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3 years ago

Carol drops a stone in a mine shaft 122.5 metres deep.How

Physics

1 answer:

Verizon [17]3 years ago

7 0

Answer:

T = 5.36 s

Explanation:

given,

depth of the mine shaft = 122.5 m

speed of the sound = 340 m/s

time taken = ?

time taken by the stone to reach at the bottom

using equation of motion

$s = u t + \dfrac{1}{2}gt^2$

initial speed , u = 0 m/s

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 122.5}{9.8}}$

t = 5 s

time taken by the sound to travel

d =v x t

$t = \dfrac{d}{v}$

$t = \dfrac{122.5}{340}$

t = 0.36 s

total time taken for the sound to reach carol after dropping the stone

T = 5 + 0.36

T = 5.36 s

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#1 Not sure where to start. This is for AP Physics!

yaroslaw [1]

First,

$\rho=\dfrac mV$

where $\rho$ is density, $m$ is mass, and $V$ is volume. We can compute the volume of the roll:

$2.7\,\dfrac{\mathrm g}{\mathrm{cm}^3}=\dfrac{1275\,\mathrm g}V$

$\implies V\approx472.22\,\mathrm{cm}^3\approx4.72\,\mathrm m^3$

When the roll is unfurled, the aluminum will be a rectangular box (a very thin one), so its volume will be the product of the given area and its thickness $x$ . Note that we're assuming the given area is not the actual total surface area of the aluminum box, but just the area of the largest face (i.e. the area of one side of the unrolled sheet of aluminum).

So we have

$V=Ax$

where $A$ is the given area, so

$4.72\,\mathrm m^3=\left(18.5\,\mathrm m^2\right)x$

$\implies x\approx0.255\,\mathrm m=255\,\mathrm{mm}$

If we're taking significant digits into account, the volume we found would have been $V=470\,\mathrm m^3$ , in turn making the thickness $x=250\,\mathrm{mm}$ .

8 0

3 years ago

Please help! thank you

BlackZzzverrR [31]

Answer:

poor, too precise

good

Explanation:

8 0

3 years ago

1. Calculati greutatea unui sac cu 5 kg de cartofi într-o zonă in care acceleratia gravitatională

harkovskaia [24]

Answer:

the weight is 49.1 N

Explanation:

The computation of the weight is shown below:

As we know that

= 5kg of potatoes × gravitational acceleration

= 5kg of potatoes × 9.82 m/s

= 49.1 N

Hence, the weight is 49.1 N

We simply applied the above formula in order to determine the weight

6 0

3 years ago

An object is acted upon by a force of 22 newtons to the right and a force of 13 newtons to the left. What is the magnitude and d

Harman [31]

The magnitude and direction of the net force is 9 newtons to the right ⇒ answer A

Explanation:

The net force is the vector sum of all the forces that act upon an object

If two forces are horizontal where,

$F_{1}$ > $F_{2}$
They are in opposite direction
Net force F = $F_{1}$ - $F_{2}$ in the direction of $F_{1}$

An object is acted upon by a force of 22 newtons to the right

and a force of 13 newtons to the left

We need to find the magnitude and direction of the net force

→ $F_{1}$ = 22 newtons to the right

→ $F_{2}$ = 13 newtons to the left

$F_{1}$ > $F_{2}$ and they are in opposite direction

Substitute these values in the rule of the net force above

→ Net force F = 22 - 13 = 9

→ The direction of the net force is the same direction of the larger

force, the direction of the larger force is to the right

→ Net force is 9 newtons to the right

The magnitude and direction of the net force is 9 newtons to the right

Learn more:

You can learn more about the net force in brainly.com/question/4033012

#LearnwithBrainly

8 0

3 years ago

A diver makes 0.50 revolutions on the way from a 9.6-m-high platform to the water. Assuming zero initial vertical velocity, find

devlian [24]

Answer:

ω = 0.36 rev/s = 2.24 rad/s

Explanation:

First, we will find the time taken by the diver to reach the water. For this we use 2nd equation of motion:

h = Vi t + (1/2)gt²

where,

h = height = 9.6 m

Vi = initial vertical velocity = 0 m/s

t = time taken = ?

g = 9.8 m/s²

Therefore,

9.6 m = (0 m/s)(t) + (1/2)(9.8 m/s²)t²

t = √1.95 s²

t = 1.4 s

Now, the average angular speed of diver will be:

ω = No. of Revolutions/t

ω = 0.5 rev/1.4 s

7 0

3 years ago