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2 years ago

4

Hazardous wastes are placed into categories to aid in _____. A. determining toxicity B. tracking the number of yearly accidents

C. determining disposal methods D. studying their side effects

2 answers:

Natasha_Volkova [10]2 years ago

6 0

The answer to this question is D. studying their side effects

Alisiya [41]2 years ago

5 0

Determining disposal methods is the most accurate answer, thanks to the comment on the first response

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Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.

Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 1 (air) E = 7.5 / 0.10*10^-3

E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

C = k*A*ε / d

k = 1 (air) C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 2.213 pF

Q = C*V

Q = 7.5*(2.213)

Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 3.13 E = 7.5 / (3.13)0.10*10^-3

E = 24,000 V/m

F) What is the capacitor's charge after the Mylar is inserted?

C = k*A*ε / d

k = 3.13 C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 6.927 pF

Q = C*V

Q = 7.5*(6.927)

Q = 52 pC

6 0

2 years ago

A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass

Vlada [557]

1) 15 / 12 = 1.25 ratio
2) to increase acceleration 1.25 times (with same F, or same engine) you have to lower mass 1.25 times
3) 1515/1.25 = 1212 kg

choose A

6 0

2 years ago

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t

Juliette [100K]

Answer:

24.348mm

Explanation:

NB: I'll be attaching pictures so as to depict missing mathematical expressions or special characters which are not easily found on keyboards

K = d / €^n

Note : d represents the greek alphabet epsilion.

K = 345 / 0.02⁰.²² = 816mPa

The true strain based upon the stress of 414mPa =

€= (€/k)^1/n = (414/816)¹/⁰.²² = 0.04576

However the true relationship between true strain and length is given by

€ = ln(Li/Lo)

Making Li the subject of formula by rearranging,

Li = Lo.e^€

Li = 520e⁰.⁰⁴⁵⁷⁶

Li = 544.348mm

The amount of elongation can be calculated from

Change in L = Li - Lo = 544.348 - 520 change in L = 24.348mm.

8 0

2 years ago

Two swift canaries fly toward each other, each moving at 15.0 m/s relative to the ground, each warbling a note of frequency 1750

lesya692 [45]

Answer:

Part a)

f = 1911.5 Hz

Part b)

$\lambda = 0.186 m$

Explanation:

Here the source and observer both are moving towards each other

so we know that the apparent frequency is given as

$f' = f_0 (\frac{v + v_o}{v - v_s})$

here we know that

$f_0 = 1750 Hz$

$v_o = 15 m/s$

$v_s = 15 m/s$

now we will have

$f' = (1750)(\frac{340 + 15}{340 - 15})$

$f' = 1911.5 Hz$

Part b)

Apparent wavelength is given by the formula

$\lambda = \frac{v_{relative}}{f_{app}}$

here we will have

$\lambda = \frac{340 + 15}{1911.5}$

$\lambda = 0.186 m$

3 0

2 years ago

A certain factory whistle can be heard up to a distance of 2.5 km. Assuming that the acoustic output of the whistle is uniform i

enyata [817]

Answer:

Emitted power will be equal to $7.85\times 10^{-5}watt$

Explanation:

It is given factory whistle can be heard up to a distance of R=2.5 km = 2500 m

Threshold of human hearing $I=10^{-12}W/m^2$

We have to find the emitted power

Emitted power is equal to $P=I\times A$

$P=I\times 4\pi R^2$

$P=10^{-12}\times 4\times 3.14\times 2500^2=7.85\times 10^{-5}watt$

So emitted power will be equal to $7.85\times 10^{-5}watt$

4 0

2 years ago