By experiment, determine what makes a force attractive or repulsive. Describe your experiments and observations with some exampl
1 answer:
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Answer:
Explanation:
Average velocity of oxygen molecule at given temperature is
now we have
M = 32 g/mol = 0.032 kg/mol
T = 27 degree C = 300 K
now we have
now for de Broglie wavelength we know that
Answer:
30.67m
Explanation:
Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;
s = ut ± ------------(i)
Where;
s = horizontal/vertical displacement
u = initial horizontal/vertical component of the velocity
a = acceleration of the projectile
t = time taken for the projectile to reach a certain horizontal or vertical position.
Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;
h = vt ± ------------(ii)
Where;
h = vertical displacement
v = initial vertical component of the velocity = usinθ
a = acceleration due to gravity (since vertical motion is considered)
t = time taken for the projectile to hit the target
<em>From the question;</em>
u = 47m/s, θ = 0.6rads
=> usinθ = 47 sin 0.6
=> usinθ = 47 x 0.5646 = 26.54m/s
t = 1.7s
Take a = -g = -10.0m/s (since motion is upwards against gravity)
Substitute these values into equation (ii) as follows;
h = vt -
h = 26.54(1.7) -
h = 45.118 - 14.45
h = 30.67m
Therefore, the vertical distance is 30.67m
In triangle ABC , using Pythagorean theorem
BC = sqrt(AB² + AC²)
r = sqrt(y² + x²) eq-1
taking derivative both side relative to "t"
dr/dt = (1/(2 sqrt(y² + x²) ) ) (2 y (dy/dt) + 2 x (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) (2 (0.5) (dy/dt) + 2 (0.5) (dx/dt))
dr/dt = (1/(2 sqrt(0.5² + 0.5²) ) ) ( v₁ + v₂)
15= (1/(2 sqrt(0.5² + 0.5²) ) ) ( - 30 + v₂)
v₂ = 51.2 m/s
