Answer:
3A
Explanation:
Using Ohms law U=I×R solve for I by I=U/R
Answer:
Molar mass -56.0774 g/mol
There is a way to calculate this
atomic number for calcium is 20
atomic number for oxygen is 8
The molar mass for calcium is 40
The molar mass for oxygen is 16
CALCIUM OXIDE
40. +. 16
=56g/mol
1 molecule = 1×56=56g/mol
2 molecules=2×56 =112g/mol
Answer:
Gravitational force (pulled downward by the Earth)
Normal force (pushed upward by the ground)
Applied force (pushed by the person)
Friction force (pulled opposite the direction of motion by the roughness of the ground)
The rovers were designed to trek up to 100 meters (about 110 yards or 328 feet) across the martian surface each martian day, though they have gone much farther. While a complete martian day (called a sol) is about 24 hours and 40 minutes long (or 24 hours 37.5 minutes if you prefer), the Sun can only provide enough power for driving during a four-hour window around high noon. That means the rovers have to be able to move quickly and effectively.
Moving safely from rock to rock or location to location is a major challenge because of the communication time delay between Earth and Mars, which is about 20 minutes on average. Unlike a remote controlled car, the drivers of rovers on Mars cannot instantly see what is happening to a rover at any given moment and they cannot send quick commands to prevent the rover from running into a rock or falling off of a cliff.
During surface operations on Mars, each rover receives a new set of instructions at the beginning of each sol. Sent from the scientists and engineers on Earth, the command sequence tells the rover what targets to go to and what science experiments to perform on Mars. The rover is expected to move over a given distance, precisely position itself with respect to a target, and deploy its instruments to take close-up pictures and analyze the minerals or elements of rocks and soil.
Answer:
Explanation:
For any large steel plate with an infinite length and width and a center through the crack under tension, the stress intensity factor can be expressed as:
where;
a = 2.5 mm/2
a = 1.25 mm
a = 1.25 × 10⁻³ m
Therefore; the maximum stress in tension capacity can be computed as;
