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AfilCa [17]
2 years ago
7

A chief finds a glittering stone which he shows to the family. The family assures him that the stone is pure gold but he doubts.

Prepare a message of what you can do with the stone to give the chief and his family the advice
Physics
1 answer:
Nina [5.8K]2 years ago
5 0

I will tell the Chief about the high worth of the Gold and the various

substances it can be used to produce.

<h3>What is Gold?</h3>

Gold is a naturally occurring element with a bright, slightly orange yellow

color. It has a high value and can be used in the production of various

substances such as:

  • Jewelries
  • Medals and awards
  • Arts etc.

The Chief should however be told about the high worth value of Gold and

the various things it can be used for so he can make a decision.

Read more about Gold here brainly.com/question/1159978

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Some drops a ball off of the top of a 125-m-tall building. In this prob-lem, you will be solving for the time it takes the ball
Nimfa-mama [501]

Answer:

t = 5.05 s

Explanation:

This is a kinetic problem.

a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m

b) in this system the equations of motion are

              y = v₀ t + ½ g t²

where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth

e)    y = 0 + ½ g t²

     t = √ (2y / g)

     t = √(2 125 / 9.8)

     t = 5.05 s

6 0
2 years ago
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Bingel [31]
<span>
At the Earth's surface, warm air expands and rises, creating
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8 0
3 years ago
Read 2 more answers
Motion maps for two objects, Y &amp; Z, are shown.
serg [7]

Answer: it’s 3 seconds

Explanation:

5 0
2 years ago
What is the function of a hypothesis in the scientific inquiry process
iris [78.8K]
You can use them by analyzing the way you can solve it.
8 0
3 years ago
If the ball shown in the figure lands in 0.5 s, about what height was it thrown from?
Ede4ka [16]

Answer: 1.22 m

Explanation:

The equation of motion in this situation is:

y=y_{o}+V_{oy}t-\frac{g}{2}t^{2} (1)

Where:

y=0 is the final height of the ball

y_{o}=h is the initial height of the ball

V_{oy}=V_{o}sin(0\°)=0 is the vertical component of the initial velocity (assuming the ball was thrown vertically and there is no horizontal velocity)

t=0.5 s is the time at which the ball lands

g=9.8 m/s^{2} is the acceleration due gravity

So, with these conditions the equation is rewritten as:

h=\frac{g}{2}t^{2} (2)

h=\frac{9.8 m/s^{2}}{2}(0.5 s)^{2} (3)

Finally:

h=1.22 m

4 0
2 years ago
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