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jeyben [28]
3 years ago
10

A research study that proposed to describe the behaviors of high school teachers would be

Physics
2 answers:
german3 years ago
5 0

Answer:

That's a question?

Explanation:

Dnndnxnskwmem fmwo

nikdorinn [45]3 years ago
5 0

Answer:

<em>The Correct Answer is </em><u><em>D</em></u>

Explanation:

d) Descriptive

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Applying the Law of Conservation of Energy. If a car was released down the track from a height what happens to the potential ene
erastova [34]

Answer:

According to the law of conservation of energy, energy cannot be created or destroyed,  although it can be changed from one form to another.    KE + PE = constant. A simple example involves a stationary car at the top of a hill.  As the car coasts down the hill, it moves faster and so it’s kinetic energy increases and it’s potential energy decreases.  On the way back up the hill, the car converts kinetic energy to potential energy.  In the absence of friction, the car should end up at the same height as it started.

This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum.  When the ball is at the top of its swing, all of the pendulum’s energy is potential energy.   When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy.   The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

4 0
2 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
3 years ago
A clamp-type measuring instrument operates on the principle of
Hitman42 [59]

<em>A clamp-type measuring instrument operates on the principle of; </em>

A. induction

8 0
3 years ago
Speedy Sue, driving at 34.0 m/s, enters a one-lane tunnel. She then observes a slow-moving van 160 m ahead traveling at 5.20 m/s
Zielflug [23.3K]

Answer:

there will be collision

Explanation:

v_{s} =  speed of sue = 34 m/s

v_{v} = speed of van = 5.20 m/s

v_{sv} = speed of sue relative to van  = v_{s} - v_{v} = 34 - 5.20 = 28.8 m/s

d_{s} = stopping distance after brakes are applied

D = distance between sue and van = 160 m

v_{f} = final speed of sue = 0 m/s

a = acceleration = - 1.80 m/s²

Using the kinematics equation

v_{f}^{2} = v_{o}^{2} + 2 a d_{s}

0^{2} = 28.8^{2} + 2 (1.80) d_{s}

d_{s} = 230.4 m

Since  d_{s} < D

hence there will be collision

7 0
3 years ago
OK brainly if you want hard here here are sixty cups on a table. If one falls down, then how many remain
vivado [14]

Answer:

If one cup falls down then there will be 59 cups left.

5 0
3 years ago
Read 2 more answers
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