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3 years ago

Which of the following is a good insulator? Wire, doorknob, plastic, nickle

2 answers:

4 0

Plastic would be a good insulator. Some other common insulators are glass, rubber, air, and wood.
Insulators usually protect us from the dangers of electric currents.
Hope that helped!

vovangra [49]3 years ago

3 0

Plastic is a good insulator.

Doorknob MAY be; depends on what it's made of.

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A 26.0 kg beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal. If

Nuetrik [128]

Answer:

Force exerted by the hinge on the beam = 109.24N

Explanation:

Weight = mg = 26 x 9.81 = 255.06 N

Vertical component = T sin θ

Horizontal component = Tcos θ

Now, there are 3 vertical forces acting on the beam. These are;

- The downward force which is the weight of the beam.

- The vertical components of the tension in the cable.

-The force that hinge exerts on the beam are the upward forces.

Hence, for the beam to remain horizontal, the sum of the upward forces must be equal to the weight of the beam.

For us to determine the vertical component of the tension in the cable, we will do a torque problem. Let the pivot point be at the hinge. Let’s assume that the length of the beam is L. The vertical component of the tension in the cable will produce clockwise torque while the weight of the beam will produce counter clockwise torque.

Tbus;

Clockwise torque = TL sin 61

Since the center of mass of beam is at the middle of the beam, the distance from the hinge to the weight of the beam is L/2.

Counter clockwise torque = WL/2

Thus;

TL sin 61 = WL/2

L will cancel out.

T sin 61 = 255.06/2

T x 0.8746 = 127.53

T = 127.53/0.8746 = 145.82 N

Now, the equation to determine the vertical component of the force that the hinge exerts on the beam is given as;

T + F = W

Thus;

145.82 + F = 255.06

F = 255.06 - 145.82 = 109.24 N

8 0

3 years ago

Which of the following is a contact force?

Nutka1998 [239]

The contact force is indeed caused by "Contact".

Air resistance is basically a type of friction, which is apparently present providing two object contact.

The rest selections are all interaction force, which is not necessarily caused by contact.

4 0

3 years ago

Read 2 more answers

The electron's velocity at that instant is purely horizontal with a magnitude of 2 \times 10^5 ~\text{m/s}2×10 5 m/s then ho

Lesechka [4]

Complete question:

At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.

[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]

Answer:

The time it will take the particle to pass through point (P) again is 1.639 ns.

Explanation:

F = qvB

Also;

$F = \frac{MV}{t}$

solving this two equations together;

$\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}$

where;

m is the mass of electron = 9.11 x 10⁻³¹ kg

q is the charge of electron = 1.602 x 10⁻¹⁹ C

B is the strength of the magnetic field = 3.47 x 10⁻³ T

substitute these values and solve for t

$t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns$

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.

5 0

3 years ago

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to c

IRINA_888 [86]

During a rodeo, a clown runs 7.7 m north, turns 49.9 degrees east of north, and runs 6.4 m. Then after waiting for the bull to come near, the clown turns due east and runs 19.8 m to exit the arena. The magnitude of the clown’s displacement is 27 m.

<u>Explanation: </u>

As the clown is running in the north direction for about 7.7 m and then he turns 49.9 degrees east of north. In the east of north, he covers a distance of 6.4 m and then turns east to exit the arena after covering a distance of 19.8 m. Let’s have a simple diagram to easily understand the problem.

In first step, the clown runs 7.7 m in north direction, so the image will be as in fig 1. Then he takes a direction of north east and covers a distance of 6.4 m, so the image will be modified as in fig 2. Then after the bull comes, he turns east and runs 19.8 m to exit the arena, so the image will be as in figure 3.

So, the extension of North line and the East line at a point shown as the dotted line in the above image, forms the total displacement as the hypotenuse of a right angled triangle. The extended dotted lines is nothing but the horizontal and vertical components of the angle 49.9 degree.

By using Pythagoras theorem, the total displacement can be found as

$\text { Total displacement }=\sqrt{(o p p)^{2}+(a d j)^{2}}$