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Andru [333]
4 years ago
11

Exercise combined with a blank can help control weight

Physics
1 answer:
SIZIF [17.4K]4 years ago
8 0

Answer:

Exercise combined with a diet can help manage weight, or you can contact a doctor

Explanation:

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two coils close to each other have a mutual inductance of 32 mh. if the current in one coil decays according to , where i0
Harlamova29_29 [7]

The EMF induced in the second coil is 43 Volts.

Michael Faraday was the first to discover electromagnetic induction back in the 1830s. Faraday discovered that moving a permanent magnet in and out of a coil or a single loop of wire caused an electromotive force, or EMF—otherwise known as a voltage—to be produced.

Changing magnetic flux results in varied currents flowing through the coil, which in turn generates its own magnetic field. This self-induced EMF opposes the change that is creating it, and the stronger the opposing EMF is, the faster the rate at which the current is changing. According to Lenz's law, this self-induced EMF will oppose the change in current in the coil, and because of its orientation, it is typically referred to as a back-EMF.

To learn more about EMF please visit-

brainly.com/question/15121836

#SPJ4

3 0
2 years ago
Suppose you design an apparatus in which a uniformly charged disk of radius R is to produce an electric field. The field magnitu
ANEK [815]

Answer:

The electric field will be decreased by 29%

Explanation:

The distance between point P from the distance z = 2.0 R

Inner radius = R/2

Outer raidus = R

Thus;

The electrical field due to disk is:

\hat {K_a} = \dfrac{\sigma}{2 \varepsilon _o} \Big( 1 - \dfrac{z}{\sqrt{z^2+R_i^2}} \Big))

\implies \dfrac{\sigma}{2 \vaepsilon _o} \Big ( 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0\ R)^2+(R)^2}} \Big)

Similarly;

\hat {K_b} = \hat {k_a} - \dfrac{\sigma}{2 \varepsilon_o} \Big( 1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ r)^2 + (\dfrac{R}{2}^2)}}\Big)

However; the relative difference is: \dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{E_a -E_a + \dfrac{\sigma}{2 \varepsilon_o  \Big[1 - \dfrac{2.0 \ R}{\sqrt{(2.0 \ R)^2 + (\dfrac{R}{2})^2}} \Big] } } { \dfrac{\sigma}{2 \varepsilon_o \Big [ 1 - \dfrac{2.0 \ R}{\sqrt{ (2.0 \ R)^2 + (R)^2}} \Big] }}

\dfrac{\hat {k_a} - \hat {k_b}}{\hat {k_a} }= \dfrac{1 - \dfrac{2.0}{\sqrt{(2.0)^2 + \dfrac{1}{4}}} }{1 - \dfrac{2.0 }{\sqrt{(2.0)^2 + 1}}}

= 0.2828 \\ \\ \mathbf{\simeq  29\%}

3 0
3 years ago
What happens to the volume the gas occupies when the pressure on a gas increases?
taurus [48]
Each time they collide with the walls they exert a force on them. More collisions mean more force, so the pressure will increase. When the volume decreases, the pressure increases. This shows that the pressure of a gas is inversely proportional to its volume.
8 0
3 years ago
Read 2 more answers
Is it likely that mesosaurus swam back and forth across the Atlantic Ocean?
Natalija [7]
No, Their arms and back legs are too short, but the mesosaurus was also a coastal animal! It's very unlikely.
3 0
3 years ago
PLZ HELP ME
bonufazy [111]

Answer:

1. The stone will strike the ground 49.46 m from the base of the cliff

2. A) Approximately 0.542 seconds

B)  Approximately 3.69 m/s

3. A) The time the ball spends in the air is approximately 4.0775 s

B) The horizontal range is approximately  141.25 m.

Explanation:

1. The time it takes the stone to land is given by the equation, t = √(h/(1/2 × g)

∴ t = √(30/(1/2 × 9.81)) ≈ 2.473 seconds

The horizontal distance covered by the stone in that time = 20 × 2.473 ≈ 49.46 m

The stone will strike the ground 49.46 m from the base of the cliff

2. A) The time the ball spends in the air = t = √(h/(1/2 × g)

∴ t = √(1.44/(1/2 × 9.81)) ≈ 0.542 seconds

B) The initial horizontal velocity, u = Horizontal distance/(Time) = 2/0.542 ≈ 3.69 m/s

The initial horizontal velocity ≈ 3.69 m/s

3. A) The time the ball spends in the air is given by the following equation;

t = 2 × u × sin(θ)/g = 2 × 40 × sin(30)/9.81 ≈ 4.0775 s

t ≈ 4.0775 s

B) The horizontal range, R, of the  ball is given by the equation for the range of a projectile as follows;

Range, R = \dfrac{u^2 \times sin (2 \cdot \theta) }{g}

Substituting the known values, gives;

Range, R = \dfrac{40^2 \times sin (2 \times 30^{\circ}) }{9.81} \approx 141.25 \ m

The horizontal range ≈ 141.25 m.

4 0
3 years ago
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