Ask question

Ask question

All categories

2 years ago

13

An electric field can be modeled as lines of force radiating out into space.

1 answer:

Hatshy [7]2 years ago

5 0

Yes. Electric field lines are the imaginary lines that represents strength & direction of electric field due to a charge. It can be modeled as lines.

In short, Your Answer would be "True"

Hope this helps!

You might be interested in

Balanced or Unbalanced?

Galina-37 [17]

Unbalanced because if it is pushing then stopping, that means that it is unbalanced.

6 0

2 years ago

A box weighs 25N. How much mass does it have?

Rudiy27

Explanation:

If box weight 25N on ground

MA=F

M(10)=25

M=2.5Kg

3 0

2 years ago

The force against a solid object moving through a gas or a liquid

boyakko [2]

This drag force is always opposite to the object's motion, and unlike friction between solid surfaces, the drag force increases as the object moves faster.

3 0

2 years ago

What is the weight of an object with a mass of 6.0 kg on Earth?

gregori [183]

<h2>since weight is measured in newtons, convert the 6 kg to newtons</h2><h3>the formula to convert is kg x 9.807 = N</h3>

6 x 9.807
= 58.842 N

hope that helps :))

3 0

2 years ago

Read 2 more answers

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

2 years ago