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GalinKa
19 days ago
11

At a given location from a source charge, the electric field______________.a. strength is dependent on the source charge and the

charge that is at the given location. points away from the source charge, regardless of the sign of the source b. points away from the source charge, if the source charge is positive. c. strength is only dependent on the source charge.d. points toward the source charge, if the source charge is negative.
Physics
1 answer:
Softa [2K]19 days ago
3 0
b. The electric field diverges from the source charge when it is positive. d. If the source charge is negative, the electric field converges towards it. In more depth: A positive source charge generates an electric field that delivers a repulsion to a positively charged test charge. Hence, the field points outward from positively charged sources. Conversely, a positively charged test charge is drawn toward a negative source charge, resulting in electric field vectors consistently pointing towards negatively charged objects. Additionally, the intensity of the electric field solely depends on the test charge. Thus, the valid responses are b and d, confirming that the electric field points away from a positive source charge and toward a negative source charge.
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Water flows without friction vertically downward through a pipe and enters a section where the cross sectional area is larger. T
kicyunya [2264]

Answer:

v_{2} will be less than v_{1} and P_{2} will be greater than P_{1}.

Explanation:

The law of conservation of mass states that the rate of fluid mass (m_{1}) entering a system equals the rate at which the fluid mass (m_{2}) exits the system.

The mass flow rate can be expressed as follows:

m = \rho A v

where \rho denotes the fluid density, A signifies the cross-sectional area through which fluid flows, and v represents the fluid's velocity.

Based on the problem conditions, as the fluid's density remains constant, we can write:

&& m_{1} = m_{2}\\&or,& \rho A_{1} v_{1} = \rho A_{2} v_{2}\\&or,& \dfrac{v_{2}}{v_{1}} = \dfrac{A_{1}}{A_{2}}

where A_{1} and A_{2} are the cross-sectional areas for the fluid flow, while v_{1} and v_{2} are the corresponding velocities across those areas.

Given the conditions in the problem, A_{2} > A_{1}, leading from the formula to v_{2} < v_{1}.

Furthermore, fluid pressure arises from the fluid's movement through any specific area. When the fluid accelerates, part of its energy increases its speed in the direction of flow, resulting in lower pressure.

Thus, in this instance, v_{2} < v_{1} the pressure in the larger cross-sectional area P_{2} will exceed the pressure P_{1} in the smaller cross-sectional area, implying

P_{2} > P_{1}.

6 0
25 days ago
The wavelength of light is 5000 angstrom. Express it in nm and m.
serg [2598]

Answer:

1 angstrom equals 0.1 nm.

To convert 5000 angstroms: 5000 angstrom = 5000 / 1 × 0.1 nm.

= 500 nm

1 \:  angstrom = 1 \times  {10}^{ - 10} m

To express 5000 angstroms in meters: 5000 angstrom = 5000 × 1 × 10^-10.

= 5 × 10^-7 m

Hope this explanation is useful for you.

7 0
1 month ago
A simple harmonic wave of wavelength 18.7 cm and amplitude 2.34 cm is propagating along a string in the negative x-direction at
Ostrovityanka [2214]

Answer

Given:

Wavelength = λ = 18.7 cm

                  = 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A)  Calculate the angular frequency.

     f = \dfrac{v}{\lambda}

     f = \dfrac{0.38}{0.187}

     f =2.03\ Hz

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

      K = \dfrac{2\pi}{\lambda}

     K= \dfrac{2\pi}{0.187}

    K =33.59\ m^{-1}

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

4 0
1 month ago
A robot probe drops a camera off the rim of a 239 m high cliff on mars, where the free-fall acceleration is −3.7 m/s2 .
Maru [2360]
<span>a. To determine the velocity at which the camera strikes the ground: v^2 = (v0)^2 + 2ay = 0 + 2ay v = sqrt{ 2ay } v = sqrt{ (2)(3.7 m/s^2)(239 m) } v = 42 m/s The camera impacts the ground with a speed of 42 m/s. b. To calculate the duration it takes for the camera to reach the bottom: y = (1/2) a t^2 t^2 = 2y / a t = sqrt{ 2y / a } t = sqrt{ (2)(239 m) / 3.7 m/s^2 } t = 11.4 seconds
       
The camera descends for 11.4 seconds before hitting the ground.</span>
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16 days ago
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Keith_Richards [2268]
The ball covers a horizontal distance of 0.902 meters. The trajectory of a kicked football adheres to a quadratic equation expressed as: f(x), where f(x) indicates the vertical distance in feet, and x signifies how far the ball travels horizontally. To compute the distance the ball will advance before striking the ground, we set the condition f(x) = 0. Upon solving this quadratic equation, we find that the horizontal distance traveled by the ball is: x = -0.902 meters, leading us to conclude that it travels 0.902 meters across the field.
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4 days ago
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