Answer:
<h2>10 kg.m/s</h2>
Explanation:
The momentum of an object can be found by using the formula
momentum = mass × velocity
From the question we have
momentum = 20 × 0.5
We have the final answer as
<h3>10 kg.m/s</h3>
Hope this helps you
Answer:
a) S = 1.69 10⁹ W/m², b) P = 5.63 Pa
, c) F = 20.6 10⁻¹² N
Explanation:
a) The intensity defined as the energy per unit area
S = U / A
Area of a circle is
W = 6.2 mw = 6.2 10-3 W
R = 1080 nm = 1080 10⁻⁹ m = 1.080 10⁻⁶ m
A = π R2
A = π (1,080 10⁻⁶)²
A = 3.66 10 -12 m²
S = 6.2 10-3 / 3.66 10-12
S = 1.69 10⁹ W / m²
b) The radiation pressure
P = 1 / c (dU / dt) / A
S = (dU / dt) / A
P = S / c
P = 1.69 10 9 / 3. 108
P = 5.63 Pa
c) the definition of pressure is force over area
P = F / A
F = P A
F = 5.63 3.66 10⁻¹²
F = 20.6 10⁻¹² N
d) for this we use Newton's second law
F = ma
a = F / m
Answer:
The velocity of Mosquito with respect to earth will be 0.302m/s
Explanation:
V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air
V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.
Velocity of Mosquito with respect to earth will be
V(me) = V(ma) + V(ae)
We need to find the mosquito’s speed with respect to Earth in the x direction.
V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )
Angle (ae) = –90.0° − 35°=−125°
V(x, me) = 1.10 + (1.4)Cos(-125)
= 1.10 + 1.4(-0.57)
= 1.10 -0.798
= 0.302
So the velocity of Mosquito with respect to earth will be 0.302m/s
