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3 years ago

Which statement is true of an object in equilibrium

2 answers:

8 0

If I remember correctly, it is a state an object is in when no force is acting on it or when there is equal amount of force in each direction is being applied to it. Basically, either the object is motionless or is traveling at a constant velocity, there is no acceleration.

Olin [163]3 years ago

6 0

The rate of the forward reaction is equal to the rate of the reverse reaction.

Equilibrium is reached when the rates of the forward and reverse reactions are the same.

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B c<br>_ - _<br>a b<br>( a=3 b=-5 c=6)

Pachacha [2.7K]

(-5)/3 - 6/(-5)
You can solve it now :)

7 0

2 years ago

Which best explains why a wood-burning fireplace represents an open system?

Alona [7]

The answer will be B

5 0

3 years ago

Three resistors of 100 W, 3900 W, and 1000 W are connected in series across a 200-V battery. What is the voltage drop across the

Gala2k [10]

Answer:

40 V

Explanation:

I will assume that the resistors are

100 and 3900 and 1000 OHMS <=====(NOT W)

In series , the resistances add together 100 + 3900 + 1000 = 5000 ohms total

V = IR

I = V / R so the total current will be 200 v / 5000 ohms = .04 amps

this is the current through all of the resistors

so for the 1000 ohm resistor V = IR .04 (1000) = 40 V

7 0

2 years ago

An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$

$V_{f1}=\frac{mV_0+m_TV_O}{m-m_O}$

$V_{f1}=\frac{(98.7)*0+5.3*15}{98.7-5.3}$

$V_{f1}=0.8511m/s$

B) $7.9Kg\rightarrow 11.2m/s$

$V_{f2}=\frac{mV_{f1}+m_TV_O}{m-m_O}$

$V_{f2}=\frac{(98.7)(0.8511)+(7.9)(11.2)}{98.7-5.3-7.9}$

$V_{f2} = 2.0173m/s$

C) $10.5Kg\rightarrow 7m/s$

$V_{f3}=\frac{mV_{f2}+m_TV_O}{m-m_O}$

$V_{f3}=\frac{(98.7)(2.0173)+(10.5)(7)}{98.7-5.3-7.9-10.5}$

$V_{f3} = 3.63478m/s$

Therefore the final velocity of astronaut is 3.63m/s

7 0

3 years ago

If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be

zalisa [80]

If Earth was twice as far from the sun, the force of gravity attracting the Earth to the sun would be only one-quarter as strong. The correct answer will be C.

6 0

2 years ago

Read 2 more answers