Answer:
the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Explanation:
The computation of the magnitude of the each force is shown below:
Provided that
Ratio of forces = 3: 5
Let us assume the common factor is x
Now
first force = 3x
And, the second force = 5x
Resultant force = 35 N
The Angle between the forces = 60 degrees
Based on the above information
Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos
35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]
35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)
35 = √49 x²
x = 5
So, the magnitude of first force = 3 × 5= 15 N
ANd, the magnitude of second force = 5 × 5 = 25 N
Answer:
71 rpm
Explanation:
Given that:
Angular momentum (L) = 0.26
Diameter = 25cm = 0.25 cm
Radius, r = (d/2) = 0.125m
Mass = 5.6 kg
Moment of inertia (I) = 2mr² / 5
I = (2 * 5.6 * 0.125^2) / 5
= 0.175
= 0.175 / 5
= 0.035 kgm²
Angular speed (w) ;
w = L / I
w = 0.26 / 0.035
= 7.4285714
= 7.429 rad/s
w = (7.429 * 60/2π)
w = 445.74 / 2π rpm
w = 70.941724
Angular speed = 70.94 rpm
= 71 rpm
200g*1 mole/ 18g=11.1 moles There are 11.1 moles of water.
Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V
Answer:
At 6 minutes, the soil temperature at a 45° angle of insolation is higher than at 0°.
At 15 minutes, the soil temperature at a 90° angle of insolation is higher than at 45°
Explanation:
