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irina [24]
3 years ago
7

When the switch in the drawing is closed, the current in the coil increases to its equilibrium value. While the current is incre

asing there is an induced current in the metal ring. The ring is free to move. What happens to the ring? a. It does not move b. It jumps upward c. It jumps downward
Physics
1 answer:
Crazy boy [7]3 years ago
3 0

Answer:

the previous correct answer is b

Explanation:

When the circuit is closed in the system, a current is induced that follows the lenz law, which opposes the change that is occurring and therefore the coil increases and the idicidal current in the ring must reach the maximum oppositing is the current of the coil, so quiet force is repulsion

Consequently, the previous correct answer is b

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The particles that are found in the nucleus of an atom are
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Answer: Electrons are the smallest of the three particles that make up atoms. Electrons are found in shells or orbitals that surround the nucleus of an atom

Explanation: hope this helps

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3 years ago
What are atoms and molecules constantly doing?
Alexus [3.1K]

A. molecules are CONSTANTLY moving.

7 0
3 years ago
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Question 2
Delvig [45]

Answer:

Approximately 73\; {\rm N}, assuming that the acceleration of this ball is constant during the descent.

Explanation:

Assume that the acceleration of this ball, a, is constant during the entire descent.

Let x denote the displacement of this ball and let t denote the duration of the descent. The SUVAT equation x = (1/2)\, a\, t^{2} would apply.

Rearrange this equation to find an expression for the acceleration, a, of this ball:

\begin{aligned} a &= \frac{2\, x}{t^{2}}\end{aligned}.

Note that x = 11\; {\rm m} and t = 1.5\; {\rm s} in this question. Thus:

\begin{aligned} a &= \frac{2\, x}{t^{2}} \\ &= \frac{2 \times 11\; {\rm m}}{(1.5\; {\rm s})^{2}} \\ &\approx 9.78\; {\rm m \cdot s^{-2}}\end{aligned}.

Let m denote the mass of this ball. By Newton's Second Law of Motion, if the acceleration of this ball is a, the net external force on this ball would be m\, a.

Since m = 7.5\; {\rm kg} and a \approx 9.78\; {\rm m\cdot s^{-2}}, the net external force on this ball would be:

\begin{aligned} (\text{net force}) &= m\, a \\ &\approx 7.5\; {\rm kg} \times 9.78\; {\rm m\cdot s^{-2}} \\ &\approx 73\; {\rm kg \cdot m \cdot s^{-2} \\ &= 73\; {\rm N} && (1\; {\rm N} = 1\; {\rm kg \cdot m\cdot s^{-2}}) \end{aligned}.

4 0
2 years ago
A certain cloud contains 220 water droplets per cubic centimeter. If 1cubic meter = 1,000,000cubic centimeters, how many drops a
lianna [129]
Wouldn't you just have to multiply 220 by 1,000,000? That would mean there are 220,000,000 water droplets in one cubic of the cloud.
3 0
3 years ago
A child whose weight is 276 N slides down a 5.90 m playground slide that makes an angle of 34.0° with the horizontal. The coeffi
garik1379 [7]

Answer:

Explanation:

Frictional force acting on the child = μ mg cosθ

, μ is coefficient of kinetic friction ,  m is mass of child θ is inclination

work done by frictional force

μ mg cosθ x d , d is displacement on inclined plane

work done = .13 x 276 x cos34 x 5.9

= 175.5 J

This work will be converted into heat energy.

b ) Initial energy of child = mgh + 1/2 m v ² , h is height , v is initial velocity

= 276 x 5.9 sin34  + 1/2 x 276 / 9.8 x .518² [ mass m = 276 / g ]

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= 914.36 J

loss of energy due to friction = 175.5

Net energy at the bottom

= 738.86 J

If v be the velocity at the bottom

1/2 m v² = 738 .86

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v² = 52.47

v = 7.24 m /s .

4 0
3 years ago
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