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3 years ago

What state of matter must ionic compounds be in to conduct electricity?

Physics

1 answer:

mixas84 [53]3 years ago

5 0

In either liquid or aqueous.

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You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa

rusak2 [61]

The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

7 0

3 years ago

What are two ways electromagnetic waves are used in a home computer scanner?

miss Akunina [59]

Answer:

A and B

Explanation:

8 0

2 years ago

Read 2 more answers

An object weighing 49 N is pushed across a floor by a force of 12 N. What is the acceleration of the object?

NISA [10]

Answer:

Explanation:

Given parameters:

Weight of object = 49N

Force applied = 12N

Unknown:

Acceleration of object = ?

Solution:

The acceleration of the object is found by dividing the force by the weight;

Acceleration = $\frac{12}{49}$ = 0.25m/s²

3 0

2 years ago

A meter stick is free to rotate about an axis through one of its end. Find the force F needed to balance this meter stick if the

Mumz [18]

Answer:

Explanation:

Component of force perpendicular to stick

= F Sin 60°

=√3 / 2 F.

Taking torque about the other end

= √3 / 2 F x 1 Nm

Weight of stick = 60 gm

= 60 x 10⁻³ kg

= 60 x 10⁻³ x 9.8 N

= .588 N

This weight will act from the middle point of stick so torque about the

other end

= .588 x 1 Nm

Balancing these two torques we have

.588 = √3 /2 F

$F=\frac{2\times0.588}{\sqrt{3} }$

F = 0.679 N

6 0

3 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

Delvig [45]

Answer:

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

Explanation:

Given that

For straight wire

Charge density= λ

For hollow metal cylinder

Charge density=2 λ

We know that electric filed for wire given as

$E_w=\dfrac{\lambda_{wire} }{2\pi \varepsilon _or}$

$E_w=\dfrac{\lambda }{2\pi \varepsilon _or}$

Now the electric filed due to hollow metal cylinder

$E_c=\dfrac{\lambda_{cylinder} }{2\pi \varepsilon _or}$

$E_c=\dfrac{2\lambda }{2\pi \varepsilon _or}$

Now by considering the Gaussian surface r<R then only electric fild due to wire will present.So

At r<R

$E=\dfrac{\lambda }{2\pi \varepsilon _or}$

5 0

3 years ago