All categories

3 years ago

A urine sample has a mass of 151 g and a density of 1.020 g / mL . What is the volume of this sample?

Physics

1 answer:

babymother [125]3 years ago

6 0

Answer:

0.15 L.

Explanation:

Density of a substance is defined as a property of a substance which is the mass of unit volume of substance. Its SI unit is kg/m3.

Density = mass/volume

Mass = 151 g

Density = 1.020 g/ml

To find volume,

= 151/1.020

= 148.04 ml

Converting ml to liters,

1000 ml = 1 liter,

148.04 ml = 0.148 l

= 0.15 L.

You might be interested in

A package of mass 5 kg sits on an airless asteroid of mass 7.6 × 1020 kg and radius 8.0 × 105 m. We want to launch the package i

Effectus [21]

Answer:

s = 1.7 m

Explanation:

from the question we are given the following:

Mass of package (m) = 5 kg

mass of the asteriod (M) = 7.6 x 10^{20} kg

radius = 8 x 10^5 m

velocity of package (v) = 170 m/s

spring constant (k) = 2.8 N/m

compression (s) = ?

Assuming that no non conservative force is acting on the system here, the initial and final energies of the system will be the same. Therefore

• Ei = Ef

• Ei = energy in the spring + gravitational potential energy of the system

• Ei = \frac{1}{2}ks^{2} + \frac{GMm}{r}

• Ef = kinetic energy of the object

• Ef = \frac{1}{2}mv^{2}

• \frac{1}{2}ks^{2} + (-\frac{GMm}{r}) = \frac{1}{2}mv^{2}

• s = $\sqrt{\frac{m}[k}(v^{2}+\frac{2GM}{r})}$

s = $\sqrt{\frac{5}[2.8 x 10^5}(170^{2}+\frac{2 x 6.67 x10^{-11} x 7.6 x 10^{20}}{8 x 10^5})}$

s = 1.7 m

7 0

3 years ago

Why is it that the weight of an object weighing 1N air, weighs more when immersed in water ?

Anni [7]

There is no "why", because that's not what happens. The truth is
exactly the opposite.

Whatever the weight of a solid object is in air, that weight will appear
to be LESS when the object is immersed in water.

The object is lifted by a force equal to the weight of the fluid it displaces.
It displaces the same amount of air or water, and any amount of water
weighs more than the same amount of air. So the force that lifts the
object in water is greater than the force that lifts it in air, and the object
appears to weigh less in the water.

4 0

3 years ago

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside thi

Zanzabum

Complete question:

Resistor is made of a very thin metal wire that is 3.2 mm long, with a diameter of 0.4 mm. What is the electric field inside this metal resistor? If the potential difference due to electric field between the two ends of the resistor is 10 V.

Answer:

The electric field inside this metal resistor is 3125 V/m

Explanation:

Given;

length of the wire, L = 3.2 mm = 3.2 x 10⁻³ m

diameter of the wire, d = 0.4 mm = 0.4 x 10⁻³ m

the potential difference due to electric field between the two ends of the resistor, V = 10 V

The electric field inside this metal resistor is given by;

ΔV = EL

where;

ΔV is change in electric potential

E = ΔV / L

E = 10 / (3.2 x 10⁻³ )

E = 3125 V/m

Therefore, the electric field inside this metal resistor is 3125 V/m

7 0

3 years ago

A toy train is pushed forward and released at Xi = 4.0 m with a speed of 2.5 m/s. It rolls at a steady speed for 2.0 s, then one

Fittoniya [83]

I don't think that 4m has anything to do with the problem. anyway. here. A___________________B_______C where A is the point that the train was released. B is where the wheel started to stick C is where it stopped From A to B, v=2.5m/s, it takes 2s to go A to B so t=2 AB= v*t = 2.5 * 2 = 5m The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m then BC= AC-AB = 7.7-5 = 2.7m now consider BC v^2=u^2+2as where u is initial speed, in this case is 2.5m/s v is final speed, train stop at C so final speed=0, so v=0 a is acceleration s is displacement, which is BC=2.7m substitute all the number into equation, we have 0^2 = 2.5^2 + 2*a*2.7 0 = 6.25 + 5.4a a = -6.25/5.4 = -1.157 so acceleration is -1.157m/(s^2)

8 0

3 years ago

Read 2 more answers

Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)

SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c² => c = 0.707 m

Charge to the right: In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x $10^{9}$ N m²/C²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below: For y direction

e = kq/r²

e = (9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²

e = 9 x $10^{5}$ N/C

Charge diagonal:

e = kq/r²

e = [(9 x $10^{9}$ )(250 x $10^{-7}$ ) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

4 0

3 years ago