Answer:
maximum speed of the car to prevent sliding is 13.1m/s
Explanation:
Given data
Radius of curve r=50m
Mass of car m=4907kg
Coefficient of friction u=0.35
Limiting for R=?
Hence limiting force R=ma
R=4907*9.81
R=48137.7N
We know that the force to overcome friction is
F=uR
Hence
F=0.35*48137.7
F=16848.2N
Centripetal force along the curve is given as
Fc=mv²/r
Fc = centripetal force
m = mass
v = velocity
r = radius
To solve for velocity we have to equate both force required to overcome friction and the centripetal force
Fc=mv²/r=F=uR
mv²/r=uR
Making velocity subject of formula we have
v²=u*r*R/m
v²=(0.35*50*48137.7)/4907
v²=842409.75/
v²=171.67
v=√171.67
v=13.1m/s
b nonelectrolyte
c liquid
d solid
Answer:
B. Nonelectrolyte.
Explanation:
Nonelectrolytes do not dissociate into ions in solution, hence, nonelectrolyte solutions don't conduct electricity.
A non-electrolyte doesn’t conduct electric current even when it forms a solution.
Answer: Option B
Explanation:
Where electrolytes are defined as the compounds that can conduct electric current with mobile ions existing in its solution, non-electrolytes are the compounds that don’t behave the same either in the aqueous solution or in the molten state.
This is all because these compounds don’t produce mobile ions to flow from one electrode to the other and hence conduct electric flow in the solution. Sugar and ethanol are the best examples of non-electrolytes that don’t induce electric current even after getting dissolved in water.
Write the velocity vectors in component form.
• initial velocity:
v₁ = 4 m/s at 45º N of E
v₁ = (4 m/s) (cos(45º) i + sin(45º) j)
v₁ ≈ (2.83 m/s) i + (2.83 m/s) j
• final velocity:
v₂ = 4 m/s at 10º N of E
v₂ = (4 m/s) (cos(10º) i + sin(10º) j)
v₂ ≈ (3.94 m/s) i + (0.695 m/s) j
The average acceleration over this 3-second interval is then
a = (v₂ - v₁) / (3 s)
a ≈ (0.370 m/s²) + (-0.711 m/s²)
with magnitude
||a|| = √[(0.370 m/s²)² + (-0.711 m/s²)²] ≈ 0.802 m/s²
and direction θ such that
tan(θ) = (-0.711 m/s²) / (0.370 m/s²) ≈ -1.92
→ θ ≈ -62.5º
which corresponds to an angle of about 62.5º S of E, or 27.5º E of S. To use the notation in the question, you could say it's E 62.5º S or S 27.5º E.
Answer:
Apparent frequency of the bell to the observer is 546.12 Hz
Explanation:
The frequency of train bell (frequency of source) = 505 Hz
The speed of train (observer) = 27.6 m/s
The speed of sound in the air is (velocity of sound) = 339 m/s
The apparent frequency of the bell to the observer is calculated as follows:
Apparent frequency of bell to the observer.
Answer:
0.25( m1m1) , 0.75( m2m2)
Explanation:
Noted the formula for kinetic energy is 1/2(Mass × Velocity).
Therefore the original value of the mass is 0.5, giving half away makes it 0.25 to another mass which is primarily 0.5. This now makes the new mass 0.25+0.5=0.75.
Thank you.
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L
The BOD concentration 50 km downstream from the wastewater treatment plant is approximately 15.865 mg/L.
To calculate the BOD concentration 50 km downstream, we need to consider the rate of dilution due to the flow of the river and the first-order reaction that destroys BOD. The concentration of BOD downstream can be calculated using the equation C2 = C1 * exp(-k * d/v), where C1 is the initial concentration, k is the rate constant, d is the distance, and v is the velocity of the river.
Plugging in the given values, we have C2 = 75 * exp(-0.05 * 50/15), which gives us a BOD concentration of approximately 15.865 mg/L 50 km downstream from the wastewater treatment plant.
Learn more about downstream here:
#SPJ3
Answer: With no friction, the box will accelerate down the ramp
Explanation: