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2 years ago

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1. The center of gravity of both cats and humans is roughly in the thoracic region. However, they each bear their weight very di

fferently. a. Which are the major, weight-bearing structures of the two skeletons

1 answer:

Mama L [17]2 years ago

6 0

Answer:

Major, weight-bearing structures are the bones of the body that are strong and dense to be able to bear the weight of the body. The major, weight-bearing structures of cat and human skeletons are :

Human skeleton: The body weight of an individual is on his pelvic girdles that are attached to the bones of lower limbs. Thigh bones, leg bones, and bones of feet comprise lower limbs The lower limbs consist of the thigh, the leg, and the foot.

Cat skeleton: cats are quadrupedal so it bears all the body weight on shoulders and legs that includes the Scapula and pelvis.

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Which equation can be used to model simple harmonic motion

mixer [17]

Answer:x(t)= Acos(wt)

Explanation:

According to Newton's 2nd law,a particle of mass m acted on by a force is given by:Fs=-kx

Where x is displacement from equilibrium

K = spring constant

Therefore X(t) = Acos(2pit/T)

X(t)= Acos(wt)

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3 years ago

How long does it take light to travel 850 km in a vacuum? Answer in ms.(Express your answer to two significant figures.)

poizon [28]

Answer:

0.002833 sec

Explanation:

Speed of light in vacuum is $3\times 10^{8}m/sec$

Given distance = 850 km = 850×1000=850000 m

We have to calculate the time that light take to travel the distance 850 km

Time $T=\frac{distance }{speed}=\frac{850000}{3\times 10^8}=2.833\times 10^{-3}sec$

So the time taken by light to travel 850 km is 0.002833 sec

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3 years ago

The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal

8090 [49]

Α $=2.4 \frac{m}{s}$

β $=1.2 \frac{m}{s^2}$

$x(t)=at$

$y(t)=3-$ β $t^2$

$Vx(t)=$ α

$Vy(t)=-2$ β $t$

$vectorV=[$ α $;-2$ β $t]$

$ax(t)=0$

$ay(t)=-2$ β $t$

$vector a [0;-2$ β $t]$

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3 years ago

In a particular case of an object in front of a spherical mirror with a focal length of +12.0 cm, the magnification is +4.00.(a)

salantis [7]

Answer:

9 cm

-36 cm

Explanation:

u = Object distance

v = Image distance

f = Focal length = 12

m = Magnification = 4

$m=-\frac{v}{u}\\\Rightarrow 4=-\frac{v}{u}\\\Rightarrow v=-4u$

Lens equation

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{12}=\frac{1}{u}+\frac{1}{-4u}\\\Rightarrow \frac{1}{12}=\frac{3}{4u}\\\Rightarrow u=9\ cm$

Object distance is 9 cm

$v=-4\times 9=-36\ cm$

Image distance is -36 cm (other side of object)

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3 years ago

Write an email to a classmate explaining why the velocity of the current in a river has no FN C02-F20-OP11USB effect on the time

anzhelika [568]

The velocity of the current in a river has no effect on the the time it takes to paddle a canoe across the river, given that the boat is pointed perpendicular to the bank of the river, because

The velocity of the river does not change the velocity and therefore the distance traveled in the direction of the boat which is directed perpendicular to it

Reason:

Let $\overset \rightarrow {v_y}$ represent the velocity of the boat across the river in the direction, $\overset \rightarrow {d_y}$ , and let, $\overset \rightarrow {v_x}$ , represent the velocity of the river, we have;

The velocity of the boat perpendicular to the direction of the river = $\overset \rightarrow {v_y}$

Therefore, the distance covered per unit time in the perpendicular direction

to the flow of the river is $\overset \rightarrow {v_y}$ , such that the time it takes to cross the river in

the perpendicular direction is the same, for every value of the velocity of

the river.

This is so because the velocity in the perpendicular direction to the flow of

the river, which is the velocity of the boat is unchanged by the velocity of

the river, because there is no perpendicular component of velocity in the

velocity of the river.

$\overset \rightarrow {v_y}$ = 3 m/s

$\overset \rightarrow {v_x}$ = 4 m/s

The width of the river, $w_y$ = 6 meters, we have;

The resultant velocity = $\sqrt{(3 \ m/s)^2 + (4 \ m/s)^2} =5 \ m/s$

The direction, θ, is given as follows;

$\theta = \arctan \left(\dfrac{4}{3} \right) \approx 53.13^{\circ}$

The length of the path of the boat, <em>l</em>, is given as follows;

$l = \dfrac{6}{cos \left(\arctan \left(\dfrac{4}{3} \right)\right)} = 10$

The length of the path the boat takes = 10 m

The time it takes to cross the river, t = $\dfrac{l}{v}$ , therefore;

$t = \dfrac{10}{5} = 2$

The time it takes to cross the river, t = 2 seconds

Considering only the y-components, we have;

$t = \dfrac{w_y}{v_y}$

Therefore;

$t = \dfrac{6 \ m}{3 \ m/s} = 2 \, s$

Which expresses that the time taken is the same and given that the

vectors of the velocities of the river and the boat are perpendicular, the

distance covered in the direction of the boat is unaffected by the velocity

of the river.

Learn more vectors here:

brainly.com/question/15907242

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3 years ago