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3 years ago

What is the center of the atom called?

Physics

1 answer:

11111nata11111 [884]3 years ago

8 0

The center of an atom is called the nucleus.

Hope I helped!
~ Zoe

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A cohesive force between the liquids molecules is responsible for the fluids is called

kiruha [24]

Answer:

static force

Explanation:

mark me brainliest

5 0

2 years ago

Read 2 more answers

How fast (in rpm) must a centrifuge rotate if a particle 7.3 cm from the axis of rotation is to experience an acceleration of 1.

guajiro [1.7K]

The formula we use here is:

radial acceleration = ω^2 * R 

110,000 * 9.81 m/s^2 = ω^2 * 0.073 m
ω^2 = 110,000 * 9.81 / 0.073
ω = 3844.76 rad/s

and since: ω = 2pi*f --> f = ω/(2pi)
f = 3844.76 / (2pi) = 611.91 rps = 611.91 * 60 rpm
= 36,714.77 rpm

5 0

3 years ago

A bucket of mass M (when empty) initially at rest and containing a mass of water is being pulled up a well by a rope exerting a

Naily [24]

Answer:

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

Explanation:

Given that

Constant rate of leak =R

Mass at time T ,m=RT

At any time t

The mass = Rt

So the total mass in downward direction=(M+Rt)

Now force equation

(M+Rt) a =P- (M+Rt) g

$a=\dfrac{P}{M+Rt}-g$

We know that

$a=\dfrac{dV}{dt}$

$\dfrac{dV}{dt}=\dfrac{P}{M+Rt}-g$

$\int_{0}^{V}V=\int_0^T \left(\dfrac{P}{M+Rt}-g\right)dt$

$V=\dfrac{P}{R}\ ln\dfrac{M+RT}{M}-gT$

$V=\dfrac{PT}{m}\ ln\dfrac{M+m}{M}-gT$

This is the velocity of bucket at the instance when it become empty.

6 0

3 years ago

a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(

horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.

Explanation:

We can find the velocity of the skier by energy conservation:

$E_{1} = E_{2}$

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

$mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}$ (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

$v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}$

$v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s$

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!

6 0

2 years ago

A playground merry-go-round has a radius of 4.6 m and a moment of inertia of 200 kg-m2 and turns with negligible friction about

tankabanditka [31]

Answer:

8050 J

Explanation:

Given:

r = 4.6 m

I = 200 kg m²

F = 26.0 N

t = 15.0 s

First, find the angular acceleration.

∑τ = Iα

Fr = Iα

α = Fr / I

α = (26.0 N) (4.6 m) / (200 kg m²)

α = 0.598 rad/s²

Now you can find the final angular velocity, then use that to find the rotational energy:

ω = αt

ω = (0.598 rad/s²) (15.0 s)

ω = 8.97 rad/s

W = ½ I ω²

W = ½ (200 kg m²) (8.97 rad/s)²

W = 8050 J

Or you can find the angular displacement and find the work done that way:

θ = θ₀ + ω₀ t + ½ αt²

θ = ½ (0.598 rad/s²) (15.0 s)²

θ = 67.3 rad

W = τθ

W = Frθ

W = (26.0 N) (4.6 m) (67.3 rad)

W = 8050 J

6 0

2 years ago