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A mass of mercury occupies 0.750 L. What volume would an equal mass of ethanol occupy? The density of mercury is 13.546 g/mL, an

melamori03 [73]

Answer:

Ve=12876.43mL : Ethanol volume

Explanation:

We apply the formula to calculate the density:

ρ= $\frac{m}{V}$ Formula (1)

Where:

ρ:Density in g/mL

m: mass in g (grams)

V= Volume in ml (milliliters)

We know the following data:

1L= 1000mL

Vm=Volume of mercury:VHg=0.750 L*1000mL/L=750mL

ρ-m=Density of mercury =13.546 g/mL

ρ-e: density of ethanol is 0.789 g/mL

$m_{m} =m_{e}$

Development of the problem:

In formula 1 We replace the known data for the mercury :

ρ-m= $\frac{m_{m} }{V_{m} }$

$13.546 \frac{g}{mL} =\frac{m_{m} }{750mL}$

$m_{m}=13.546 \frac{g}{mL}*750mL$

$m_{m} =10159.5 g$

We know that $m_{e} =m_{m} =10159.5 g$ , then, We replace the known data for the ethanol In formula 1:

ρ-e= $\frac{m_{e} }{V_{e} }$

$0.789 \frac{g}{mL} =\frac{10159.5g}{V_{e} }$

$Ve=\frac{10159.5 g}{0.789 \frac{g}{mL}}$

$Ve=12876.43mL$

8 0

3 years ago

Going around a ferris wheel without stopping is an example of _____ circular motion<br>(apex)

victus00 [196]

going around a ferris wheel without stopping is an example of uniform circular motion (apex).

3 0

3 years ago

Read 2 more answers

A rope horizontally pulls a massive object lying on a surface with friction with a constant

SVEN [57.7K]

Answer:

Equal to the frictional force

Explanation:

<u>Question</u>; The options given with regards to a similar question posted online are;

A. Equal (equivalent) to the frictional force

B. Larger than the frictional force

C. Equal to the object's weight

D. More than the object's weight

Explanation

According to Newton's first law of motion, every object shall remain at rest or continue moving with uniform (constant speed) motion unless there is a net force acting on the object

Given that the velocity of the massive block, lying on the surface that has friction, being pulled by the rope = Constant

Therefore;

The net force acting on the moving block while being pulled by the rope = 0

From which we have;

The pulling force = The resistive force

Where;

The pulling force = The (pulling) force (applied) on the rope

The resistive force = The frictional force of the surface which tends to prevent the motion of the block

Therefore, given that the net force acting on the block = 0

The force on the rope = The frictional force (of the surface)

The correct option is option A. Equal to the frictional force.

5 0

3 years ago

You get pulled over for running a red light. You explain to the police officer that the light appeared green to you due to the D

Ludmilka [50]

Answer:

$ 3085713685.71

Explanation:

$\lambda_0$ = Actual wavelength = 700 nm

$\lambda$ = Changed wavelength = 500 nm

Let the wavelength of red color be 700 nm and green be 500 nm

Change in wavelength is

$\Delta \lambda=700-500\\\Rightarrow \theta\lambda=200\ nm$

We have the relation

$\dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}\\\Rightarrow v=\dfrac{\Delta\lambda}{\lambda_0}c\\\Rightarrow v=\dfrac{200}{700}\times 3\times 10^8\\\Rightarrow v=85714285.7143\ m/s$

The speed of the vehicle is 85714285.7143 m/s

$\dfrac{85714285.7143\times 3600}{1000}=308571428.571\ km/h$

By how much was the car speeding

$308571428.571-60=308571368.571\ km/h$

The number of 10 km/h in the above speed

$308571368.571\times\dfrac{1}{10}=30857136.8571$

Cost of the ticket

$30857136.8571\times 100=\$ 3085713685.71$

The cost of the ticket is $ 3085713685.71

8 0

3 years ago

Read 2 more answers

A particle initially located at the origin has an acceleration of = 1.00ĵ m/s2 and an initial velocity of i = 6.00î m/s. (a) Fin

Ira Lisetskai [31]

Answer:

a) d = (6.00 t i ^ + 0.500 t²) m , b) v = (6.00 i ^ + 1.00 t j ^) m / s

c) d = (24.00 i ^ + 8.00 j^ ) m , d) v = (6.00 i ^ + 5 j^ ) m/s

Explanation:

This exercise is about kinematics in two dimensions

a) find the position of the particle on each axis

X axis

Since there is no acceleration on this axis, we can use the relation of uniform motion

v = x / t

x = v t

we substitute

x = 6.00 t

Y Axis

on this axis there is an acceleration and there is no initial speed

y = v₀ t + ½ a t²

y = ½ at t²

we substitute

y = ½ 1.00 t²

y = 0.500 t²

in vector position is

d = x i ^ + y j ^

d = (6.00 t i ^ + 0.500 t²) m

b) x axis

as there is no relate speed is concatenating

vₓ = v₀

vₓ = 6.00 m / s

y Axis

there is an acceleration and the initial speed is zero

$v_{y}$ = v₀ + a t

v_{y} = a t

v_{y} = 1.00 t

the velocity vector is

v = vₓ i ^ + v_{y} j ^

v = (6.00 i ^ + 1.00 t j ^) m / s

c) the coordinates for t = 4 s

d = (6.00 4 i ^ + 0.50 4 2 j⁾

d = (24.00 i ^ + 8.00 j^ ) m

x = 24.0 m

y = 8.00 m

d) the velocity of for t = 4 s

v = (6 i ^ + 1 5 j ^)

v = (6.00 i ^ + 5 j^ ) m/s

7 0

3 years ago