Let us use the formula for Newton's Second Law of Motion:
Net force = Mass*Acceleration
Net force = Applied Force - μ*Normal Force
where μ is the coefficient of kinetic friction
Normal Force = Force due to gravity = mass*gravity
Normal Force = (210 kg)(9.81 m/s²) =<em> 2,060.1 N</em>
Then,
Net force = 4100 - 0.38*2060.1 = 3317.162 N
3317.162 N = (210 kg)(a)
Solving for acceleration,
<em>a = 15.796 m/s²</em>
Answer:
b) 252 Hz or 260 Hz
c) 0.25 s
Explanation:
b) The frequency of the beats is 4 Hz, and one tuning fork has a frequency of 256 Hz. Therefore, the second tuning fork is either 4 Hz lower or 4 Hz higher.
f = 252 Hz or 260 Hz
c) Period is the inverse of frequency.
T = 1/f
T = 1 / (4 Hz)
T = 0.25 s
The basic formula for power is P=W/t meaning it is the rate at which work is accomplished.
Answer:
The potential difference across a given given wire of an circuit is directly proportional to the current flowing through it provided it's temperature remains the same. This is known as ohm's law
V is proportional to I
V/I = constant
V/I = R
V = IR where R is the constant of proportionality and resistance of the wire
Hope this helps!
Answer:
F = 3.6 kN, direction is 9.6º to the North - East
Explanation:
The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.
Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.
Wind
X axis
F₁ = 2.50 kN
Tide
cos 30 = F₂ₓ / F₂
sin 30 = F_{2y} / F₂
F₂ₓ = F₂ cos 30
F_{2y} = F₂ sin 30
F₂ₓ = 1.20cos 30 = 1.039 kN
F_{2y} = 1.20 sin 30 = 0.600 kN
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = 2.50 +1.039
Fₓ = 3,539 kN
F_y = F_{2y}
F_y = 0.600
to find the vector we use the Pythagorean theorem
F = 
F = 
F = 3,589 kN
the address is
tan θ = F_y / Fₓ
θ = tan⁻¹
θ = tan⁻¹
0.6 / 3.539
θ = 9.6º
the resultant force to two significant figures is
F = 3.6 kN
the direction is 9.6º to the North - East