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1 year ago

A receptor that contains many mechanically-gated ion channels would function BEST as a ____________.

1 answer:

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A receptor that contains many mechanically-gated ion channels would function BEST as a tactile receptor.

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A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing

Damm [24]

Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft, $\tau _{shaft}$ = 30 MPa

Shearing stress of key, $\tau _{key}$ = 240 MPa

width of key, w = $\frac{d}{4}$

d is shaft diameter

Solution:

Torque, T = $\frac{P}{\omega }$

where,

$\omega = \frac{2\pi N}{60}$

$T = \frac{7460}{\frac{2\pi (1200 )}{60}}$ = 59.365 N-m

Now,

$\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}$

$30\times 10^{6} = \frac{2\times 59.365}{\pi (\frac{d}{2})^{3}}$

d = 0.0216 m

Now,

w = $\frac{d}{4}$ = $\frac{0.02116}{4}$ = 5.4 mm

Now, for shear stress in key

$\tau _{key}$ = $\frac{F}{wl}$

we know that

T = $F \times r$ = F. $\frac{d}{2}$

⇒ $\tau _{key}$ = $\frac{\frac{T}{\frac{d}{2}}}{wl}$

⇒ $240\times 10^{6}$ = $\frac{\frac{59.365}{\frac{0.0216}{2}}}{0.054l}$

length of the rectangular key, l = 4.078 mm

7 0

1 year ago

Read 2 more answers

A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30

ElenaW [278]

Answer:

70 cm

Explanation:

0.5 kg at 20 cm

0.3 kg at 60 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

$0.5(50-20)=0.3(60-50)+0.6x\\\Rightarrow x=\dfrac{0.5(50-20)-0.3(60-50)}{0.6}\\\Rightarrow x=20\ cm$

The position of the third mass of 0.6 kg is at 20+50 = 70 cm

7 0

2 years ago

Read 2 more answers

Explica la relación entre momento de torsión y aceleración angular mencionando tres ejemplos Una varilla uniforme delgada mide 1

sukhopar [10]

Answer:

1) τ = I α whereby the torque is provided by the angular acceleration

2) L = 216 Kg m² / s

Explanation:

1) Let's start with Newton's second law

F = m a

multiply by the arm or perpendicular distance

F r = m a r

if the distance is not perpendicular a way of realizing the relations using the vector product

τ = F r = F x r

the bold are vectors. The angular and linear acceleration are related

a = α r=

τ = m (α r) r

τ = (m r²) α

the inertia of the rotational motion is

I = m r²

we substitute

τ = I α

whereby the torque is provided by the angular acceleration.

As an example we have:

* a spinning disk

* a ball rotating in the air

* a pulley

2) The rotational momentum is

L = I w

the moment of inertia of a rod that through its center

I = m L²

we substitute

L = m L² w

let's calculate

L = 6 1.5 2 16

L = 216 Kg m² / s

7 0

1 year ago

Floor lamps usually have a base with large inertia, while the long body and top have much less inertia. Part A If you want to sh

Novay_Z [31]

Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia and this mathematically given as

$I = mr^2$

Where m is the mass

r is the rotation radius

For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

Explanation:

5 0

1 year ago

Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of th

Citrus2011 [14]

Answer:

(a) Rm = 268.4 m

(b) f = 6

Explanation:

The horizontal range of a projectile is given by the following formula:

R = V₀² Sin 2θ/g

(a)

For moon:

R = Range on moon = Rm

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on moon = (9.8 m/s²)/6 = 1.63 m/s²

Therefore,

Rm = (28 m/s)²Sin (2*17°)/(1.63 m/s²)

Rm = 268.4 m

(b)

For Earth:

R = Range on Earth = Re

V₀ = Launch Speed = 28 m/s

θ = Launch Angle = 17°

g = acceleration due to gravity on Earth = 9.8 m/s²

Therefore,

Re = (28 m/s)²Sin (2*17°)/(9.8 m/s²)

Re = 44.7 m

Therefore.

f = Rm/Re = 268.4 m/44.7 m

f = 6

3 0

1 year ago