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
polet [3.4K]
3 years ago
9

an ampere is the si unit of which of the following? electric power, electric current, resistance, electric potential difference

Physics
1 answer:
Ivan3 years ago
3 0
The Ampere is the SI unit of electric current. This unit is often shortened to "amp" and is abbreviated "A" when placed at the end of a measurement. The name Ampere comes from <span>André-Marie </span><span>Ampère, and is equal to the flow of one coulomb per second.</span>
You might be interested in
In Which senecio is an animal doing work
Arisa [49]

Answer:

A horse pulls a wagon along a road

4 0
3 years ago
In each of two coils the rate of change of the magnetic flux in a single loop is the same. The emf induced in coil 1, which has
irinina [24]

Answer:

 Coil 2 have  235  loops

Explanation:

Given  

The number of loops in coil 1 is n ₁= 159

The emf induced in coil 1 is  ε ₁ = 2.78 V

The emf induced in coil 2 is  ε ₂ = 4.11 V

Let

n ₂  is the number of loops in coil 2.

Given, the emf in a single loop in two coils are same. That is,

ϕ ₁/n ₁= ϕ ₂ n ₂⟹ 2.78/159 = 4.11/ n ₂

n₂=\frac{159 * 4.11}{2.78}

n₂=235

Therefore, the coil 2 has  n ₂= 235  loops.

4 0
3 years ago
Read 2 more answers
the chef was careless and put the aluminum cookie sheet directly on the hot stove, which melted 0.05 kg of the aluminum. how muc
sergejj [24]

Heat required to melt 0.05 kg of aluminum is 28.7 kJ.

<h3>What is the energy required to melt 0.05 kg of aluminum?</h3>

The heat energy required to melt 0.05 kg of aluminum is obtained from the heat capacity of aluminum and the melting point of aluminum.

The formula to be used is given below:

  • Heat required = mass * heat capacity * temperature change

Assuming the aluminum sheet was at room temperature initially.;

Room temperature = 25 °C

Melting point of aluminum = 660.3 °C

Temperature difference = (660.3 - 25) = 635.3 903

Heat capacity of aluminum = 903 J/kg/903

Heat required = 0.05 * 903 * 635.3

Heat required = 28.7 kJ

In conclusion, the heat required is obtained from the heat change aluminum and the mass of the aluminum melted.

Learn more about heat capacity at: brainly.com/question/21406849

#SPJ1

3 0
2 years ago
What phase difference between two otherwise identical harmonic waves, moving in the same direction along a stretched string, wil
lora16 [44]

Answer:

\theta=145

Explanation:

The amplitude of he combined wave is:

B=2Acos(\theta/2)\\

A, is the amplitude from the identical harmonic waves

B, is the amplitude of the resultant wave

θ, is the phase, between the waves

The amplitude of the combined wave must be 0.6A:

0.6A=2Acos(\theta/2)\\ cos(\theta/2)=0.3\\\theta/2=72.5\\\theta=145

5 0
3 years ago
You have a summer job at a company that developed systems to safely lower large loads down ramps. Your team is investigating a m
Fofino [41]

Answer:

Note that the emf induced is

emf = B d v cos (A)

---> v = emf / [B d cos (A)]

where

B = magnetic field

d = distance of two rails

v = constant speed

A = angle of rails with respect to the horizontal

Also, note that

I = emf/R

where R = resistance of the bar

Thus,

I = B d v cos (A) / R

Thus, the bar experiences a magnetic force of

F(B) = B I d = B^2 d^2 v cos (A) / R, horizontally, up the incline.

Thus, the component of this parallel to the incline is

F(B //) = F(B) cos(A) = B I d = B^2 d^2 v cos^2 (A) / R

As this is equal to the component of the weight parallel to the incline,

B^2 d^2 v cos^2 (A) / R = m g sin (A)

where m = the mass of the bar.

Solving for v,

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]   [ANSWER, the constant speed, PART A]

******************************

v = [R m g sin (A) / B^2 d^2 cos^2 (A)]

Plugging in the units,

m/s = [ [ohm * kg * m/s^2] / [T^2 m^2] ]

Note that T = kg / (s * C), and ohm = J * s/C^2

Thus,

m/s = [ [J * s/C^2 * kg * m/s^2] / [(kg / (s * C))^2 m^2] ]

= [ [J * s/C^2 * kg * m/s^2] / [(kg^2 m^2) / (s^2 C^2)]

As J = kg*m^2/s^2, cancelling C^2,,

= [ [kg*m^2/s^2 * s * kg * m/s^2] / [(kg^2 m^2) / (s^2)]

Cancelling kg^2,

= [ [m^2/s^2 * s * m/s^2] / [(m^2) / (s^2)]

Cancelling m^2/s^2,

= [s * m/s^2]

Cancelling s,

=m/s   [DONE! WE SHOWED THE UNITS ARE CORRECT! ]

8 0
3 years ago
Other questions:
  • NEED ANSWER ASAP PLEASE!!!
    8·2 answers
  • Which is NOT a property of magnetic field lines?
    6·2 answers
  • A probe from earth, with a mass of 106 kg, is in orbit around planet Pud in another solar system. Pud has a radius of 106 m and
    14·1 answer
  • If you jump off a 7 m diving platform, how long will it take you to hit the
    6·1 answer
  • A 1.72-kg block of soft wood is suspended by two strings from the ceiling. The wood is free to rotate in pendulum-like fashion w
    8·1 answer
  • Match each measurement tool with what it measures.
    14·1 answer
  • Missy's favorite ride at the Topsfield Fair is the rotor, which has a radius of 4.0 m. The ride
    5·1 answer
  • Saturn isn’t the only planet with rings, but they are the most beautiful ones. <br> Fact or Opinion?
    8·1 answer
  • General Circulation Models (GCM) :_________
    13·1 answer
  • 7. The Cuyahoga River in Ohio has caught fire
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!