Given that,
Power = 60 W
Distance = 1.0 m
Distance between speakers = 1.5 m
We need to calculate the intensity
Using formula of intensity
Put the value into the formula
We need to calculate the intensity
Using formula of intensity
Put the value into the formula
We need to calculate the intensity of the sound waves produced by four speakers
Using formula for intensity
Put the value into the formula
Hence, The intensity of the sound waves produced by four speakers is 13.78 W/m².
Answer:
q = 8.85 x 10⁻¹¹ C
Explanation:
given,
Electric field, E = 1.18 N/C
distance, r = 0.822 m
Charge magnitude = ?
using formula of electric field.
k is the coulomb constant
q = 8.85 x 10⁻¹¹ C
The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C
wireequal to the strength of the Earth's magnetic field of about
5.0 x10^5 T?
Answer:
The distance is 2 cm
Solution:
According to the question:
Magnetic field of Earth, B_{E} =
Current, I = 5.0 A
We know that the formula of magnetic field is given by:
where
d = distance from current carrying wire
Now,
d = 0.02 m 2 cm
Answer:
m2=3.2722lbm/s
Explanation:
Hello!
To solve this problem follow the steps below
1. Find water densities and entlapies in all states using thermodynamic tables.
note Through laboratory tests, thermodynamic tables were developed, which allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy, etc.)
through prior knowledge of two other properties, such as pressure and temperature.
D1=Density(Water;T=50;x=0)=62.41 lbm/ft^3
D2=Density(Water;T=120;x=0)=61.71 lbm/ft^3
D3=Density(Water;T=80;x=0)=62.21 lbm/ft^3
h1=Enthalpy(Water;T=50;x=0)=18.05 BTU/lbm
h2=Enthalpy(Water;T=120;x=0)=88 BTU/lbm
h3=Enthalpy(Water;T=80;x=0)=48.03 BTU/lbm
2. uses the continuity equation that states that the mass flow that enters a system is the same as the one that must exit
m1+m2=m3
3. uses the first law of thermodynamics that states that all the flow energy entering a system is the same that must come out
m1h1+m2h2=m3h3
18.05(m1)+88(m2)=48.03(m3)
divide both sides of the equation by 48.03
0.376(m1)+1.832(m2)=m3
4. Subtract the equations obtained in steps 3 and 4
m1 + m2 = m3

0.376m1 + 1.832(m2) =m3

0.624m10.832m2=0
solving for m2
(0.624/0.832)m1=m2
0.75m1=m2
5. Mass flow is the product of density by velocity across the crosssectional area
m1=(D1)(A)(v1)
internal Diameter for 2" Sch 40=2.067in=0.17225ft
m1=(62.41 lbm/ft^3)(0.0233ft^2)(3ft/S)=4.3629lbm/s
6.use the equation from step 4 to find the mass flow in 2
0.75m1=m2
0.75(4.3629)=m2
m2=3.2722lbm/s
Answer:
Average speed
Explanation:
We have given current through power i = 20 A
Diameter d = 1 cm = 0.01 m
So radius r = 0.005 m
So area
Charge on electron e =
We know that current is given by , here n is nuber density of free electron, e is charge on electron, A is area and is average speed
We know that for copper n =
So average speed
Answer:
The volume flow rate is 3.27m³/s
Diameter at the refinery is 88.64cm
Explanation:
Given
At the wellhead
Pipes diameter, d2 = 59.1cm = 0.591m
Flow speed of petroleum f2 = 11.9m/s
At the refinery,
Pipes diameter, d1 = ? Unknown
Flow speed of petroleum, f1 = 5.29m/s
Calculating the volume flow rate of petroleum along the pipe.
Volume flow rate = Flow rate * Area along the pipe
V = 11.9 * πd²/4
V = 11.9 * 22/7 * 0.591²/4
V = 3.265778m³/s
The volume flow rate is 3.27m³/s  Approximated
Since it's not stated if the flowrate is uniform throughout the pipe, we'll assume that flow rate is the same through out...
Using V1A1 = V2A2, where V1 & V2 Volume flow rate at both ends and area = Area of pipes at both ends
This gives;
V1A1 = V1A2
V1*πd1²/4 = V2 * πd2²/4  Divide through by π/4
So, we are left with
V1d1² = V2d2²
5.29 * d1²= 11.9 * 59.1²
d1² = 11.9 * 59.1²/5.29
d1² = 7857.172
d1 = √7857.172
d1 = 88.6406904305240618
d1 = 88.64cm  Approximated
The value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.
Given data:
The mass of sphere is, m = 10 kg.
The radius of sphere is, r = 0.5 m.
The density of air is, .
The drag coefficient of object is, .
The altitude is, h = 5000 m.
The density of air at altitude is, .
The mathematical expression for the terminal velocity of an object is,
here,
g is the gravitational acceleration.
A is the area of sphere.
Solving as,
Now, the terminal velocity at the altitude of 5000 m is given as,
Solving as,
Thus, we can conclude that the value of terminal velocity of object is 7.99 m/s and the value of terminal velocity at an altitude of 5000 m is 10.30 m/s.
Learn more about the terminal velocity here:
Answer:
The terminal velocity at sea level is 7.99 m/s
The terminal velocity at an altitude of 5000 m is 10.298 m/s
Explanation:
mass of sphere m = 10 kg
radius of sphere r = 0.5 m
air density at sea level p = 1.22 kg/m^3
drag coefficient Cd = 0.8
terminal velocity = ?
Area of the sphere A = = 4 x 3.142 x = 3.142 m^2
terminal velocity is gotten from the relationship
where g = acceleration due to gravity = 9.81 m/s^2
imputing values into the equation
= 7.99 m/s
If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3
= 10.298 m/s