(a) Differentiate the position vector to get the velocity vector:
<em>r</em><em>(t)</em> = (3.00 m/s) <em>t</em> <em>i</em> - (4.00 m/s²) <em>t</em>² <em>j</em> + (2.00 m) <em>k</em>
<em>v</em><em>(t)</em> = d<em>r</em>/d<em>t</em> = (3.00 m/s) <em>i</em> - (8.00 m/s²) <em>t</em> <em>j</em>
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(b) The velocity at <em>t</em> = 2.00 s is
<em>v</em> (2.00 s) = (3.00 m/s) <em>i</em> - (16.0 m/s) <em>j</em>
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(c) Compute the electron's position at <em>t</em> = 2.00 s:
<em>r</em> (2.00 s) = (6.00 m) <em>i</em> - (16.0 m) <em>j</em> + (2.00 m) <em>k</em>
The electron's distance from the origin at <em>t</em> = 2.00 is the magnitude of this vector:
||<em>r</em> (2.00 s)|| = √((6.00 m)² + (-16.0 m)² + (2.00 m)²) = 2 √74 m ≈ 17.2 m
(d) In the <em>x</em>-<em>y</em> plane, the velocity vector at <em>t</em> = 2.00 s makes an angle <em>θ</em> with the positive <em>x</em>-axis such that
tan(<em>θ</em>) = (-16.0 m/s) / (3.00 m/s) ==> <em>θ</em> ≈ -79.4º
or an angle of about 360º + <em>θ</em> ≈ 281º in the counter-clockwise direction.
Answer:
0.175 m/s²
Explanation:
Given:
v₀ = 35 m/s
v = 42 m/s
t = 40 s
Find: a
v = at + v₀
42 m/s = a (40 s) + 35 m/s
a = 0.175 m/s²
Answer:
50kg.m/s
Explanation:
In order to find momentum you must use the formula P=mv
p= momentum
m=mass
v= velocity
so in other words, momentum= mass times velocity
or in this case, momentum= 10 times 5 :)
The string moves to the right, as it restores its original position with the median plane of the bow. As a result, the string "pulls" on the arrow with a force F2. 2. The tip of the arrow T moves slightly to the left.
pls thank me and brainliest me
Answer:
(A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
Explanation:
Given that,
Voltage = 8.2 V
Inductor = 38 mH
Resistance = 150 Ω
Time t = 0.110 ms
The battery has negligible internal resistance, so that the total resistance in the circuit is 150 ohms. Then use this equation for current at time t in terms of inductance
We need to calculate the current
Using formula of current
Put the value into the formula
(B). We need to calculate the store energy in the inductor
Using formula of energy
Put the value into the formula
{tex]E=7.04\ \mu J[/tex]
Hence, (A). The current in the circuit is 19.25 mA.
(B). The store energy in the inductor is 7.04 μJ.
