Answer:
a) vₓ = 6,457 m / s
, v_{y} = 0.518 m / s
, b) v = 6.478 m / s, θ = 4.9°
Explanation:
a) This is a kinematic problem, let's use trigonometry to find the components of acceleration
sin 31 = / a
cos 31 = aₓ = a
a_{y} = a sin31
aₓ = a cos 31
Now let's use the kinematic equation for each axis
X axis
vₓ = v₀ₓ + aₓ (t-t₀)
vₓ = v₀ₓ + a cos 31 (t-t₀)
vₓ = 2.6 + 0.45 cos 31 (20-10)
vₓ = 6,457 m / s
Y Axis
v_{y} = v_{oy} + a_{y} t
v_{y} = v_{oy} + a_{y} sin31 (t-to)
v_{y} = -1.8 + 0.45 sin31 (20-10)
v_{y} = 0.518 m / s
b) let's use Pythagoras' theorem to find the magnitude of velocity
v = √ (vₓ² + v_{y}²)
v = √ (6,457² + 0.518²)
v = √ (41.96)
v = 6.478 m / s
We use trigonometry for direction
tan θ = v_{y} / vₓ
θ = tan⁻¹ v_{y} / vₓ
θ = tan⁻¹ 0.518 / 6.457
θ = 4.9°
c) let's look for the vector at the initial time
v₁ = √ (2.6² + 1.8²)
v₁ = 3.16 m / s
θ₁ = tan⁻¹ (-1.8 / 2.6)
θ₁ = -34.7
We see that the two vectors differ in module and direction, and that the acceleration vector is responsible for this change.
a = (v₂ -v₁) / (t₂-t₁)
An object with volume greater than it mass. to account less density and thus it will float
Answer:
Options d and e
Explanation:
The pendulum which will be set in motion are those which their natural frequency is equal to the frequency of oscillation of the beam.
We can get the length of the pendulums likely to oscillate with the formula;
where g=9.8m/s
ω= 2rad/s to 4rad/sec
when ω= 2rad/sec
L = 2.45m
when ω= 4rad/sec
L = 9.8/16
L=0.6125m
L is between 0.6125m and 2.45m.
This means only pendulum lengths in this range will oscillate.Therefore pendulums with length 0.8m and 1.2m will be strongly set in motion.
Have a great day ahead
Explanation:
A crystalline solid that conducts current under certain conditions