All categories

3 years ago

PLEASE HELP MEEE!!!

Physics

1 answer:

IRINA_888 [86]3 years ago

4 0

Answer:

Option C

Explanation:

Answer C is the correct option. water can be written as H₂O, which means that there are 2 Hydrogen atoms for every oxygen atom, therefore it will occupy more space than oxygen and push more. there is also one more possibility, if the splitting takes place in Hoffman's Voltameter then the Hydrogen will be close to the cathode as hydrogen is positive. Otherwise, option C is correct answer. Hope this Helps you!

You might be interested in

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

Steam enters a one-inlet, two-exit control volume at location (1) at 360°C, 100 bar, with a mass flow rate of 2 kg/s. The inlet

yaroslaw [1]

Answer:

The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

Explanation:

Given that,

Mass flow rate = 2 kg/s

Diameter of inlet pipe = 5.2 cm

Fifteen percent of the flow leaves through location (2) and the remainder leaves at (3)

The mass flow rate is

$m_{2}=0.15\times2$

We need to calculate the mass flow rate at reach exit

Using formula of mass

$m_{3}=m_{1}-m_{2}$

$m_{3}=2-0.15\times2$

$m_{3}=1.7\ kg/s$

We need to calculate the inlet velocity

Using formula of velocity

$v=\dfrac{m}{\rho A}$

Put the value into the formula

$v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}$

$v=21.9\ m/s$

Hence, The inlet velocity is 21.9 m/s.

The mass flow rate at reach exit is 1.7 kg/s.

7 0

4 years ago

The flow of electricity can be compared of water in

Anna007 [38]

The flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

<h3>How we compare the flow of electricity to water?</h3>

Water flowing in pipes is like flowing of electricity in a circuit. A battery is like a pump from where electricity comes and moves in the circuit. Electrons flowing through wires are like water molecules flowing through pipes. So in comparison between water and electricity, both water and electricity are similar to each other in flowing and movement.

So we can conclude that the flow of electricity can be compared of water in the pipes because both water and electricity moves in the channel.

Learn more about electricity here: brainly.com/question/776932

#SPJ1

7 0

2 years ago

A rock is dropped from the top of a building. after 4 seconds it is moving at 40 m/s. what was the acceleration of the rock

Marta_Voda [28]

Its moving 10 m/s every sec.

6 0

3 years ago

If a girl is running along a straight road with a uniform velocity 1.5 m /s find her acceleration

Phoenix [80]

Answer:

As velocity is uniform then acceleration will be zero.

Explanation:

6 0

3 years ago