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Butoxors
3 months ago
13

Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele

ctric field of N/C at point P. Determine the work required of an external agent to move a 3.0-μC charge along the x axis from infinity to point P without any net change in the kinetic energy of the particle.
Physics
1 answer:
ValentinkaMS [3.4K]3 months ago
4 0

Answer:

0.018 J

Explanation:

The work required to bring the charge from infinity to the point P is equal to the change in its electric potential energy. This can be expressed as

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C represents the charge's magnitude

and \Delta V = 6.0 kV = 6000 V signifies the potential difference between point P and infinity.

After substituting into the formula, we arrive at

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

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A new roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radius of the loop i
Sav [3153]

Respuesta:

11.4 m/s

Explicación:

La fórmula para la aceleración centrípeta es:

a=\frac{v^2}{R}

donde, a es la aceleración, v la velocidad alrededor de la circunferencia y R el radio del círculo.

En este problema,

a = g = aceleración debida a la gravedad en la cima = 9.81\ m/s^2

v = ?

R = 13.2 m

Por lo tanto,

9.81=\frac{v^2}{13.2}

v^2=9.81\times {13.2}

v = 11.4 m/s

8 0
3 months ago
A wildlife researcher is tracking a flock of geese. The geese fly 4.0 km due west, then turn toward the north by 40° and fly ano
serg [3582]

Answer:

(a). The distance traveled is 7.06 km towards the west.

(b). Their displacement magnitude is 7.51 km.

Explanation:

The information given states that,

The geese initially move 4.0 km directly west, then alter course to the north at a 40° angle, covering an additional 4.0 km.

Based on the diagram,

(a). To determine the distance

We will apply the distance formula

AD=AB+BD

Insert the values into the equation

D= 4+4.0\cos40^{\circ}

D=7.06\ km

The resultant distance is 7.06 km westward.

(b). We will find the total displacement's magnitude

Using the displacement formula

AC=\sqrt{(CD)^2+(AD)^2}

Insert the values into the equation

AC=\sqrt{(4.0\sin40)^2+(7.06)^2}

AC=7.51

The total displacement magnitude is 7.51 km.

In conclusion, (a). The traveled distance is 7.06 km towards the west.

(b). The magnitude of their total displacement is 7.51 km.

7 0
2 months ago
The acceleration of segment D is m/s2. Rank segments A, B , C from least accelerations to greatest acceleration. Least
Ostrovityanka [3204]

Answer:

D, C, B, A

Explanation:

The derivative from a velocity-time graph provides the acceleration value.

Segment A

\frac{dy}{dx} = \frac{15m/s}{1s} = 15m/s^2

Segment B

\frac{dy}{dx} = \frac{5m/s}{1s} = 5m/s^2

Segment C

\frac{dy}{dx} = \frac{0m/s}{2s} = 0m/s^2

Segment D

\frac{dy}{dx} = \frac{-20m/s}{1s} = -20m/s^2

Sorted from the lowest to the highest acceleration:

D, C, B, A

8 0
2 months ago
Read 2 more answers
A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate power by striking the buckets lo
Ostrovityanka [3204]

Response:

216000 W or 216 kW

Details:

Power: This refers to the rate at which energy is utilized or consumed. The standard unit for power is the Watt (W).

In general,

Power = Energy/time

P = E/t........................ Equation 1.

However,

E = 1/2mv²..................... Equation 2

with m representing mass, and v indicating velocity.

Substituting equation 2 into equation 1,

P = 1/2mv²/t...................... Equation 3

Assuming flow rate (Q) = m/t,

Q = m/t................ Equation 4

Insert equation 4 into equation 3,

P = Qv²/2........................ Equation 5

Where Q stands for flow rate, v is velocity, and P denotes power.

Given: Q = 120 kg/s, v = 60 m/s

Plug these values into equation 5,

P = 120(60)²/2

P = 60(60)²

P = 60×3600

P = 216000 W.

Thus, the potential power generation of the water jet is 216000 W or 216 kW.

5 0
3 months ago
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