All categories

2 years ago

A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?

Physics

1 answer:

Natali [406]2 years ago

7 0

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒ $Density = \frac{Mass}{Volume}$

or,

⇒ $Volume=\frac{Mass}{Density}$

On substituting the values, we get

⇒ $=\frac{6.3 \ g}{19.3 \ g/mL}$

⇒ $=0.32 \ mL$

You might be interested in

Which of the following is not an example of approximate simple harmonic motion? A. a ball bouncing on the floor

Reptile [31]

<h2>Answer:</h2>

Answer to this question is (A)

<h2>Explanation</h2>

A ball bouncing on the floor is not the example of simple harmonic motion. SHM is the special kind of to and fro motion in which a particle oscillate about its mean position in a straight line. The acceleration of the particle is always directed towards its mean position and is directly proportional to its displacement from its mean position.

In case of a ball bouncing on the ground, the motion of the ball is not SHM, as neither it’s a to and fro motion nor the acceleration is proportional to its displacement from its mean position.

3 0

3 years ago

How do i solve the ones in red? A student fires a cannonball horizontally with a speed of 24.0m/s from a height of 51.0m. Neglec

Rus_ich [418]

Horizontal speed = 24.0 m/s

height of the cliff = 51.0 m

For the initial vertical speed will are considering the vertical component. Therefore,

Since the student fires the canonical ball at the maximum height of 51 m, the initial vertical velocity will be zero. This means

$\text{ Initial vertical velocity = 0 m/s}$

let's find how long the ball remained in the air.

$\begin{gathered} 0=51-\frac{1}{2}(9.8)t^2 \\ 4.9t^2=51 \\ t^2=\frac{51}{4.9} \\ t^2=10.4081632653 \\ t=\sqrt[]{10.4081632653} \\ t=3.22 \\ t=3.22\text{ s} \end{gathered}$

Finally, let's find the how far from the base of the building the ball landed(horizontal distance)

$\begin{gathered} S_x=u_xt+\frac{1}{2}a_xt^2 \\ S_x=24\times3.23 \\ S_x=\text{horizontal distance=77.28 m} \\ \text{note} \\ a_x=0m/s^2 \end{gathered}$

6 0

1 year ago

A book rest on a table which it's face having a sides 30cm by 25cm . if it exerts apressure of 200pa then determine the mass of

Simora [160]

Answer:

Approximately $1.5\; \rm kg$ . (Assuming that this table is level, and that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ .

Explanation:

Convert the dimensions of this book to standard units:

$\displaystyle 30\; \rm cm = 30\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.30\; \rm m$ .

$\displaystyle 25\; \rm cm = 25\; \rm cm \times \frac{1\; \rm m}{100\; \rm cm} = 0.25\; \rm m$ .

Calculate the surface area of this book:

$0.30\; \rm m \times 0.25\; \rm m = 0.075\; \rm m^{2}$ .

Pressure is the ratio between normal force and the area over which this force is applied.

$\displaystyle \text{Pressure} = \frac{\text{normal Force}}{\text{contact Area}}$ .

Equivalently:

$(\text{normal Force}) = \text{Pressure} \cdot (\text{contact Area})$ .

In this question, $\text{Pressure} = 200\; \rm Pa = 200\; \rm N \cdot m^{-2}$ .

It was found that $(\text{contact Area}) = 0.075\; \rm m^{2}$ (assuming that the entire side of this book is in contact with the table.

Hence:

$\begin{aligned}& (\text{normal Force}) \\ &= \text{Pressure} \cdot (\text{contact Area}) \\ &= 200\; \rm N \cdot m^{-2} \times 0.075\; \rm m^{2} \\ &= 15\; \rm N \end{aligned}$ .

If that the table is level, this normal force would be equal to the weight of this book:

$\text{weight} = 15\; \rm N$ .

Assuming that the gravitational field strength is $g = 9.8\; \rm N \cdot kg^{-1}$ . The mass of this book would be:

$\begin{aligned}\text{mass} &= \frac{\text{weight}}{g} \\ &= \frac{15\; \rm N}{9.8\; \rm N \cdot kg^{-1}}\approx 1.5\; \rm kg\end{aligned}$ .

4 0

2 years ago

A cook preparing a meal for a group of people is an example of

Brums [2.3K]

Is this question about chemical/ physical change?
If it is the the answer is "chemical change"

6 0

3 years ago

Read 2 more answers

What is needed to give a large boulder a large acceleration?

jok3333 [9.3K]

More force needs to be applied

6 0

2 years ago

Read 2 more answers