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nikklg [1K]
3 years ago
12

Is prey abiotic or biotic?

Physics
1 answer:
Ray Of Light [21]3 years ago
8 0
Biotic because of the prey of the reasoning
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Inside a 30.2 cm internal diameter stainless steel pan on a gas stove water is being boiled at 1 atm pressure. If the water leve
dybincka [34]

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

<u>Q = 20.22 x 10³ W = 20.22 KW</u>

8 0
3 years ago
A nonconducting sphere is made of two layers. The innermost section has a radius of 6.0 cm and a uniform charge density of −5.0C
Leno4ka [110]

Answer:

a) E =0, b)   E = 1,129 10¹⁰ N / C , c)    E = 3.33 10¹⁰ N / C

Explanation:

To solve this exercise we can use Gauss's law

        Ф = ∫ E. dA = q_{int} / ε₀

Where we must define a Gaussian surface that is this case is a sphere; the electric field lines are radial and parallel to the radii of the spheres, so the scalar product is reduced to the algebraic product.

           E A = q_{int} /ε₀

The area of ​​a sphere is

          A = 4π r²

         E = q_{int} / 4πε₀ r²

         k = 1 / 4πε₀

         E = k q_{int} / r²

To find the charge inside the surface we can use the concept of density

        ρ = q_{int} / V ’

         q_{int} = ρ V ’

         V ’= 4/3 π r’³

Where V ’is the volume of the sphere inside the Gaussian surface

 Let's apply this expression to our problem

a) The electric field in center r = 0

     Since there is no charge inside, the field must be zero

          E = 0

b) for the radius of r = 6.0 cm

In this case the charge inside corresponds to the inner sphere

        q_{int} = 5.0  4/3 π 0.06³

         q_{int} = 4.52 10⁻³ C

        E = 8.99 10⁹  4.52 10⁻³ / 0.06²

         E = 1,129 10¹⁰ N / C

c) The electric field for r = 12 cm = 0.12 m

In this case the two spheres have the charge inside the Gaussian surface, for which we must calculate the net charge.

     The charge of the inner sphere is q₁ = - 4.52 10⁻³ C

The charge for the outermost sphere is

       q₂ =  ρ 4/3 π r₂³

       q₂ = 8.0 4/3 π 0.12³

       q₂ = 5.79 10⁻² C

The net charge is

     q_{int} = q₁ + q₂

     q_{int} = -4.52 10⁻³ + 5.79 10⁻²

     q_{int} = 0.05338 C

The electric field is

        E = 8.99 10⁹ 0.05338 / 0.12²

        E = 3.33 10¹⁰ N / C

8 0
3 years ago
What are the 2 forms of memory retrieval?
Alika [10]

recall and recognition

5 0
4 years ago
Is cordyceps fungi a predator? If not, what is it?
ASHA 777 [7]

Answer:

It is a predator.

Explanation:

They eat other insects so therefore I think they're a predator.

5 0
3 years ago
3. A crew team rows a boat at a rate of 20 km/h in still water. Beginning at time = 0 minutes, the team rows for 30 minutes up a
Sindrei [870]
Going upstream against the current gives a net speed equivalent to the speed at still water minus the speed of the current. Consequently, the speed downstream gives a net speed equal to the speed at still water plus the speed of the current, making it travel faster. The solution is:

UPSTREAM

v = 20 - 1.5 = 18.5 km/h
t = 30 mins or 0.5 hours
distance = 18.5km/h (0.5 h) = 9.25 km

DOWNSTREAM
for the same distance of 9.25 km:

v = 20 + 1.5 = 21.5 km/h
t = 9.25km / 21.5 km/h = 0.43 hours or 25.8 mins  =  26 mins --> FINAL ANS.
4 0
3 years ago
Read 2 more answers
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