All categories

3 years ago

Match each statement with the characteristic of scientists.

Physics

2 answers:

Anarel [89]3 years ago

8 0

Answer: B, A, D, C,

Explanation: i got it correct :))

Dimas [21]3 years ago

5 0

Answer:

Options B, A, D, C

Explanation:

When a scientists, let's say Roberto wonders if the presence of other elements also affects the color of a flame, he can decide to prove this through a study. Therefore, in chemistry class, Roberto sees that traces of lithium makes a flame appear bright red. Subsequently, Roberto designs an experiment to test flame color in the presence of different elements and finally Roberto's friend tells him the color of a flame cannot be changed, but Roberto is still unsure.

You might be interested in

Mbbnmzmzkdjhxhxndmfmfmckcjcjcncncngmgkgjcjcj

Amiraneli [1.4K]

Gmsisient djaektnfidiaus wjrnrksiauenrnsiaur msiwlwkrjsuakeurjrndkaiakejrjd laismrnff

8 0

3 years ago

Read 2 more answers

after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF

Snezhnost [94]

Answer:

The time is $110.16\times10^{-3}\ sec$

Explanation:

Given that,

Capacitor = 120 μF

Voltage = 150 V

Resistance = 1.8 kΩ

Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

$i_{0}=\dfrac{V_{0}}{R}$

Put the value into the formula

$i_{0}=\dfrac{150}{1.8\times10^{3}}$

$i_{0}=83.3\times10^{-3}\ A$

We need to calculate the time

Using formula of current

$i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}$

Put the value into the formula

$50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}$

$\dfrac{50}{83.3}=e^{\frac{-t}{RC}}$

$\dfrac{-t}{RC}=ln(0.600)$

$t=0.51\times1.8\times10^{3}\times120\times10^{-6}$

$t=110.16\times10^{-3}\ sec$

Hence, The time is $110.16\times10^{-3}\ sec$

4 0

3 years ago

Which of the following statements are true? Check all that apply.

lidiya [134]

We have 6 levels and we will verify which of them is true.

A ) Electric charge is conserved

According to the conservation of charge, an electric charge can neither be created nor be destroyed.

Therefore this statement is TRUE.

B ) A positive charge and negative charge attract each other

According to the properties of charges, the opposite charges attract each other and similar charges repel each other

Therefore this statement is TRUE

C) Two positive charges attract each other.

Using the previous law, we can observe that this statement is false, because equal charges repel each other.

Therefore this statement is FALSE.

D ) Two negative charges repel each other.

Using the same principle as the two previous statements, two equal charges repel each other.

Therefore this statement is TRUE

E) Electric charge is quantized.

According to the quantization of charge, charges can take values only in integral multiples of a fundamental unit of charge.

Therefore this statement is TRUE.

F) A neutral object contains no charge.

A neutral object contains equal amount of negative and positive charge. Thus, in totality a neutral object contains no charge.

Therefore this statement is TRUE.

8 0

3 years ago

Answer fast !!!!!!! pleaseee

wariber [46]

Answer: Magnetic Fields

Explanation:

5 0

3 years ago

Read 2 more answers

A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location

amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}$

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

$f=-16.0 cm$ (the focal length is negative for a diverging lens)

$p=10.0 cm$ is the distance of the object from the lens

Solvign the equation for q, we find

$\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}$

$q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm$

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0

3 years ago