Answer:
Yes it will move and a= 4.19m/s^2
Explanation:
In order for the box to move it needs to overcome the maximum static friction force
Max Static Friction = μFn(normal force)
plug in givens
Max Static friction = 31.9226
Since 36.6>31.9226, the box will move
Mass= Wieght/g which is 45.8/9.8= 4.67kg
Fnet = Fapp-Fk
= 36.6-16.9918
=19.6082
=ma
Solve for a=4.19m/s^2
If its asking the distance for the 65 db then use a proportion, if otherwise pleas clarify. It sounds like a pretty juicy conversation.
Answer: Newton, the unit of force, is defined based on Newton's Second Law (F=ma), as the force required to give a mass of one kilogram an acceleration of 1 meter/second2. Thus, it is derived from these other units.
Explanation:
Answer:
10.6 mA
Explanation:
t = time interval = 1.00 s
q = magnitude of charge on each ion = 1.6 x 10⁻¹⁹ C
n₁ = number of Na⁺ ions = 2.68 x 10¹⁶
q₁ = charge due to Na⁺ ions = n₁ q = (2.68 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.004288 C
n₂ = number of Cl⁻ ions = 3.92 x 10¹⁶
q₂ = charge due to Cl⁻ ions = n₂ q = (3.92 x 10¹⁶) (1.6 x 10⁻¹⁹) = 0.006272 C
i₁ = Current due to Na⁺ ions = 
= 
= 0.004288 A
i₂ = Current due to Cl⁻ ions = 
= 
= 0.006272 A
Current passing between the electrodes is given as
i = i₁ + i₂
i = 0.004288 + 0.006272
i = 0.01056 A
i = 10.6 x 10⁻³ A
i = 10.6 mA
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
