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2 years ago

PLS HELP PLS

Physics

2 answers:

igor_vitrenko [27]2 years ago

6 0

Good morning love ya good morning Mrs thanks love you bye bye thanks again for your birthday wishes bye

zhuklara [117]2 years ago

6 0

Answer:

sorry di ko alam

Explanation:

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Based on its orbit, which planet behaves the least like the others?

Gekata [30.6K]

I believe it's Mercury, because the only other option would be Pluto and it's not even considered a planet anymore
Hope this helps

3 0

3 years ago

What tripositive ion has the electron configuration [kr] 4d3 ? what neutral atom has the electron configuration [kr] 5s 2 4d2 ?

tigry1 [53]

(a)

Electronic configuration is given as follows:

$[Kr]4d^{3}$

Since, this is the electronic configuration of ion with+3 that means 3 electrons are removed. On adding the 3 electrons, the electronic configuration of neutral atom can be obtained.

Thus, electronic configuration of neutral atom is $[Kr]4d^{5}5s^{1}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+6=42.

This corresponds to element: molybdenum. Thus, the tripositive atom will be $Mo^{3+}$ .

(b) The given electronic configuration is $[Kr]5s^{2}4d^{2}$ .

The atomic number of Kr is 36, thus, total number of electrons become 36+4=40.

This corresponds to element zirconium, represented by symbol Zr.

6 0

3 years ago

A 6.0 g marble is fired vertically upward using a spring gun. The spring must be compressed 9.4 cm if the marble is to just reac

RoseWind [281]

Answer:

a) $\Delta U_{g} = 12.945\,J$ , b) $\Delta U_{k} = 12.945\,J$ , c) $k = 2930.059\,\frac{N}{m}$

Explanation:

a) The change in the gravitational potential energy of the marble-Earth system is:

$\Delta U_{g} = (0.06\,kg)\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (22\,m)$

$\Delta U_{g} = 12.945\,J$

b) The change in the elastic potential energy of the spring is equal to the change in the gravitational potential energy, then:

$\Delta U_{k} = 12.945\,J$

c) The spring constant of the gun is:

$\Delta U_{k} = \frac{1}{2} \cdot k \cdot x^{2}$

$k = \frac{2\cdot \Delta U_{k}}{x^{2}}$

$k = \frac{2\cdot (12.945\,J)}{(0.094\,m)^{2}}$

$k = 2930.059\,\frac{N}{m}$

4 0

2 years ago

A 200-gram liquid sample of Alcohol Y is prepared at -6°C. The sample is then added to 400 g of water at 20°C in a sealed styrof

Vinil7 [7]

The specific heat capacity of the alcohol will be 3.72 kJ/kg°C.

<h3>What is the specific heat capacity?</h3>

The amount of heat required to increase a substance's temperature by one degree Celsius is known as its "specific heat capacity."

Similarly, heat capacity is the relationship between the amount of energy delivered to a substance and the increase in temperature that results.

Given data;

Mass of liquid sample of Alcohol m₁ = 200-gram

The temperature of alcohol, T₁ = -6°C.

Mass of liquid sample of water m₂ = 400-gram

The temperature of the water, T₂= 20°C.

The specific heat capacity of the alcohol, S₁=?

The specific heat capacity of water is, S₂=4.19 kJ/kg.°C

As we know that;

<h3 />

$\rm Q_{gain}= Q{loss} \\\\ Q_{alcohol} =Q_{water} \\\\\ m_1s_1\triangle T_1 = m_2S_2 \triangle S_2 \\\\ 200 \times 10^{-3} \times S_1 [ (12-(-6) ] = 40 \times 10^{-3} \times 4.19 \times 10^{-3} \times (20-12)\\\\S_1 = 2 \times 4.19 \times 10^3 \times \frac {8}{18} \\\\ S_1 = 3.72 \ kJ /kg ^0 C$

Hence the specific heat capacity of the alcohol will be 3.72 kJ/kg°C.

To learn more about the specific heat capacity, refer to the link brainly.com/question/2530523.

#SPJ1

6 0

2 years ago

The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit

const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

W = ΔK

Where work is defined as the product of force by distance

W = fr x cos 180

The angle is because the friction force opposes the movement

Δk = $K_{f}$ –K₀

ΔK = 0 - ½ m v₀²

We substitute

- fr x = - ½ m v₀²

fr = ½ m v₀²/x

8 0

3 years ago