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3 years ago

If you were to look for a cut on the palmar surface of a dog's leg where would you look?

Physics

1 answer:

nika2105 [10]3 years ago

4 0

Answer:

If you were to look for a cut on the palmar surface of a dog's leg then you should look at the back area of the front leg below the carpus.

Explanation:

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A 150 kg safe on frictionless casters is being raised at 1.20m to the bed of a truck using planks 4m long. The force needed to p

Vesnalui [34]

The only thing you need to know in order to solve this task is that <span>plank length (which is force x), should equal the increase in potential energy, so what we have now : (mass)* g * (height).
It has to look like that: </span>
<span>F * 3.0 = 150 x 9.81 x 1.20
Then solve for F, the result should be in newtones = 588N

Do hope it makes sense.</span>

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3 years ago

The image shows street lights powered by solar panels. Which sequence shows the energy transformations taking place in these lig

yawa3891 [41]

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3 years ago

Read 2 more answers

A lone neutron spontaneously decays into a proton plus an _______

frosja888 [35]

Answer:

A lone neutron spontaneously decays into a proton plus an electron.

Explanation:

In an atom, nuclei contain protons and neutrons, which are the fundamental particles of an atom. Neutrons are stable and uncharged particles inside a nucleus.

For 15 times during its lifetime, a free neutron decays and breaks down into more smaller particles.This breakdown causes problems in nuclear reactors, as they start decaying and emit radiations of different wavelengths.

A neutron undergoes the decaying process to produce an electron, a proton, and energy.

The reaction of neutron decay:

n0 → p+ + e− + νe

5 0

3 years ago

A rock is dropped from a vertical cliff. The rock takes 3.00 s to reach the ground below the cliff. A second rock is thrown vert

Phantasy [73]

Answer:

Velocity v= 12.25 $\frac{m}{s}$

Explanation:

The first rock dropped give the distance Y in meters

$Y_{f}=Y_{o}+v_{o}*t +\frac{1}{2}*a*t^{2}\\ Y_{f}=\frac{1}{2}* 9.8 \frac{m}{s^{2} }* 3^{2} \\Y_{f}=44.1 m$

Now the motion of the second rock the time change so to know the velocity

$Y_{f}= Y_{o} +v_{o}*t +\frac{1}{2}*a *t^{2} \\v_{o}*t= -Y_{f} +\frac{1}{2} *a*t^{2} \\v_{o} =\frac{-44.1 +0.5 * 9.8* s^{2} }{2} \\v_{o}= 12.25 \frac{m}{s}$

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3 years ago

1. A 14-cm tall object is placed 26 cm from a converging lens that has a focal length of 13 cm.

AURORKA [14]

Answer:

a) Please find attached the required drawing of light passing through the lens

By the use of similar triangles;

The image distance from the lens = 26 cm

The height of the image = 14 cm

c) The image distance from the lens = 26 cm

The height of the image = 14 cm

Explanation:

Question;

a) Determine the image distance and the height of the image

b) Calculate the image position and height

The given parameters are;

The height of the object, h = 14 cm

The distance of the object from the mirror, u = 26 cm

The focal length of the mirror, f = 13 cm

The location of the object = 2 × The focal length

Therefore, given that the center of curvature ≈ 2 × The focal length, we have;

The location of the object ≈ The center of curvature of the lens

The diagram of the object, lens and image created with MS Visio is attached

From the diagram, it can be observed, using similar triangles, that the image distance from the lens = The object distance from the lens = 26 m

The height of the image = The height of the object - 14 cm

b) The lens equation is used for finding the image distance from the lens as follows;

$\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

Where;

v = The image distance from the lens

We get;

$v = \dfrac{u \times f}{u - f}$

Therefore;

$v = \dfrac{26 \times 13}{26 - 13} = \dfrac{26 \times 13}{13} = 26$

The distance of the image from the lens, v = 26 cm

The magnification, M =v/u

∴ M = 26/26 = 1, therefore, the object and the image are the same size

Therefore;

The height of the image = The height of the object = 14 cm.

5 0

3 years ago