Answer:
The answer to your question is: 15 m/s2
Explanation:
Equation x = at3 - bt2 + ct
a = 4.1 m/s3
b = 2.2 m/s2
c = 1.7 m/s
First we find x at t = 4.1 s
x = 4.1(4.1)3 - 2.2(4.1)2 + 1.7(4.1)
x = 4.1(68.921) - 2.2(16.81) + 6.97
x = 282.58 - 36.98 + 6.98
x = 252.58 m
Now we find speed
v = x/t = 252.58/ 4.1 = 61.6 m/s
Finally
acceleration = v/t = 61.6/4.1 = 15 m/s2
Answer:
6.05 cm
Explanation:
The given equation is
2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)
The initial head velocity V₀ₓ =11 m/s
The final head velocity Vₓ is 0
The accelerationis given by =1000 m/s²
the stopping distance = x-x₀=?
So we can wind the stopping distance by following formula
2 (-1000)(x-x₀)=[
]
x-x₀=6.05*
m
=6.05 cm
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
h = v₀² / 2g
, h = k/4g x²
Explanation:
In this exercise we can use the law of conservation of energy at two points, the lowest, before the shot and the highest point that the mouse reaches
Starting point. Lower compressed spring
Em₀ = K = ½ m v²
Final point. Highest on the path
= U = mg h
As or no friction the energy is conserved
Em₀ = Em_{f}
½ m v₀²² = m g h
h = v₀² / 2g
We can also use as initial energy the energy stored in the spring that will later be transferred to the mouse
½ k x² = 2 g h
h = k/4g x²
