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3 years ago

How do hormones affect only certain cells in the body but not others?

1 answer:

4 0

If you have studied enzymes its a similar concept. Cells have proteins on the surface of their cell which hormones bind to (called receptors) The receptor must be a complimentary shape to the hormone for it to bind. Only target cells have the receptor with the complimentary shape so only these cells will be affected.

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A ball is launched from ground level at 20 m/s at an angle of 40° above the

DedPeter [7]

(a) The ball's height y at time t is given by

y = (20 m/s) sin(40º) t - 1/2 g t ²

where g = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve y = 0 for t :

0 = (20 m/s) sin(40º) t - 1/2 g t ²

0 = t ((20 m/s) sin(40º) - 1/2 g t )

t = 0 or (20 m/s) sin(40º) - 1/2 g t = 0

The first time refers to where the ball is initially launched, so we omit that solution.

(20 m/s) sin(40º) = 1/2 g t

t = (40 m/s) sin(40º) / g

t ≈ 2.6 s

(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration g. So

0² - ((20 m/s) sin(40º))² = 2 (-g) y

where y in this equation refers to the maximum height of the ball. Solve for y :

y = ((20 m/s) sin(40º))² / (2g)

y ≈ 8.4 m

8 0

3 years ago

Voices of swimmers at a pool travel 400 m/s through the air and 1,600 m/s underwater. The wavelength changes from 2 m in the air

frosja888 [35]

The frequency of the wave has not changed.

In fact, the frequency of a wave is given by:

$f=\frac{v}{\lambda}$

where v is the wave's speed and $\lambda$ is the wavelength.

Applying the formula:

- In air, the frequency of the wave is:

$f=\frac{400 m/s}{2 m}=200 Hz$

- underwater, the frequency of the wave is:

$f=\frac{1600 m/s}{8 m}=200 Hz$

So, the frequency has not changed.

3 0

3 years ago

Read 2 more answers

The resistance of resistor is greater for:

Monica [59]

Answer:

c: long and thin resistor.

Explanation:

The resistance of a resistor is given by:

R = ρ*L/A

where:

R = resistance

ρ = resistivity (depends on the material)

L = length of the material

A = cross-sectional area of the material

We can see that the length is on the numerator, which means that if we increase the length, then the resistance is increased.

We also can see that the cross-sectional area is on the denominator, then if we increase the area (for example, with a ticker resistor) the resistance decreases.

Then if we want to maximize the resistance, we need to have a long and thin resistor, so the correct answer is c.

8 0

3 years ago

Carbon dioxide in a piston-‐‐cylinder is expanded in a polytropic manner. The initialtemperature and pressure are 400 K and 550

Arisa [49]

Answer:

q_poly = 14.55 KJ/kg

Explanation:

Given:

Initial State:

P_i = 550 KPa

T_i = 400 K

Final State:

T_f = 350 K

Constants:

R = 0.189 KJ/kgK

k = 1.289 = c_p / c_v

n = 1.2 (poly-tropic index)

Find:

Determine the heat transfer per kg in the process.

Solution:

-The heat transfer per kg of poly-tropic process is given by the expression:

q_poly = w_poly*(k - n)/(k-1)

- Evaluate w_poly:

w_poly = R*(T_f - T_i)/(1-n)

w_poly = 0.189*(350 - 400)/(1-1.2)

w_poly = 47.25 KJ/kg

-Hence,

q_poly = 47.25*(1.289 - 1.2)/(1.289-1)

q_poly = 14.55 KJ/kg

4 0

3 years ago

Un proyectil es lanzado horizontalmente desde una altura de 12 metro con una velocidad de 80 m/sg. a.Calcular el tiempo de vuelo

Naily [24]

Answer:

t= 1,56 s , x= 124,8 m , v = (80 i^ - 15,288 j ) m/s

Explanation:

Este es un ejercicio de lanzamiento proyectiles, comencemos por encontrar el tiempo que tarda en llegar al piso

y = y₀ + $v_{oy}$ t – ½ g t²

en este caso la altura inicial es y₀= 12 m y llega a y=0 , como es lanzado horizontalmente la velocidad vertical es cero (v_{oy}=0)

0 = y₀ – ½ g t²

t= √ (2 y₀/g)

calculemos

t= √ ( 2 12 / 9,8)

t= 1,56 s

El alcance del proyectil es la distancia horizontal recorrida

x = v₀ₓ t

x = 80 1,56

x= 124,8 m

La velocidad de impacto cuando toca el suelo

vx = v₀ₓ = 80 ms

$v_{y}$ = v_{oy} – gt

v_{y} = - 9,8 1,56

v_{y} = - 15,288 m/s

la velocidad es

v = (80 i^ - 15,288 j ) m/s

Traducttion

This is a projectile launching exercise, let's start by finding the time it takes to reach the ground

y = y₀ + v_{oy} t - ½ g t²

in this case the initial height is i = 12 m and it reaches y = 0, as it is thrown horizontally the vertical speed is zero ( $v_{oy}$ = 0)

0 =y₀I - ½ g t²

t = √ (2y₀ / g)

let's calculate

t = √ (2 12 / 9.8)

t = 1.56 s

Projectile range is the horizontal distance traveled

x = v₀ₓ t

x = 80 1.56

x = 124.8 m

Impact speed when it hits the ground

vₓ = v₀ₓ = 80 ms

v_{y} = v_{oy} - gt

v_{y} = - 9.8 1.56

v_{y} = - 15,288 m / s

the speed is

v = (80 i ^ - 15,288 j) m / s

7 0

3 years ago