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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

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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A cube and a square pyramid were joined to form the composite solid. A cube with side lengths of 12 inches. A square pyramid wit

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Answer:

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3 years ago

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A 60 cm diameter wheel accelerates from rest at a rate of 7 rad/s2. After the wheel has undergone 14 rotations, what is the radi

Mamont248 [21]

Answer:

$a=368.97\ m/s^2$

Explanation:

Given that,

Initial angular velocity, $\omega=0$

Acceleration of the wheel, $\alpha =7\ rad/s^2$

Rotation, $\theta=14\ rotation=14\times 2\pi =87.96\ rad$

Let t is the time. Using second equation of kinematics can be calculated using time.

$\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\\t=\sqrt{\dfrac{2\theta}{\alpha }} \\\\t=\sqrt{\dfrac{2\times 87.96}{7}} \\\\t=5.01\ s$

Let $\omega_f$ is the final angular velocity and a is the radial component of acceleration.

$\omega_f=\omega_i+\alpha t\\\\\omega_f=0+7\times 5.01\\\\\omega_f=35.07\ rad/s$

Radial component of acceleration,

$a=\omega_f^2r\\\\a=(35.07)^2\times 0.3\\\\a=368.97\ m/s^2$

So, the required acceleration on the edge of the wheel is $368.97\ m/s^2$ .

6 0

3 years ago

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3 years ago

When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec

igomit [66]

Answer:

The maximum electric power output is $P_{max} =1.339*10^{9} \ W$

Explanation:

From the question we are told that

The capacity of the hydroelectric plant is $\frac{V}{t} = 690 \ m^3 /s$

The level at which water is been released is $h = 220 \ m$

The efficiency is $\eta =$ 0.90

The electric power output is mathematically represented as

$P = \frac{PE_l - PE _o}{t}$

Where $PE_l$ is the potential energy at level h which is mathematically evaluated as

$PE_l = mgh$

and $PE_o$ is the potential energy at ground level which is mathematically evaluated as

$PE_o = mg(0)$

$PE_o = 0$

So

$P = \frac{mgh}{t}$

here $m = V * \rho$

where V is volume and $\rho$ is density of water whose value is $\rho = 1000 kg/m^3$

So

$P = \frac{(\rho * V) * gh}{t}$

$P = \frac{V}{t} * gh \rho$

substituting values

$P =690 * 9.8 * 220 * 1000$

$P =1.488*10^{9} \ W$

The maximum possible electric power output is

$P_{max} = P * \eta$

substituting values

$P_{max} =1.488*10^{9} * 0.90$

$P_{max} =1.339*10^{9} \ W$

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3 years ago

What would you do to improve the precision of an experiment?

GenaCL600 [577]

Explanation:

Precision represents that how close the different measurements of the sample one take are to one another.

One can increase the precision in lab by paying attention to each and every detail.
Usage of the equipment properly and also increasing the sample size.
Ensuring that the equipment is calibrated properly. They should be clean and functioning. Using equipment which is not functioning correctly can cause results to swing wildly and also bits of the debris stuck to the equipment can influence the measurements of the mass and the volume.
Each measurement must be taken multiple times, especially if experiments in which combining of the substances in specific amounts is involved.

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3 years ago

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