8

Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (HGE). In a study of ticks in the Midwest, it was fou

tiny-mole [99]

Answer:

a)0.024

b)0.148

Explanation:

Let 's represent the set of deer ticks Carrying Lyme disease with L and the set of deer ticks carrying Human Granulocytic Ehrlichiosis with H

Given:

P(L) = 0.16

P(H) = 0.10

P(L n H) = 0.1 ·P( L u H )

Hence, P( L u H) = 10 ·P( L nH)

(a)

Hence. using the equation. P(L U H) = P(L) + P(H) - P(L n H)

Hence, 10 · P(L n H ) = 0.16 + 0.1 - P(L n H )

Hence, 11 · P(L n H) = 0.16 + 0.1 = 0.26

Hence, P(L n H) = 0.26/11=0.024

(b)

We know that condition probability P(H ║ L) = p(L n H)/P(L)

hence, P(H ║ L) =(0.26/11)/0.16 =0.148

3 0

2 years ago

When should you use a piece of pipe as a leverage extension on the handle on a wrench?

Vesnalui [34]

Answer:

D

Explanation:

We must never use a piece of pipe as a leverage extension on the handle on a wrench.

Hence option d is correct.

7 0

3 years ago

If a conducting loop of radius 10 cm is onboard an instrument on Jupiter at 45 degree latitude, and is rotating with a frequency

Pepsi [2]

Answer:

a) fem = - 2.1514 10⁻⁴ V, b) I = - 64.0 10⁻³ A, c) P = 1.38 10⁻⁶ W

Explanation:

This exercise is about Faraday's law

fem = $- \frac{ d \Phi_B}{dt}$

where the magnetic flux is

Ф = B x A

the bold are vectors

A = π r²

we assume that the angle between the magnetic field and the normal to the area is zero

fem = - B π 2r dr/dt = - 2π B r v

linear and angular velocity are related

v = w r

w = 2π f

v = 2π f r

we substitute

fem = - 2π B r (2π f r)

fem = -4π² B f r²

For the magnetic field of Jupiter we use the equatorial field B = 428 10⁻⁶T

we reduce the magnitudes to the SI system

f = 2 rev / s (2π rad / 1 rev) = 4π Hz

we calculate

fem = - 4π² 428 10⁻⁶ 4π 0.10²

fem = - 16π³ 428 10⁻⁶ 0.010

fem = - 2.1514 10⁻⁴ V

for the current let's use Ohm's law

V = I R

I = V / R

I = -2.1514 10⁻⁴ / 0.00336

I = - 64.0 10⁻³ A

Electric power is

P = V I

P = 2.1514 10⁻⁴ 64.0 10⁻³

P = 1.38 10⁻⁶ W

6 0

2 years ago

An insect 5.00 mm tall is placed 20.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ

jenyasd209 [6]

Answer:

a) i = -9.63 cm , h ’= .0.24075 cm erect

b) i = 259.74 cm ,

Explanation:

For this exercise let's start by finding the focal length of the lens

1 / f = (n-1) (1 / R₁ - 1 / R₂)

1 / f = (1.70 -1)) 1 / ∞ - 1/13)

1 / f = 0.0538

f = - 18.57 cm

Now we can use the constructor equation

1 / f = 1 / o + 1 / i

1 / i = 1 / f - 1 / o

1 / i = -1 / 18.57 -1/20

1 / i = -0.1038 cm

I = -9.63 cm

For the height of the

image let's use magnification

m = h '/ h = - i / o

h ’= -h i / o

h ’= - 0.5 (-9.63) / 20

h ’= .0.24075 cm

b) we invert the lens

The focal length is

1 / f = (1.70 -1) (1/13 - 1 / int)

1 / f = 0.0538

f = 18.57 cm

1 / i = 1 / f -1 / o

1 / I = 1 / 18.57 - 1/20

1 / I = 3.85 10-3

i = 259.74 cm

h ’= - 0.5 259.74 / 20

h ’= 6.4935 cm

7 0

2 years ago

To test the performance of its tires, a car

velikii [3]

Answer:

The coefficient of static friction between the tires and the road is 1.987

Explanation:

Given:

Radius of the track, r = 516 m

Tangential Acceleration $a_r$ = 3.89 m/s^2

Speed,v = 32.8 m/s

To Find:

The coefficient of static friction between the tires and the road = ?

Solution:

The radial Acceleration is given by,

$a_{R = \frac{v^2}{r}$

$a_{R = \frac{(32.8)^2}{516}$

$a_{R = \frac{(1075.84)}{516}$

$a_{R = 2.085 m/s^2$

Now the total acceleration is

$\text{ total acceleration} = \sqrt{\text{(tangential acceleration)}^2 +{\text{(Radial acceleration)}^2$

=> $= \sqrt{ (a_r)^2+(a_R)^2}$

=> $\sqrt{ (3.89 )^2+( 2.085)^2}$

=> $\sqrt{ (15.1321)+(4.347)^2}$

=> $19.4791 m/s^2$

The frictional force on the car will be f = ma------------(1)

And the force due to gravity is W = mg--------------------(2)

Now the coefficient of static friction is

$\mu =\frac{f}{W}$

From (1) and (2)

$\mu =\frac{ma}{mg}$

$\mu =\frac{a}{g}$

Substituting the values, we get

$\mu =\frac{19.4791}{9.8}$

$\mu =1.987$

8 0

3 years ago