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3 years ago

One degree Celsius indicates the same temperature change as?

1 answer:

6 0

If you've ever seen the formula from Celsius to Fahrenheit...

$C=\frac59(F-32)$

As you can see, if you put in some number of Celsuis C your answer is going to be multiplied by 5/9...meaning that 1°C = 5/9°F.

As for Kelvin, it's literally just Celsuis, but about 273.15° more.

B. 5/9°F

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What does Coulombs law state?

dedylja [7]

Coulomb's law states<span> that: The magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distance between them.
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3 years ago

Post Test: Forces and Motion

Len [333]

Answer: B to C

Explanation: The line is curving inwards, practically calculating the stance that it had went down. If it went straight across, it stayed the same till a specific point, furthermore calculating the bent line bending upwards is actually a partial-raise, conclude points B to C is most likely an un-even balance, meaning it had went down; or decreasing. B to C is the decreasing segment of this equation/problem (question).

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3 years ago

Physics formula for Celsius to Fahrenheit

worty [1.4K]

F=9/5C+32 is the formula for Celsius to Fahrenheit.

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3 years ago

Read 2 more answers

The Mars Curiosity rover was required to land on the surface of Mars with a velocity of 1 m/s. Given the mass of the landing veh

Aliun [14]

Answer:

The value is $A = 39315 \ m^2$

Explanation:

From the question we are told that

The velocity which the rover is suppose to land with is $v = 1 \ m/s$

The mass of the rover and the parachute is $m = 2270 \ kg$

The drag coefficient is $C__{D}} = 0.5$

The atmospheric density of Earth is $\rho = 1.2 \ kg/m^3$

The acceleration due to gravity in Mars is $g_m = 3.689 \ m/s^2$

Generally the Mars atmosphere density is mathematically represented as

$\rho_m = 0.71 * \rho$

=> $\rho_m = 0.71 * 1.2$

=> $\rho_m = 0.852 \ kg/m^3$

Generally the drag force on the rover and the parachute is mathematically represented as

$F__{D}} = m * g_{m}$

=> $F__{D}} = 2270 * 3.689$

=> $F__{D}} = 8374 \ N$

Gnerally this drag force is mathematically represented as

$F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}$

Here A is the frontal area

$A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }$

=> $A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }$

=> $A = 39315 \ m^2$

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3 years ago

(Mass vs. Weight) HELP PLZ!!

Lina20 [59]

$\huge \boxed { \sf{Answers}}$

c. The weight of an object on the moon will be the same as its weight on Earth. It is false because the weight of an on the moon will be 1/6 th times its weight on Earth.
d. The weight of an object is its mass multiplied by the force of gravity. The statement is false because the formula of weight is mass × acceleration due to gravity, not force of gravity.
e. The mass and weight of an object are the same thing. The statement is false because mass means a body of matter. While weight of an object is its mass multiplied by the force of gravity.
f. The mass of an object is the force of gravity acting upon an object. It is false because it will be the weight of the object not mass.
So, the answers are c, d, e and f.

Hope you could understand.

If you have any query, feel free to ask.

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2 years ago