Orbiting satellites use geothermal energy panels. True False

A sample of gold has a volume of 2 cm3 and a mass of 38.6 grams. What would be the density, and three other properties of the sa

lukranit [14]

The density of the gold is calculated to be "19,300 kg/m³".

Explanation:

Given:

Volume = 2cm³

Mass = 38.6 grams.

To Find:

Density of the gold = ?

Solution:

Density is obtained by dividing mass of the sample by its volume and it is given in the units of kg/m³.

Mass in grams is converted into kg as,

1 g = 0.001 kg

38. 6 g = $\frac{38.6}{1000} = 0.0386 kg$

Now we have to convert cm³ to m³ as,

1 cm³ = 10⁻⁶ m³

2 cm³ = 2 × 10⁻⁶ m³

So Density = $\frac{0.0386 k g}{2 \times 10^{-6} m^{3}}=19,300 \mathrm{kg} / \mathrm{m}^{3}$

Physical properties of Gold:

Gold is a heavy metal.

It is Malleable and ductile.

It is a corrosion resistant element.

3 0

3 years ago

Subduction occurs at which of the following tectonic plate boundaries?

hjlf

I think the answer is:
B) oceanic crust-continental crust

8 0

3 years ago

Read 2 more answers

A particle moves along a straight path through displacement while force acts on it. (Other forces also act on the particle.) Wha

allsm [11]

a) c = 1.85

b) c = 0.8

c) c = 2.33

Explanation:

a)

The displacement of the particle is given by

$d=2.2i+cj$

While the force applied on the particle is

$F=3.2i-3.8 j$

So we have a problem in 2-dimensions.

The work done on the particle is given by the scalar product between force and displacement:

$W=F\cdot d$ (1)

Here the work done on the particle is zero, so

W = 0

Therefore from eq(1) we find:

$0=(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=7.04\\c=\frac{7.04}{3.8}=1.85$

b)

In this problem, the work done on the particle is

$W=4.0 J$

The force and displacement are still

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

Therefore, by calculting the scalar product between force and displacement and equating it to the work done (4.0 J), we find:

$W=F\cdot d$

$4.0 =(3.2i-3.8j)\cdot (2.2i+cj)=7.04-3.8c\\3.8c=3.04\\c=\frac{3.04}{3.8}=0.8$

c)

In this problem instead, the work done on the particle is negative:

$W=-1.8 J$

As before, the force and displacement are

$d=2.2i+cj$ (displacement)

$F=3.2i-3.8 j$ (force)

And so again, we calculate the scalar product between force and displacement and we equate it to the work done on the particle, -1.8 J.

Doing so, we find:

$W=F\cdot d$

$-1.8=(3.2i-3.8j)\cdot (2.2i+c)=7.04-3.8c\\3.8c=8.84\\c=\frac{8.84}{3.8}=2.33$

7 0

3 years ago

a man stands on weighing scale in a lift which carries him upwards with acceleration, what happen please tell

Lapatulllka [165]

He gets heavier due to the upward acceleration of the lift

5 0

3 years ago

Force f⃗ =−10j^n is exerted on a particle at r⃗ =(7i^+5j^)m. part a what is the torque on the particle about the origin?

cluponka [151]

Answer:

Torque, $\tau=0i+0j-70k$

Explanation:

It is given that,

Force acting on the particle, $F=-10j\ N$

Position of the particle, $r=(7i+5j)\ m$

We need to find the torque on the particle about the origin. It is equal to the cross product of position and the force. Its formula is given by :

$\tau=r\times F$

$\tau=(7i+5j)\times (-10j)$

The cross product of vectors is given by :

$\tau=\begin{pmatrix}0&0&-70\end{pmatrix}$

or

$\tau=0i+0j-70k$

So, the torque on the particle about the origin $0i+0j-70k$ . Hence, this is the required solution.

6 0

3 years ago