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Levart [38]
3 years ago
8

Which of the following statements about electric field lines are true? (choose all that are true) a) They are only defined for p

ositive charges.
b) They are always tangent to electric field vectors.
c) They are always perpendicular to charged surfaces.
d) They are a simple way to visualize the electric field vectors. e) None of the above.
Physics
1 answer:
cupoosta [38]3 years ago
8 0

Answer:  b) TRUE and d) TRUE

Explanation:  a) FALSE the electric field lines are used to represent the charges postives and negatives.

b) TRUE it is the definition of  electric field lines , they are tangent to the electric field vector.

c) FALSE  the electric field lines only for conductor are perpendicular to the surface in other any situation there is a tangencial electric field components.

d) TRUE since it is a way to describe and imagine the effect of vectorial fields.

e) FALSE

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A 10.0 L tank contains 0.329 kg of helium at 28.0 ∘C. The molar mass of helium is 4.00 g/mol . Part A How many moles of helium a
nadya68 [22]

Answer:

82.25 moles of He

Explanation:

From the question given above, the following data were obtained:

Volume (V) = 10 L

Mass of He = 0.329 Kg

Temperature (T) = 28.0 °C

Molar mass of He = 4 g/mol

Mole of He =?

Next, we shall convert 0.329 Kg of He to g. This can be obtained as follow:

1 Kg = 1000 g

Therefore,

0.329 Kg = 0.329 Kg × 1000 g / 1 Kg

0.329 Kg = 329 g

Thus, 0.329 Kg is equivalent to 329 g.

Finally, we shall determine the number of mole of He in the tank. This can be obtained as illustrated below:

Mass of He = 329 g

Molar mass of He = 4 g/mol

Mole of He =?

Mole = mass / molar mass

Mole of He = 329 / 4

Mole of He = 82.25 moles

Therefore, there are 82.25 moles of He in the tank.

8 0
3 years ago
Kellie found a sample of a pure element. She carefully studied the sample and made a list of its properties.
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1. (C.) -Si- is silicone which at room temp is a solid and it's a hard crystalline Brittle rock basically...

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2 years ago
What is shown in the figure above
Firdavs [7]

A single magnetic field is shown.

4 0
2 years ago
A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the
sp2606 [1]
From the question,
u = 0m {s}^{ - 1}
v = 6m {s}^{ - 1}

t = 3s
F=12N



Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

Ft =m(v-u)


12(3.0)=m(6.0- \: 0)
This implies that

36.0 = 6m
m =  \frac{36.0}{6.0}
\therefore \: m = 6.0kg


You can also use the equation of linear motion,
v = u + at
6 = 0 + a(3)
6 = 3a
a =  \frac{6}{3}

a = 2 {ms}^{ - 2}
But
F=ma
12 = m(2)
12 = 2m
\frac{12}{2}  = m
\therefore \: m = 6kg
4 0
3 years ago
A typical jet airliner has a cruise airspeed of 900 km/h , which is its speed relative to the air through which it is flying. If
Luba_88 [7]

Explanation:

It is given that,

Speed of the jet airplane with respect to air, v_{PA} = 900\ km/h          

If the wind at the airliner’s cruise altitude is blowing at 100 km/h from west to east, v_{AG}= 100\ km/h

(A) Let v_{PG} is the speed of the airliner relative to the ground if the airplane is flying from west to east,

v_{PG}=900-100=800\ km/h

(B) Let v'_{PG} is the speed of the airliner relative to the ground if the airplane is flying from east to west,

v'_{PG}=900+100=1000\ km/h

Hence, this is the required solution.                                            

8 0
3 years ago
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