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Alex_Xolod [135]
3 years ago
15

HELP PLZZZZZ Consider the following data:

Physics
2 answers:
Montano1993 [528]3 years ago
8 0
It’s A I’m sorry if it’s wrong
marishachu [46]3 years ago
4 0

Answer:

I think the answer is As you increase the current the resistance increases too.

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A feather is dropped on the moon from the height of 1.4m. The acceleration of gravity on the moon is 1.67ms-1. Determine the tim
Zinaida [17]

1.3s

Explanation:

Given parameters:

Height = 1.4m

Gravity on moon = 1.67ms⁻¹

Unknown:

Time for feather to fall = ?

Solution:

To solve this problem, we are going to use one of the motion equation that relates time, gravity and height.

    H = ut + \frac{1}{2} g t^{2}

Sine the body was dropped from rest, initial velocity is zero;

 H = height

  u = initial velocity

  t = time

  g = acceleration due to gravity

since u = 0;

H = \frac{1}{2} g t^{2}

 1.4 = \frac{1}{2} x 1.67 x t²

  t = 1.3s

learn more:

Gravity brainly.com/question/10934170

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8 0
3 years ago
3 track athletes run the 400 meter dash, Jackie 51 seconds and 1500 joules
Aleksandr [31]

Answer:

ish all usual isaiah or easier fall iF.Hg. diff potty off radial ish usually appraisal

7 0
3 years ago
The energy of the electron in Hydrogen atom can be shown to be given by En = -13.6/n^2 eV, where n represents the principal quan
Tpy6a [65]

Answer:

This represents radiation in ultra-violet region .

Explanation:

Energy of the orbit where n = 3 is given as follows

E_3 =- \frac{13.6 }{3^2}

E_3 =- \frac{13.6 }{9} = -1.511 eV

Energy of the orbit where n = 1 is given as follows

E_1 =- \frac{13.6 }{1^2}

E_1 =- \frac{13.6 }{1} = 13.6 eV

Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6

= 12.089 eV.

The wavelength of light having this energy in nm is given by the expression as follows

Wavelength in nm = 1244 / energy in eV

= 1244 / 12.089

= 102.90 nm

This represents radiation in ultra-violet region .

8 0
3 years ago
1 lb equals how many grams
shusha [124]

Answer:

1 lb. is 453.592 grams

Explanation:

1lb is 453.592 grams

8 0
3 years ago
Read 2 more answers
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
slega [8]
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
where dm =M(b*dr)/(ab)
Now we find the moment of inertia by integrating from -a/2 to a/2
The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



4 0
3 years ago
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