The maximum force of static friction is the product of normal force (P) and the coefficient of static friction (c). In a flat surface, normal force is equal to the weight (W) of the body.
P = W = mass x acceleration due to gravity
P = (0.3 kg) x (9.8 m/s²) = 2.94 kg m/s² = 2.94 N
Solving for the static friction force (F),
F = P x c
F = (2.94 N) x 0.6 = 1.794 N
Therefore, the maximum force of static friction is 1.794 N.
Answer:
If it is not an object in motion, all forces are balanced.
Answer:
» An electron is lighter than a proton.
<u>explanation</u><u>:</u>
hence it's mass number is zero
hence it's mass number is 4
<u>Therefore</u><u>,</u><u> </u><u>proton</u><u> </u><u>is</u><u> </u><u>heavier</u><u> </u><u>than</u><u> </u><u>electron</u>
» An electron has a small charge magnitude than a proton.
<u>Explanation</u><u>:</u>
An electron has charge of -1 while proton has charge of +2, therefore electron is less deflected by any energetic fields than a proton
Answer:
a) m =1 θ = sin⁻¹ λ / d, m = 2 θ = sin⁻¹ ( λ / 2d)
, c) m = 3
Explanation:
a) In the interference phenomenon the maxima are given by the expression
d sin θ = m λ
the maximum for m = 1 is at the angle
θ = sin⁻¹ λ / d
the second maximum m = 2
θ = sin⁻¹ ( λ / 2d)
the third maximum m = 3
θ = sin⁻¹ ( λ / 3d)
the fourth maximum m = 4
θ = sin⁻¹ ( λ / 4d)
b) If we take into account the effect of diffraction, the intensity of the maximums is modulated by the envelope of the diffraction of each slit.
I = I₀ cos² (Ф) (sin x / x)²
Ф = π d sin θ /λ
x = pi a sin θ /λ
where a is the width of the slits
with the values of part a are introduced in the expression and we can calculate intensity of each maximum
c) The interference phenomenon gives us maximums of equal intensity and is modulated by the diffraction phenomenon that presents a minimum, when the interference reaches this minimum and is no longer present
maximum interference d sin θ = m λ
first diffraction minimum a sin θ = λ
we divide the two expressions
d / a = m
In our case
3a / a = m
m = 3
order three is no longer visible
Answer:
2.4 m/s". 1
Explanation:
A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1
