Which describes a function of the thalamus?

Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

lina2011 [118]

Answer:

-6.0 m/s, 10.4 m/s

Explanation:

To find the x- and y- components, we have to apply the formulas:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v = 12.0 m/s is the magnitude of the vector

$\theta$ is the angle between the direction of the vector and the positive x-axis

Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is

$\theta=180^{\circ} - 60^{\circ} = 120^{\circ}$

So, the two components are:

$v_x = (12.0 m/s) cos 120^{\circ}=-6.0 m/s$

$v_y = (12.0 m/s) sin 120^{\circ}=10.4 m/s$

5 0

3 years ago

How many atoms of hydrogen are represented by the formula: A) 1 B) 4 C) 8 D) 28

Keith_Richards [23]

It takes only one atom to make hydrogen; however, hydrogen is commonly found as H2, meaning two atoms make up hydrogen. Because of this, hydrogen is known as a diatomic element

The answer is prob A.1 but you didnt' give me a formula.

8 0

4 years ago

Read 2 more answers

In a car lift used in a service station, compressed air exerts a force on a small piston of circular cross section having a radi

Vitek1552 [10]

Answer:

(a) 1,569.63 N

(b) 195,933.99 Pa

(c) As pressure and volume are equal for each piston, workdone must also be equal

(d) 1,647.47 kg

Explanation:

Let the cross-sectional area (CSA) of small piston = A₁

Let the cross-sectional area (CSA) of the bigger piston = A₂

Let the Force applied at the smaller piston = F₁

Let the Force applied at the bigger piston = F₂

The principle of hydraulic lift assumes the that the fluid is in-compressible, resulting to a constant pressure system.

F₁/A₁ =F₂/A₂----------------------------------------------------------- (1)

(a) F₁= F₂ xA₁ /A₂

F₂ = 13,300 N

A₁ = π r₁²

=π x (0.0505)²

= 0.008011 m²

A₂= π r₂²

=π x (0.147)²

= 0.06788 m²

Substituting into (1)

F₁ = 13,300 x 0.008011/0.06788

= 1,569.6272

≈ 1,569.63 N

(b) Air pressure = Force/Area

= F₁/A₁

= 1,569.6272/ 0.008011

= 195,933.99 Pa

(c) The pressure is constant for both pistons according to Pascal Law.

Workdone = force x distance----------------------------------------- (2)

force = pressure × area

distance = volume/area from

Substituting into (2)

Workdone = pressure × volume.

As pressure and volume are equal for each piston, work must also be equal

(d) F₂ = F₁ x A₂/ A₁---------------------------------------------------- (3)

A₁ = π r₁²

=π x (0.079)²

= 0.01960 m²

A₂= π r₂²

=π x (0.353)²

= 0.3914 m²

Substituting into (2)

F₂= 825 x 0.3914/ 0.01960

= 16,474.7448

≈ 16,474.74 N

= 1,647.47 kg

3 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

Luden [163]

Answer:

The ball will be attracted to the negatively charged plate. It'll touch and pick up some electrons from the plate so that the ball becomes negatively charged. Immediately the ball is repelled by the negative plate and is attracted to the positive plate. The ball gives up electrons to the positive plate so that it is positively charged and suddenly attracts to the negative plate again, flies over to it and picks up enough electrons to be repulsed by negative plate and again to the positive plate and that continues.

8 0

3 years ago

A giant wall clock with diameter d rests vertically on the floor. The minute hand sticks out from the face of the clock, and its

Katyanochek1 [597]

Answer:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

Explanation:

We can try writing the equation of the horizontal component of the length of the minute hand in terms of distance and the angle, that depends of time in this particular case.

The x-component of the length of the minute hand is:

$d_{x}(t)=dcos(\theta (t))$ (1)

d is the length of the minute hand (d=D/2)
D is the diameter of the clock
t is the time (min)

Now, using the angular kinematic equations we can express the angle in term of angular velocity and time. As we know, the minute hand moves with a constant angular velocity, so we can use this equation:

$\theta (t)=\omega *t$ (2)

Also we know, that the minute hand moves 90 degrees or π/2 rad in 15 min, so using the definition of angular velocity, we have:

$\omega=\frac{\Delta \theta}{\Delta t}=\frac{\theta_{f}-\theta_{i}}{t_{f}-t{i}}=\frac{\pi/2-0}{15-0}=\frac{\pi}{30}$

Now, let's put this value on (2)

$\theta (t)=\frac{\pi}{30}*t$

Finally the length x(t) of the shadow of the minute hand as a function of time t, will be:

$d_{x}(t)=(D/2)cos(\frac{\pi}{30}*t)$

I hope it helps you!

6 0

3 years ago