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2 years ago

Plzzzzz helppppp plzzzzzzzzzzzzzzzzzzzzzzz

Physics

2 answers:

QveST [7]2 years ago

8 0

Answer:

A. Refraction

Explanation:

Helen [10]2 years ago

6 0

Answer:

Explanation:

ap3x

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What is the mass number of an atom with 3 protons, 4 neutrons, and 3 electrons

snow_tiger [21]

Mass number = no. of protons + no. of neutrons

so, it would be, 3+4 = 7

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3 years ago

A hurricanes energy comes from?

Shalnov [3]

Answer:

a or b

Explanation:

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2 years ago

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Consider an electron that is 10-10 m from an alpha particle (9 = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electri

Studentka2010 [4]

Answer:

a) $E=2.88*10^{11}N/C$

b) $E=1.44*10^{11}N/C$

c) $F=4.61*10^{-8}N$

Explanation:

We use the definition of a electric field produced by a point charge:

$E=k*q/r^2$

a)Electric Field due to the alpha particle:

$E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(10^{-10})^2=2.88*10^{11}N/C$

b)Electric Field due to electron:

$E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(10^{-10})^2=1.44*10^{11}N/C$

c)Electric Force on the alpha particle, on the electron:

The alpha particle and electron feel the same force but with opposite direction:

$F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(10^{-10})^2=4.61*10^{-8}N$

4 0

3 years ago

PL-1) A spring that hangs vertically is 25 cm long when no weight is attached to its lower end. Steve adds 250 g of mass to the

Anuta_ua [19.1K]

Answer:

20.42 N/m

Explanation:

From hook's law,

F = ke ......................... Equation 1

Where F = Force applied to the spring., k = spring constant, e = extension.

Make k the subject of the equation,

k = F/e ................. Equation 2

Note: The force on the spring is equal to the weight of the mass hung on it.

F = W = mg.

k = mg/e................ Equation 3

Given: m = 250 g = 0.25 kg, e = 37-25 = 12 cm = 0.12 m.

Constant: g = 9.8 m/s²

Substitute into equation 3

k = (0.25×9.8)/0.12

k = 20.42 N/m.

Hence the spring constant = 20.42 N/m

7 0

3 years ago

A 2.0 kg particle moves in a circle of radius 3.1 m. As you look down on the plane of its orbit, the particle is initially movin

Ghella [55]

Answer

given,

L(t) = 10 - 3.5 t

mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

τ = $\dfrac{dL}{dt}$

τ = $\dfrac{d}{dt}(10 - 3.5 t)$

τ = -3.5 N.m

Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.

L decreases the angular acceleration upward. so, net torque is upward.

b) Moment of inertia of the particle

I = m R^2

I = 2 x 3.1²

I = 19.22 kg.m²

L = I ω

ω = $\dfrac{L}{I}$

ω = $\dfrac{10 - 3.5 t}{19.22}$

ω = $0.520 - 0.182 t$

A = 0.52 rad/s B = -0.182 rad/s²

5 0

3 years ago

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