Which of these objects has the greatest inertia?

daser333 [38]

C Weight is the gravitational pull on an object

8 0

3 years ago

Heptane and water do not mix, and heptane has a lower density (0.684 g/mL) than water (1.00 g/mL). A graduated cylinder contains

lakkis [162]

Given that the density of heptane is

$d_h=\frac{0.684g}{mL}$

The mass of heptane is

$m_h=31\text{ g}$

The density of water is

$d_w=\frac{1g}{mL}$

The mass of water is

$m_w=37\text{ g}$

The volume of heptane will be

$\begin{gathered} V_h=\frac{m_h}{d_h} \\ =\frac{31}{0.684} \\ =45.32\text{ mL} \end{gathered}$

The volume of water will be

$\begin{gathered} V_w=\frac{m_w}{d_w} \\ =\frac{37}{1} \\ =37\text{ mL} \end{gathered}$

Thus, the volume of heptane is 45.32 mL and the volume of water is 37 mL.

The total volume of liquid in the cylinder will be

$\begin{gathered} V=V_h+V_w \\ =45.32+37 \\ =82.32\text{ mL} \end{gathered}$

The total volume of liquid in the cylinder will be 82.32 mL.

7 0

1 year ago

13) The condenser pressure is 417.4 psig for r-410A and the condenser outlet temperature is 108f. how much subcooling is there i

Tanzania [10]

Answer:

12°F

Explanation:

Calculation for how much subcooling is there in the condenser

Since the CONDENSING TEMPERATURE for 417.4 psig discharge pressure is 120 degrees (120°) which means that the amount of subcooling that is there in the condenser will be calculated using this formula

Amount of Condenser subcooling= Condensing Temperature discharge pressure -Condenser outlet temperature

Let plug in the formula

Amount of Condenser subcooling=120°-108f

Amount of Condenser subcooling=12°F

Therefore the amount of subcooling that is there in the condenser will be 12°F

8 0

3 years ago

The gravitational force between two objects is 2400 N. What will be the gravitational force between the objects if the mass of o

motikmotik

Answer:

4800N

Explanation:

Lets assume,

Mass of first object = m₁

Mass of second object = m₂

Distance between the two objects = r

Thus the force between the two objects will be

$F = \frac{G\times m_{1}\times m_{2}}{r^{2}}$

where, G = Universal gravitational constant

Given, F = 2400N

New mass of second object = 2m₂

Now, the force will be

$F_{2} = \frac{G\times m_{1}\times 2m_{2}}{r^{2}}$

$F_{2}= 2\frac{G\times m_{1}\times m_{2}}{r^{2}}$

$F_{2}= 2F$

$F_{2}= 2\times2400$

Thus, F₂ = 4800N

6 0

3 years ago

A traveling wave on a string can be described by the equation : y = (5.26 ~\text{m}) \cdot \sin \big( (1.65 ~\frac{\text{rad}}{\

zloy xaker [14]

Answer:

t = 1.77 s

Explanation:

The equation of a traveling wave is

y = A sin [2π (x /λ -t /T)]

where A is the oscillation amplitude, λ the wavelength and T the period

the speed of the wave is constant and is given by

v = λ f

Where the frequency and period are related

f = 1 / T

we substitute

v = λ / T

let's develop the initial equation

y = A sin [(2π / λ) x - (2π / T) t +Ф]

where Ф is a phase constant given by the initial conditions

the equation given in the problem is

y = 5.26 sin (1.65 x - 4.64 t + 1.33)

if we compare the terms of the two equations

2π /λ = 1.65

λ = 2π / 1.65

λ = 3.81 m

2π / T = 4.64

T = 2π / 4.64

T = 1.35 s

we seek the speed of the wave

v = 3.81 / 1.35

v = 2.82 m / s

Since this speed is constant, we use the uniformly moving ratios

v = d / t

t = d / v

t = 5 / 2.82

t = 1.77 s

3 0

3 years ago