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Nastasia [14]
3 years ago
10

How do I convert 14.8 cm into MEters?

Physics
2 answers:
stellarik [79]3 years ago
6 0
Multiply it by a fraction equal to ' 1 ', like this:

(14.8 cm) x (1 meter/100 cm) = 14.8/100 = 0.148 meter
irinina [24]3 years ago
5 0
1 cm = 0.01 m = 1 \cdot 10^{-2} m
We can write it in this form:
14 cm = 0.14 m = 14 \cdot 10^{-2} m
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A 2.0-kg mass is oscillating about the origin at 24 rad/s. The amplitude of the oscillations is 0.040 m. At what position is the
Darya [45]

Answer:

0.0327 m

Explanation:

m = 2 kg

ω = 24 rad/s

A = 0.040 m

Let at position y, the potential energy is twice the kinetic energy.

The potential energy is given by

U = 1/2 m x ω² x y²

The kinetic energy is given by

K = 1/2 m x ω² x (A² - y²)

Equate both the energies as according to the question

1/2 m x ω² x y² = 2 x 1/2 m x ω² x (A² - y²)

y² = 2 A² - 2 y²

3y² = 2A²

y² = 2/3 A²

y = 0.82 A = 0.82 x 0.040 = 0.0327 m

4 0
3 years ago
Fill in the blanks to complete the sentence.
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The answer is:

Fill in the blanks to complete the sentence.

Light acts like a  PARTICAL  when it bounces off surfaces,

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4 0
3 years ago
Read 2 more answers
An unstable particle is created in the upper atmosphere from a cosmic ray and travels straight down toward the surface of the ea
german

Answer:

Q1)  Time taking by particle to travel the 40 km wrt. earth = 1.34\times10^{-6} sec.

Q2) The distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

Q3) The time taking by particle to travel from where it is created to the surface of the earth = 1.285\times10^{-5} sec.

Explanation:

Given :

Speed of particle wrt. earth v=0.99537c

Distance between where particle is created and earth surface = 40 km

we know that,

⇒       v = \frac{x}{t}

Where x = 40\times10^{3} m, v = 0.99537c, we know speed of light c = 3 \times10^{8}

∴      t = \frac{x}{v}

         = \frac{40 \times10^{3} }{0.99537\times3\times10^{8} }

      t = 1.34\times10^{-6} sec

∴ Thus, time taking by particle to travel the 40 km wrt. earth t = 1.34\times10^{-6} sec

According to the lorentz transformation,

⇒    l = l_{o} \sqrt{1-\frac{v^2}{c^2} }

Where l = improper length, l_{o} =proper length (distance measured wrt. rest frame) = 40 km

     l = 40 \sqrt{1-\frac{v^2}{c^2 }

     l = 40 \times 0.096

     l = 3.84 km

∴ Thus, the distance traveled by particle where the particle created to surface of earth wrt. particle's frame = 3.84 km.

According to the time dilation,

   \Delta t = \frac{\Delta t_{o} }{\sqrt{1-\frac{v^2}{c^2} } }

Where \Delta t = improper time (wrt. earth frame time) =1.34\times10^{-6} sec ,  \Delta t _{o} = proper time (wrt. particle frame).

 1.34\times10^{-6} = \frac{ \Delta t_{o}}{0.096}

 \Delta t_{o} = 1.285 \times10^{-5} sec

Thus, the time taking by particle to travel from where it is created to the surface of the earth = 1.285 \times10^{-5} sec.

6 0
2 years ago
. A 40.0-kg child standing on a frozen pond throws a 0.500-kg stone to the east with a speed of 5.00 m/s. Neglecting friction be
tamaranim1 [39]

Answer:

Explanation:

A 40kg child throw stone of 0.5kg

At a direction of 5m/s

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m_1•v_1 = -m_2•v_2

The negative sign show that the momentum of the boy is directed oppositely to that of the stone

m_1 Is mass of boy

v_1 is the recoil velocity of the boy

m_2 is mass of stone

v_2 is the velocity of stone

Then,

m_1•v_1 = -m_2•v_2

40•v_1 = -0.5 × 5

40•v_1 = -2.5

v_1 = -2.5 / 40

v_1 = -0.0625 m/s

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A cylinder has a height of 5.3 cm and a diameter of 3.0 cm. what is its volume?
anzhelika [568]
The cylinder has a volume of 37.46 cubic cm
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2 years ago
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