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
abruzzese [7]
2 years ago
11

When is a model used in science?

Physics
1 answer:
Mamont248 [21]2 years ago
8 0
C. Both of the above
You might be interested in
The faintest sound an ear can hear (20 micro-pascals) is roughly ______ times smaller than atmospheric pressure at sea level.
gtnhenbr [62]

Answer: 5.06\times10^{9}

At sea level, there is one standard atmospheric pressure which is equal to 101.325 kilopascals.

The pressure of faintest sound that a human ear can hear is 20 micro-pascals.

taking the ratio of two:

\frac {101.325 kilopascals. }{20 micropascals} = \frac {101.325 \times 10^{3}Pa}{20 \times 10^{-6}Pa}

=5.06\times10^{9}

Hence, the atmospheric pressure at sea level is 5.06\times10^{9} times greater than the faintest sound that a human ear can hear.



6 0
2 years ago
29. If you traveled to Mars, your mass would be<br> than your mass on Earth. *
marin [14]
Yes it would be the same your weight would change:)
8 0
3 years ago
Read 2 more answers
While running these tests, crall notices a similarity in the velocity measured at the ground independent of whether the tennis b
lara [203]

At the ground the ball will always have velocity along the direction of gravity. If upward motion is taken positive it will always have negative velocity at the ground because, if the ball was given an initial upward velocity then gravity will decelerate it and bring it down with a negative final velocity. If the ball is given an initial downward velocity then the ball will be further accelerated by gravity in the downward direction only, again maintaining negative direction. The magnitude however in both cases will be different. the final velocity at the ground will have higher magnitude in case of elevator moving downwards.

6 0
3 years ago
Read 2 more answers
Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 60.7 A. The resistanc
lisov135 [29]

Answer:

a) The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b)The mass per unit length \lambdaλ for a copper cable is 0.757kg/m

c)The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d)The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

Explanation:

The expression for electric field of conductor is,

E =  \frac{V}{L}

The general equation of voltage is,

V = iR

The expression for current density in term of electric field is,

J = \frac{E}{p}

Substitute (V/L)  for E in the above equation of current density.

J = \frac{V}{pL} ------(1)

Substitute iR for V in equation (1)

J = \frac{iR}{pL} ------(2)

Substitute 1.69 × 10⁸ Ω .m for p

50A for i

0.200Ω.km⁻¹ for (R/L) in eqn (2)

J = \frac{(50) (0.200\times 10^-^3) }{1.69 \times 10^-^8 } \\\\= 5.91 \times 10^5A.m^-^2

The magnitude JJ of the current density for a copper cable is 5.91 × 10⁵A.m⁻²

b) The expression for resistivity of the conductor is,

p = \frac{RA}{L}

A = \frac{pL}{R}

The expression for mass density of copper is,

m = dV

where, V is the density of the copper.

Substitute AL for V in equation of the mass density of copper.

m=d(AL)

m/L = dA

λ is use for (m/L)

substitute,

pL/R for A  and λ is use for (m/L) in the eqn above

\lambda = d\frac{p}{\frac{R}{L} } ------(3)

Substitute 0.200Ω.km⁻¹ for (R/L)

8960kgm⁻³  for d and 1.69 × 10⁸ Ω .m

\lambda = (8960) \frac{(1.69 \times 10^-^8 }{0.200\times 10^-^3} \\\\= 0.757kg.m^-^1

c) Using the equation (2) current density for aluminum cable is,

J = \frac{iR}{pL}

p is the resistivity of the aluminum cable.

Substitute 2.82 × 10⁻⁸Ω.m for p ,

50A for i and 0.200Ω.km⁻¹ for (R/L)

J = \frac{(50)(0.200\times10^-^3) }{2.89\times 10^-^8} \\\\= 3.5 \times10^5A/m^2

The magnitude J of the current density for an aluminum cable is 3.5 × 10⁵A/m²

d) Using the equation (3) mass per unit length for aluminum cable is,

\lambda = d\frac{p}{\frac{R}{L} }

p is the resistivity and is the density of the aluminum cable.

Substitute 0.200Ω.km⁻¹ for (R/L), 2700 for d and 2.82 × 10⁻⁸Ω.m for p

\lambda = (2700) \frac{(2.82 \times 10^-^8) }{(0.200 \times 10^-^3) } \\\\= 0.380kg/m

The mass per unit length \lambdaλ for an aluminum cable is 0.380kg/m

7 0
3 years ago
Read 2 more answers
A homeowner wants to reduce the environmental damage of the energy source used to heat her home.
liberstina [14]

Answer:

wind

Explanation:

wind is the only one of those onto that list that is a renewable source

4 0
3 years ago
Read 2 more answers
Other questions:
  • Plyometric helps strengthen your bones true or false
    11·1 answer
  • Why is energy important to motion?
    5·2 answers
  • Is a roller coaster moving downhill is a potential or kinetic energy?
    7·1 answer
  • Please help
    15·2 answers
  • What is the French word for shepherds crook
    10·2 answers
  • By exerting a force on the astronaut, the vehicle in which they orbit experiences an equal and opposite force. If the mass of th
    8·1 answer
  • 1. If point Q is reflected across x = 1, what are the coordinates of its reflection image?
    12·2 answers
  • Flow of electrical current in a wire is analogous to the flow of water in a pipe. True False
    8·2 answers
  • A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?
    6·1 answer
  • If the displacement of a horizontal mass-spring system was doubled, the elastic potential energy in the system would change by a
    8·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!