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3 years ago

Three balls with the same radius 21 cm are in water. Ball 1 floats, with

Physics

1 answer:

Masteriza [31]3 years ago

7 0

Answer:

Explanation:

A )

The ball floats with half of it exposed above the water level . So it must have density half that of water . In other words its density must have been 500 kg / m³

B )

Tension in the ball will be equal to net force acting on the ball

Net force on the ball = buoyant force - weight .

4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1000 - 893 )

= 40.65 x 10⁻⁶ N .

C )Tension in the 3 rd ball will be equal to net force acting on the ball

Net force on the ball = weight - buoyant force

= 4/3 x π x .21³ x 10⁻⁶ x 9.8 ( 1320 - 1000 )

= 121.6 x 10⁻⁶ N .

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Differences between weightlessness in space and weightlessness in earth

zhuklara [117]

Answer:

it depends on a person's own weight

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3 years ago

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A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime

Olin [163]

Answer:

The leaves of the electroscope move further apart.

Explanation:

This is what happens; when the positive object is brought near the top, negative charges migrating from the gold leaves to the top. This is because the negative charges in the gold are attracted by the positive charge. Thus, it leaves behind a net positive charge on the leaves, though the scope remains neutral overall. To that effect, the leaves repel each other and move apart. If a finger touches the top of the electroscope at the moment when the positive object remains near the top, it basically grounds the electroscope and thus the net positive charge in the leaves flows to the ground through the finger. However, the positive object continues to "hold" negative charges in place at the top. Ar this moment the gold leaves have lost their net positive charge, so they no longer repel, and they move closer together. If the positive object is moved away, the negative charges at the top are no longer attracted to the top, and they redistribute themselves throughout the electroscope, moving into the leaves and charging them negatively.

Thus, the leaves move apart from each other again and we now have a negatively charged electroscope. If a negatively charged object is now brought close to the top, but without touching, the negative charges already in the electroscope will be repelled down toward the leaves, thereby making them more negative, causing them to repel more, and hence move even further apart.

So, the leaves move further apart.

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3 years ago

A planet's moon travels in an approximately circular orbit of radius 7.0 ✕ 10⁷ m with a period of 6 h 38 min. Calculate the mass

natka813 [3]

Answer:

3.56×10²⁶ Kg.

Explanation:

Note: The gravitational force is acting as the centripetal force.

Fg = Fc........................... Equation 1

Where Fg = gravitational Force, Fc = centripetal force.

Recall,

Fg = GMm/r²......................... Equation 2

Fc = mv²/r............................. Equation 3

Where M = mass of the planet, m = mass of the moon, r = radius of the orbit and G = Universal gravitational constant.

Substituting equation 2 and 3 into equation 1

GMm/r² = mv²/r

Simplifying the equation above,

M = v²r/G .............................. Equation 4.

The period of the moon in the orbit

T = 2πr/v

Making v the subject of the equation,

v = 2πr/T............................. Equation 5

where r = 7.0×10⁷ m, T = 6 h 38 min = (6×3600 + 38×60) s = (21600+2280) s

T = 23880 s, π = 3.14

v = (2×3.14×7.0×10⁷ )/23880

v = 18409 m/s

Also Given: G = 6.67×10⁻¹¹ Nm²/kg²

Also substituting into equation 4

M = 18409²×7.0×10⁷ /(6.67×10⁻¹¹)

M = 3.56×10²⁶ Kg.

Thus the mass of the planet = 3.56×10²⁶ Kg.

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3 years ago

Global climate change has caused ocean temperatures to increase. Which evidence supports this claim?

Anuta_ua [19.1K]

D. Sea level has risen more than 300 centimeters. This is true because as the earth gets warmer, ice will melt and turn into water which goes into the ocean making it rise.

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3 years ago

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A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci

Vlad [161]

Answer:

Approximately $\rm 2.5\; m \cdot s^{-1}$ .

Explanation:

Let the increase in the rocket's velocity be $\Delta v$ . Let $v_0$ represent the initial velocity of the rocket. Note that for this question, the exact value of $v_0$ doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

Mass of the rocket with the 5 kg of fuel: $1000$ .
Initial velocity of the rocket and the fuel: $v_0$ .
Hence the initial momentum of the rocket: $1000\,v_0$ .

Mass of the rocket without that 5 kg of fuel: $1000 - 5 = 995$ .
Final velocity of the rocket: $v_0 + \Delta v$ .
Hence the final momentum of the rocket: $995\,(v_0 + \Delta v)$ .

Mass of the 5 kg of fuel: $5$ .
Final velocity of the fuel: $v_0 - 500$ (assuming that the the 500 m/s in the question takes the rocket as its reference.)
Hence the final momentum of the fuel: $5\,(v_0 - 500)$ .