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aleksklad [387]
3 years ago
9

a sphere of mass 5kg and volume 2×10-5completely immersed in water find the buoyant force exerted water​

Physics
1 answer:
gregori [183]3 years ago
4 0

Answer:

Buoyant force exerted water​ =  0.196 Newton

Explanation:

Given:

Mass of sphere ball = 5 kg

Volume = 2 x 10⁻⁵

Find:

Buoyant force exerted water​

Computation:

Buoyant force exerted water​ = Gravity due to acceleration x volume of object x density of given liquid

Buoyant force exerted water​ = 9.8 x 2 x 10⁻⁵ x 1000

Buoyant force exerted water​ =  0.196 Newton

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Explanation:

Velocity=√(centripetal acceleration×radius)

Velocity=√(13.33×30)

Velocity=√(399.9)

Velocity=19.997

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Explanation:

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2 years ago
The specific heat of water is 4.18 J/gºC. How many Joules of
densk [106]

Answer:

Q = 59565 [J]

Explanation:

In order to calculate the amount of thermal energy needed we must use the following equation.

Q=m*C_{p}*(T_{i}-T_{f})

where:

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What level of intensity is playing golf while pulling or carrying a set of clubs?
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In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. P
tatiyna

Answer:

Time period for first satellites 24.46 days and for second satellites 37.67 days

Explanation:

Given :

Distance of first satellites r_{sat1} = 48000 \times 10^{3} m

Distance of second satellites r _{sat2} = 64000 \times 10^{3} m

Distance of charon r_{c} = 19600 \times 10^{3} m

Time period of charon T_{c} = 6.39 days

From the kepler's third law,

Square of the time period is proportional to the cube of the semi major axis.

   T^{2} = r^{3}

   \frac{T}{r^{\frac{3}{2} } } = constant

For first satellites,

  \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat1} }{r_{sat1} ^{\frac{3}{2} }  }

{T_{sat1} } = 6.39 \times \frac{(48000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat1} = 24.46 days

For second satellites,

   \frac{T_{c} }{r_{c} ^{\frac{3}{2} }  }  = \frac{T_{sat2} }{r_{sat2} ^{\frac{3}{2} }  }

{T_{sat2} } = 6.39 \times \frac{(64000 \times 10^{3} )^{\frac{3}{2} } }{(19600\times 10^{3} )^{\frac{3}{2} }}

T_{sat2} = 37.67 days

Therefore, time period for first satellites = 24.46 days and for second satellites 37.67 days

8 0
3 years ago
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