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ASHA 777 [7]
3 years ago
7

6

Physics
1 answer:
Tcecarenko [31]3 years ago
8 0

Explanation:

Melting point

Boiling point

Does the compound

dissolve in water?

Can the compound

conduct electricity?

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Does anyone know this or done it? It’s for life skills and about MyPlate:)
vivado [14]
The answer is no I’ve not done that
6 0
2 years ago
Use this free body diagram to help you find the magnitude of the force F1 needed to keep this block in static equilibrium 15.3 N
bija089 [108]

Do you have a picture of the diagram?

5 0
3 years ago
How much thermal energy is needed to melt 1.25 kg of water at its melting point? Use Q = masslaten heat of fusion.
Amanda [17]

Answer:

Latent heatnof fusion = 417.5 J

Explanation:

Specific latent heat of fusion of water is 334kJ.kg-1.

The heat required to melt water when it's ice I called latent heat because there is no temperature change, the only change observed is change in physical structure.

The amount of heat required to change 1 kg of solid to its liquid state (at its melting point) at atmospheric pressure is called Latent heat of Fusion.

Latent heat = ML

Latent heat= 1.25 kg * 334kJ.kg-1

Latent heat = 1.25*334 *(J/kg)*kg

Latent heat = 417.5 J

8 0
3 years ago
A piston-cylinder device initially contains 0.08 m3 of nitrogen gas at 150 kPa and 200°C. The nitrogen is now expanded to a pres
Lemur [1.5K]

Answer:

V_2 = 0.125 m^3

Work done =  = 5 kJ

Explanation:

Given data:

volume of nitrogen v_1 = 0.08 m^3

P_1 = 150 kPa

T_1 = 200 degree celcius = 473 Kelvin

P_2 = 80 kPa

Polytropic exponent n = 1.4

\frac{T_2}{T_1} = [\frac{P_2}{P_1}]^{\frac{n-1}{n}

putting all value

\frac{T_2}{473} = [\frac{80}{150}]^{\frac{1.4-1}{1.4}

\frac{T_2} = 395.23 K = 122.08 DEGREE \ CELCIUS

polytropic process is given as

P_1 V_1^n = P_2 V_2^n

150\times 0.08^{1.4} = 80 \times V_2^{1.4}

V_2 = 0.125 m^3

work done = \frac{P_1 V_1 -P_2 V_2}{n-1}

= \frac{150 \times 0.8 - 80 \times 0.125}{1.4-1}

                  = 5 kJ

4 0
3 years ago
You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. Th
Rama09 [41]

Answer:

 I = 8.75 kg m

Explanation:

This is a rotational movement exercise, let's start with kinetic energy

        K = ½ I w²

They tell us that K = 330 J, let's find the angular velocity with kinematics

      w² = w₀² + 2 α θ

as part of rest w₀ = 0

      w = √ 2α θ

let's reduce the revolutions to the SI system

      θ = 30.0 rev (2π rad / 1 rev) = 60π rad

let's calculate the angular velocity

      w = √(2  0.200  60π)

      w = 8.683 rad / s

we clear from the first equation

        I = 2K / w²

let's calculate

        I = 2 330 / 8,683²

        I = 8.75 kg m

4 0
2 years ago
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