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dimulka [17.4K]

3 years ago

8

The electric potential at a point equidistant from two particles that have charges +Q and –Q is larger than zero. a. smaller tha

n zero.equal to zero.c. equal to the average of the two distances times the charges.d. equal to the net electric field.

1 answer:

Marizza181 [45]3 years ago

8 0

Answer:

idk, idk cause i'm steppin on my toes and i can't stop i make flips ou of my flops

Explanation:

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What is buoyant force?

Nostrana [21]

This is the upthrust on an object which is placed inside a fluid

This force act upwards and always push upwards

so the correct answer is given as

D. A force within a fluid that pushes upward

this force is always due to pressure difference at two levels of

at lower level since pressure is more that is why the force is upwards and this upthrust is known as Buoyancy

3 0

3 years ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

5 0

3 years ago

The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .

ASHA 777 [7]

Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

- The external force, whose work is $W=1.20 \cdot 10^{-3}J$

- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: $W_e=q\Delta V$

where

q is the charge

$\Delta V$ is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

$K_f - K_i = W + W_e = W+q\Delta V$

and since the charge starts from rest, $K_i = 0$ , so the formula becomes

$K_f = W+q\Delta V$

In this problem, we have

$W=1.20 \cdot 10^{-3}J$ is the work done by the external force

$q=-6.70 \mu C=-6.7\cdot 10^{-6}C$ is the charge

$K_f = 4.72\cdot 10^{-4}J$ is the final kinetic energy

Solving the formula for $\Delta V$ , we find

$\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V$

4 0

2 years ago

Read 2 more answers

A car accelerates uniformly from +10.0 m/s to +40.0 m/s over a distance of 125 m. How long did it take to go that distance? Show

Marizza181 [45]

Answer:

Explanation:

$v^2-u^2=2as\\40^2-10^2=2a*125\\1600-100=250a\\a=\frac{1500}{250}=6~m/s^2\\v=u+at\\40=10+6t\\6t=30\\ t=\frac{30}{6}=5 ~s$

7 0

3 years ago

An electron in a region where there is an electric field experiences a force of magnitude 3.5 10-16 n. what is the magnitude of

natima [27]

<span>The force that an object feels in an electric field is the magnitude of the electric field E times the charge on the object Q. We know that the magnitude of the charge of an electron is 1.6 x 10^{-19} C F = E x Q E = F / Q E = (3.5 x 10^{-16} N) / (1.6 x 10^{-19} C) E = 2.19 x 10^3 N/C The magnitude of the electric field at the location of the electron is 2.19 x 10^3 N/C</span>

8 0

2 years ago