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1 month ago

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A ball collides elastically with an immovable wall fixed to the earth’s surface. Which statement is false? 1. The ball's speed i

s unchanged by the collision.2. The ball does no net work on the wall.3. The ball's kinetic energy is unchanged by the collision.4. the wall doos no net work on the ball.5. the ball's size and shape are unchanged by the collision.6. All of these statements are true7. The ball's momentum is unchanged by the collision

1 answer:

Maru [3.3K]1 month ago

4 0

Answer:

Statements 4, 6 & 7 are incorrect.

Explanation:

In any elastic collision, the overall momentum vector sum of the system remains zero.

In this scenario, an elastic collision occurs between the ball and a stationary wall. The ball's velocity will consistently revert after the impact, leading to a change in direction of momentum.

The initial momentum of the ball is represented as:

$p=m.v$

where:

m = mass of the ball

v = initial velocity of the body

post-collision for the elastic interaction:

$p=m.(-v)$

Here, the momentum changes solely in direction, thus contradicting statement 7.

During the impact, both the ball and the wall exert forces on each other that are equal and opposite. The wall remains motionless, while the ball is influenced by the wall's reaction force, performing work on it, which contradicts statement 4.

Given that this collision is elastic, the ball's form and dimensions do not alter.

The previous points clearly indicate that not all provided statements hold true, thus violating statement 6.

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Two large insulating parallel plates carry charge of equal magnitude, one positive and the other negative, that is distributed u

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Answer:

The correct choice is C: points 1, 4, and 5 are equal, followed by 2 and 3 being equal.

Explanation:

Here’s the breakdown:

The electric field from the positive sheets E₁ = б/2E₀

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At points 1, 4, and 5, the electric fields created by the sheets oppose each other.

At point 1, the total field is calculated as -E₁ + E₂ = 0.

Similarly, at point A, the total field results in -E₁ - E₂ = 0.

However, at any point in between the plates, the electric field is directed consistently in one way.

At points 2 and 3, the field is directed to the right.

Thus, we have:

E net = E₁ + E₂

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=б/E₀

Note: Please refer to the attached document for the full question accompanying this solution.

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It's a snowy day and you're pulling a friend along a level road on a sled. You've both been taking physics, so she asks what you

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Answer:

0.0984

Explanation:

The first diagram below illustrates a free body diagram that will aid in resolving this problem.

According to the diagram, the force's horizontal component can be expressed as:

$F_X = F_{cos \ \theta}$

Substituting 42° for θ and 87.0° for $F$

$F_X =87.0 \ N \ *cos \ 42 ^\circ$

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Meanwhile, the vertical component is:

$F_Y = Fsin \ \theta$

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$F_Y =87.0 \ N \ *sin \ 42 ^\circ$

$F_Y =58.21 \ N$

In resolving the vector, let A denote the components in mutually perpendicular directions.

The magnitudes of both components are illustrated in the second diagram provided and can be represented as A cos θ and A sin θ

The frictional force can be expressed as:

$f = \mu \ N$

Where;

$\mu$ is the coefficient of friction

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Also, the normal reaction (N) is calculated as mg - F sin θ

Substituting $F_Y \ for \ F_{sin \ \theta}$ . Normal reaction becomes:

$N = mg \ - \ F_Y$

By balancing the forces, the horizontal component of the force equals the frictional force.

The horizontal component is described as follows:

$F_X = \mu \ ( mg - \ F_Y)$

Rearranging the equation above to isolate $\mu$ leads to:

$\mu \ = \ \frac{F_X}{mg - F_Y}$

Substituting in the following values:

$F_X \ = \ 64.65 \ N$

m = 73 kg

g = 9.8 m/s²

$F_Y = \ 58.21 N$

Thus:

$\mu \ = \ \frac{64.65 N}{(73.0 kg)(9.8m/s^2) - (58.21 \ N)}$

$\mu = 0.0984$

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Answer:

The equivalent distance in kilometers is 4012 × $10^{-6}$ km.

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