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3 years ago

Which vertebrae are normally fused together to form a solid mass of bone? A. Sacral B. Thoracic C. Coccygeal D. Cervical

Physics

2 answers:

Nikolay [14]3 years ago

6 0

A) sarcral is the answer to this question

UNO [17]3 years ago

5 0

A. Sacral is the answer.

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What force works against friction? <br> Inertia<br> Potential <br> Kinetic

Oksanka [162]

Kinetic, as inertia means to remain unchanged.

6 0

4 years ago

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A boat crossing a 153.0 m wide river is directed so that it will cross the river as quickly as possible. The boat has a speed of

Lynna [10]

We have the relation

$\vec v_{B \mid E} = \vec v_{B \mid R} + \vec v_{R \mid E}$

where $v_{A \mid B}$ denotes the velocity of a body A relative to another body B; here I use B for boat, E for Earth, and R for river.

We're given speeds

$v_{B \mid R} = 5.10 \dfrac{\rm m}{\rm s}$

$v_{R \mid E} = 3.70 \dfrac{\rm m}{\rm s}$

Let's assume the river flows South-to-North, so that

$\vec v_{R \mid E} = v_{R \mid E} \, \vec\jmath$

and let $-90^\circ < \theta < 90^\circ$ be the angle made by the boat relative to East (i.e. -90° corresponds to due South, 0° to due East, and +90° to due North), so that

$\vec v_{B \mid R} = v_{B \mid R} \left(\cos(\theta) \,\vec\imath + \sin(\theta) \, \vec\jmath\right)$

Then the velocity of the boat relative to the Earth is

$\vec v_{B\mid E} = v_{B \mid R} \cos(\theta) \, \vec\imath + \left(v_{B \mid R} \sin(\theta) + v_{R \mid E}\right) \,\vec\jmath$

The crossing is 153.0 m wide, so that for some time $t$ we have

$153.0\,\mathrm m = v_{B\mid R} \cos(\theta) t \implies t = \dfrac{153.0\,\rm m}{\left(5.10\frac{\rm m}{\rm s}\right) \cos(\theta)} = 30.0 \sec(\theta) \, \mathrm s$

which is minimized when $\theta=0^\circ$ so the crossing takes the minimum 30.0 s when the boat is pointing due East.

It follows that

$\vec v_{B \mid E} = v_{B \mid R} \,\vec\imath + \vec v_{R \mid E} \,\vec\jmath \\\\ \implies v_{B \mid E} = \sqrt{\left(5.10\dfrac{\rm m}{\rm s}\right)^2 + \left(3.70\dfrac{\rm m}{\rm s}\right)^2} \approx 6.30 \dfrac{\rm m}{\rm s}$

The boat's position $\vec x$ at time $t$ is

$\vec x = \vec v_{B\mid E} t$

so that after 30.0 s, the boat's final position on the other side of the river is

$\vec x(30.0\,\mathrm s) = (153\,\mathrm m) \,\vec\imath + (111\,\mathrm m)\,\vec\jmath$

and the boat would have traveled a total distance of

$\|\vec x(30.0\,\mathrm s)\| = \sqrt{(153\,\mathrm m)^2 + (111\,\mathrm m)^2} \approx \boxed{189\,\mathrm m}$

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2 years ago

An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple

Anuta_ua [19.1K]

Answer:6.71 m/s

Explanation:

Given

Apple fall from a height of $h=2.3 m$

We need to find the impact speed of apple which can be given by using

$v^2-u^2=2gh$

where v=final velocity

u=initial velocity

h=Displacement

Assuming initial velocity to be zero

substituting the value we get

$v^2-0=2\times 9.8\times 2.3$

$v=\sqrt{2\times 9.8\times 2.3}$

$v=6.71\ m/s$

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3 years ago

Find the volume of a box with length 25 cm, height 25 cm and width 1.0 m.

emmasim [6.3K]

Before finding the volume, it is always good to change the units as same.
length 25cm
height 25cm
width 100cm

volume
length * height * width
25 * 25 * 100
62500 cm^3

5 0

4 years ago

An object's motion changes

Strike441 [17]

Answer:A.

only when a nonzero net force is applied to the object.

Explanation:

3 0

3 years ago

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