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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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Light with a wavelength of about 490 nm is made to pass through a diffraction grating. the angle formed between the path of the

polet [3.4K]

The number of lines per mm in the diffraction grating is 326.

<h3>What is diffraction grating?</h3>

A diffraction grating is a type of optical instrument obtained with a continuous pattern. The pattern of the diffracted light by a grating depends on the structure and number of elements present.

The given data in the problem is

$\rm \theta$ is the angle formed between the path of the incident light and the diffracted light = 9. 2°

λ is the wavelength of the light=490nm=4.9

N is the number of lines per mm in the diffraction grating=?

n is ordered = 1

The formula for the diffraction grating is;

$n \lambda = d sin\theta \\\\ d = \frac{n \lambda}{sin \theta } \\\\ d = \frac{1 \times 4.90 \times 10^{-7}}{sin 9.2^0 } \\\\ d=3.06 \times 10^{-6} \\\\ d=3.06 \times 10^{-3} \ mm$

The number of lines per mm is found as;

$\rm N= \frac{1}{d} \\\\ N= \frac{1}{3.06 \TIMES 10^{-3}} \\\\ N=326.8 /mm$

Hence the number of lines per mm in the diffraction grating is 326.

To learn more about diffraction grating refer to the link;

brainly.com/question/1812927

5 0

3 years ago

A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci

Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = $\frac{C_{3}C_{4} }{C_{3} + C_{4} }$

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = $\frac{4.25\times3.98 }{4.25 + 3.98 }$

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3 = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = $\frac{Q}{C_{4} }$ = $\frac{35.46\times10^{-6} }{3.98\times10^{-6}}$ = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0

3 years ago

Which describes Einstein’s second postulate about the special theory of relativity? The speed of light in a vacuum is constant,

crimeas [40]

Answer:A

Explanation:

5 0

3 years ago

Read 2 more answers

An Iowa class warship holds the record for the fastest warship. If the ship accelerates uniformly from rest at 0.15 m/s² for 2 m

Mumz [18]

The equation to be used is the derived formulas for rectilinear motion at a constant acceleration. The formula for acceleration is

a = (v - v₀)/t
where
v and v₀ are the initial and final velocities, respectively
t is the time
a is the acceleration

Since it started from rest, v₀ = 0. Using the formula:

0.15 m/s² = (v - 0)/[2 minutes*(60 s/1 min)]
Solving for v,
v = 18 m/s

3 0

3 years ago

Barack is playing basketball in his back yard. He takes a shot 7.0 m from the basket (measured along the ground), shooting at an

VMariaS [17]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given $\Delta x=7\ m$

$\theta=45 \°$

Also, $\Delta y=(3.5-2)=1.5\ m$

$a_x=0\ and\ a_y=-9.81\ m/s^2$

Let us say the velocity in the x-direction is $v_x$ and in the y-direction is $v_y$ . And acceleration in the x-direction is $a_x$ and in the y-direction is $a_y$ .

Also, $\Delta x\ and\ \Delta y$ is distance covered in x and y direction respectively. And $t$ is the time taken by the ball to hit the backboard.

We can write $v_x=v_0cos(45)\ and\ v_y=v_0sin(45)$ . Where $v_0$ is velocity of ball.

Now,

$\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt$

$\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}$

Also,

$\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s$ .

Plugging this value in

$t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}$

$t=1.06\ seconds$

So, the time of flight of the ball is 1.06 seconds.

6 0

4 years ago