1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Artist 52 [7]
3 years ago
10

28. Which of the following statements is true? a. An ideal metal conducts because each electron ‘belongs’ to every nucleus in th

e crystal lattice, while in an ideal insulator, each electron is localized to a single nucleus. b. Metals conduct much better than insulators because, per gram of material, metals have orders of magnitude more electrons. c. At sufficiently low temperatures, (intrinsic) semi-conductors are effectively insulators.
Physics
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:768

Explanation:

You might be interested in
An initially stationary object experiences an acceleration of 6 m/s2 for a time of 15 s. How far will it travel during that time
Arisa [49]

Explanation:

Distance = (intial speed)X(Time) + 1/2(acceleration)X(Time) [Third equation of motion]

As initial speed is zero, therefore;

Distance = 1/2(acceleration)X(Time)

              = 1/2 (6 X 15)

              = 1/2 (90)

              = 45 meters

Hence, the object traveled 45 meters.

4 0
4 years ago
A uniform rod of mass 3.30×10−2 kg and length 0.450 m rotates in a horizontal plane about a fixed axis through its center and pe
Alex73 [517]

(a) 2.75 rev/min

The moment of inertia of the rod rotating about its center is:

I_R=\frac{1}{12}ML^2

where

M=3.30\cdot 10^{-2} kg is its mass

L = 0.450 m is its length

Substituting,

I_R=\frac{1}{12}(3.30\cdot 10^{-2})(0.450)^2=5.57\cdot 10^{-4} kg m^2

The moment of inertia of the two rings at the beginning is

I_r = 2mr^2

where

m = 0.200 kg is the mass of each ring

r=5.20\cdot 10^{-2} m is their distance from the center of the rod

Substituting,

I_r=2(0.200)(5.20\cdot 10^{-2})^2=1.08\cdot 10^{-3} kg m^2

So the total moment of inertia at the beginning is

I_1=I_R+I_r = 5.57\cdot 10^{-4}+1.08\cdot 10^{-3}=1.64\cdot 10^{-3}kg m^2

The initial angular velocity of the system is

\omega_1 = 35.0 rev/min

The angular momentum must be conserved, so we can write:

L=I_1 \omega_1 = I_2 \omega_2 (1)

where I_2 is the moment of inertia when the rings reach the end of the rod; in this case, the distance of the ring from the center is

r=\frac{0.450 m}{2}=0.225 m

so the moment of inertia of the rings is

I_r=2(0.200)(0.225)^2=0.0203 kg m^2

and the total moment of inertia is

I_2 = I_R + I_r =5.57\cdot 10^{-4} + 0.0203 = 0.0209 kg m^2

Substituting into (1), we find the final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{0.0209}=2.75 rev/min

(b) 103.0 rev/min

When the rings leave the rod, the total moment of inertia is just equal to the moment of inertia of the rod, so:

I_2 = I_R = 5.57\cdot 10^{-4}kg m^2

So using again equation of conservation of the angular momentum:

L=I_1 \omega_1 = I_2 \omega_2

We find the new final angular speed:

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(1.64\cdot 10^{-3})(35.0)}{5.57\cdot 10^{-4}}=103.0 rev/min

7 0
3 years ago
According to the chart, one gram of copper and
ad-work [718]

Answer:

C) three

Explanation:

Let gram of gold required be m . Let temperature change in both be Δ t .

heat absorbed = mass x specific heat x change in temperature

for copper

heat absorbed = 1 x .385 x Δt

for gold

heat absorbed = m x .129 x Δt

So

m x .129 x Δt = 1 x .385 x Δt

m = 2.98

= 3 g approximately .

4 0
3 years ago
Kinetic Energy - What does it depend on?
Greeley [361]

Answer:

faster; more kinetic energy

Explanation:

7 0
3 years ago
A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One
Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\

To get the required work done, we will divide the mass by the speed of one knot to have:

w=\frac{7230}{0.51}\\w= 14,176.47Joules

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0
3 years ago
Other questions:
  • a 95.00 kg man on ice skates catches a moving ball at 18 m/s. the man is initially at rest. the man and ball move together after
    13·1 answer
  • if an observer is hearing a rise in pitch that means the frequency of the waves in their position are falling
    5·1 answer
  • All digits shown on the measuring device, plus one estimated digit are considered____
    5·1 answer
  • the four forces shown have the same strength. which force would be most effective in opening the door?
    9·1 answer
  • A mass is connected to a spring on a horizontal frictionless surface. The potential energy of the system is zero when the mass i
    7·1 answer
  • Pilots often take advantage of the ____,which are highly-speed winds between 7km and 16 km above earths surface.
    9·1 answer
  • A person who is heavier than the standard for the person’s height is _____
    15·2 answers
  • Help Please! a lot of points for 3 questions
    14·1 answer
  • I’m not sure what this is… help please
    6·1 answer
  • If a child of mass 60.6 kg sits in a seat, what is
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!