Answer: 4.64 × 10^7 mg
Explanation:
Given:
T =1299g + 45.1kg
Convert kg to g
T = 1299g + 45.1kg ×1000g/kg
T = 1299g + 45100g
T = 46399g
T= 4.64 × 10^4 g
Convert to milligram
T = 4.64 × 10^4 g × 1000mg/g
T = 4.64 × 10^7 mg
Therefore, 1299g + 45.1kg = 4.64 × 10^7 mg
I think the answer is:
Velocity
Answer:
θ = 21.8º
Explanation:
We can solve this exercise using Newton's conditions for equilibrium, in the attached we can see a diagram of the forces.
The most used coordinate system is an axis parallel to the plane (x axis) and an axis perpendicular to the plane (y axis), let's write Newton's equations on this axes
Y Axis
N- 
= 0
N =
X axis
fr - Wₓ = 0 (1)
Let's use trigonometry to find the normal ones
sin θ = Wₓ / W
cos θ = / W
Wx = W sin θ
= W cos θ
The friction force has the formula
fr = μ N
fr = μ (W cos θ)
We substitute in 1
μ mg cosθ = mg sin θ
μ cos θ = sin θ
tan θ = μ
θ = tan⁻¹ μ
calculate
θ = tan⁻¹ 0.40
θ = 21.8º
Answer: 4.17m
Explanation:
The observer at C will hear a sound on no sound upon whether the interference is constructive or destructive.
If the listeners hears sounds it is caled constructive interference but if he hears no sound its called destructive interference.
So
d2 - d1 = (n *lamba)/ 2
Where n=1,3,5
lamda=v/f =349/62.8
lamda=5.56m
d2= d1 + nlamda/2
d2= 1 + 5.56/2
d2= 3.78m
X'= 1 cos 60= 0.5m
Y= 1 sin60= 0.866m
X"^2 + Y^2 =d2^2
X" =√(y^2 - d2^2)
X"=√(3.78^2 - 0.886^2)
X"= 3.67m
So therefore the closest that speaker A can be to speaker B so the listener does not hear any sound is X' + X"= 0.5 + 3.67
4.17m
