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Viktor [21]
2 years ago
7

It wont let me make account i tried twice this the only way i hope i can get to you to tell you this.I also just picked a random

subject so this can send
Physics
2 answers:
Mandarinka [93]2 years ago
5 0

Answer:

if you doing it on a computer i recommend to do it better on your phone and download the app, I should let you create one.

Explanation:

Alex777 [14]2 years ago
4 0
Installing on a mobile device is a lot easier then a pc or laptop
You might be interested in
Which is NOT one of the four principals of Dalton’s atomic theory?
Gre4nikov [31]

Answer:

Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>

Explanation:

Hope it helps..

Just correct me if I'm wrong, okay?

But ur welcome!!

(;ŏ﹏ŏ)(◕ᴗ◕✿)

8 0
2 years ago
Read 2 more answers
4. Una cuerda de acero de piano mide 1.60 m de longitud y 0.20 cm de diámetro. ¿Cuál es la tensión en la cuerda si se estira 0.2
bazaltina [42]

Answer:

1030.83\ \text{N}

Explanation:

\Delta L = Cambio en la longitud de la cuerda = 0.25 cm

T = tensión en cuerda

A = Área de la cadena = \dfrac{\pi}{4}d^2

d = Diámetro de la cuerda = 0.2 cm

L = Longitud original de la cuerda = 1.6 m

El cambio de longitud de una cuerda viene dado por

\Delta L=\dfrac{TL}{AE}\\\Rightarrow T=\dfrac{\Delta LAE}{L}\\\Rightarrow T=\dfrac{0.25\times 10^{-2}\times \dfrac{\pi}{4}(0.2\times 10^{-2})^2\times 210\times 10^9}{1.6}\\\Rightarrow T=1030.84\ \text{N}

La tensión en la cuerda es 1030.84\ \text{N}.

8 0
2 years ago
Under ideal conditions (no atmospheric interference of any kind), if I hit a golf ball at an angle of 25 degrees at an initial s
g100num [7]

Answer:

The required angle is (90-25)° = 65°

Explanation:

The given motion is an example of projectile motion.

Let 'v' be the initial velocity and '∅' be the angle of projection.

Let 't' be the time taken for complete motion.

Let 'g' be the acceleration due to gravity

Taking components of velocity in horizontal(x) and vertical(y) direction.

v_{x} =  v cos(∅)

v_{y} =  v sin(∅)

We know that for a projectile motion,

t =\frac{2vsin(∅)}{g}

Since there is no force acting on the golf ball in horizonal direction.

Total distance(d) covered in horizontal direction is -

d = v_{x}×t = vcos(∅)×\frac{2vsin(∅)}{g} = \frac{v^{2}sin(2∅) }{g}.

If the golf ball has to travel the same distance 'd' for same initital velocity v = 23m/s , then the above equation should have 2 solutions of initial angle 'α' and 'β' such that -

α +β = 90° as-

d = \frac{v^{2}sin(2α) }{g} = \frac{v^{2}sin(2[90-β]) }{g} =\frac{v^{2}sin(180-2β) }{g} = \frac{v^{2}sin(2β) }{g} .

∴ For the initial angles 'α' or 'β' , total horizontal distance 'd' travelled remains the same.

∴ If α = 25° , then

     β = 90-25 = 65°

∴ The required angle is 65°.

5 0
3 years ago
Read 2 more answers
A circular 10-turn coil with a radius of 5.0 cm carries a current of 5 A. It lies in the xy plane in a uniform magnetic field =
Tju [1.3M]

Answer:

U = – 0.12J

Explanation:

Given N = 10 turns, I = 5A, r = 5×10-²m

B^ = 0.05 T iˆ+ 0.3 T kˆ

Magnitude of the magnetic field vector B = √(0.05²+0.3²) = 0.304T

Area = πr² = π(5×10-²)² = 7.85×10-³m²

Magnetic moment μ = NIA

μ = 10×5×7.85×10-³ = 0.3925Am²

U = -μ•B = –0.3925×0.304 = –0.12J

The sign is negative because the magnetic moment is aligned with the magnetic field.

3 0
2 years ago
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
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