Answer:
the nucleus-containing central part of a neuron exclusive of its axons and dendrites that is the major structural element of the gray matter of the brain and spinal cord, the ganglia, and the retina. — called also perikaryon, soma.
Answer:
The distance traveled by Tina before passing David is 900 m
Given:
Initial speed of David,
Acceleration of Tina,
Solution:
Now, as per the question, we use 2nd eqn of motion for the position of David after time t:
where
s = distance covered by David after time 't'
a = acceleration of David = 0
Thus
Now, Tina's position, s' after time 't':
where
, initially at rest
(1)
At the instant, when Tina passes David, their distances are same, thus:
s = s'
t = 30 s
Now,
The distance covered by Tina before she passes David can be calculated by substituting the value t = 30 s in eqn (1):
= 900 m
Answer:
calculate each person's relative strength
Explanation:
The answer is:
-when Jackhammer is a huge hammer used to breaking up the hard rocks and the hard materials to smaller pieces.
-and the metamorphic rocks are type of rocks which change to metamorphic rock by heating and high pressure so, the chemical and physical properties of the rock will be changed.
-and here in cause of using the Jackhammer there is no change in the chemical or physical properties of the original rock and no high temp(more than 200°C ) or high pressure to change to metamorphic rock . It is just a breaking up of the rock to small parts.
The time taken for the train to stop is 8.9 seconds.
The distance traveled by the train during acceleration is 9,984 m.
The distance traveled by the train during deceleration is 17.8 m.
<h3>
Time taken for the train to stop</h3>
Use the following kinematic equation to determine the time taken for the train to stop.
v = u + at
where;
- v is the final velocity of the train = 0
- u is the initial velocity = 4 m/s
- a is the deceleration = -0.45 m/s²
- t is time of motion
0 = 4 - 0.45t
0.45t = 4
t = 4/0.45
t = 8.89 seconds = 8.9 seconds
<h3>Distance traveled by the train during acceleration</h3>
s = ut + ¹/₂at²
where;
- t is time of motion = 8 min = 480 seconds
s = (4)(480) + ¹/₂(0.07)(480)²
s = 9,984 m
<h3>Distance traveled by the train during deceleration</h3>
s = ut + ¹/₂at²
s = (4 x 8.9) + ¹/₂(-0.45)(8.9)²
s = (4 x 8.9) - ¹/₂(0.45)(8.9)²
s = 17.8 m
Learn more about deceleration here: brainly.com/question/75351
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