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3 years ago

Please help me .finish this paper

Physics

1 answer:

Evgesh-ka [11]3 years ago

7 0

Solution-1:-

$\boxed{\sf \dfrac{10\times 1000}{60\times 60}}$

Solution:-2

$\boxed{\sf Sodium\:and\:Potassium}$

Solution:-3

$\boxed{\sf 320m}$

Solution:-4

$\boxed{\sf Rough\:tiles\:are\:used\:in\:bathroom}$

Solution:-5

$\boxed{\sf Mg_3N_2}$

Solution:-6

$\boxed{\sf Grapes\:and\:Rambutan}$

Solution:-7

$\boxed{\sf {}^{}_{}N}$

Solution:-8

$\boxed{\sf Galactuse}$

Solution:-9

$\boxed{\sf Y-X}$

Solution:-10

$\boxed{\sf Cell\:wall}$

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A charged particle having mass 6.64 x 10-27 kg (that of a helium atom) moving at 8.70 x 105 m/s perpendicular to a 1.30-T magnet

Fiesta28 [93]

Answer:

the charge of the particle is 2.47 x 10⁻¹⁹ C

Explanation:

Given;

mass of the particle, m = 6.64 x 10⁻²⁷ kg

velocity of the particle, v = 8.7 x 10⁵ m/s

strength of the magnetic field, B = 1.3 T

radius of the circle, r = 18 mm = 1.8 x 10⁻³ m

The magnetic force experienced by the charge is calculated as;

F = ma = qvB

where;

q is the charge of the particle

a is the acceleration of the charge in the circular path

$a = \frac{v^2}{r} \\\\ma = qvB\\\\q = \frac{ma}{vB} \\\\q = \frac{mv^2}{rvB} = \frac{mv}{rB} \\\\q = \frac{(6.64\times 10^{-27} ) \times (8.7\times 10^5)}{(1.8\times 10^{-2}) \times (1.3)} \\\\q = 2.47 \ \times 10^{-19} \ C$

Therefore, the charge of the particle is 2.47 x 10⁻¹⁹ C

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3 years ago

Who was the first woman to compete and win a championship in the Olympics

natita [175]

Hélène de Pourtalès she was the first women

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3 years ago

The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it

Sindrei [870]

Answer with Explanation:

We are given that

Speed , $v_0$ =107257 kmph

Angular velocity, $\omega=7.292\times 10^{-5} rad/s$

Radius of earth,r= $6371 km=6371000 m$

1 km=1000 m

Linear velocity, $v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s$

Linear velocity, $v=464.57\times \frac{18}{5}=1672.46 km/h$

Velocity at point A, $v_A=-vi+v_0 j=-1672.46 i+107257j kmph$

Velocity at point B, $v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph$

Velocity at point C, $v_C=v_0j+vi=1672.46 i+107257j kmph$

Velocity at point D, $v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph$

3 0

3 years ago

A ball is dropped from rest. how fast is the ball going after 3 seconds

Airida [17]

Answer:

v = 29.4m/s

Explanation:

Since the ball is dropped at rest,

u = 0m/s

a = 9.81m/s²

Using

v = u + at

After 3 seconds,

v = 0 + (9.81)(3)

v = 29.4m/s

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3 years ago

A rope, under a tension of 221 N and fixed at both ends, oscillates in a second-harmonic standing wave pattern. The displacement

mamaluj [8]

Answer:

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Explanation:

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2 years ago

Please help me .finish this paper​

Please help me .finish this paper