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alexandr402 [8]
3 years ago
12

Please help me I’ll mark brainless .

Physics
1 answer:
Ivan3 years ago
7 0
The mass is 10.811 hope this helps
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How do latitude and proximity to bodies of water relate to the temperature?
SSSSS [86.1K]

Answer:

The answer is below.

Explanation:

According to geographical coordinates and related information, when the latitude is lower, that is closer to the equator the temperature becomes warmer. While the far away from the equator the latitude is the temperature decreases.

Similarly, the water bodies have a relative effect that is considered little or less severe on the temperature throughout any of the seasonal period or day and night.

7 0
3 years ago
What is the potential difference VB – VA when the I= 1.5 A in the circuit segment below?​
son4ous [18]

Answer:

Dahil to sa pilipinas para Hindi I to masakof ng chinaPara lingtas ang mga tao sa pilipinas at is a pa Wang mag kalat sa ebang ebang dumiy

Explanation:

bakit lahat tayo ay ekonomista

6 0
3 years ago
Estimate the Joule-Thomson coefficient of refrigerant-134a at 0.70 MPa and 50°C. Assume the second state will be selected for a
leva [86]

Answer:

\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

Explanation:

The Joule-Thomson coefficient is the ratio of the change of temperature to the change of pressure under isoenthalpic conditions:

\mu = \left(\frac{\Delta T}{\Delta P}\right)_{h}

Initial and final properties are:

T_{1} = 50^{\textdegree}C, P_{1}=700\,kPa, h_{1}=288.54\,\frac{kJ}{kg}. Superheated Vapor.

T_{2} = 48.186^{\textdegree}C, P_{2}=600\,kPa, h_{2}=288.54\,\frac{kJ}{kg}. Superheated Vapor.

The Joule-Thomson coefficient is approximately:

\mu = \frac{50^{\textdegree}C-48.186^{\textdegree}C}{700\,kPa-600\,kPa}

\mu = 0.018\,\frac{^{\textdegree}C}{kPa}

5 0
4 years ago
Which formula represents the law of conservation of energy?
Amiraneli [1.4K]

Answer:

KE+PE

Explanation:

4 0
3 years ago
What is the speed of a proton whose kinetic energy is 3.4 kev ?
Andreas93 [3]
The kinetic energy of the proton is 3.4 kev
1 kev = 1.602 × 10^-16 joules
therefore 3.4 kev is equivalent to;
3.4 ×  (1.602 ×10^-16)= 5.4468 × 10^-16 J
Kinetic energy is calculated by the formula 1/2mv² where m is the mass and v is the velocity.
Therefore V = √((2 × ( 5.4468×10^-16))/ (1.67 ×10^-27))
                    = 8.077 × 10^5 m/s

8 0
3 years ago
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