Ask question

Ask question

All categories

3 years ago

8

What was the role of reformers in American life? a. to arrest political bosses b. to organize immigrants into labor unions c. to

take photographs of social abuses d. to point out social ills and help solve them

1 answer:

mash [69]3 years ago

7 0

It seems like d is the only possibility :)

You might be interested in

41. Which of these is NOT a major influence on Public opinion

Furkat [3]

I think it is B I apologize if I am incorrect

3 0

3 years ago

Please help please??????????????

malfutka [58]

Hey the answer is C . Buffalos

3 0

3 years ago

Read 2 more answers

What was murphy's law?

tamaranim1 [39]

I believe it was, "<span>Anything that can go wrong, will go wrong." But I'm not quite sure. </span>

4 0

2 years ago

Read 2 more answers

As part of the fastest growing type of tourism activity, more and more tourists attend the opera or ballet performances at their

Roman55 [17]

Answer:

Cultural tourism

Explanation:

According to sociology, there are different ways in which people do tourism. One of these types of tourism is the cultural tourism

Cultural tourism refers to the tourism that a traveler does when they engage with the lifestyle of the people from the areas they visit, the art, history, and everything that has to do with the culture of the area.

Therefore, tourists that attend the opera or ballet performances, museums, art galleries, dance ceremonies are engaging in forms of culture related to the area they visit and therefore this is an example of cultural tourism.

7 0

2 years ago

Suppose you find a rock that contains 10 micrograms of radioactive potassium-40, which has a half-life of 1.25 billion years. By

attashe74 [19]

Answer:

The rock is 3.75 billion years ago.

Explanation:

Given that:

Half life = 1.25 billion years

1 billion years = 10⁹ years

So,

Half life = 1.25 × 10⁹ years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac {ln\ 2}{t_{1/2}}$

$k=\frac {ln\ 2}{1.25\times 10^9}\ year^{-1}$

The rate constant, k = 5.5452 × 10⁻¹⁰ years⁻¹

Initial amount [A₀] = 80 micrograms

Final amount $[A_t]$ = 10 micrograms

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

So,

$10=80\times e^{-5.5452\times 10^{-10}\times t}$

$80e^{-5.5452\times \:10^{-10}x}=10$

$\frac{80e^{-5.5452\times \:10^{-10}x}}{80}=\frac{10}{80}$

$\ln \left(e^{-5.5452\times \:10^{-10}x}\right)=\ln \left(\frac{1}{8}\right)$

$x=\frac{10^{10}\times \:3\ln \left(2\right)}{5.5452}=3.75\times 10^9$

So,

<u>The rock is 3.75 billion years ago.</u>

8 0

3 years ago