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algol13
3 years ago
12

Thing 1 and Thing 2 push with a force of 2000 N to move a rather small 500 kg elephant 30 m across the kitchen floor. How much w

ork did they do?
Physics
1 answer:
Amanda [17]3 years ago
3 0

Answer:

60000 J

Explanation:

Assuming the force is applied parallel to the displacement of the elephant, the work done to move it across the floor is

W=Fd

where

F = 2000 N is the force applied

d = 30 m is the displacement of the elephant

Substituting the numbers into the formula, we find

W=(2000 N)(30 m)=60,000 J

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balu736 [363]

Answer:

23N

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Franny drew a diagram to compare images produced by concave and convex lenses.
yanalaym [24]

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virtual

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Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0
3 years ago
Assortative mating changes ______ frequencies but does not change ______ frequencies.
andrew-mc [135]

Allele frequencies are unaffected by assortative mating, but genotype frequencies .

<h3>Assortative mating: </h3>

Individuals with similar phenotypes and genotypes mate with others more frequently than is anticipated under a random mating pattern in assortative mating, which is a mating pattern and a type of sexual selection.

<h3>Frequencies of genotypes:</h3>

A population's genotype frequency is calculated by dividing the number of people having a particular genotype by the overall population size. The genotype frequency in population genetics is the frequency or ratio (i.e., 0 f 1) among genotypes inside a population.

<h3>The frequency for alleles in biology:</h3>

The term "allele frequency" describes the prevalence of an allele in a population. It is calculated by calculating the number of times the allele occurs in the population and dividing by the sum of all the gene copies.

To know more about Assortative mating visit:

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4 0
1 year ago
an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu
Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

E1= k*q1/(h1-h2)..............(1)

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1


[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2\\

E1=[tex]E= 359502+359502\\E=719004 V/mV/m[/tex]

similarly electric feild due to negative charge q2=-40

[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2\\

E2=359502V/m

Total electric feild E at the aircraft is given as

E= E1+ E2\\...............(2)

Put values of  E1 and E2 in equation2

\\E=359502+359502\\E= 719004V/m\\

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0
3 years ago
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