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kolbaska11 [484]
3 years ago
12

A closed loop conductor that forms a circle with a radius of 2.0 m is located in a uniform but changing magnetic field. If the m

aximum emf Induced in the loop is 5.0 V what is the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies
Physics
1 answer:
Vladimir79 [104]3 years ago
5 0

Answer:

The maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

Explanation:

Given;

radius of the circular loop, r = 2.0 m

maximum induced emf, E = 5.0 V

The emf induced in a magnetic field is given as;

emf = \frac{d\phi}{dt} \\\\\phi = AB\\\\emf = A\frac{dB}{dt} \\\\\frac{dB}{dt}  = \frac{emf}{A} \\\\where;\\A \ is \ the \ area \ circular \ l00p = \pi r^2 = \pi (2)^2 = 4\pi \ m^2\\\\\frac{dB}{dt}  = \frac{5}{4\pi} \\\\\frac{dB}{dt}  = 0.398 \ T/s

Therefore, the maximum rate at which the magnetic field strength is changing if the magnetic field is oriented perpendicular to the plane in which the loop lies 0.398 T/s.

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You fill two balloons with gas, one with hydrogen and one with carbon dioxide. You hold a match to each balloon. The hydrogen ba
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Answer:E. Hydrogen was able to participate in an exergonic reaction and carbon dioxide couldn't

Explanation:

An exergonic reaction releases energy to the environment. The combustion of hydrogen contained in the balloon is a chemical reaction. The reaction can take place because hydrogen combines with oxygen in air, that is, the gas is combustible. CO2 does not support combustion, it does not combine with oxygen in air and it is also denser than air, hence does not participate in the exergonic reaction.

4 0
3 years ago
An object initially at rest experiences an acceleration of 1.90 ­m/s² for 6.60 s then travels at that constant velocity for anot
borishaifa [10]

Answer:

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

Explanation:

Hi there!

The average velocity is calculated as the displacement of the object divided by the time it takes the object to do that displacement.

The displacement is calculated as the distance between the final position of the object and the initial position. <u>In this problem</u>, the displacement is equal to the traveled distance because the object travels only in one direction:

a.v = Δx/t

Where:

a.v = average velocity.

Δx = displacement = final position - initial position

t = time

So, let's find the distance traveled while the object was accelerating. For that, we will use this equation:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the object at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

In this case, since the object is initially at rest, v0 = 0. If we place the origin of the frame of reference at the point where the object starts moving, then x0 = 0. So, the equation of the position of the object after a time t will be:

x = 1/2 · a · t²

x = 1/2 · 1.90 m/s² · (6.60 s)²

x = 41.4 m

The object traveled 41.4 m during the first 6.60 s.

Now, let's find the rest of the traveled distance.

When the velocity is constant, a = 0. Then, the equation of position will be:

x = x0 + v · t

Let's place now the origin of the frame of reference at the point where the object starts traveling at constant velocity so that x0 = 0:

x = v · t

The velocity reached by the object during the acceleration phase is calculated as follows:

v = v0 + a · t   (v0 = 0 because the object started from rest)

v = 1.90 m/s² · 6.60 s

v = 12.5 m/s

Then, the distance traveled by the object at a constant velocity will be:

x = 12.5 m/s · 8.50 s

x = 106 m

The total traveled distance in 15.1 s is (106 m + 41.4 m) 147 m.

Then the displacement will be:

Δx = final position - initial position

Δx = 147 m - 0 = 147 m

and the average velocity will be:

a.v = Δx/t

a.v = 147 m / 15.1 s

a.v = 9.74 m/s

The magnitude of the object's average velocity is 9.74 m/s (9.80 m/s without any intermediate rounding).

8 0
3 years ago
A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
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Answer:

a. A = 0.735 m

b. T = 0.73 s

c. ΔE = 120 J decrease

d. The missing energy has turned into interned energy in the completely inelastic collision

Explanation:

a.

4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax

vmax = 4.0 m/s

¹/₂ * m * v²max = ¹/₂ * k * A²

m * v² = k * A²  ⇒ 10 kg * 4 m/s = 100 N/m * A²

A = √1.6 m ² = 1.26 m

At = 2.0 m - 1.26 m = 0.735 m

b.

T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s

T = 2π *√ 10 / 100 *s² = 1.99 s

T = 1.99 s -1.26 s = 0.73 s

c.

E = ¹/₂ * m * v²max =

E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J

E₂ = ¹/₂ * 10 * 4² = 80 J

200 J - 80 J  = 120 J decrease

d.

The missing energy has turned into interned energy in the completely inelastic collision

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