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2 years ago

An empty container has a mass of 3 g. When it is filled with 5 cm3 of a liquid,

Physics

1 answer:

IrinaK [193]2 years ago

3 0

Answer:

THE MASS OF THE LIQUID IS 22.5 g

Explanation:

Density = 0.180 g/cm3

Side length = 5 cm

Mass = unknown

To calculate the mass of the liquid, we use the formula:

Mass = density * volume

Volume of a cube or cuboid container = l^3

Volume = 5 ^3 = 125 cm3

So therefore, the mass of the liquid is:

Mass = 0.180 * 125

Mass = 22.5 g

In conclusion, the mass of the liquid in the container is 22.5 g

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An object is moving initially with a velocity of 4.7 m/s . After 3.9 s the object's velocity is -2.1 m/s . What is the object's

IgorC [24]

Answer: The acceleration of the object is 0.67m/s^2 west.

Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.

a=dv/t

a=vf-vi/t

a=-2.1-4.7/3.9

a= 0.66m/s^2 west

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2 years ago

Clutter is _______. annoying frustrating death

Sindrei [870]

Answer: _____= beautiful, yet annoying frustrating death

Explanation:

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2 years ago

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The body contains many small currents caused by the motion of ions in the organs and cells. Measurements of the magnetic field a

RoseWind [281]

Answer:

270 μA

Explanation:

Use the magnetic field due to long, straight wire and solve for current I.

$B= \frac{\mu_0I}{2\pi r}$

$I= \frac{2\pi r B}{\mu_0}$

plug in the values

$I= \frac{2\pi(5.40\times10^{-2})\times10\times10^{-6}\times10^{-4}}{4\pi\times10^{-7}}$

= 2.7×10^{-4)×10^6

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2 years ago

The a992 steel rod bc has a diameter of 50 mm and is used as a strut to support the beam. determine the maximum intensity w of t

EastWind [94]

Answer:

w = 11.211 KN/m

Explanation:

Given:

diameter, d = 50 mm

F.S = 2

L = 3

Due to symmetry, we have:

$Ay = By = \frac{w * 6}{2} = 3w$

$P_c_r = 3w * F.S = 3w * 2.0 = 6w$

$I = \frac{\pi}{64} (0.05)^4 = 3.067*10^-^7$

To find the maximum intensity, w, let's take the Pcr formula, we have:

$P_c_r = \frac{\pi^2 E I}{(KL)^2}$

Let's take k = 1

$E = 200*10^9$

Substituting figures, we have:

$6w = \frac{\pi^2 * 200*10^9 * 3.067*10^-^7}{(1 * 3)^2}$

Solving for w, we have:

$w = \frac{67266.84}{6}$

w = 11211.14 N/m = 11.211 KN/m

Since Area, A= pi * (0.05)²

$\sigma _c_r = \frac{w}{A}$

$\sigma _c_r = \frac{11.211}{\pi (0.05)^2} = 1.4 MPA < \sigma y$ . This means it is safe

The maximum intensity w = 11.211KN/m

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2 years ago

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B if not please comment back .

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