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2 years ago

8

An explosion inside a nuclear plant resulted in an exposure of 250 Rem/hr. (2 miles away). How far will you have to move away to

decrease your exposure to 2 Rem/hr.?

1 answer:

makkiz [27]2 years ago

7 0

Answer:

f something happens to go wrong at a nuclear reactor, anyone living in a 10-mile radius of the plant may have to evacuate. This map also shows a 50-mile evacuation zone, the safe distance that the U.S. government recommended to Americans who were near

because

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a) A Kuiper Belt Object orbits the Sun with an average orbital distance of 40 AU. What is its orbital period? Round your answer

denpristay [2]

Answer:

253 years

Explanation:

a = Distance in Astronomical units from the sun = 40 AU

P = Orbital period in years

From Kepler's law we have

$P^2=a^3$

$\Rightarrow P^2=40^3$

$\Rightarrow P^2=64000$

$\Rightarrow P=\sqrt{64000}$

$\Rightarrow P=252.982212813$

$\Rightarrow P\approx 253\ years$

The orbital period of the Kuiper belt object is 253 years.

7 0

2 years ago

A 200N force pushes forward on the 20kg truck. The truck pulls two crates connected together by ropes, as shown. The gravitation

devlian [24]

Force on left crate, right crate and truck we need to draw here

First we will write the equation for left crate

$T_{left} = m*a$

$T_{left} = 20*a$

Now similarly for right crate

$T_{right} - T_{left} = 20*a$

now for truck

$F - T_{right} = 20*a$

Now we know that F = 200 N

$200 - T_{right} = 20*a$

now add all three equation

$T_{left} + T_{right} - T_{left} + 200 - T{right} = 20a + 20 a + 20a$

$200 = 60 a$

$a = 10/3 m/s^2$

now we will find all tension using this value of acceleration

$T_{left} = 20 * \frac{10}{3} = \frac{200}{3} N$

$T_{right} - \frac{200}{3} = 20*\frac{10}{3}$

$T_{right} = \frac{400}{3} N$

6 0

2 years ago

What is the rhyme scheme in Emily Dickinson’s poem, “Hope is the thing with feathers”?

kaheart [24]

Answer is B. ABAB. Hope it helped you, and have a great day.
-Charlie

5 0

2 years ago

A spaceship moves radially away from Earth with acceleration 29.4 m/s 2 (about 3g). How much time does it take for sodium street

Gemiola [76]

Answer:

doppler shift's formula for source and receiver moving away from each other:

<em>λ'=λ°√(1+β/1-β)</em>

Explanation:

acceleration of spaceship=α=29.4m/s²

wavelength of sodium lamp=λ°=589nm

as the spaceship is moving away from earth so wavelength of earth should increase w.r.t increasing speed until it vanishes at λ'=700nm

using doppler shift's formula:

<em>λ'=λ°√(1+β/1-β)</em>

putting the values:

700nm=589nm√(1+β/1-β)

after simplifying:

<em>β=0.17</em>

by this we can say that speed at that time is: v=0.17c

to calculate velocity at an acceleration of a=29.4m/s²

we suppose that spaceship started from rest so,

<em>v=v₀+at</em>

where v₀=0

so<em> v=at</em>

as we want to calculate t so:-

<em>t=v/a</em> v=0.17c ,c=3x10⁸ ,a=29.4m/s²

putting values:

=0.17(3x10⁸m/s)/29.4m/s²

<em>t=1.73x10⁶</em>

7 0

2 years ago

The circuits, P and Q, show two different ammeter-voltmeter methods of measuring resistance. Suppose the ammeter has a resistanc

qaws [65]

Answer:

Uncorrected values for

For circuit P

R = 2.4 ohm

For circuit Q

R = 2.4 ohm

Corrected values

for circuit P

R = 12 OHM

For circuit Q

R = 2.3 ohm

Explanation:

Given data:

Ammeter resistance 0.10 ohms

Resister resistance 3.0 ohms

Voltmeter read 6 volts

ammeter reads 2.5 amp

UNCORRECTED VALUES FOR

1) circuit P

we know that IR =V

$R = \frac{6}{2.5} - 2.4 ohm$

2) circuit Q

R = 2.4 ohm as no potential drop across ammeter

CORRECTED VALUES FOR

1) circuit p

IR = V

$\frac{3R}{R+3} \times 2.5 = 6$

R= 12 ohm

2) circuit Q

$I\times (R+0.1) =V$

$R+0.1 =\frac{6}{2.5}$

R = 2.3 ohm

5 0

2 years ago