All categories

3 years ago

What is the density of a 2 gallon milk jug that has a mass of 2.0 kg? Answer should be in g/ml.

Physics

1 answer:

kirill [66]3 years ago

8 0

volume of milk is given as

$V = 2 Gallon$

we will convert it into mL unit

$V = 2 Gal = 7570.82 ml$

mass of the milk m = 2kg

m = 2000 g

now for the density we can use

$\rho = \frac{m}{V}$

$\rho = \frac{2000}{7570.82}$

$\rho = 0.264 g/mL$

<em>so the density is 0.264 g/mL for above sample</em>

You might be interested in

A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference

Andre45 [30]

a) $400 \Omega$

b) 0.43 V

c) 0.44 %

Explanation:

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

$V=E-Ir$ (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when $R_1=550 \Omega$ , $V_1=0.25 V$

Using Ohm's Law, the current is:

$I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A$

2) when $R_2=1000 \Omega$ , $V_2=0.31 V$

Using Ohm's Law, the current is:

$I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A$

Now we can rewrite eq.(1) in two forms:

$V_1 = E-I_1 r$

$V_2=E-I_2 r$

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

$V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega$

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

$V=E-Ir$

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

$V=0.25 V$ is the voltage

$I=4.5\cdot 10^{-4}A$ is the current

$r=400\Omega$ is the internal resistance

Solving for E,

$E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V$

In this part, we are told that the area of the cell is

$A=4.0 cm^2$

While the intensity of incoming radiation (the energy received per unit area) is

$Int.=5.5 mW/cm^2$

This means that the power of the incoming radiation is:

$P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W$

This is the power in input to the resistor.

The power in output to the resistor can be found by using

$P'=I^2R$

where:

$R=1000 \Omega$ is the resistance of the resistor

$I=3.1\cdot 10^{-4} A$ is the current on the resistor (found in part A)

Susbtituting,

$P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W$

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

$\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%$

7 0

3 years ago

Que propiedad de la Luz se produce cuando ves tu cara reflejada en la cuchara?

kramer

Answer:

refracción

Explanation:

I don't speak spanish but that's the answer

5 0

3 years ago

A position vector in the first quadrant has an x-component of 18 m and a magnitude of 30 m. What is the value of its y-component

Snezhnost [94]

Answer:

The value is 24meters

Explanation:

Using

r= xi+yj

To get the magnitude of vector x

We say

/r/= √x²+y²

30²= √18² + y²

y= √576

Y= 24m

7 0

2 years ago

A ball is thrown straight up with a speed of 30 m/s, and air resistance is negligible. How long does it take the ball to reach t

PolarNik [594]

Answer:

3 seconds

Explanation:

Applying,

v = u±gt................ Equation 1

Where v = final velocity, u = initial velocity, t = time, g = acceleration due to gravity.

From the question,

Given: v = 0 m/s ( at the maximum height), u = 30 m/s

Constant: g = -10 m/s

Substitute these values into equation 1

0 = 30-10t

10t = 30

t = 30/10

t = 3 seconds

6 0

2 years ago

(a) What is the fluid speed in a fire hose with a 9.00-cm diameter carrying 80.0 L of water per second? (b) What is the flow rat

son4ous [18]

Answer:

12.5752053801 m/s

$80\times 10^{-3}\ m^3/s$

No.

Explanation:

Q = Volume flow rate = $80\ L/s=80\times 10^{-3}\ m^3/s$

d = Diameter of pipe = 9 cm

A = Area = $\dfrac{\pi}{4}d^2$

Volume flow rate is given by

$Q=Av\\\Rightarrow v=\dfrac{Q}{A}\\\Rightarrow v=\dfrac{80\times 10^{-3}}{\dfrac{\pi}{4} (9\times 10^{-2})^2}\\\Rightarrow v=12.5752053801\ m/s$

Velocity of fluid is 12.5752053801 m/s

The volume flow rate in m³/s is $80\times 10^{-3}\ m^3/s$

The flow of fluid does not depend on the type of water used. Hence the answers would be same. If Q is constant v will be the same irrespective of the type of water used.

8 0

3 years ago