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Answer:
The distance from the building is 11.41 m.
Explanation:
Given that,
Speed = 30.0 m/s
Angle = 45.0°
Distance = 10.0 m
We need to calculate the vertical component
Using formula of vertical component
We need to calculate the time
Using equation of motion
Put the value into the formula
On neglecting of higher value of t
Then use the horizontal component
We need to calculate the distance
Put the value into the formula
Hence, The distance from the building is 11.41 m.
Answer:
Explanation:
If I am not wrong
current = charge/time
All you have to take care of is the units should be in the same system
so
current = 12/(2*60) --------- 2 min = 2*60 sec
current = 12/120 = 0.1 amp
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)