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Tasya [4]
4 years ago
7

Consider what happens at the moment when the block leaves the surface of the globe. Which of the following statements are correc

t?
- The net acceleration of the block is directed straight down.
- The component of the force of gravity toward the center of the globe is equal to the magnitude of the normal force.
- The force of gravity is the only force acting on the block.
Physics
2 answers:
Musya8 [376]4 years ago
8 0

Answer:

Option (3) is correct.

Explanation:

when the block leaves the surface, the normal reaction acting on the block is zero and the acceleration is acceleration due to gravity which acts towards the centre of globe.

So, only the force of gravity acts on the block which acts towrads the centre of the globe.

Option (3) is correct.

erma4kov [3.2K]4 years ago
6 0

Answer:

Statement 1 and 3 are correct.

Explanation:

1. The mass moves  downward, so the net acceleration of the block is straight downward.

2.The mass is sliding through the globe, so only the force of gravity is acting on the mass which pulls it in downward direction. The force of gravity has two components [mg sin∅] and [mg cos∅].

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Light shined through a single slit will produce a diffraction pattern. Green light (565 nm) is shined on a slit with width 0.210
kondor19780726 [428]

Answer:(a)9.685 mm

(b)4.184 mm

Explanation:

Given

Wavelength of light (\lambda )=565nm \approx 565\times 10^{-9}m

Width of slit(b)=0.210

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=\frac{2\lambda L}{b}

=\frac{2\times 565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

=9.685 mm

(b)Width of the first order bright fringe

Y_1=\frac{\lambda \times L}{b}

Y_1=\frac{565\times 10^{-9}\times 1.8}{0.210\times 10^{-3}}

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3 years ago
Two identical balls are thrown vertically upward. the second ball is thrown with an initial speed that is twice that of the firs
Temka [501]
The motion of the ball on the vertical axis is an accelerated motion, with acceleration 
a=g=-9.81 m/s^2
The following relationship holds for an uniformly accelerated motion:
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If we take the moment the ball reaches the maximum height (let's call this height h), then at this point of the motion the vertical velocity is zero:
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So we can rewrite the equation as
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h= \frac{v_i^2}{19.62} (1)

Now let's assume that v_i is the initial velocity of the first ball. The second ball has an initial velocity that is twice the one of the first ball: 2v_i. So the maximum height of the second ball is
h= \frac{(2v_i)^2}{19.62}= \frac{4v_i^2}{19.62} (2)

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Answer:

<h2>A. Nearsightedness</h2>

Explanation:

A nearsightedness is an eye defect that occurs when someone is only able to see close ranged object but not far distance object. According to the question, if the length of my eye decreases slightly as I age, this means there is a possibility that I will find it difficult to view a far distance object as I age.

At 70, once my eyes had decreased slightly in length, this means I will only be able to see close ranged object but not far distant object, showing that I am now suffering from nearsightedness according to its definition above.

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