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Artemon [7]
3 years ago
9

Please help answer much as possible there's 4 questions

Physics
2 answers:
frozen [14]3 years ago
8 0
6.A
7. A
9.B
That is the answer btw
Not sure tho sorry if wrong!!!
katovenus [111]3 years ago
3 0

Answer:

6. A

7. A

9. B

hope dis helps ^-^

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A child goes down the slide,starting from rest. If the length of the slide is 2m and it takes the child 3 seconds to go down the
lapo4ka [179]

Answer:

0.44 m/s^2

Explanation:

We have the following data:

- distance covered by the child: d = 2 m (length of the slide)

- time taken to cover this distance: t = 3 s

- initial velocity of the child: 0 m/s (he starts from rest)

So we can find the acceleration by using the equation:

d=ut+\frac{1}{2}at^2

Where a is the acceleration.

Substituting the values and solving for a,

a=\frac{2d}{t^2}=\frac{2(2)}{3^2}=0.44 m/s^2

3 0
3 years ago
Problem #2: An apple is thrown upward with an initial velocity of +24.0 m/s. a. Sketch the apple's trip and label what you know.
bogdanovich [222]

Answer:

The answer is below

Explanation:

a) The initial velocity (u) = 24 m/s

We can solve this problem using the formula:

v² = u² - 2gh

where v = final velocity, g= acceleration due to gravity = 9.8 m/s², h = height.

At maximum height, the final velocity = 0 m/s

v² = u² - 2gh

0² = 24² - 2(9.8)h

2(9.8)h = 24²

2(9.8)h = 576

19.6h = 576

h = 29.4 m

b) The time taken to reach the maximum height is given as:

v = u - gt

0 = 24 - 9.8t

9.8t = 24

t = 2.45 s

The total time needed for the apple to return to its original position = 2t = 2 * 2.45 = 4.9 s

4 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
Free tingsssssssssss
il63 [147K]

Explanation:

yayyy thx u soo muchihuhuhu

5 0
2 years ago
Read 2 more answers
Have an electrical charge of +1, while
Scrat [10]

Answer:

B

Explanation:

Protons have a positive electrical charge of +1,

Electrons have a negative charge of -1,

Neutrons have a neutral charge of about 0.

4 0
3 years ago
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