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LUCKY_DIMON [66]
3 years ago
5

A typical ceiling fan running at high speed has an airflow of about 1.85 ✕ 103 ft3/min, meaning that about 1.85 ✕ 103 cubic feet

of air move over the fan blades each minute. Determine the fan's airflow in m3/s.
Physics
1 answer:
densk [106]3 years ago
7 0

Answer:

0.8726  m^3/s

Explanation:

We are to convert 1.85 x 10^3 ft^3/min to m^3/s

First, let us convert the numerator from ft3 to m3

1 ft3 = 0.0283 m3

Hence,

1.85 x 10^3 ft3 = 1.85 x 10^3 x 0.0283 m3

     = 52.355 m3

Now, let us convert the denominator from minutes to seconds

1 min = 60 sec

Therefore;

1.85 x 10^3 ft^3/min = 52.355/60  m^3/s

        = 0.8726  m^3/s

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What is the forms of energy given below.
Damm [24]

Answer:

Lightening of the table lamp

Explanation:

Energy has a different form of energy. In physics, the capacity of the form of energy is work. The energy can exist in the form of thermal, potential, kinetic, chemical and electrical, and nuclear. There are other forms of energy such as work and heat.

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3 0
3 years ago
.Find the uncertainty in a calculated electrical potential difference from the measurements of current and resistance. Electric
maw [93]

Answer:

a) The uncertainty in calculated V, ΔV = 25.3

b) The uncertainty in calculated v, Δv = 0.41 m/s

c) The uncertainty in calculated V, ΔV = 22.2 V

Explanation:

We'll use Upper-Lower Bounds method of uncertainty to estimate the uncertainties.

a) I = 5.1 A, ΔI = 0.3 A

I = (5.1 ± 0.3) A

R = 77.5 ohms, ΔR = 0.4 ohms

R = (77.5 ± 0.4) ohms

V = IR = 5.1 × 77.5 = 395.25 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 5.1 - 0.3 = 4.8 A

Rₗ = 77.5 - 0.4 = 77.1 ohms

Vₗ = 4.8 × 77.1 = 370.08 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 5.1 + 0.3 = 5.4 A

Rᵤ = 77.5 + 0.4 = 77.9 ohms

Vᵤ = 5.4 × 77.9 = 420.66 V

The average of the differences from the mean voltage/true value is 25.3 V

V = 395.25 V, Δ = 25.3V

V = (395.25 ± 25.3) V

b) x = 2.9 m, Δx = 0.3 m

x = (2.9 ± 0.3) m

t = 4.4 s, Δt = 1.8 s

t = (4.4 ± 1.8) ohms

v = x/t = 2.9/4.4 = 0.659 m/s

The lower bound for average speed will be calculated using the lower bounds for distance and upper bounds for time.

xₗ = 2.9 - 0.3 = 2.6 m

tᵤ = 4.4 + 1.8 = 6.2 s

vₗ = 2.6/6.2 = 0.419 m/s

The upper bound for the average speed will be calculated using the upper bound for the distance and lower bound for time

xᵤ = 2.9 + 0.3 = 3.2 m

tₗ = 4.4 - 1.8 = 2.6 s

vᵤ = 3.2/2.6 = 1.231 m/s

The average of the differences from the mean average speed/true value is 0.41 m/s

v = 0.659 m/s, Δv = 0.41 m/s

v = (0.659 ± 0.41) m/s

c) ) I = 9.8 A, ΔI = 0.5 A

I = (9.8 ± 0.5) A

R = 40.5 ohms, ΔR = 0.2 ohms

R = (40.5 ± 0.2) ohms

V = IR = 9.8 × 40.5 = 396.9 V

The lower bound for the voltage will be calculated using the lower bounds for the current and resistance

Iₗ = 9.8 - 0.5 = 9.3 A

Rₗ = 40.5 - 0.2 = 40.3 ohms

Vₗ = 9.3 × 40.3 = 374.79 V

The upper bound for the voltage will be calculated using the upper bounds for the current and resistance

Iᵤ = 9.8 + 0.5 = 10.3 A

Rᵤ = 40.5 + 0.2 = 40.7 ohms

Vᵤ = 10.3 × 40.7 = 419.21 V

The average of the differences from the mean voltage/true value is 22.2 V

V = 396.9 V, Δ = 22.2 V

V = (396.9 ± 22.2) V

7 0
3 years ago
A 30.0 kg dolphin decelerates from 12.0 to 7.00 m/s in 2.60 s to join another dolphin in play. What average force (in N) was exe
Crazy boy [7]

Answer:

Force is 57.69 N to the opposite direction of motion of dolphin.

Explanation:

We have force is the product of mass and acceleration.

That is    

                 Force = Mass x Acceleration

                         F = ma

Mass of dolphin, m = 30 kg

We have equation of motion, v = u + at

Final velocity, v = 7 m/s

Initial velocity, u = 12 m/s

Time, t = 2.60 s

Substituting

                   7 = 12 + a x 2.6

                    a = -1.92 m/s²

Force,   F = 30 x -1.92 = -57.69 N

So the force is 57.69 N to the opposite direction of motion of dolphin.

7 0
2 years ago
A car moves with constant acceieration of -0.s m/s? on a straight portion of the road. Att- 0s the car has a velocity of 69 mph,
Crazy boy [7]

Answer:

b) d = 0.71 Km

Explanation:

Car kinematics

Car 1 moves with uniformly accelerated movement

v_f^2=v_o^2+2a*d Formula (1)

d: displacement in meters (m)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

Equivalences:

1mile = 1609.34 meters

1 hour = 3600s

1km = 1000m

Known data

v_o = 69\frac{mile}{hour} *1609.34\frac{meter}{mile} *\frac{1}{3600}\frac{hour}{s}=30.8 \frac{m}{s}

v_f= \frac{69}{2} \frac{mile}{hour} =34.5\frac{mile}{hour}=15.4 \frac{m}{s}

a = -0.5 m/s²

Distance calculation

We replace data in the Formula (1)

15.4^2 = 30.8^2+2(-0.5)*d

2(0.5)*d = 30.8^2 - 15.4^2

d =\frac{ 30.8^2 - 15.4^2}{2(0.5) }= 717.6m

d = 717.6 m\frac{1km}{1000m} = 0.7176Km

8 0
3 years ago
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