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BARSIC [14]
3 years ago
15

Find the potential energy of a 2 kg ball 15 m in the air.

Physics
1 answer:
mart [117]3 years ago
8 0

Answer:

294.3 Joules

Explanation:

2kg*9.81m/s^2*Δ15=294.3J

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The mass number of a nucleus (except Hydrogen) is:
netineya [11]
The mass number of a nucleus (except Hydrogen) is: number of protons + number of neutrons.

A=Z+N

A=mass number=protons + neutrons.
Z=atomic number=number of protons.
N=number of neutrons. 

In the case of Hydrogen it depends of isotope of Hydrogen . 
the hydrogen has three isotopes, 
protium : A=Z, because N=0
deuterium:  A=Z+N;   N=1
 tritium: A=Z+N;    N=2 
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If an object has a constant negative acceleration, what can we assume about the forces on it?
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Its slowing down I believe
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A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr
nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

 a= μ g                          ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g  =ω²X

We know for spring mass system natural frequency given as

\omega=\sqrt{\dfrac{k}{M+m}}

By putting the values

\omega=\sqrt{\dfrac{150}{5.67+1}}

ω = 4.47 rad/s

μ g  =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2

kX^2=(m+M)v^2

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

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3 years ago
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a particle is moving along a circular path having a radius of 4 in such that its position as a function of time is given by thet
ANTONII [103]

Answer:

Explanation:

Given

radius of circular path r=4\ in.

Position is given by

\theta =\cos 2t---1

Differentiate 1  to angular velocity we get

\frac{\mathrm{d} \theta }{\mathrm{d} t}=\omega =-2\sin 2t----2

Differentiate 2 to get angular acceleration

\frac{\mathrm{d} \omega }{\mathrm{d} t}=-2^2\cos 2t ---3

Net acceleration is the vector summation of tangential and centripetal force

a_t=\alpha \times r

a_t=-4\cos 2t\times 4=-16\cos 2t

a_r=\omega ^2\cdot r

a_r=(-2\sin 2t)^2\cdot 4

a_r=16\sin^2(2t)

a_{net}=\sqrt{a_r^2+a_t^2}

a_{net}=\sqrt{(16\sin ^2(2t)+(-16\cos 2t)^2}

a_{net}=\sqrt{256\cos ^2(2t)+256\sin ^4(2t)}                                                    

6 0
3 years ago
There is a moon orbiting an Earth-like planet. The mass of the moon is 9.58 × 1022 kg, the center-to-center separation of the pl
kaheart [24]

Answer:

= 4.38 × 10³⁴kgm²/s

Explanation:

Given that,

mass of moon m = 9.5 × 10²²kg

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Angular momentum of the moon about the planet

L = mvr

L = mr²w

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