Where is information first stored in a human brain

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Answer 1

Answer:

Answer:

when u learn something new it goes to ur short term memory

Related Questions

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The atmosphere on Venus consists mostly of CO2. The density of the atmosphere is 65.0 kg/m3 and the bulk modulus is 1.09 x 107 N/m2. A pipe on a lander is 75.0 cm long and closed at one end. When the wind blows across the open end, standing waves are caused in the pipe (like blowing across the top of a bottle). a) What is the speed of sound on Venus? b) What are the first three frequencies of standing waves in the pipe?

Answers

Answer:

a. 409.5 m/s b. f₁ = 136.5 Hz, f₂ = 409.5 Hz, f₃ = 682.5 Hz

Explanation:

a. The speed of sound v in a gas is v = √(B/ρ) where B = bulk modulus and ρ = density. Given that on Venus, B = 1.09 × 10⁷ N/m² and ρ = 65.0 kg/m³

So, v = √(B/ρ)

= √(1.09 × 10⁷ N/m²/65.0 kg/m³)

= √(0.01677 × 10⁷ Nm/kg)

= √(0.1677 × 10⁶ Nm/kg)

= 0.4095 × 10³ m/s

= 409.5 m/s

b. For a pipe open at one end, the frequency f = nv/4L where n = mode of wave = 1,3,5,.., v = speed of wave = 409.5 m/s and L = length of pipe = 75.0 cm = 0.75 m

Now, for the first mode or frequency, n = 1

f₁ = v/4L

= 409.5 m/s ÷ (4 × 0.75 m)

= 409,5 m/s ÷ 3 m

= 136.5 Hz

Now, for the second mode or frequency, n = 2

f₂ = 3v/4L

= 3 ×409.5 m/s ÷ (4 × 0.75 m)

= 3 × 409,5 m/s ÷ 3 m

= 3 × 136.5 Hz

= 409.5 Hz

Now, for the third mode or frequency, n = 5

f₃ = 5v/4L

= 5 × 409.5 m/s ÷ (4 × 0.75 m)

= 5 × 409,5 m/s ÷ 3 m

= 682.5 Hz

Find a glass jar with a screw-top metal lid. Close the lid snugly and put the jar into the refrigerator. Leave it there for about 10 minutes and then take the jar out and try to open the lid. (a) Did the lid become tighter or looser? Explain your observation.

Answers

Answer:

The lid becomes tighter

It becomes tighter because metals have a lower heat capacity than glass meaning their temperature drops (or increases) much faster than glass for the same energy change. So in this example, the metal will contract faster than the glass causing it to be more tighter around the glass.

A 110 kg ice hockey player skates at 3.0 m/s toward a railing at the edge of the ice and then stops himself by grasping the railing with his outstretched arms. During the stopping process, his center of mass moves 30 cm toward the railing. (a) What is the change in the kinetic energy of his center of mass during this process? (b) What average force must he exert on the railing?

Answers

Answer:

(a). The change in the kinetic energy of his center of mass during this process is -495 J.

(b). The average force is 1650 N.

Explanation:

Given that,

Mass = 110 kg

Speed = 3.0 m/s

Distance = 30 cm

(a). We need to calculate the change in the kinetic energy of his center of mass during this process

Using formula of kinetic energy

$\Delta K.E=K.E_(2)-K.E_(1)$

$\Delta K.E=(1)/(2)mv_(f)^2-(1)/(2)mv_(i)^2$

Put the value into the formula

$\Delta K.E=(1)/(2)*110*0^2-(1)/(2)*110*(3.0)^2$

$\Delta K.E=-495\ J$

(b). We need to calculate the average force must he exert on the railing

Using work energy theorem

$W=\Delta K.E$

$Fd=\Delta K.E$

$F=(\Delta K.E)/(d)$

Put the value into the formula

$F=(-495)/(30*10^(-2))$

$F=-1650\ N$

The average force is 1650 N.

Hence, (a). The change in the kinetic energy of his center of mass during this process is -495 J.

(b). The average force is 1650 N.

Answer

given,

mass of ice hockey player = 110 Kg

initial speed of the skate = 3 m/s

final speed of the skate = 0 m/s

distance of the center of mass, m = 30 cm = 0.3 m

a) Change in kinetic energy

$\Delta KE = (1)/(2)mv_f^2 - (1)/(2)mv_i^2$

$\Delta KE = (1)/(2)m(0)^2 - (1)/(2)* 110 * 3^2$

$\Delta KE = - 495\ J$

b) Average force must he exerted on the railing

using work energy theorem

W = Δ KE

F .d = -495

F x 0.3 = -495

F = -1650 N

the average force exerted on the railing is equal to 1650 N.

Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?
(b) Find the ball's speed at impact.
(c) Find the horizontal range of the ball.

Answers

Answer:

Explanation:

Recall your experimental setup from Lab 05A: a constant force was applied to a disc by attaching a mass to a light string wrapped around a mass-less pulley and hanging the mass over the edge of the apparatus. In the lab, you used energy conservation arguments to derive an expression for the angular velocity of the disc after the mass had fallen a distance x . Your goal now is to use kinematics and dynamics to confirm your expression. Use the following symbols throughout this question: m is the mass of the hanging mass, M is the mass of the disc, r is the radius of the pulley, R is the radius of the disc, x is the distance the mass has fallen, and g is the acceleration due to gravity. What is the linear acceleration of the mass after it has fallen a distance x

Answers

Answer:

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

Explanation:

For this exercise let's start by applying Newton's second law to the mass with the string

W - T = m a

In this case, as the system is going down, we will assume the vertical directional down as positive.

T = W - m a

Now we apply Newton's second law for rotational motion to the pulley of radius r. We will assume the positive counterclockwise rotations

∑ τ = I α

T r = I α

the moment of inertia of the disk is

I = ½ M R²

angular and linear acceleration are related

a = α r

we substitute

T r = (½ m R²) (a / r)

T = ½ m ( $(R)/(r)$ )² a

we write our two equations

T = W - m a

T = ½ m ( $(R)/(r)$ )² a

we solve the system of equations

W - m a = ½ m (\frac{R}{r} )² a

m g = m a [ 1 + ½ (\frac{R}{r} )² ]

a = $(g)/( 1 + (1)/(2) ((R)/(r))^2 )$

this acceleration is constant throughout the trajectory, so with the angular and lineal kinematics relations

w² = w₀² + 2 α θ

v² = v₀² + 2 a y

as the system is released its initial angular velocity is zero

w² = 0 + 2 α θ

v² = 0 + 2 a y

we look for the angular acceleration

a =α r

α = a / r

α = $(g)/(r (1 + (1)/(2) ((R)/(r))^2 )$

we look for the angle, remember that they must be measured in radians

θ = s / r

in this case we approximate the arc to the distance

s = y

θ = y / r

we substitute

w = $\sqrt{2 (g)/( r( (1)/(2) ((R)/(r))^2 ) (y)/(r) }$

w = $\sqrt{(2gy)/(r^2 + (1)/(2) R^2 ) }$

for the simple case where r = R

w = $\sqrt{ (2gy)/( (3)/(2) R^2 ) }$

w = $\sqrt{ (4)/(3) (gy)/(R^2) }$

A helicopter is hovering above the ground. Jim reaches out of the copter (with a safety harness on) at 180 m above the ground. A package is launched upward, from a point on a roof 10 m above the ground. The initial velocity of the package is 50.5 m/s. Consider all quantities as positive in the upward direction. Does Jim Bond have a chance to catch the package? (calculate how high will it go)

Answers

Answer:

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

Explanation:

Hi there!

The equation of height and velocity of the package are the following:

h = h0 + v0 · t + 1/2 · g · t²

v = v0 + g · t

Where:

h = height of the package at time t.

h0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.81 m/s² because we consider the upward direction as positive).

v = velocity of the package at a time t.

First, let´s find the time it takes the package to reach the maximum height. For this, we will use the equation of velocity because we know that at the maximum height, the velocity of the package is zero. So, we have to find the time at which v = 0:

v = v0 + g · t

0 = 50.5 m/s - 9.8 m/s² · t

Solving for t:

-50.5 m/s / -9.81 m/s² = t

t = 5.15 s

Now, let´s find the height that the package reaches in that time using the equation of height. Let´s place the origin of the frame of reference on the ground so that the initial position of the package is 10 m above the ground:

h = h0 + v0 · t + 1/2 · g · t²

h = 10 m + 50.5 m/s · 5.15 s - 1/2 · 9.81 m/s² · (5.15 s)²

h = 140 m

The maximum height of the package is 140 m above the ground. Jim Bond will not catch the package.

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