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Tems11 [23]
3 years ago
8

A child sits 1.6 m from the center of a merry go round that makes one complete revolution in 4.7 s. What is his angular accelera

tion?
Physics
1 answer:
olga2289 [7]3 years ago
7 0

Answer: angular acceleration = 2.86\ rad/sec^{2}

Given:

Distance from center of axis = 1.6 m

Time taken to complete one revolution = 4.7 sec

Therefore, we can evaluate the angular acceleration using the following formula:

\alpha = r\times \omega^{2}

\alpha = r\times (\frac{2\pi}{T})^{2}

\alpha = 1.6\times (\frac{2\times3.14}{4.7})^{2}

\alpha = 2.86\ rad/sec^{2}

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What is the increase in pressure required to decrease volume of mercury by 0.001%
REY [17]

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

3 0
2 years ago
What does the geology of the two continents indicate about past events in Earth history?
Nikolay [14]

Answer:

Explanation:

Rocks tell us a great deal about the Earth's history. Igneous rocks tell of past volcanic episodes and can also be used to age-date certain periods in the past. Sedimentary rocks often record past depositional environments (e.g deep ocean, shallow shelf, fluvial) and usually contain the most fossils from past ages.

6 0
3 years ago
A thin, light wire 75.2 cm long having a circular cross section 0.560 mm in diameter has a 25.2 kg weight attached to it, causin
blsea [12.9K]

Answer:

The stress is calculated as 1.003\times 10^{9}\ Pa

Solution:

As per the question:

Length of the wire, l = 75.2 cm = 0.752 m

Diameter of the circular cross-section, d = 0.560 mm = 0.560\times 10^{- 3}\ m

Mass of the weight attached, m = 25.2 kg

Elongation in the wire, \Delta l = 1.10\ mm = 1.10\times 10^{- 3}\ m

Now,

The stress in the wire is given by:

Stress,\ \sigma = \frac{Force,\ F}{Area,\ A}          (1)

Now,

Force is due to the weight of the attached weight:

F = mg = 25.2\times 9.8 = 246.96\ N

Cross  sectional Area, A = \pi (\frac{d}{2})^{2} = \pi (\frac{0.560\times 10^{- 3}}{2})^{2} = 2.46\times 10^{- 7}\ m^{2}

Using these values in eqn (1):

\sigma = \frac{246.96}{2.46\times 10^{- 7}} = 1.003\times 10^{9}\ Pa  

8 0
2 years ago
A particle is projected from a point on a horizontal plane and has an initial velocity of 28root 3 m/s at an angle of 60 degree
Lelechka [254]

Answer:

uujjjjjctc7tox7txr9ll8rz8lr5xl8r6l8dl85x8rl5x8rl5x8rl5xrx8l58rk5xr8l5xr6l8xr68lc

4 0
3 years ago
Can y’all help me with 5 plsssss
Alona [7]

Answer:

Bar graph

Explanation:

each day collects data so a bar graph would work.

4 0
2 years ago
Read 2 more answers
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