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alina1380 [7]
3 years ago
14

PLEASE HELP! FOR BRAINLIEST!!!

Physics
2 answers:
Stells [14]3 years ago
7 0

Answer:

I think it's C I'm so sorry if I'm wrong.

Explanation:

SVETLANKA909090 [29]3 years ago
7 0

Answer:

<h2><em>A or</em> <em>Alex > Chai > Jimmy > Sanni</em></h2>

Explanation:

__________________________________________________________

<em>Here's why this is the answer,</em>

<em>__________________________________________________________</em>

<em>Alex's Range = 3.20 - 0.90 = 2.30</em>

<em>Sanni's Range = 1.35 - 1.30 = 0.05</em>

<em>Chai's Range = 2.10 - 1.05 = 1.05</em>

<em>Jimmy's Range = 2.20 - 1.55 = 0.65</em>

__________________________________________________________

<u>Greatest to Least</u>

<em>Alex > Chai > Jimmy > Sanni</em>

<em>Here's why this is correct.</em>

<em>Alex = 2.30</em>

<em>Chai = 1.05</em>

<em>Jimmy = 0.65</em>

<em>Sanni = 0.05</em>

<em>I do not see the answer I have on this question so I think it is A.</em>

<em>__________________________________________________________</em>

<em>Hope this helps! <3</em>

<em>__________________________________________________________</em>

<em></em>

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In a playground, there is a small merry-go-round of radius 1.20 m and mass 160 kg. Its radius of gyration is 91.0 cm. (Radius of
aksik [14]

Answer:

a) 145.6kgm^2

b) 158.4kg-m^2/s

c) 0.76rads/s

Explanation:

Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation 

(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and

(c) the angular speed of the merry-go-round and child after the child has jumped on.

a) From I = MK^2

I = (160Kg)(0.91m)^2

I = 145.6kgm^2

b) The magnitude of the angular momentum is given by:

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L = r × mc

L = (1.20m)(44.0kg)(3.0m/s)

L = 158.4kg-m^2/s

c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:

L = Iw

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7 0
3 years ago
When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5
Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C   → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C   → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C   → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C   → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first  charge , 2nd to 5 th charges can be  written as 2e,  3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of  1.6 x 10⁻¹⁹ C  exists.

3 0
3 years ago
A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat
UNO [17]
<h2>Spring constant is 14.72 N/m</h2>

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Here force is weight of mass

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Elongation, x  = 36 cm = 0.36 m

Substituting

           F = kx

           5.3 = k x 0.36

             k = 14.72 N/m

Spring constant is 14.72 N/m

6 0
3 years ago
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