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3 years ago

Which particles are heavy and which are light

Physics

2 answers:

VashaNatasha [74]3 years ago

3 0

Answer:

Protons and neutrons are heavy, Electrons are extremely light

Explanation:

Protons and neutrons are heavier than electrons and reside in the nucleus at the center of the atom. Electrons are extremely lightweight and exist in a cloud orbiting the nucleus.

aleksklad [387]3 years ago

3 0

Protons- heavy
Electrons- light

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A piece of wood 400N in weight and 50cm X 30cm X 20cm in size lies on 50cm X 20cm face. Calculate the pressure exerted by it.

Serga [27]

Answer:

P = 4000 [Pa]

Explanation:

Pressure is defined as the relationship between Force and the area where the body rests.

The support area is equal to:

$A=50*20=1000[cm^{2} ]$

But we must convert from square centimeters to square meters.

$1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]$

And the pressure is:

$P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]$

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2 years ago

A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.2 rad/s in 3.07 s. (a) f

Hitman42 [59]

(a) The angular acceleration of the wheel is given by
$\alpha = \frac{\omega_f - \omega_i }{t}$
where $\omega_i$ and $\omega_f$ are the initial and final angular speed of the wheel, and t the time.

In our problem, the initial angular speed is zero (the wheel starts from rest), so the angular acceleration is
$\alpha = \frac{(11.2 rad/s) - 0}{3.07 s} =3.65 rad/s^2$

(b) The wheel is moving by uniformly rotational accelerated motion, so the angle it covered after a time t is given by
$\theta (t) = \omega_i t + \frac{1}{2} \alpha t^2$
where $\omega_i = 0$ is the initial angular speed. So, the angle covered after a time t=3.07 s is
$\theta= \frac{1}{2} \alpha t^2 = \frac{1}{2}(3.65 rad/s^2)(3.07 s)^2 = 17.2 rad$

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2 years ago

You have a horizontal grindstone (a disk) that is 86 kg, has a 0.33 m radius, is turning at 92 rpm (in the positive direction),

Rudiy27

Answer:

a) α = 0.338 rad / s² b) θ = 21.9 rev

Explanation:

a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque

τ = I α

fr r = I α

Now we write the translational Newton equation in the radial direction

N- F = 0

N = F

The friction force equation is

fr = μ N

fr = μ F

The moment of inertia of a saying is

I = ½ m r²

Let's replace in the torque equation

(μ F) r = (½ m r²) α

α = 2 μ F / (m r)

α = 2 0.2 24 / (86 0.33)

α = 0.338 rad / s²

b) let's use the relationship of rotational kinematics

w² = w₀² - 2 α θ

0 = w₀² - 2 α θ

θ = w₀² / 2 α

Let's reduce the angular velocity

w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s

θ = 9.634 2 / (2 0.338)

θ = 137.3 rad

Let's reduce radians to revolutions

θ = 137.3 rad (1 rev / 2π rad)

θ = 21.9 rev

7 0

3 years ago