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dedylja [7]
3 years ago
12

2. How does inherent value differ from instrumental value? *

Physics
1 answer:
s344n2d4d5 [400]3 years ago
4 0

Answer:

The difference between inherent and instrumental value is that inherent value is valued for its own sake, whereas instrumental value is valued for the end results gained from it.

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Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?
Orlov [11]
  • Force=10N
  • Displacement=30cm=0.3m

\\ \sf\bull\longmapsto Wd=Force(Displacement)

\\ \sf\bull\longmapsto Work=10(0.3)

\\ \sf\bull\longmapsto Work=3J

5 0
3 years ago
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Josh and Jake are both helping to build a brick wall which is 6 meters in height. They each lay 250 bricks, but Josh finishes th
Molodets [167]
<span>Each laid 250 bricks but while Jake was still working, Josh was lounging in the shade. Josh has more power but that power was only on for 3 hours out of 4.5. Obviously Josh could get more done is less time as long as he keeps working. Jake will get the hang of it soon.</span>
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3 years ago
ne true-breeding line of mice is obese and dark, and the other true-breeding line is lean and light. Dark is dominant to light,
LenKa [72]

Answer:

 \frac{3}{8}

Explanation:

Let the allele for dark color be represented by "D" and the allele for light color be represented by "d"

Also D is dominant over d

Similarly, let the allele for obese trait in mice be represented by "O" and the allele for lean trait in mice be represented by "L"

Obese and lean exhibit incomplete dominance

Genotype of true breeding dark and obese mice be DDOO

Genotype of true-breeding lean and light mice is ddLL

Offspring from F1 cross -  

DdOL

F2 cross is between DdOL * DdOL

DO         DL         dO          dL

DO DDOO DDOL DdOO DdOL

DL DDOL DDLL DdOL DdLL

dO DdOO DdOL ddOO ddOL

dL DdOL DdLL ddOL ddLL

dark and intermediate between obese and lean offspring genotype –  

2 DDOL, 4 DdOL  

So total six out of sixteen offspring are dark and intermediate between obese and lean -  

 \frac{6}{16} \\\frac{3}{8}

8 0
4 years ago
Which three forms of light are invisible light?
Flauer [41]
Radio waves
infrared
x-rays
and red and blue light are not a thing, their on a color spectrum called visible light.
3 0
3 years ago
Read 2 more answers
I NEED PHYSICS HELP? THE QUESTION IS IN THE PIC
Marina CMI [18]

Speed of the block at the bottom of the incline: 5.42 m/s

The first part of the problem can be solved by using the law of conservation of energy. Since the ramp is frictionless, the initial gravitational potential energy of the block at the top of the ramp is converted into kinetic energy at the bottom:

mgh = \frac{1}{2}mv^2 (1)

where

m is the mass of the block

g = 9.8 m/s^2 is the acceleration of gravity

h is the initial height of the block

v is the speed of the block at the bottom

The initial height of the block is equal to the height of the ramp, so

h=L sin \theta (2)

where

L = 3.00 m is the length of the ramp

\theta=30^{\circ} is the angle of the ramp

Substituting (2) into (1) and re-arranging the equation, we  find the speed

2gL sin \theta = v^2

v=\sqrt{2gL sin \theta}=\sqrt{2(9.8)(3.00)sin 30^{\circ}}=5.42 m/s

Coefficient of kinetic friction between the floor and the block: 0.3

In the second part of the motion, the block is slowed down by friction along the flat surface. According to the work-energy theorem, the work done by friction is equal to the change in kinetic energy of the block:

W=\Delta K=K_f -K_i

where

W is the work done by friction

Kf is the final kinetic energy of the block, which is zero since the block comes to rest

K_i = \frac{1}{2}mv^2 is the initial kinetic energy of the block, where

m = 10.0 kg is the mass of the block

v = 5.42 m/s is its initial speed

Substituting into the equation, we find

W=-\frac{1}{2}mv^2=-\frac{1}{2}(10.0)(5.42)^2=-146.9 J

and the work is negative, since the direction of the force of friction is opposite to the direction of motion of the block.

Now we can rewrite the work as the product between the force of friction and the displacement of the block:

W=-F_f d = - \mu mg d

where

\mu is the coefficient of friction

d = 5.00 m is the displacement of the block

Solving for \mu,

\mu = - \frac{W}{mgd}=-\frac{-146.9}{(10.0)(9.8)(5.00)}=0.3

4 0
3 years ago
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