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mojhsa [17]
3 years ago
8

Rewrite the false statements correctly

Physics
1 answer:
FrozenT [24]3 years ago
5 0

Explanation:

1. if an object sjnks in one liquid and floats on another liquid it implies that the density of second liquid is greater than the density of first liquid

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A leaky faucet drips once each second. at this rate, it takes 1 hour to fill a 1-liter graduated cylinder to the 200-cm3 mark. w
UNO [17]
Define
v = volume of a drop per second, cm³/s

The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.

Therefore,
\frac{200 \,  cm^{3}}{v \, cm^{3}/s}  = 3600 \, s \\ v =  \frac{200}{3600} =0.0556 \, cm^{3}

The average volume of a single drop is approximately 0.0556 cm³.

Answer: 0.0556 cm³

8 0
3 years ago
What is the ratios of sodium hydroxide
Katena32 [7]
Clarify what you mean by ratios?
8 0
3 years ago
In the following figure, electric field at y axis will be maximum at y=?
Strike441 [17]

Because of symmetry electric field component in the x axis cancels out. Now just use electric field formula and slap that sine of theta cause you want the vertical component of electric field and multiply that by two since there’s two charges. I’ve shown my work. Hope it helps✌

5 0
3 years ago
Read 2 more answers
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
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
The power of the kettle was 2.6 kW
faltersainse [42]

Answer:

Cp = 4756 [J/kg*°C]

Explanation:

In order to calculate the specific heat of water, we must use the equation of energy for heat or heat transfer equation.

Q = m*Cp*(T_f - T_i)/t

where:

Q = heat transfer = 2.6 [kW] = 2600[W]

m = mass of the water = 0.8 [kg]

Cp = specific heat of water [J/kg*°C]

T_f  = final temperature of the water = 100 [°C]

T_i = initial temperature of the water = 18 [°C]

t = time = 120 [s]

Now clearing the Cp, we have:

Cp = Q*t/(m*(T_f - T_i))

Now replacing

Cp = (2600*120)/(0.8*(100-18))

Cp = 4756 [J/kg*°C]

7 0
3 years ago
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