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3 years ago

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2. A hammer hits a nail with a force of 50 N into some wood. The area of the point of the nail is 0.02 cm2. What is the pressure

the nail puts on the wood?
What is the pressure the nail puts on the wood?

2 answers:

Levart [38]3 years ago

8 0

I hope this helps.......

almond37 [142]3 years ago

6 0

Answer:

hmmdndbdjjdjdhdjejjejdjdhdudududjhdhdhdhd

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How is the direction of light changed when it travels from an optically denser medium to an optically rarer medium????? please a

ANTONII [103]

Answer:

The light bends away from the normal

Explanation:

We can solve the problem by using Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where:

$n_1$ is the index of refraction of the first medium

$n_2$ is the index of refraction of the second medium

$\theta_1$ is the angle of incidence (angle between the incoming ray and the normal to the interface)

$\theta_2$ is the angle of refraction (angle between the outcoming ray and the normal to the interface)

We can rearrange the equation as

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

In this problem, light travels from an optically denser medium to an optically rarer medium, so

$n_1 > n_2$

Therefore, the term $\frac{n_1}{n_2}$ is greater than 1, so

$sin \theta_2 > sin \theta_1\\\rightarrow \theta_2 > \theta_1$

which means that the angle of refraction is greater than the angle of incidence, and so the light will bend away from the normal.

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3 years ago

Identify the phyla of the organisms on the basis of following distinct characteristics:

MrRissso [65]

Answer:

Phylum Annelida commonly referred as segmented worms possess long , cylindrical and segmented body .

Phylum Aschelminthes commonly referred as round worms possess long , cylindrical , unsegmented body and show sexual dimorphism .

Phylum Echinodermata which includes star fish have tube feet as locomotory organ .

Phylum Porifera commonly referred as pore bearing animals and are diploplastic which includes euspongia etc.

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2 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

$0=F-FrictionAB$

$F=FrictionAB=Nab*$ μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

$0=Nab-W$

$Nab=W=mg$

replacing this we have:

$F=$ μs $*Nab=$ μs* $mg=41.6N$

and μs $=41.6N/(mg)$

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

$F-Friction1(ground)-Friction2(AB)=0$

This is the new external F force that we are looking for:

$F=Friction1(ground)+Friction2(AB)$

we know Friction2(AB) because we found that in the previous block so:

$F=Friction1(ground)+mg$ *μs

for the other friction we have to use the equation:

$Friction(ground)=N(ground)$ *μs

from y axis we have:

$N(ground)-w-Normal(AB)=0$

$N(ground)=w+Normal(AB)$

we found the value of Normal(AB) with the previous block so:

$N(ground)=mg+mg=2mg$

and:

$Friction(ground)=2mg$ *μs

$F=Friction(ground)+mg$ *μs

$F=2mg$ *μs+μs* $mg=3mg$ *μs

and since: μs* $mg=41.6N$

the new F force would be:

$F=3mg$ *μs $=41.6*3=124.8N$

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3 years ago

Joe balances a stationary coin on the the tip of his finger 20 cm from the top of the table. How much work is Joe doing?

adell [148]

The work done by Joe is 0 J.

<u>Explanation</u>:

When a force is applied to an object, there will be a movement because of the applied force to a certain distance. This transfer of energy when a force is applied to an object that tends to move the object is known as work done.

The energy is transferred from one state to another and the stored energy is equal to the work done.

W = F . D

where F represents the force in newton,

D represents the distance or displacement of an object.

Force = 0 N, D = 20 cm = 0.20 m

W = 0 $\times$ 0.20 = 0 J.

Hence the work done by Joe is 0 J.

7 0

2 years ago

Which is the correct answer?

Nezavi [6.7K]

Answer:

Point A

Explanation:

The work done by stretching or compressing a spring is given by E=1/2kx²

The potential energy is numerically equal to the work done.

This means that the higher the bigger the value of the extension, x, the higher the energy contained.

In this scenario the modulus of x is considered.

Among the given values of x the modulus of -5 is the largest.

thus it gives the highest value of energy.

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3 years ago