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3 years ago

A current relationship is one that

Physics

2 answers:

Alex3 years ago

8 0

C is the best answer
hope it helped

Alecsey [184]3 years ago

5 0

The Answer To This Question Is C (Active Now)
Hope This Helps!

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Describe the electron transfers that occur in the formation of calcium fluoride from elemental calcium and elemental fluorine.

boyakko [2]

Answer:

Check explanation

Explanation:

In the formation of calcium fluoride we take calcium and fluorine.

in elemental form calcium exist in solid form and fluorine in gaseous form.

formation of compound takes place to complete their octet, in case of calcium need to remove two electron and need to add one elecron in fluorine to complete their octet so two electron will ransferred from calcium to two fluorine atom.

3 0

2 years ago

What is the modern periodic table of elements?

Kamila [148]

C is the correct answer

3 0

2 years ago

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A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left

Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

Length of the massless beam; $L = 4.00m$

Distance of support from the left end; $x = 3.00m$

First mass; $m1 = 31.3 kg$

Distance of beam from the left end( m₁ is attached to ); $x_1 = ?$

Second mass; $m_2 = 61.7 kg$

Distance of beam from the right of the support( m₂ is attached to ); $x_1 = 0.273m$

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, $m_1g( x-x_1) = m_2gx_2$

we divide both sides by $g$

$m_1( x-x_1) = m_2x_2$

Next, we make $x_1$ , the subject of the formula

$x_1 = x - [ \frac{m_2x_2}{m_1} ]$

We substitute in our given values

$x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]$

$x_1 = 3.00m - 0.538m$

$x_1 = 2.46m$

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0

2 years ago

A ball is thrown up at a speed of 20 m/s.

ioda

Given:

(Initial velocity)u=20 m/s

At the maximum height the final velocity of the ball is 0.

Also since it is a free falling object the acceleration acting on the ball is due to gravity g.

Thus a=- 9.8 m/s^2

Now consider the equation

v^2-u^2= 2as

Where v is the final velocity which is measured in m/s

Where u is the initial velocity which is measured in m/s

a is the acceleration due to gravity measured in m/s^2

s is the displacement of the ball in this case it is the maximum height attained by the ball which is measured in m.

Substituting the given values in the above formula we get

0-(20x20)= 2 x- 9.8 x s

s= 400/19.6= 20.41m

Thus the maximum height attained is 20.41 m by the ball

6 0

2 years ago

A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Pig. P7.41). (a) How fast

Leona [35]

Answer:

a)Velocity of car =v=16 m/s

b)Force against the track at point B=1.15* $10^{4}$ N

Explanation:

Given mass of roller coaster=m=350 kg

Position of A=Ha=25 m

Position of B=Hb=12 m

Net potential energy=mg(ha-hb)

Net potential energy=(350)(9.80)(25-12)

Net potential energy=44590 J

Using energy conservation

net kinetic energy=net potential energy

(1/2)mv^2=mg(ha-hb)

m=350

velocity=v=16 m/s

b)There two force acting,centripetal force upward and gravity downward.

Thus net force acting will be

Net force=(mv^2/r)-mg

Net force=14933.33-3430

Net force=1.15* $10^{4}$ N

3 0

2 years ago

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