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3 years ago

T=2pi square root 1/g solve for g. Explanation would be really helpful.

Physics

1 answer:

Natalija [7]3 years ago

7 0

I added individual steps for clarity. Note that g must be positive if the solution is to be real.

$T=2\pi \sqrt{\frac{1}{g}}=2\pi g^{-\frac{1}{2}}\\g^{-\frac{1}{2}} = \frac{T}{2\pi}\\(g^{-\frac{1}{2}})^{-2} = (\frac{T}{2\pi})^{-2}\\g = \frac{4\pi^2}{T^2}\,\,\,, g>0}$

Let me know if you have any questions.

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) The magnitude of scalar product of two unit vectors perpendicular to each other is

fenix001 [56]

This question involves the concept of the scalar product.

The magnitude of the scalar product will be "0".

<h3>SCALAR PRODUCT</h3>

The scalar product, also known as the dot product of the two vectors is given by the following formula:

$A.B = |A||B|Cos\theta$

where,

A.B = Scalar product = ?
|A| = Mangnitude of vector A = 1 unit
|B| = Magnitude of Vector B = 1 unit
θ = Angle between vectors = 90°

Therefore,

$A.B = (1)(1)Cos90^o = (1)(1)(0)$

A.B = 0

Learn more about scalar product here:

brainly.com/question/6849226

5 0

2 years ago

0.403 L is equal to: <br> 4.03 mL <br> 40.3 mL <br> 403 mL <br> 4,030 mL

Natalija [7]

The answer is 403 ml coz when you divide 0.403 with 1000 ml you get 403 ml

3 0

3 years ago

Read 2 more answers

Helppp meeee

worty [1.4K]

T = (v-0)/a
t = (45.5)/(9.8)
= 4.64m/s.

hope this helps :)

7 0

2 years ago

n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.

9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0

3 years ago

You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what

pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector $F_{net}$ = vector Fe + vector Fn,