PLEASE HELP!!!!!

1 answer:

7 0

Answer: The electroscope had a negative charge before the experiment started. When the negatively charged rod comes near the electroscope more electrons are repelled down to the metal pieces and the metal pieces move farther away from each other.

Step-by-step explanation: First of all, how we know the answer is not A or B is because unlike charges attract each other which means the metal pieces would not spread out. Now that we narrowed it down to C and D we can find the answer, In answer choice D it states that the negatively charged rod attracted the electrons from the electroscope, if that happened then something would happen to the rod, not the electroscope. In answer choice C it says that the rod repels the electrons downward on to the metal pieces, which would cause the metal pieces to spread out since they are both negative. Sorry, I'm not the best with explaining, I hope you understand.

You might be interested in

What's the answers to the hear questions??

scoundrel [369]

a.129.6+16.2
= 145.8
b.129.6×16.2
= 2099.52
c.129.6÷16.2
= 8
d.129.6−16.2
= 113.4
e.192.4+14.8
= 207.2

8 0

3 years ago

Can someone pls help with this?

telo118 [61]

Person above is correct:)

3 0

3 years ago

Two boats start their journey from the same point A and travel along directions AC and AD, as shown below

adell [148]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Using Newton's Law of Cooling. You are a medical examiner called to an abandoned house in which a dead body has been discovered.

Artyom0805 [142]

A) The differential equation comes from the fact that the rate of temperature change is proportional to the difference in temperatures.
$\frac{dT}{dt} = \alpha(T -52)$

B) Find general solution by separating variables and integrating $\int\frac{dT}{T-52} = \alpha \int dt \\ ln (T-52) = \alpha t + C \\ T = C e^{\alpha t} +52$ :

C) Initial condition is t=0, T = 87
$87 = C +52 \\ C = 35$

D) Total time elapsed is 10 minutes, new temperature is 84
$84 = 35 e^{10\alpha} +52$
solve for alpha
$e^{10\alpha} =\frac{32}{35} \\ \alpha = \frac{ln(32/35)}{10} \\ \alpha = -.00896$

E) Temperature function is:
$T = 35 e^{-.00896 t} +52$
solving for t
$t = \frac{ln (\frac{T-52}{35})}{-.00896}$
Plug in T = 98.6
$t = -31.95$

This is approximately 32 minutes before he arrived or about 1:20 AM

3 0

3 years ago

Please find the vertex to this quadratic

Fudgin [204]

Is = 2 ………………………………..,…..

3 0

3 years ago