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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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Where do we hear loud sound, at antinode or node?

tester [92]

Answer:

node

Explanation:

on the graph node is higher than antinode

so it can get or hear loud sounds faster

4 0

3 years ago

What is the net torque on the square plate, with sides 0.2 m, from each of the three forces? F1=18 N, F2=26 N, and F3=14 N. Use

marshall27 [118]

Answer:

The net torque on the square plate is 2.72 N-m.

Explanation:

Given that,

Side = 0.2 m

Force $F_{1}=18\ N$

Force $F_{2}=26\ N$

Force $F_{3}=14\ N$

We need to calculate the torque due to force F₁

Using formula of torque

$\tau_{1}=-F_{1}d_{1}$

$\tau_{1}=-F_{1}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{1}=-18\times\dfrac{0.2}{2}$

$\tau_{1}=-1.8\ N-m$

We need to calculate the torque due to force F₂

Using formula of torque

$\tau_{2}=F_{2}d_{2}$

$\tau_{2}=F_{2}\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{2}=26\times\dfrac{0.2}{2}$

$\tau_{2}=2.6\ N-m$

We need to calculate the torque due to force F₃

Using formula of torque

$\tau_{3}=F_{3}d_{3}$

$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

Put the value into the formula

$\tau_{3}=0.1(14\sin45+14\cos45)$

$\tau_{3}=1.92\ N-m$

We need to calculate the net torque on the square plate

$\tau=\tau_{1}+\tau_{2}+\tau_{3}$

$\tau=-1.8+2.6+1.92$

$\tau=2.72\ N-m$

Hence, The net torque on the square plate is 2.72 N-m.

3 0

3 years ago

A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th

fiasKO [112]

Let's use the mirror equation to solve the problem:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$
where f is the focal length of the mirror, $d_o$ the distance of the object from the mirror, and $d_i$ the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for $d_i$ by using the numbers given in the text of the problem:
$\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}$
$\frac{1}{d_i}= -\frac{7}{60 cm}$
$d_i = -8.6 cm$
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.

6 0

3 years ago

Read 2 more answers

The five general principles from the APA are meant to __________. A. be enforceable rules B. be posted in every office C. guide

Paha777 [63]

The answer is C guide and inspire good conduct

7 0

2 years ago

Two test tubes are filled with a solution of bromthymol blue. A student exhales through a straw into each tube, and the bromthym

saw5 [17]

Answer:

oxygen was produced by phosynthesis

Explanation:

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3 years ago