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tangare [24]
2 years ago
5

A screw having 50% efficiency is driven by a rod and 25 cm. The pitch of the screw is 1/10cm Calculate velocity ratio and mechan

ical advantage of the screw. ​
Physics
1 answer:
neonofarm [45]2 years ago
4 0

(a) The velocity ratio of the screw is 1570.8.

(b) The mechanical advantage of the screw is 785.39.

<h3>Velocity ratio of the screw</h3>

The velocity ratio of the screw is calculated as follows;

V.R = 2πr/P

where;

  • P is the pitch = 1/10 cm = 0.1 cm = 0.001 m
  • r is radius = 25 cm = 0.25 m

V.R = (2π x 0.25)/(0.001)

V.R = 1570.8

<h3>Mechanical advantage of the screw</h3>

E = MA/VR x 100%

0.5 = MA/1570.8

MA = 785.39

Learn more about mechanical advantage here: brainly.com/question/18345299

#SPJ1

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3 years ago
A softball player moving 3.89 m/s
Ahat [919]

Answer:

0.119 s

Explanation:

Given that

U=3.89\ m/s\\a=-1.44\ m/s^2\\S=4.8\ m

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We know that

V^2=U^2+2aS\\V^2=3.89^2-2\times 1.44\times 4.8\\V=3.7\ m/s

Also We know that

V=U+at\\

Putting the value of V ,U and a in the previous equation  We get

3.7=3.89-1.44\times t\\t=0.119\ s

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3 0
3 years ago
An electron, moving south, enters a uniform magnetic field. Because of this uniform magnetic field, the electron curves upward.
Flura [38]

Answer:

Towards the west

Explanation:

Magnetic force is the interaction between a moving charged particle and a magnetic field.

Magnetic force is given as

F = q (V × B)

Where F is the magnetic force

q is the charge

V is the velocity

B is the magnetic field

V×B means the cross product of the velocity and the magnetic field

NOTE:

i×i=j×j×k×k=0

i×j=k.  j×i=-k

j×k=i.  k×j=-i

k×i=j.  i×k=-j

So, if the electron is moving southward, then, it implies that the velocity of it motion is southward, so the electron is in the positive z-direction

Also, the electron is curved upward due to the magnetic field, this implies that the force field is directed up in the positive y direction.

Then,

V = V•k

F = F•j

Then, apply the theorem

F •j = q ( V•k × B•x)

Let x be the unknown

From vector k×i =j.

This shows that x = i

Then, the magnetic field point in the direction of positive x axis, which is towards the west

You can as well use the Fleming right hand rule

The thumb represent force

The index finger represent velocity

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3 0
3 years ago
Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the su
yKpoI14uk [10]

Complete Question

Red blood cells can be modeled as spheres of 6.53 μm diameter with −2.55×10−12 C excess charge uniformly distributed over the surface. Find the electric field at the following locations, with radially outward defined as the positive direction and radially inward defined as the negative direction. The permittivity of free space ????0 is 8.85×10−12 C/(V⋅m). What is the electric field

E⃗ 1 inside the cell at a distance of 3.05 μm from the center?

E⃗ 2 Just inside the surface of the cell

E⃗ 3 Just outside the surface of the cell

E⃗ 4 At a point outside the cell 3.05 μm from the surface

Answer:

E⃗ 1

      0 V/m

E⃗ 2

      0 V/m

E⃗ 3

         E_3 =  2.153 *10^{9} \  V/m

E⃗ 4

E_4 =  5.754 *10^ {8} \  V/m

Explanation:

From the question we are told that

The diameter is d =  6.53 \mu m  = 6.53*10^{-6}\  m

The charge is Q =  -.2.55 *10^{-12} \  C

The permittivity of free space is \epsilon_o  =  8.85* 10^{-12}\  C / V.m

The distance considered is d =  3.05 \mu m  =  3.05 *10^{-6} \ m

Generally the electric field inside the cell at a distance of 3.05 μm from the center is

0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just inside the surface of the cell is 0 V/m

This because there is no electric field felt inside the cell according Gauss the cell is taken as a point charge

Generally the electric field just outside the cell is mathematically represented as

E_3 =  \frac{ k  *  |Q|}{ r^2 }

Here k is the coulomb constant with value

k  =   9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}

r is the radius of the sphere which is mathematically as

r =  \frac{d}{2} =   \frac{6.53*10^{-6}}{2}  = 3.265 *10^{-6} \  m

E_3 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [3.265 *10^{-6} ]^2 }

E_3 =  2.153 *10^{9} \  V/m

Generally the electric field at a point outside the cell 3.05 μm from the surface is mathematically represented as

E_4 =  \frac{ k  *  |Q|}{ R^2 }

Here R is mathematically represented as

R  =  3.265 *10^{-6} +  3.05 *10^{-6}

=>       R  =  6.315 *10^{-6}

So

E_4 =  \frac{ 9*10^{9}  *  |-2.55 *10^{-12} |}{ [ 6.315 *10^{-6} ]^2 }

E_4 =  5.754 *10^ {8} \  V/m

3 0
3 years ago
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