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3 years ago

What does the spinal cord on a sheep brain do

2 answers:

7 0

Cerebellum: controls balance and muscle coordination; located caudal to the cerebrum in the sheep brain. ... Gray matter: areas of the brain and spinal cord containing neuronal cell bodies, dendrites, and unmyelinated axons. Found in the cerebral cortex of the brain and inner area of the spinal cord

german3 years ago

3 0

Answer:

It carries nerve impulses between the brain and body

Explanation:

buzz

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Fossil A is found in a rock layer above a layer containing Fossil B. Which fossil is probably older? A) Fossil A B) Fossil B C)

luda_lava [24]

Fossil B is older the lower they found the fossil the older it is

8 0

2 years ago

Read 2 more answers

Saturn moves in an orbit around the Sun with radius 10 AU. How many degrees does it move on the Celestial in one year? (Hint: Ca

Lana71 [14]

Answer:

B. About 12 degrees

Explanation:

The orbital period is calculated using the following expression:

T = 2π*( $\sqrt{\frac{r^3}{Gm}}$ )

Where r is the distance of the planet to the sun, G is the gravitational constant and m is the mass of the sun.

Now, we don't actually need to solve the values of the constants, since we now that the distance from the sun to Saturn is 10 times the distance from the sun to the earth. We now this because 1 AU is the distance from the earth to the sun.

Now, we divide the expression used to calculate the orbital period of Saturn by the expression used to calculate the orbital period of the earth. Notice that the constants will cancel and we will get the rate of orbital periods in terms of the distances to the sun:

$\frac{Tsaturn}{Tearth}$ = $\sqrt{\frac{rSaturn^3}{rEarth^3} } = \sqrt{10^3}}$

Knowing that the orbital period of the earth is 1 year, the orbital period of Saturn will be $\sqrt{10^3}}$ years, or 31.62 years.

We find the amount of degrees it moves in 1 year:

$1year * \frac{360degrees}{31.62years} = 11.38 degrees$

or about 12 degrees.

6 0

3 years ago

Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

$sin\theta - \mu_k cos\theta = 1$

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

$F_f = \mu Mgcos\theta$

now if it is moving down the inclined plane at constant speed

so we will have

$Mgsin\theta - T - \mu mgcos\theta = 0$

on other side the mass "m" will go up at constant speed

so we have

$T - mg = 0$

so we have

$Mgsin\theta = \mu Mgcos\theta + mg$

so we have

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

for special case when m = M

then we have

$sin\theta - \mu_k cos\theta = 1$

5 0

2 years ago

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

We will use the second equation of motion to find out the initial speed of the arrow.

$h= v_it + \frac{1}{2}gt^2\\$

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}$

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0

1 year ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

3 years ago