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weeeeeb [17]
3 years ago
6

A fire hose sends 2429 gallons of water per minute against a burning building. The water strikes the building at 66 m/s and does

not bounce back. There are 3.785 L/gal Liters in a gallon, and the density of water is 1 kg/L.
a) What is the magnitude of the rate of change of momentum of the water? Answer in units of N.
b) The rate of change of momentum of the water 1. is negative. 2. cannot be determined. or 3. is positive?
c ) What force does the water exert on the building in 1.2 minutes? Answer in units of N.
Physics
1 answer:
Anastasy [175]3 years ago
5 0

Answer:

a)   Δp = - 10113.2 Kg m / s, b) he rate of change is negative, c)    F = 140.46N

Explanation:

a) For this part let's analyze a water particle, it has a velocity of 66 m / s and when it collides with the building its velocity changes to zero, so the change in moment is

           Δp = mv_f - m v₀

           Δp = -m v₀                              (1)

the change of the moment in a second is

           

if 2429 gallons arrive in in minute (60s) in a second how many gallons arrive      

            c_agua = 2429/60

            c_water = 40,483 gallon/ s

let's use the concept of density to find the mass

            ρ = m / V

            m = ρ V

let's reduce gallons to liters

             c_water = 40,483 gal (3,785 l / 1 gal) = 153.23 l

           

            m = 1 153.23

            m 153.23 kg

           

we substitute in 1

            Δp = - 153.23 66

            Δp = - 10113.2 Kg m / s

b) From the previous result the rate of change is negative

c) Let's use the impulse ratio

             I = f t = Δp

             F t = Δp

               

              F = Δp / t

              F = \frac{10113.2}{1.2 \ 60}

              F = 140.46 N

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3 0
3 years ago
The magnetic field perpendicular to a single 16.7-cm-diameter circular loop of copper wire decreases uniformly from 0.750 T to z
sammy [17]

Answer:

1.24 C

Explanation:

We know that the magnitude of the induced emf, ε = -ΔΦ/Δt where Φ = magnetic flux and t = time. Now ΔΦ = Δ(AB) = AΔB where A = area of coil and change in magnetic flux = Now ΔB = 0 - 0.750 T = -0.750 T, since the magnetic field changes from 0.750 T to 0 T.

The are , A of the circular loop is πD²/4 where D = diameter of circular loop = 16.7 cm = 16.7 × 10⁻²m

So, ε = -ΔΦ/Δt = -AΔB/Δt= -πD²/4 × -0.750 T/Δt = 0.750πD²/4Δt.

Also, the induced emf ε = iR where i = current in the coil and R = resistance of wire = ρl/A where ρ = resistivity of copper wire =1.68 × 10⁻⁸ Ωm, l = length of wire = πD and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.25 mm = 2.25 × 10⁻³ m.

So, ε = iR = iρl/A = iρπD/πd²/4 = 4iρD/d²

So,  4iρD/d² = 0.750πD²/4Δt.

iΔt = 0.750πD²/4 ÷ 4iρD/d²

iΔt = 0.750πD²d²/16ρ.

So the charge Q = iΔt

= 0.750π(Dd)²/16ρ

= 0.750π(16.7 × 10⁻²m 2.25 × 10⁻³ m)²/16(1.68 × 10⁻⁸ Ωm)

= 123.76 × 10⁻² C

= 1.2376 C

≅ 1.24 C

4 0
3 years ago
What is the magnitude of the applied electric field inside an aluminum wire of radius 1.2 mm that carries a 3.0-a current? [ σal
galben [10]
Hello

1) First of all, since we know the radius of the wire (r=1.2~mm=0.0012~m), we can calculate its cross-sectional area
A=\pi r^2 = 3.14 \cdot (0.0012~m)^2=4.5\cdot10^{-6}~m^2

2)  Then, we can calculate the current density J inside the wire. Since we know the current, I=3~A, and the area calculated at the previous step, we have
J= \frac{I}{A}= \frac{3~A}{4.5\cdot10^{-6}~m^2} = 6.63\cdot10^5 ~A/m^2

3) Finally, we can calculate the electric field E applied to the wire. Given the conductivity \sigma=3.6\cdot10^7~ \frac{A}{Vm} of the aluminium, the electric field is given by
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4 0
3 years ago
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lidiya [134]

Answer:

The samples specific heat is 14.8 J/kg.K

Explanation:

Given that,

Weight = 28.4 N

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Temperature = 18°C

We need to calculate the samples specific heat

Using formula of specific heat

Q=mc\Delta T

c=\dfrac{Q}{m\Delta T}

Where, m = mass

c = specific heat

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Put the value into the formula

c=\dfrac{1.25\times10^{4}}{\dfrac{28.4}{9.8}\times(18+273)}

c=14.8\ J/kg. K

Hence, The samples specific heat is 14.8 J/kg.K

8 0
3 years ago
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