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3 years ago

Gold has a density of 19.3 grams per cubic centimeter. what is the density of gold in metric tons per cubic meter?

1 answer:

6 0

In order to determine this, we will first need some conversions. We will need to convert metric tons and grams into one another and also cubic centimeters to cubic meters into one another.

1 metric ton = 1000 kg

1 kg = 1000 grams

1 metric ton = 10⁶ grams

So 10⁶ grams / metric ton

1 meter = 100 cm
1 m³ = (100)³ cm³
1 m³ = 10⁶ cm³

So 10⁶ cm⁶ / m³

Now, we manipulate the given value:

(19.3 grams / cm³) * (1 metric ton / 10⁶ grams) * (10⁶ cm³ / m³)
= 19.3 metric tons / m³

The density of gold is 19.3 metric tons meter meter cubed.

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Answer:

14 m/s

Explanation:

Using the principle of conservation of energy, the potential energy is converted to kinetic energy, assuming any losses.

Kinetic energy is given by ½mv²

Potential energy is given by mgh

Where m is the mass, v is the velocity, g is acceleration due to gravity and h is the height.

Equating kinetic energy to be equal to potential energy then

½mv²=mgh

Making v the subject of the formula

v=√(2gh)

Substituting 9.81 m/s² for g and 10 m for h then

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4 years ago

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True, an object at rest stays and rest and an object in motion stays in motion

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3 years ago

Read 2 more answers

An object is moving east, and its velocity changes from 66 mvs to 26 mvs in 10 seconds. Which describes the

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Answer:

negative acceleration of 4m/s²

Explanation:

Given parameters:

Initial velocity = 66m/s

Final velocity = 26m/s

Time taken = 10s

Unknown:

Description of the acceleration = ?

Solution:

Acceleration is the rate of change of velocity with time.

It is mathematically expressed as;

acceleration = $\frac{v - u}{t}$

v is the final velocity

u is the initial velocity

t is the time taken

Now insert the parameters and solve;

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3 years ago

3. a car takes off with initial velocity of 20m/s and move with uniform acceleration of 12m/s2 for 20s. It maintained a constant

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The total distance traveled by the car at the given velocity and time is 900 m.

The given parameters:

initial velocity of the car, u = 20 m/s
acceleration of the car, a = 12 m/s²
time of motion of the car, t = 20 s
final time = 30 s
final acceleration = 2 m/s²

The final time of motion of car before coming to rest is calculated as follows;

$v_f = v_0 -at\\\\0 = 20 - 2t\\\\t = 10 \ s$

The graph of the car's motion is in the image uploaded.

The total distance traveled by the car is calculated as follows;

$total \ distance = A \ + B \ + C\\\\total \ distance = (\frac{1 }{2} \times 20 \times 20) \ + (20 \times 30) \ + (\frac{1}{2}\times 10 \times 20)\\\\total \ distance = 900 \ m$

Thus, the total distance traveled by the car at the given velocity and time is 900 m.

Learn more about velocity-time graph here: brainly.com/question/24874645

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