Answer:
The initial acceleration of the 59g particle is 
Explanation:
Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:
(1)
We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:
with q1 and q2 the charge of the particles, r the distance between them and k the constant 
. So:
Using that value on (1) and solving for a
Half mass car because it's traveling faster
ANSWER:
The easiest way to get a fairly accurate measure of your water flow rate is to time yourself filling up a bucket. So for example if you fill up a 10 litre bucket in 1.5 minutes, then your flow rate will be: 10/1.5 = 6.66 Litres per minute.
The acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².
<h3>How to calculate acceleration?</h3>
The acceleration of a freight train can be calculated using the following formula:
Force = mass × acceleration
According to this question, a 600,000kg freight train can produce 100,000N of force. The acceleration is as follows:
100,000 = 600,000 × a
100,000 = 600,000a
a = 0.167m/s²
Therefore, the acceleration of a 600,000 kg freight train, if each of its three engines can provide 100,000N of force is 0.167m/s².
Learn more about acceleration at: brainly.com/question/12550364
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The kayaker has velocity vector
<em>v</em> = (2.50 m/s) (cos(45º) <em>i</em> + sin(45º) <em>j</em> )
<em>v</em> ≈ (1.77 m/s) (<em>i</em> + <em>j</em> )
and the current has velocity vector
<em>w</em> = (1.25 m/s) (cos(315º) <em>i</em> + sin(315º) <em>j</em> )
<em>w</em> ≈ (0.884 m/s) (<em>i</em> - <em>j</em> )
The kayaker's total velocity is the sum of these:
<em>v</em> + <em>w</em> ≈ (2.65 m/s) <em>i</em> + (0.884 m/s) <em>j</em>
That is, the kayaker has a velocity of about ||<em>v</em> + <em>w</em>|| ≈ 2.80 m/s in a direction <em>θ</em> such that
tan(<em>θ</em>) = (0.884 m/s) / (2.65 m/s) → <em>θ</em> ≈ 18.4º
or about 18.4º north of east.
