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3 years ago

When energy is added to an object, it will vibrate at it's...

Physics

1 answer:

sergiy2304 [10]3 years ago

6 0

Answer:

I think D

Explanation:

if energy is added,it wont stay the same,I beleive D is the only one that makes since,sorry if wrong

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The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

3 years ago

A 3.0 m tall, 40 cm diameter concrete column supports a 235,000 kg load. by how much is the column compressed? assume young's mo

statuscvo [17]

The Young modulus E is given by:
$E= \frac{F L_0}{A \Delta L}$
where
F is the force applied
A is the cross-sectional area perpendicular to the force applied
$L_0$ is the initial length of the object
$\Delta L$ is the increase (or decrease) in length of the object.

In our problem, $L_0 = 3.0 m$ is the initial length of the column, $E=3.0 \cdot 10^{10}N/m^2$ is the Young modulus. We can find the cross-sectional area by using the diameter of the column. In fact, its radius is:
$r= \frac{d}{2}= \frac{40 cm}{2}=20 cm=0.2 m$
and the cross-sectional area is
$A=\pi r^2 = \pi (0.20 m)^2=0.126 m^2$
The force applied to the column is the weight of the load:
$W=mg=(235000 kg)(9.81 m/s^2)=2.305 \cdot 10^6 N$

Now we have everything to calculate the compression of the column:
$\Delta L = \frac{F L_0}{EA}= \frac{(2.305\cdot 10^6 N)(3.0 m)}{(3.0\cdot 10^{10}N/m^2)(0.126 m^2)} =1.83\cdot 10^{-3}m$
So, the column compresses by 1.83 millimeters.

3 0

3 years ago

Read 2 more answers

Which of the following is not a true statement?<br> v=at<br> O art

shepuryov [24]

Answer:

t = Δa / v

Explanation:

To know which option is not true, we shall fine a relationship between acceleration (a), velocity (v), time (t) and radius (r). This is illustrated below:

Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as shown below:

Acceleration = change in velocity / time

a = Δv / t ..... (1)

But

Δv = v₂ – v₁

Substitute the value of Δv into equation (1)

a = Δv / t

a = v₂ – v₁ / t ....... (2)

From equation (1), make Δv the subject of the equation.

a = Δv / t

Cross multiply

Δv = at .... (3)

From equation (1), make t the subject of the equation.

a = Δv / t

Cross multiply

at = Δv

Divide both side by a

t = Δv /a ...... (4)

From circular motion, centripetal's force is given by:

F = mv²/r

F = ma꜀

Therefore,

ma꜀ = mv²/r

Cancel out m

a꜀ = v²/r

SUMMARY:

a = Δv / t

a = v₂ – v₁ / t

Δv = at

t = Δv /a

a꜀ = v²/r

Considering the options given in question above, t = Δa / v is not a true statement.

8 0

3 years ago

How Do You Appreciate the role of spherical mirrors in daily life ?

marin [14]

I appreciate the role of spherical mirrors in daily life. THey are used in corridors to see who comes form other sides from sideways.
THey are used in car drivers and motor cyclists, to view who comes behind them.
they are used by dentists.
they are used in long terrestrial Telescopes.

5 0

3 years ago

Which of the following BEST explain why a book lying on a table does not move, even though the force of gravity is acting on it

muminat

It would not move because while gravity is pushing it down, the table is also pushing it up therefor canceling out

8 0

3 years ago