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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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Two blocks of the same mass but made of different material slide across a horizontal, rough surface and eventually come to rest.

Akimi4 [234]

Answer:

a. $\mathbf{F_{f_2} = 2 F_{f1}}$ , $\mathbf { since \ the \ force \ o f \ friction \ is \ represented \ as \ the \ slope \ for \ each \ of \ the \ two \ curves.}$

Explanation:

From the information given;

Using the work-energy theorem

ΔKE = W = $\mathbf{ F_f \times r}$

K = $\mathbf{ F_f \times r}$

∴

$\dfrac{K_1}{K_2} = \dfrac{F_{f1}}{F_{f2}} (\dfrac{r_1}{r_2})$

Since $K_1 = K_2$ and r_1 = 4, and r_2 = 2 (from the missing diagram which is attached below)

Then;

$1 = \dfrac{F_{f1}}{F_{f2}} (\dfrac{4 \ m}{2 \ m})$

$\mathbf{F_{f_2} = 2 F_{f1}}$

5 0

2 years ago

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.50 mm. A 25

torisob [31]

Answer:

The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

Explanation:

Given that,

Area =7.70 cm²

Distance = 1.50 mm

Potential difference = 25.0 V

Suppose we find the electric field between the plates, the surface charge density, the capacitance and the charge on each plates.

We need to calculate the electric field

Using formula of electric field

$E=\dfrac{V}{d}$

Put the value into the formula

$E=\dfrac{25.0}{1.50\times10^{-3}}$

$E=16666.66\ V/m$

We need to calculate the charge density

Using formula of charge density

$\sigma=E\times\epsilon_{0}$

Put the value into the formula

$\sigma=16666.66\times8.85\times10^{-12}$

$\sigma=1.474\times10^{-7}\ C/m^2$

We need to calculate the capacitance

Using formula of capacitance

$C=\dfrac{\epsilon_{0}A}{d}$

Put the value into the formula

$C=\dfrac{8.85\times10^{-12}\times7.60\times10^{-4}}{1.50\times10^{-3}}$

$C=4.484\times10^{-12}\ F$

We need to calculate the charge

Using formula of charge

$q=CV$

Put the value into the formula

$q=4.484\times10^{-12}\times25.0$

$q=112.1\times10^{-12}\ C$

Hence, The electric field is 16666.66 V/m.

The surface charge density is $1.474\times10^{-7}\ C/m^2$

The capacitance is $4.484\times10^{-12}\ F$

The charge on each plate is $112.1\times10^{-12}\ C$

8 0

3 years ago

Plz help... asap I need it

riadik2000 [5.3K]

8000 J is the answer

3 0

2 years ago

Read 2 more answers

An alien spacecraft flies directly over a football stadium at a speed of 0.50c. if the proper length of the field is 100 yards,

Ivahew [28]

Go on way and you will get your answer

6 0

3 years ago

The question is on the picture

Firlakuza [10]

Answer:

<em> think 2 also if not im so sorry but i think it is :)</em>

7 0

2 years ago