How is item A different from Item B?

Physics

2 answers:

iris [78.8K]3 years ago

8 0

Explanation:

well there is nothing there and it could be different by diffrent objects, idk

Rudik [331]3 years ago

4 0

Answer:

I have deduced that the answer is C) "The crust in Item A is found partially below sea level as compared to the crust in Item B."

Explanation:

This is because the crust in picture A has its crust slightly underneath sea level compared to picture B. I'd get more in-depth, but, I'm a bit too lazy. I believe this is to be learnt about in the future anyhow.(I believe)

You might be interested in

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor

const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is $\tau = 34.3 \ N\cdot m$

Explanation:

From the question we are told that

The mass of the steel ball is $m = 3.0 \ kg$

The length of arm is $l = 70 \ cm = 0.7 \ m$

The mass of the arm is $m_a = 4.0 \ kg$

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

$r = \frac{l}{2}$

=> $r = \frac{ 0.7}{2}$

=> $r = 0.35 \ m$

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

$\tau = m_a * g * r + m * g * L$

=> $\tau = 4 * 9.8 * 0.35 + 3 * 9.8 * 0.70$

=> $\tau = 34.3 \ N\cdot m$

5 0

3 years ago

A volumen constante un gas ejerce una presión de 880 mmHg a 20o Celsius dentro de una olla a presión ¿Qué temperatura habrá si e

jasenka [17]

Answer: Hence, the final temperature is 350 K

Explanation :

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$