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Harlamova29_29 [7]

3 years ago

8

A stone is dropped from from rest at the top of a mine shaft. It takes 95 seconds for the stone to fall to the bottom of the min

e shaft. How deep is the mine shaft? The acceleration of gravity is 9.8 m/s^2. Answer in units of m

1 answer:

Marianna [84]3 years ago

6 0

Distance of fall from rest,
without air resistance = (1/2) (gravity) (time)²

   = (1/2) (9.8 m/s²) (95 sec)²

   = (4.9 m/s²) (9,025 sec²)

   = 44,222.5 meters .

The depth of the mine shaft is five times the height of Mt. Everest !

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The density of a glass is 2.5 g/cm3. if the glass breaks, what is the density of the smaller pieces

sp2606 [1]

Answer:

See the explanation below.

Explanation:

Density will remain the same since density is the relationship between mass and volume. As we can see in the equation below.

$Ro=m/V$

where:

Ro = density = 2.5 [g/cm³]

m = mass [g]

V = volume [cm³]

In such a way that when the glass is broken the small fragments retain the same density ratio. That is, each fragment has a small mass and a small volume. That's why the density remains the same.

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a passing warm front changes air pressure. does a passing warm front increase or decrease air pressure

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A group of big city tenants were upset about rising rent for their apartments, and decided to play a prank on the Building Manag

Naya [18.7K]

Answer:

t = 6 s

Explanation:

This is a free fall exercise

y = y₀ + v₀ t - ½ g t²

If the balloon is released its initial velocity is zero, when it reaches the floor its height is also zero, we substitute

0 = y₀ + 0 - ½ g t2

t = √(2yo / g)

let's calculate

t = √ (2 176.4 / 9.8)

t = 6 s

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Comentarios

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3 years ago

The universal force that is effective over the longest distance between very massive objects like planets is

fenix001 [56]

The universal force that is effective over the longest distance between very massive objects like planets is Gravity

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3 years ago

A block of unknown mass is attached to a spring with a spring constant of 7.00 N/m 2 and undergoes simple harmonic motion with a

KatRina [158]

Answers:

a) $0.80 kg$

b) $2.12 s$

c) $1.093 m/s^{2}$

Explanation:

We have the following data:

$k=7 N/m$ is the spring constant

$A=12.5 cm \frac{1 m}{100 cm}=0.125 m$ is the amplitude of oscillation

$V=32 cm/s=0.32 m/s$ is the velocity of the block when $x=\frac{A}{2}=0.0625 m$

Now let's begin with the answers:

<h3>a) Mass of the block</h3>

We can solve this by the conservation of energy principle:

$U_{o}+K_{o}=U_{f}+K_{f}$ (1)

Where:

$U_{o}=k\frac{A^{2}}{2}$ is the initial potential energy

$K_{o}=0$ is the initial kinetic energy

$U_{f}=k\frac{x^{2}}{2}$ is the final potential energy

$K_{f}=\frac{1}{2} m V^{2}$ is the final kinetic energy

Then:

$k\frac{A^{2}}{2}=k\frac{x^{2}}{2}+\frac{1}{2} m V^{2}$ (2)

Isolating $m$ :

$m=\frac{k(A^{2}-x^{2})}{V^{2}}$ (3)

$m=\frac{7 N/m((0.125 m)^{2}-(0.0625 m)^{2})}{(0.32 m/s)^{2}}$ (4)

$m=0.80 kg$ (5)

<h3>b) Period</h3>

The period $T$ is given by:

$T=2 \pi \sqrt{\frac{m}{k}}$ (6)

Substituting (5) in (6):

$T=2 \pi \sqrt{\frac{0.80 kg}{7 N/m}}$ (7)

$T=2.12 s$ (8)

<h3>c) Maximum acceleration</h3>

The maximum acceleration $a_{max}$ is when the force is maximum $F_{max}$ , as well :

$F_{max}=m.a_{max}=k.x_{max}$ (9)

Being $x_{max}=A$

Hence:

$m.a_{max}=kA$ (10)

Finding $a_{max}$ :

$a_{max}=\frac{kA}{m}$ (11)

$a_{max}=\frac{(7 N/m)(0.125 m)}{0.80 kg}$ (12)

Finally:

$a_{max}=1.093 m/s^{2}$

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3 years ago