Decompose the forces acting on the block into components that are parallel and perpendicular to the ramp. (See attached free body diagram. Forces are not drawn to scale)
• The net force in the parallel direction is
∑ <em>F</em> (para) = -<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
• The net force in the perpendicular direction is
∑ <em>F</em> (perp) = <em>n</em> - <em>mg</em> cos(21°) = 0
Solving the second equation for <em>n</em> gives
<em>n</em> = <em>mg</em> cos(21°)
<em>n</em> = (0.200 kg) (9.80 m/s²) cos(21°)
<em>n</em> ≈ 1.83 N
Then the magnitude of friction is
<em>f</em> = <em>µn</em>
<em>f</em> = 0.25 (1.83 N)
<em>f</em> ≈ 0.457 N
Solve for the acceleration <em>a</em> :
-<em>mg</em> sin(21°) - <em>f</em> = <em>ma</em>
<em>a</em> = (-0.457N - (0.200 kg) (9.80 m/s²) sin(21°))/(0.200 kg)
<em>a</em> ≈ -5.80 m/s²
so the block is decelerating with magnitude
<em>a</em> = 5.80 m/s²
down the ramp.
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The force on a charged particle in a magnetic field is given by
the speed of the charged particle = 10842 m/s.
Explanation:
F= q V B sinθ
F=force=3.5 x 10⁻²N
q= charge= 8.4 x 10⁻⁴ C
B= magnetic field= 6.7 x 10⁻³ T
θ=35⁰
Thus the velocity is given by V=
V=(3.5 x 10⁻²)/[(8.4 x 10⁻⁴)(6.7 x 10⁻³)(sin35)]
V=10842 m/s
The strength of the gravitational field is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
.
Noble gases are not highly reactive
