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2 years ago

A(n ___ is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions.

1 answer:

5 0

A stereotaxic atlas is used to determine where an implanted electrode is to be placed in a brain in 3-dimensions. A stereotaxic atlas is used for doing a stereotactic surgery in order to cure the disorder occurred in the brain. This tool increase the accuracy, reliability, and the measurement of the brain because it gives the accurate brain map<span>.</span>

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If you want the ball to land in your hand when it comes back down, should you toss the ball straight upward, in a forward direct

denpristay [2]

Straight upward

the ball moves in the forward direction with your walking speed at all times. If you want the ball to land in your hand when it comes back down, you should toss the ball straight upward.

<h3>What is Projectile motion ?</h3>

Projectile motion is the motion of an object thrown (projected) into the air.

After the initial force that launches the object, it only experiences the force of gravity. The object is called a projectile, and its path is called its trajectory

A projectile can be a thrown ball, a bullet or a springboard diver ... Except for air resistance, the forward velocity of any projectile is constant and is equal to the initial velocity when it was released.

Learn more about Projectile motion here:

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7 0

1 year ago

Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha

AlexFokin [52]

1) Electric potential inside the sphere: $\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

The electric field inside the sphere is given by

$E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}$

where

$\epsilon_0=8.85\cdot 10^{-12}F/m$ is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

$E(r)=-\frac{dV(r)}{dr}$

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

$V(R)-V(r)=-\int\limits^R_r E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)$

The potential at the surface, V(R), is that of a point charge, so

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore we can find the potential inside the sphere, V(r):

$V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})$

At the center,

r = 0

Therefore the potential at the center of the sphere is:

$V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}$

On the other hand, the potential at the surface is

$V(R)=\frac{Q}{4\pi \epsilon_0 R}$

Therefore, the ratio V(center)/V(surface) is:

$\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}$

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is $\frac{3}{2}V(R)$ , as found in part b) (where $V(R)=\frac{Q}{4\pi \epsilon_0 R}$ )

- Between r and R, the potential decreases as $-\frac{r^2}{R^2}$

- Then at r = R, the potential is $V(R)$

- Between r = R and r = 3R, the potential decreases as $\frac{1}{R}$ , therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 ( $\frac{1}{3}V(R)$ )

Learn more about electric fields and potential:

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7 0

3 years ago

An electric motor rotating a workshop grinding wheel at a rate of 151 rev/min is switched off. Assume constant angular decelerat

tamaranim1 [39]

Answer:

Explanation:

Initial angular velocity ω₀ = 151 x 2π / 60

= 15.8 rad /s

final velocity = 0

Angular deceleration α = 2.23 rad / s

ω² = ω₀² - 2 α θ

0 = 15.8² - 2 x 2.23 θ

= 55.99 rad

one revolution = 2π radian

55.99 radian = 55.99 / 2 π no of terns

= 9 approx .

8 0

2 years ago

Why is the ocean warmer than the air at night

Sliva [168]

You see, during the day the ocean collects heat from the sun. So the air above the ocean get warm at night, but the rest of the air on the land gets cooler because water has the ability to collect energy from the Sun.

4 0

2 years ago

An empty beaker is placed on a top-pan balance. Some water is now poured into the beaker.What is the weight of the water? A. 0.0

mario62 [17]

Answer:

A. 0.044 kg

Explanation:

We need to subtract the sum of (beaker+water - empty beaker) which is 0.106 kg - 0.062 kg = 0.044 kg. The answer will not be written in Newton because this unit is used for force only and in this question w have to find the weight.

Hope it is enough.

Please mark me as brainliest.

6 0

2 years ago