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madreJ [45]
3 years ago
12

Everything is in the pic.

Physics
1 answer:
nikitadnepr [17]3 years ago
3 0
The answer is A

Good luck!
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I can not find the answer to the one after potassium ?
Zepler [3.9K]

Answer:

it’s halogen

Explanation:

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3 years ago
The process of conversion of solid into gas and again gas into solid is known as
pantera1 [17]
Sublimation is when a solid becomes a gas.  And gas to solid is deposition.
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A 1.5 kg ball is dropped from a height of 2.Gm. Assuming energy is
Vikki [24]

Answer:

Plug in the given values and solve for the final velocity. Remember, when the ball is on the ground it has a height of zero.

Explanation:

6 0
2 years ago
Thermal conductivity of a material is given as 129Btuft–¹ h–¹°F–¹.Calculate this thermal conductivity in Jm–¹s–¹°C–¹(Given: 1Btu
ss7ja [257]

Answer:

223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

Explanation:

The thermal conductivity of an object is defined as the measure or the ability of the object to transfer heat or conduct heat through its body.

In the context, the thermal conductivity of the material is given as

$=129 \text{ Btu ft}^{-1}\text{h}^{-1}^\circ\text{F}^{-1}$

And it is given that :

1 Btu = 1055 J

1 ft = 0.3048 m

$1^\circ F = \frac{5}{9}^\circ C$

We know that 1 h = 3600 s

So the thermal conductivity of the material in $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$ is  :

Thermal conductivity :

$=\frac{129 \text{ Btu}}{1 \text{ ft }\times \text{1 h}\times 1^\circ\text{F}}$

$=\frac{129 \times 1055 \text{ J}}{0.3048 \text{ m} \ \times 3600 \text{ s}\ \times \frac{5}{9}^\circ \text{C}}$

=  223.25 $\text{Jm}^{-1}\text{s}^{-1}^\circ\text{C}^{-1}$

3 0
3 years ago
A rocket sled with an initial mass of 3 metric tons, invluding 1 ton of fuel, rests on a level section of track. At t=0, the sol
Katena32 [7]

Answer:

v = 719.2 m / s and     a = 83.33 m / s²

Explanation:

This is a rocket propulsion system where the system is made up of the rocket plus the ejected mass, where the final velocity is

           v - v₀ = v_{e} ln (M₀ / M)

where v₀ is the initial velocity, v_{e} the velocity of the gases with respect to the rocket and M₀ and M the initial and final masses of the rocket

In this case, if fuel burns at 75 kg / s, we can calculate the fuel burned for the 10 s

            m_fuel = 75 10

            m_fuel = 750 kg

As the rocket initially had a mass of 3000 kg including 1000 kg of fuel, there are still 250 kg, so the mass of the rocket minus the fuel burned is

              M = 3000 -750 = 2250 kg

let's calculate

            v - 0 = 2500 ln (3000/2250)

            v = 719.2 m / s

To calculate the acceleration, let's use the concept of the rocket thrust, which is the force of the gases on it. In the case of the rocket, it is

             Push = v_{e} dM / dt

let's calculate

             Push = 2500  75

             Push = 187500 N

            If we use Newton's second law

             F = m a

             a = F / m

let's calculate

              a = 187500/2250

              a = 83.33 m / s²

7 0
3 years ago
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