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4 years ago

13

Which of the following does not represent a significant negative environmental impact of using nuclear fuel? a. air pollution b.

mining c. radiation d. waste production Please select the best answer from the choices provided A B C D

1 answer:

sasho [114]4 years ago

3 0

The answer is a. mining.

You might be interested in

A certain humidifier operates by raising water to the boiling point and then evaporating it. Every minute 30 g of water at 20◦ C

Sveta_85 [38]

Answer:

The value of total energy needed per minute for the humidifier = 77.78 KJ

Explanation:

Total energy per minute the humidifier required = Energy required to heat water to boiling point) + Energy required to convert liquid water into vapor at the boiling point) ----- (1)

Specific heat of water = 4190 $\frac{J}{kg k}$

The heat of vaporization is = 2256 $\frac{KJ}{kg}$

Mass = 0.030 kg

Energy needed to heat water to boiling point = $m c ( T_{2} - T_{1} )$

Energy needed to heat water to boiling point = 0.030 × 4.19 × (100 - 20)

Energy ( $E_{1}$ ) = 10.08 KJ

Energy needed to convert liquid water into vapor at the boiling point

$E_{2}$ = 0.030 × 2256 = 67.68 KJ

Thus the total energy needed E = $E_{1}$ + $E_{2}$

E = 10.08 + 67.68

E = 77.78 KJ

This is the value of total energy needed per minute for the humidifier.

9 0

4 years ago

Estimate how fast your hair grows, in units of m/s, assuming it takes 30 days for your hair to grow 1 inch. note that 1 inch =2.

VashaNatasha [74]

Answer:

1. 9.8 x 10^-7 cm/s

2. 50970 L

Explanation:

1.

time, t = 30 days = 30 x 24 x 60 x 60 seconds = 2592000 seconds

length of hair, d = 1 inch = 2.54 cm

rate of growth = length of hair grows per second = 2.54 / 2592000

= 9.8 x 10^-7 cm/s

Thus, the grown rate of hair is 9.8 x 10^-7 cm/s.

2. length of the pool, l = 15 feet

width of the pool, w = 15 feet

height of pool, h = 8 feet

Volume of the pool, V = length x width x height

V = 15 x 15 x 8 = 1800 ft^3

To convert ft^3 into m^3 , we use

1 ft = 0.3048 m

so, 1 ft^3 = (0.3048)^3 m^3 = 0.0283 m^3

So, 1800 ft^3 = 1800 x 0.0283 = 50.97 m^3

now, 1 m^3 = 1000 L

So, 50.97 m^3 = 50.97 x 1000 L = 50970 L

3 0

3 years ago

Can uh help in in this question step by step

Luda [366]

Initial velocity=u=72km/h

Convert to m/s

$\\ \sf \longmapsto 72\times \dfrac{5}{18}=5(4)=20m/s$

Final velocity=v=0m/s
Time=2s=t

$\\ \sf \longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf \longmapsto Acceleration=\dfrac{0-20}{2}$

$\\ \sf \longmapsto Acceleration=\dfrac{-20}{2}$

$\\ \sf \longmapsto Acceleration=a=-10m/s^2$

Distance be s

Using second equation of kinematics

$\\ \sf \longmapsto s=ut+\dfrac{1}{2}at^2$

$\\ \sf \longmapsto s=20(2)+\dfrac{1}{2}(-10)(2)^2$

$\\ \sf \longmapsto s=40+(-20)$

$\\ \sf \longmapsto s=40-20$

$\\ \sf \longmapsto s=20m$

Now

Mass=m=5000kg

Using newtons second law

$\\ \sf \longmapsto Force=ma$

$\\ \sf \longmapsto Force=5000(-10)$

$\\ \sf \longmapsto Force=-50000N$

Force is in opposite direction so its negative

$\\ \sf \longmapsto Force=50kN$

7 0

3 years ago

Read 2 more answers

Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Zigmanuir [339]

Answer:

10.1 m/s

Explanation:

By Newton's third law, the force on the squid and that due to the water expelled form an action reaction pair.

And by the law of conservation of momentum,

initial momentum of squid + expelled water = final momentum of squid + expelled water.

Now, the initial momentum of the system is zero.

So, 0 = final momentum of squid + expelled water

0 = MV + mv where M = mass of squid = 6.50kg, V = velocity of squid = 2.40m/s, m =mass of water in cavity = 1.55 kg and v = velocity of water expelled

So, MV + mv = 0

MV = -mv

v = -MV/m

= -6.50 kg × 2.40 m/s ÷ 1.55 kg

= -15.6 kgm/s ÷ 1.55 kg

= -10.1 m/s

So, speed must it expel this water to instantaneously achieve a speed of 2.40 m/s to escape the predator is 10.1 m/s

4 0

3 years ago

A girl of mass 50.6 kg stands on the edge of a frictionless merry-go-round of mass 827 kg and radius 3.72 m that is not moving.

GaryK [48]

Answer:

(a). The value of angular speed of the merry-go-round $\omega$ = - 5.82 × $10^{-3}$ $\frac{rad}{s}$

(b). The linear speed of the girl after the rock is thrown V = -1.89 × $10^{-2}$ $\frac{m}{s}$

Explanation:

Given data

Mass of the girl $m_{g}$ = 50.6 kg

Mass of merry-go-round $m_{m}$ = 827 kg

Radius r = 3.72 m

The speed of the rock relative to the ground $V_{r}$ = 7.82 $\frac{m}{s}$

(a). The angular speed of the merry-go-round is given by

$\omega = - [\frac{m_{r}v_{r} }{r} ] \frac{2}{m_{m} + 2m_{g} }$

Put all the values in above formula

$\omega = \frac{(1.13)(7.82)}{3.27} \frac{2}{827 + (2)50.6}$

$\omega$ = - 5.82 × $10^{-3}$ $\frac{rad}{s}$

This is the value of angular speed of the merry-go-round.

(b). The liner speed of the girl is given by

⇒ V = r × $\omega$

⇒ V = - 3.72 × 5.82 × $10^{-3}$

⇒ V = -1.89 × $10^{-2}$ $\frac{m}{s}$

This is the linear speed of the girl after the rock is thrown.

7 0

3 years ago