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2 years ago

An electron is released from rest in a region of space with a nonzero electric field.

Physics

1 answer:

enyata [817]2 years ago

3 0

Answer:

Decreases

Explanation:

When an electron is released from rest in a region of space with nonzero electric field. And when the electrons move the electric potential energy of the system decreases because the kinetic energy increases and for the total energy to remain constant the potential energy must reduce.

An electron is accelerated in an electric field meaning its kinetic energy increases.

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Which is more work, pushing with 115 N over 15m or lifting 20N over 10 m?

Nikitich [7]

Work = (force) x (distance)

You could look at the two cases, and see right away that
the first one has more force acting through more distance,
so it must be more work. But since I just gave you the formula
for Work, let's calculate the amount of it for both cases:

First case: Work = (115 N) x (15 m) = 1,725 joules

Second case: Work = (20 N) x (10 m) = 200 joules

The first case involves 8.625 times as much work as the second case.

3 0

2 years ago

Read 2 more answers

The acceleration of a baseball after it is hit by a bat depends on the mass of the ball and the net force on the

Scrat [10]

Answer:

B. Newton's second law of motion

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Net \; Force = mass * acceleration$

Making acceleration the subject of formula, we have;

$Acceleration = \frac {Net \; Force}{Mass}$

In this scenario, the acceleration of a baseball after it is hit by a bat depends on the mass of the ball and the net force on the ball. Thus, this example best illustrates Newton's second law of motion.

7 0

2 years ago

Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long

NNADVOKAT [17]

Answer:

a) 578.0 cm²

b) 25.18 km

Explanation:

We're given the density and mass, so first calculate the volume.

D = M / V

V = M / D

V = (6.740 g) / (19.32 g/cm³)

V = 0.3489 cm³

a) The volume of any uniform flat shape (prism) is the area of the base times the thickness.

V = Ah

A = V / h

A = (0.3489 cm³) / (6.036×10⁻⁴ cm)

A = 578.0 cm²

b) The volume of a cylinder is pi times the square of the radius times the length.

V = πr²h

h = V / (πr²)

h = (0.3489 cm³) / (π (2.100×10⁻⁴ cm)²)

h = 2.518×10⁶ cm

h = 25.18 km

3 0

2 years ago

During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

a tank circuit contains a capacitor and an inductor that produce 30 of reactance at the resonant frequency. the inductor has a q

Romashka-Z-Leto [24]

The total circuit current at the resonant frequency is 0.61 amps

What is a LC Circuit?

A capacitor and an inductor, denoted by the letters "C" and "L," respectively, make up an LC circuit, also referred to as a tank circuit, a tuned circuit, or a resonant circuit.
These circuits are used to create signals at particular frequencies or to receive signals from more complicated signals at particular frequencies.

Q =15 = (wL)/R

wL = 30 ohms = Xl

R = 2 ohms

Zs = R + jXl = 2 +j30 ohms where Zs is the series LR impedance

| Zs | = 30.07 <86.2° ohms

Xc = 1/(wC) = 30 ohms

The impedance of the LC circuit is found from:

Zp = (Zs)(-jXc)/( Zs -jXc)

Zp = (2+j30)(-j30)/(2 + j30-j30) = (900 -j60)2 = 450 -j30 = 451 < -3.81°

I capacitor = 277/-j30 = j9.23 amps

I Zs = 277/(2 +j30) = (554 - j8,310)/904 = 0.61 - j9.19 amps

I net = I cap + I Zs = 0.61 + j0.04 amps = 0.61 < 3.75° amps

Hence, the total circuit current at the resonant frequency is 0.61 amps

To learn more about LC Circuit from the given link

brainly.com/question/29383434

#SPJ4

5 0

11 months ago