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3 years ago

A system absorbs 194 kj of heat and the surroundings do 120 kj of work on the system. internal eneergy change

1 answer:

6 0

We can solve the problem by using the first law of thermodynamics, which states that:
$\Delta U = Q-W$
where
$\Delta U$ is the change in internal energy of the system
Q is the heat absorbed by the system
W is the work done by the system

In our problem, the heat absorbed by the system is Q=+194 kJ, while the work done is W=-120 kJ, where the negative sign means the work is done by the surroundings on the system. Therefore, the variation of internal energy is
$\Delta U= Q-W=+194 kJ - (-120 kJ)=+314 kJ$

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A 15x10^-6c charge is placed at the origin and a 9x10^-6C charge is placed on the x-axis at x=1.00m. where, on the x-axis is the

Harlamova29_29 [7]

The Electric field is zero at a distance 2.492 cm from the origin.

Let A be point where the charge $15\times10^-6$ C is placed which is the origin.

Let B be the point where the charge $9\times 10^-6$ C is placed. Given that B is at a distance 1 cm from the origin.

Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.

i.e., at distance 'x' from B.

Using Coulomb's law, $\frac{kQ_A^2}{d_A^2} = \frac{kQ_B^2}{d_B^2}$ ,

$Q_A$ = $15\times 10^-6 C$

$Q_B=9\times10^-6C$

$d_A = 1+x cm$

$d_B=x cm$

k is the Coulomb's law constant.

On substituting the values into the above equation, we get,

$\frac{(15\times10^-6)^2}{(1+x)^2} =\frac{(9\times10^-6)^2}{x^2}$

Taking square roots on both sides and simplifying and solving for x, we get,

1.67x = 1+x

Therefore, x = 1.492 cm

Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.

Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926

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3 years ago

An archer defending a castle is on an 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take

kodGreya [7K]

Answer:

about 602 milliseconds

Explanation:

The motion can be approximated by the equation ...

y = -4.9t^2 -22.8t +15.5

where t is the time since the arrow was released, and y is the distance above the ground.

When y=0, the arrow has hit the ground.

Using the quadratic formula, we find ...

t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))

= (22.8 ± √823.64)/(-9.8)

The positive solution is ...

t ≈ 0.60195193

It takes about 602 milliseconds for the arrow to reach the ground.

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