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nataly862011 [7]
3 years ago
5

Electric field lines always begin at _______ charges (or at infinity) and end at _______ charges (or at infinity). One could als

o say that the lines we use to represent an electric field indicate the direction in which a _______ test charge would initially move when released from rest. Which of the following fills in the three missing words correctly?
a. (positive; negative; negative)
b. (positive; negative; positive)
c. (negative; positive; negative)d. (negative; positive; positive)
Physics
1 answer:
lesantik [10]3 years ago
3 0

Answer:

b)

Explanation:

By convention, the electric field lines (which are tangent to the direction of the electric field at a given point) always begin at positive charges, and finish at negative charges.

This is a consequence of the convention that states that the electric field has the direction of the trajectory of a positive test charge when released from rest in an electric field.

(As the positive charge would move away from positive charges and would  be attracted by negative ones).

So, the combination of answers that is true is b) (positive, negative, positive).

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A passenger in a helicopter traveling upwards at 15 m/s accidentally drops a package out the window. If it takes 15 seconds to r
Alexeev081 [22]

Answer:

The helicopter was 1103.63 meters high when the package was dropped.

Explanation:

We consider positive speed as a downward movement

y: height (m)

t: time (s)

v₀: initial speed (m/s)

Δy = v₀t + \frac{1}{2}gt²

Δy= 15\frac{m}{s}×15 s + \frac{1}{2}×9.81\frac{m}{s^{2} }×(15 s)²

Δy= 1103.63 m

8 0
3 years ago
How do we determine the conditions that existed in the very early universe? A We can only guess at the conditions, since we have
lakkis [162]

Answer:

D By looking all the way to the cosmological horizon, we can see the actual conditions that prevailed all the way back to the first instant of the Big Bang.

Explanation:

Astrophysicists are able to determine the conditions that existed in the early universe, by using instruments such as telescopes to observe and study cosmic horizons. More ideas about the early universe can be found from the thermal light present in cosmic backgrounds.

Scientists study these details that provide an insight into the conditions that existed so many years ago. They have been able to determine that the Big Bang involved so many collisions from these observations.

5 0
3 years ago
Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass
Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

\Delta W = \Delta PE + \Delta KE

Where,

PE = Potential Energy

KE = Kinetic Energy

\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

P = \frac{\Delta W}{\Delta t}

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

The rate of mass flow is,

\frac{\Delta m}{\Delta t} = \rho_w Av

Where,

\rho_w = Density of water

A = Area of the hose \rightarrow A=\pi r^2

The given radius is 0.83cm or 0.83 * 10^{-2}m, so the Area would be

A = \pi (0.83*10^{-2})^2

A = 0.0002164m^2

We have then that,

\frac{\Delta m}{\Delta t} = \rho_w Av

\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)

\frac{\Delta m}{\Delta t} = 1.16856kg/s

Final the power of the pump would be,

P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)

P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)

P = 57.1192W

Therefore the power of the pump is 57.11W

6 0
3 years ago
State examples of a transverse wave. ​
laiz [17]

ripples on the surface of water.

vibrations in a guitar string.

a Mexican wave in a sports stadium.

electromagnetic waves – eg light waves, microwaves, radio waves.

seismic S-waves.

6 0
3 years ago
Read 2 more answers
The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is
garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom =  a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the  water = 1000 kg/m³

The total pressure at the bottom of the pool =  (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a =  P A

ρ g h a =  P A - P a

h=\dfrac{P ( A-a)}{\rho g a}

h=\dfrac{P ( 1.5 a-a)}{\rho g a}

h=\dfrac{P ( 1.5- 1)}{\rho g}

h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m

h=5.15 m

The depth is 5.15 m.

7 0
3 years ago
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