(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:
69.74 N
Explanation:
We are given that
Weight of sled=49 N
Coefficient of kinetic friction
Weight of person=585 N
Total weight==mg=49+585=634 N
We know that
Force needed to pull the sled across the snow at constant speed,F=Kinetic friction
Where N= Normal=mg
Hence, the force is needed to pull the sled across the snow at constant speed=69.74 N
Carbon is found in the solid form in geosphere of our earth. Coal and oil are some of the examples of materials containing carbon in the geosphere. when the coal or oil is burnt, carbon dioxide is formed and released in atmosphere. This carbon dioxide is absorbed by the water of the hydrosphere with the help of algae and plankton. The water turns acidic in nature. This way carbon is transferred from geosphere to hydrosphere.
1). the product of the two masses being gravitationally attracted to each other
2). the distance between their centers of mass
And that's IT. The gravitational force between them depends on
only those two things, nothing else.
Answer:
The maximum potential difference is 186.02 x 10¹⁵ V
Explanation:
formula for calculating maximum potential difference
where;
Ke is coulomb's constant = 8.99 x 10⁹ Nm²/c²
k is the dielectric constant = 2.3
b is the outer radius of the conductor = 3 mm
a is the inner radius of the conductor = 0.8 mm
λ is the linear charge density = 18 x 10⁶ V/m
Substitute in these values in the above equation;
Therefore, the maximum potential difference this cable can withstand is 186.02 x 10¹⁵ V
