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3 years ago

Can you describe Saturn?

1 answer:

6 0

Saturn is the sixth planet from the Sun and the second-largest in the Solar System, after Jupiter. It is a gas giant with an average radius about nine times that of Earth.

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A sample of an unknown substance has a mass of 89.5 g. If 345.2 J of heat are required to heat the substance from 285 K to 305 K

Zinaida [17]

Heat gained in a system can be calculated by multiplying the given mass to the specific heat capacity of the substance and the temperature difference. It is expressed as follows:

Heat = mC(T2-T1)
345.2 = 89.5(C)(305 - 285)
C = 0.1928 J/g•K

7 0

3 years ago

Read 2 more answers

A car moves with a speed of 30 metres per second calculate the distance travelled in 30 seconds

Leni [432]

30x30=900

The answer is 900 meters after 30 seconds

7 0

3 years ago

As a physics instructor hurries to the bus stop, her bus passes her, stops ahead, and begins loading passengers. She runs at 6.0

Alika [10]

velocity of the physics instructor with respect to bus

$v = 6 m/s$

acceleration of the bus is given as

$a = 2 m/s^2$

acceleration of instructor with respect to bus is given as

$a = -2 m/s^2$

now the maximum distance that instructor will move with respect to bus is given as

$v_f^2 - v_i^2 = 2 a d$

$0 - 6^2 = 2(-2)(d)$

$-36 = - 4 d$

$d = 9 m$

so the position of the instructor with respect to door is exceed by

$\delta x = 9 - 6 = 3 m$

so it will be moved maximum by 3 m distance

7 0

3 years ago

Analyze how some insects are able to move around on the surface of a lake or pond

koban [17]

It is called surface tension it is the elastic personality of some liquids as they pull together to take up as little surface area as possible. the water molecules would rather stay together than be pulled apart

3 0

3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

$Fn_{3} = \sqrt{405^{2}+701.5^{2} }$

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0

3 years ago