All categories

3 years ago

All of the following activities produce quantitative data exept

Physics

1 answer:

Keith_Richards [23]3 years ago

7 0

Answer:

The answer to your question are A and C

Explanation:

Quantitative data are quantities, something that we get after measuring something.

A. Measuring the rate of gas production from a chemical. This example is a quantitative measure, because we are measuring the rate.

B. Describing the clarity of water in a sample If we are describing something, means that we are not measuring anything, so this is not a quantitative measure.

C. Calculating the energy released from an electrochemical reaction If we are not measuring but we are using the data somebody else got to calculate energy, them this is a quantitative data.

You might be interested in

A 0.00275 kg air‑inflated balloon is given an excess negative charge q1 =−3.50×10−8 C by rubbing it with a blanket. It is found

kati45 [8]

Answer:

1) $\rm q_2$ is<u> positive.</u>

<u></u>

2) $\rm q_2=4.56\times 10^{-10}\ C.$

Explanation:

<u></u>

The charged rod is held above the balloon and the weight of the balloon acts in downwards direction. To balance the weight of the balloon, the force on the balloon due to the rod must be directed along the upwards direction, which is only possible when the rod exerts an attractive force on the balloon and the electrostatic force on the balloon due to the rod is attractive when the polarities of the charge on the two are different.

Thus, In order for this to occur, the polarity of charge on the rod must be positive, i.e., $\rm q_2$ is <u>positive.</u>

<u></u>

<u></u>

<u>Given:</u>

Mass of the balloon, m = 0.00275 kg.
Charge on the balloon, $\rm q_1 = -3.50\times 10^{-8}\ C.$
Distance between the rod and the balloon, d = 0.0640 m.
Acceleration due to gravity, $\rm g = 9.81\ m/s^2.$

In order to balloon to be float in air, the weight of the balloom must be balanced with the electrostatic force on the balloon due to rod.

Weight of the balloon, $\rm W = mg = 0.00275\times 9.81=2.70\times 10^{-2}\ N.$

The magnitude of the electrostatic force on the balloon due to the rod is given by

$\rm F_e = \dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}.$

$\rm \dfrac{1}{4\pi \epsilon_o}$ is the Coulomb's constant.

For the elecric force and the weight to be balanced,

$\rm F_e = W\\\dfrac{1}{4\pi \epsilon_o}\dfrac{|q_1||q_2|}{d^2}=W\\8.99\times 10^9\times \dfrac{3.50\times10^{-8}\times |q_2| }{0.0640^2}=2.70\times 10^{-2}\\|q_2| = \dfrac{2.70\times 10^{-2}\times 0.00640^2}{8.99\times 10^9\times 2.70\times 10^{-7}}=4.56\times 10^{-10}\ C.$

3 0

3 years ago

qual o nome da posição que o planeta tem maior velocidade? este ponto é mais próximo ou mais afastado do sol?

klio [65]

Em inglês, esse ponto é 'perihelion' ... 'periélio'. É o mais próximo que a Terra ou o planeta chega do sol.

7 0

3 years ago

During combustion, a fuel’s electromagnetic energy is converted into thermal energy.

swat32

True I believe..................

6 0

3 years ago

The generator at a power plant produces AC at 20,000 V. A transformer steps this up to 355,000 V for transmission over power lin

Masja [62]

Answer:

Number of coil in the output is 39938

Explanation:

We have given a step up transformer

Input voltage of transformer, that is primary voltage $v_p=20000volt$

Output voltage, that is secondary voltage $v_s=355000volt$

Number of turns in primary $N_p=2250$

For transformer we know that $\frac{V_p}{V_s}=\frac{N_p}{N_s}$

$\frac{20000}{355000}=\frac{2250}{N_s}$

$N_s=39937.5$

As the number of turns can not be in fraction so number of turns in the output coil is 39938

7 0

3 years ago

A high jumper jumps over a bar that is 2 m above the mat. With what velocity does the jumper strike the mat in the landing area?

docker41 [41]

Answer:

The velocity with which the jumper strike the mat in the landing area is 6.26 m/s.

Explanation:

It is given that,

A high jumper jumps over a bar that is 2 m above the mat, h = 2 m

We need to find the velocity with which the jumper strike the mat in the landing area. It is a case of conservation of energy. let v is the velocity. it is given by :

$v=\sqrt{2gh}$

g is acceleration due to gravity

$v=\sqrt{2\times 9.81\ m/s^2\times 2\ m}$

v = 6.26 m/s

So, the velocity with which the jumper strike the mat in the landing area is 6.26 m/s. Hence, this is the required solution.

8 0

3 years ago