 2 k  \ Ze^2}{ 2m}

with this expression we can find the closest approach distance (r)

3 0
3 years ago
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