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3 years ago

Do other planets aside from Saturn have rings?

Physics

2 answers:

alex41 [277]3 years ago

5 0

Other gas giants such as Jupiter, Uranus and Neptune also have rings.

andreyandreev [35.5K]3 years ago

3 0

Yes For example Neptune

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A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

Kruka [31]

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0

3 years ago

How much energy ( in joules ) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature ?

aleksley [76]

Answer:

B. 17,705.1 J

Explanation:

The hear released when the mercury condenses into a liquid is given by:

$Q=m \lambda_v$

where

m = 0.06 kg is the mass of the mercury

$\lambda_v$ is the latent heat of vaporization

For mercury, the latent heat of vaporization is $\lambda_v = 296 kJ/kg$ , so the heat released during the process is:

$Q=(0.06 kg)(296 kJ/kg)=17.76 kJ = 17,760 J$

So, the closest option is

B. 17,705.1 J

5 0

3 years ago

Read 2 more answers

During a long jump, an Olympic champion's center of mass rose about 1.2 m from the launch point to the top of the arc. 1) What m

Tanzania [10]

Answer: Minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

Explanation:

Velocity is only in horizontal direction at the top most point which is similar to the velocity in the horizontal direction at the time of launch.

Now, according to the law of conservation of energy the formula used is as follows.

$mgh = \frac{1}{2} mv^{2}_{y}\\v_{y} = \sqrt{2gh}\\= \sqrt{2 \times 9.8 m/s^{2} \times 1.2}\\= 4.85 m/s$

As speed at which the person is travelling was 6.8 m/s. Hence, the initial velocity will be calculated as follows.

$v = \sqrt{v^{2}_{x} + v^{2}_{y}}\\= \sqrt{(6.8)^{2} + (4.85 m/s)^{2}}\\= 11.65 m/s$

Thus, we can conclude that minimum speed needed by the Olympic champion at launch if he was traveling at 6.8 m/s at the top of the arc is 11.65 m/s.

6 0

3 years ago

In what stage do children see the world in an egocentric way?

zheka24 [161]

Answer: Preoperational stage (second stage).

Explanation: In the second stage of Child's development i.e., from the age of 2 years to 7 years approximately, which is called as Preoperational stage, a child develops a strong sense of Egocentrism. It is a situation where a child is unable to see thing from other peoples' eye and feels that everyone's reaction to something should be same as the child's. At this stage, a child is busy and completely trapped in his own dreamy world and is not ready to understand things properly.

Hence, the stage is Preoperational stage.

3 0

3 years ago

In his experiment, Rutherford shot alpha particles at which type of metal?

astra-53 [7]

B.gold also known as the famous gold foil experiment

4 0

3 years ago

Read 2 more answers