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3 years ago

1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.

Physics

1 answer:

Kisachek [45]3 years ago

7 0

Answer:

1. Density = 1200[kg/m^3]; 2. Volume= 0.005775[m^3], mass= 15.59[kg]

Explanation:

1. We know that the density is defined by the following expression.

$Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]$

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

long = 11 [dm] = 11 decimeters = 11/10 = 1.1[m]

Thick = 15[mm] = 15/1000 = 0.015[m]

Now we can find the density using the expression for the density.

$density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]$

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1. 20.0 mL of a liquid has a mass of 58.2 g. What is the density of the liquid?

skelet666 [1.2K]

Answer:

use the formula mass over volume

7 0

2 years ago

Read 2 more answers

La superficie de unas botas suman 400 cm2 y la persona que las usa tiene 45 kg de masa, calcula la presión que ejerce sobre el p

liubo4ka [24]

Answer:

11025 N / m²

Explanation:

Los siguientes datos se obtuvieron de la pregunta:

Área (A) = 400 cm²

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Presión (P) =?

A continuación, determinaremos la fuerza aplicada. Esto se puede obtener de la siguiente manera:

Masa (m) = 45 Kg

Aceleración por gravedad (g) = 9,8 m / s²

Fuerza (F) =.?

F = m × g

F = 45 × 9,8

F = 441 N

A continuación, convertiremos 400 cm² a m². Esto se puede obtener de la siguiente manera:

1 cm² = 0,0001 m²

Por lo tanto,

400 cm² = 400 cm² × 0,0001 m² / 1 cm²

400 cm² = 0,04 m²

Por tanto, 400 cm² equivalen a 0,04 m².

Finalmente, determinaremos la presión ejercida de la siguiente manera:

Área (A) = 0.04 m².

Fuerza (F) = 441 N

Presión (P) =?

P = F / A

P = 441 / 0,04

P = 11025 N / m²

Por tanto, la presión ejercida es 11025 M / m²

4 0

2 years ago

A 20 kg wagon is pulled along the level ground by a rope inclined at 30 degree above the horizontal. A friction force of 30 N op

Elan Coil [88]

(a) 34.6 N

To solve the problem, we have to analyze the forces acting along the horizontal direction.

We have:

- Forward: the component of the pull parallel to the ground, which is

$F cos \theta$

where

F is the magnitude of the pull

$\theta=30^{\circ}$ is the angle

- Backward: the force of friction, which is

$F_f = 30 N$

So, the equation of motion is

$F cos \theta - F_f = ma$

where

m = 20 kg is the mass of the wagon

a is the acceleration

In this part, the wagon is moving at constant speed, so a =0 and the equation becomes

$F cos \theta - F_f = 0$

Therefore, we can find the pulling force:

$F=\frac{F_f}{cos \theta}=\frac{30}{cos 30}=34.6 N$

(b) 43.9 N

In this case, the acceleration is

$a=0.40 m/s^2$

So, the equation of motion in this case is

$F cos \theta - F_f = ma$

So this time we have to take into account the term (ma).

Using the same data as before:

m = 20 kg

$\theta=30^{\circ}$

$F_f = 30 N$

We find the new magnitude of F:

$F=\frac{ma+F_f}{cos \theta}=\frac{(20)(0.40)+30}{cos 30}=43.9 N$

6 0

2 years ago

Problem 6.056 Air enters a compressor operating at steady state at 15 lbf/in.2, 80°F and exits at 275°F. Stray heat transfer and

vova2212 [387]

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

Here

P = Pressure

T = Temperature

$\gamma =$ The ratio of specific heats. For air normally is 1.4.

Our values are given as,

$P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K$

Therefore replacing we have,

$(\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}$

$(\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

Solving for $P_2,$

$P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}$

$P_2 = 44.15Lbf/in^2$

Therefore the maximum theoretical pressure at the exit is $44.15Lbf/in^2$

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3 years ago

Experiment: Group 1 drinks 500 mL of coffee a day.

slega [8]

Answer:

How does the drink content affect an individual's blood pressure?

Explanation:

In every experiment using the scientific method, an observation lays the foundation of that experiment. A problem must be observed, which then leads to asking a SCIENTIFIC QUESTION in order to investigate. A scientific question must include the variable being changed called INDEPENDENT VARIABLE and the variable being measured called DEPENDENT VARIABLE.

In this experimental procedure or set up,

- Group 1 drinks 500 mL of coffee a day.

- Group 2 drink 500 mL of tea a day,

- Group 3 is a control group i.e no drink

At the end of 60 days all participants

blood pressure is tested.

This set up indicates that the variable being changed (independent) is the DRINK CONTENT while the variable being measured (dependent) is the BLOOD PRESSURE. Hence, these variables serve as the template to ask a scientific question which goes thus:

HOW DOES THE DRINK CONTENT AFFECT AN INDIVIDUAL'S BLOOD PRESSURE?

This scientific question relates how the independent variable (drink) causes the dependent variable to respond (blood pressure).

3 0

2 years ago