Answer:
utrasonic
Explanation:
these are sounds beyond our hearing capacity range of 20-20kHz
It is made up of mostly water and salt. Cytoplasm<span> is present within the cell membrane of all cell types and contains all organelles and cell parts.
The cytoplasm is like a </span><span>bathtub water because it holds a kind of jelly fluid just like a bathtub
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eukaryote<span> is an </span>organism<span> with complex cells, or a single cell with a complex structure. </span><span>
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Apparent magnitude depends mainly on the brightness of the object as seen from an observer on Earth. This is taken into account without the effects of the atmosphere.
Answer:
The resultant force would (still) be zero.
Explanation:
Before the 600-N force is removed, the crate is not moving (relative to the surface.) Its velocity would be zero. Since its velocity isn't changing, its acceleration would also be zero.
In effect, the 600-N force to the left and 200-N force to the right combines and acts like a 400-N force to the left.
By Newton's Second Law, the resultant force on the crate would be zero. As a result, friction (the only other horizontal force on the crate) should balance that 400-N force. In this case, the friction should act in the opposite direction with a size of 400 N.
When the 600-N force is removed, there would only be two horizontal forces on the crate: the 200-N force to the right, and friction. The maximum friction possible must be at least 200 N such that the resultant force would still be zero. In this case, the static friction coefficient isn't known. As a result, it won't be possible to find the exact value of the maximum friction on the crate.
However, recall that before the 600-N force is removed, the friction on the crate is 400 N. The normal force on the crate (which is in the vertical direction) did not change. As a result, one can hence be assured that the maximum friction would be at least 400 N. That's sufficient for balancing the 200-N force to the right. Hence, the resultant force on the crate would still be zero, and the crate won't move.
Answer:
1.8m
Explanation:
Let the Elastics of the steel ASTM-36 
The strain of the bar when subjected to 150 MPa is
Therefore, if the bar elongates by 1.35 mm, then the original length L would be:
or 1.8m
