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jeyben [28]
3 years ago
10

Two 1.1 kg masses are 1 m apart (center to center) on a frictionless table. Each has +10 JC of charge. What is the initial accel

eration (in m/s2) of this massif it is released and allowed to move?
Physics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

acceleration = 0.8181 m/s²

Explanation:

given data

mass = 1.1 kg

apart d = 1 m

charge q = 10 μC

to find out

What is the initial acceleration

solution

we know that acceleration is

acceleration = \frac{force}{mass}   .................1

here force = k \frac{q1q2}{r^2}

here q1 q2 is charge and r is distance and Coulomb constant k = 9 × 10^{9} Nm²/C²

force = 9*10^{9} \frac{(10*10^{-6})^2}{1^2}

force = 0.9 N

so  from equation 1

acceleration = \frac{0.9}{1.1}

acceleration = 0.8181 m/s²

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snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
If you are six feet tall how far back from a 3 foot mirror do you have to stand in order to see yourself completely?
OverLord2011 [107]

Answer:

you would have to stand 6 ft back

Explanation:

7 0
3 years ago
Help meee please fill in the blanks !!
Ratling [72]

The atomic number is the simply the number of protons in the atom. So in the first row with atomic number 2, the number of protons is 2.

If the atom has no charge, which I think you can assume for all of these, the number of electrons is equal to the number of protons. So the number of electrons is also 2.

The number of neutrons (which are the particles with no charge in the nucleus) is simply the mass number minus the atomic number i.e. 4 - 2 = 2.

The isotopic symbol is the symbol which is found on the periodic table of elements. There are 2 numbers associated which each element on the table. The larger is the mass number and the smaller is the atomic number. The atomic number or number of protons is what identifies the element. Looking at the periodic table ( https://sciencenotes.org/wp-content/uploads/2015/01/PeriodicTableOfTheElementsBW.pdf or https://simple.wikipedia.org/wiki/Periodic_table_(big) ), it can be seen that the element on the first row above with an atomic number of 2 is Helium with a symbol He. The number that is included with the name is simply the mass number which is 4 in this case, which tells us that this type of helium has 2 neutrons.

Another type (or isotope) of helium is Helium-3 which has one neutron.

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3 0
3 years ago
If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m
algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

H=ut+\dfrac{1}{2}at^2

Here, a = g

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H=\dfrac{1}{2}\times 9.8\times (2.261)^2

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

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3 years ago
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