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Colt1911 [192]
2 years ago
12

A fan is to accelerate quiescent air to a velocity of 12.5 m/s at a rate of 9 m3/s. Determine the minimum power that must be sup

plied to the fan. Take the density of air to be 1.18 kg/m3.
Physics
1 answer:
Reika [66]2 years ago
5 0

Answer:

= 829.69 Watt

≅ 830 Watt

Explanation:

Given that,

Velocity of air flow = 12.5m/s

Rate of flow of air = 9m³/s

Density of air = 1.18kg/m³

power by kinetic energy = 1/2(mv²)

mass = density × volume

m = 1.18 × 9

  = 10.62 kg/s

power = 1/2 mV²

           = 1/2 (10.62 × 12.5²)

           = 829.69 Watt

           ≅ 830 Watt

Flow rate  

u

=

9

 

m

3

/

s

Velocity of the air  

V

=

8

 

m/s

Density of the air  

ρ

=

1.18

 

kg

/

m

3

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The spectrum of Star A has an absorption line of hydrogen at 660.0 nm. The spectrum of Star B has an absorption line at 666 nm.
rjkz [21]

Answer:

The stars are moving away from us.

Explanation:

The observed wavelengths of hydrogen transition for stars A and B (660.0 nm and 666 nm respectively) are greater than that observed in the laboratory (656.2 nm). The observed long wavelengths for the stars means that the light from the stars is red-shifted.

According to the Doppler effect, red-shifted light means that the source is moving a way from the observer; therefore, we arrive at the conclusion that the stars A and B are moving away from us.

6 0
2 years ago
1 lb equals how many grams
shusha [124]

Answer:

1 lb. is 453.592 grams

Explanation:

1lb is 453.592 grams

8 0
2 years ago
Read 2 more answers
What is the formula that can be used to find velocity if kinetic energy and mass are known?
viva [34]
The formula for kinetic energy is \frac{1}{2}m\Delta v^2. Thus, the equation for velocity is v=   \sqrt{ \frac{2TotalKineticEnergy}{m} }. 
6 0
2 years ago
Read 2 more answers
A particle with a charge of -1.24 x 10"° C is moving with instantaneous velocity (4.19 X 104 m/s)î + (-3.85 X 104 m/s)j. What is
astra-53 [7]

Answer:

(a) F= 6.68*10¹¹⁴ N (-k)

(b) F =( 6.68*10¹¹⁴ i  + 7.27*10¹¹⁴ j  ) N

Explanation

To find the magnetic force in terms of a fixed amount of charge q that moves at a constant speed v in a uniform magnetic field B we apply the following formula:

F=q* v X B Formula (1 )

q: charge (C)

v: velocity (m/s)

B: magnetic field (T)

vXB : cross product between the velocity vector and the magnetic field vector

Data

q= -1.24 * 10¹¹⁰ C

v= (4.19 * 10⁴ m/s)î + (-3.85 * 10⁴m/s)j

B  =(1.40 T)i  

B  =(1.40 T)k

Problem development

a) vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)i =

            = - (-3.85*1.4) k = 5.39* 10⁴ m/s*T (k)

1T= 1 N/ C*m/s

We apply the formula (1)

  F= 1.24 * 10¹¹⁰ C*  5.39* 10⁴ m/s* N/ C*m/s (-k)

   F= 6.68*10¹¹⁴ N (-k)

a)  vXB = (4.19 * 10⁴ m/s)î + (-3.85* 10⁴m/s)j X (1.40 T)k =

             =( - 5.39* 10⁴i - 5.87* 10⁴j)m/s*T

1T= 1 N/ C*m/s

We apply the formula (1)

F= 1.24 * 10¹¹⁰ C*  (  5.39* 10⁴i + 5.87* 10⁴j) m/s* N/ C*m/s

F =( 6.68*10¹¹⁴  i  + 7.27*10¹¹⁴  j  ) N

8 0
2 years ago
A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul
Rashid [163]
The Young modulus is given by:
E= \frac{F /A}{\Delta L / L_0}
where
F is the force applied
L_0 is the initial length of the wire
A is the cross-sectional area of the wire
\Delta L is the stretch of the wire

The wire in the problem stretches by 0.5% of its length, this means 
\frac{\Delta L}{L_0}  = 0.005

We can also calculate the area of the wire; its radius is in fact half the diameter:
r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m
and so the area is
A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

E=  \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2
3 0
2 years ago
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