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cricket20 [7]
2 years ago
5

Si usted trata de contrabandear lingotes de oro llenando su mochila, cuyas dimensiones son de 56 cm * 28 cm * 22 cm, ¿cuál sería

su masa?
Physics
1 answer:
kati45 [8]2 years ago
5 0

Answer:

Mass, m = 665.77 kg

Explanation:

The question says that "If you try to smuggle gold bars by filling your backpack, whose dimensions are 56 cm * 28 cm * 22 cm, what would be its mass? "

The dimension of gold bar is 56 cm×28 cm×22 cm

We know that the density of gold is 19.3 g/cm³

The mas per unit volume of a material is called its density. It can be given by :

d=\dfrac{m}{V}\\\\m=d\times V\\\\m=(19.3\ g/cm^3)\times (56\times 28\times 22\ cm^3)\\\\m=665772.8\ g\\\\m=665.77\ kg

So, the mass of the gold bar is 665.77 kg.

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Fiberglass, an insulator, can be found in the wals and roofs of some houses and buildings. Why would an insulator be needed insi
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It keeps the heat inside of the building

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Imagine that you are going to visit your friend. Before you get there, you decide to stop at the variety store. If you walk 200
SashulF [63]

Answer:

400m

Explanation:

Brainliest? :))

Let your initial displacement from your home to the store be

Dd

>

1 and your displacement from the store to your friend’s house

be Dd

>

2.

Given: Dd

>

1 = 200 m [N]; Dd

>

2 = 600 m [S]

Required: Dd

>

T

Analysis: Dd

>

T 5 Dd

>

1 1 Dd

>

2

Solution: Figure 6 shows the given vectors, with the tip of Dd

>

1

joined to the tail of Dd

>

2. The resultant vector Dd

>

T is drawn in red,

from the tail of Dd

>

1 to the tip of Dd

>

2. The direction of Dd

>

T is [S].

Dd

>

T measures 4 cm in length in Figure 6, so using the scale of

1 cm : 100 m, the actual magnitude of Dd

>

T is 400 m.

Statement: Relative to your starting point at your home, your

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6 0
2 years ago
This diagram shows two different forces acting on a skateboarder. The
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Answer:

B

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7 0
3 years ago
A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
expeople1 [14]

Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

\theta=10.47^{\circ}

Hence, this is the required solution.

7 0
3 years ago
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