Answer:
4. 5g
Explanation:
F=ma so, m=Fa. All you have to do is 10/2. Don't be confused by the units. Mass will normally be in grams.
Answer:
92.7 km
Explanation:
Since the magnetic field due to a solenoid is given by B = μ₀Ni/L where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, N = number of turns of solenoid, L = length of cardboard tube = 58 cm = 0.58 m, , i = current in wire = 2.5 A and l = length of wire.
So, N = BL/μ₀i
Since B = 2.0 kG = 2.0 × 10³ G = 2.0 × 10³ × 10⁻⁴ T = 2.0 × 10⁻¹ T = 0.2 T
So, substituting the variables into the equation, we have
N = BL/μ₀i
N = 0.2 T × 0.58 m/(4π × 10⁻⁷ H/m × 2.5 A)
N = 1.16 Tm/(31.416 × 10⁻⁷ HA/m)
N = 0.0369 × 10⁷ turns
N = 0.0369 × 10⁷ turns
N = 3.69 × 10⁵ turns
length of wire l = NC where N = number of turns and C = circumference of tube = πD where D = diameter of tube = 8.0 cm = 0.08 m
So, l = NC
= NπD
= πND
= π × 3.69 × 10⁵ turns × 0.08 m
= 0.9274 × 10⁵ m = 9.274 × 10⁴ m
= 92.74 × 10³ m
= 92.74 km
≅ 92.7 km
Speed = (distance) / (time)
Speed = (2.3 m) / (3 sec)
Speed = (2.3/3) (m/s)
<em>Speed = 0.766... m/s</em>
Answer: reaction force = -558N
Explanation:
w = f = 558N
since action force and reaction force are equal in magnitude and opposite in direction,
reaction force = -(f)
reaction force = -558N
if that helps.
Given:
Area of pool = 3m×4m
Diameter of orifice = 0.076m
Outlet Velocity = 6.3m/s
Accumulation velocity = 1.5cm/min
Required:
Inlet flowrate
Solution:
The problem can be solved by this general formula.
Accumulation = Inlet flowrate - Outlet flowrate
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
First, we need to convert the units of the accumulation velocity into m/s to be consistent.
Accumulation velocity = 1.5cm/min × (1min/60s)×(1m/100cm)
Accumulation velocity = 0.00025 m/s
We then calculate the area of the pool and the area of the orifice by:
Area of pool = 3 × 4 m²
Area of pool = 12m²
Area of orifice = πd²/4 = π(0.076m)²/4
Area of orifice = 0.00454m²
Since we have all we need, we plug in the values to the general equation earlier
Accumulation velocity × Area of pool = Inlet flowrate - Outlet velocity × Area of orifice
0.00025 m/s × 12m² = Inlet flowrate - 6.3m/s × 0.00454m²
Transposing terms,
Inlet flowrate = 0.316 m³/s