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Xelga [282]
3 years ago
13

How did ernest rutherford discover the nucleus of the atom

Physics
1 answer:
motikmotik3 years ago
3 0

Answer:

THE ANSWER IS:  Rutherford, Marsden and Geiger discovered the dense atomic nucleus by bombarding a thin gold sheet with the alpha particles emitted by radium

Explanation:

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A slender rod is 80.0 cm long and has mass 0.390 kg . A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.05
Lilit [14]

Answer:

v = 1.08 m/s

Explanation:

What is the linear speed of the 0.0500-kg sphere as its passes through its lowest point?

The decrease in PE is

d = 80.0cm * 1 / 1000m = 0.80m

h = 0.80 m /2 = 0.40 m

ΔPE = m*g*h

ΔPE = (0.0500 - 0.0200)kg * 9.8m/s² * 0.400 m

ΔPE = 0.1176 J

The moment of inertia of the assembly is

I = 1/12*m*L² + (m1 + m2)*(L/2)²

I = 1/12*0.390kg*(0.800m)² + 0.0700kg*(0.400m)²

I = 0.032 kg·m²

KE = ½Iω²

0.1176 J = ½ * 0.032kg·m² * ω²

ω = 2.71 rad/s

v = ωr = 2.71 rad/s * 0.400m

The linear velocity

v = 1.08 m/s

3 0
3 years ago
What had to be invented before we could discover Saturn and Neptune?
Ann [662]

The discovery of Uranus, Neptune, most asteroids, the moons of
any planet other than Earth, Pluto, and all the other dwarf planets,
required the invention and use of telescopes.  Those objects are all
too dim to be seen with a bare naked human eye.

Saturn is bright and plainly visible with your eyes, if you know when
and where to look, and what you're looking at.

4 0
3 years ago
Read 2 more answers
7.According to the periodic table, how many total particles are there in the nucleus of a Titanium, Ti, atom?
FromTheMoon [43]

Answer:

there are 22 massive, positively charged, fundamental particles in the element's nucleus. Given this, the element is unquestionably titanium.

3 0
3 years ago
A series RLC circuit consists of a 52.0 Ω resistor, a 4.80 mH inductor, and a 330 nF capacitor. It is connected to an oscillator
Tju [1.3M]

Answer:

(G) 75.11 ohm

(H) 0.08 A

(I) 46.2 degree

Explanation:

R = 52 ohm

L = 4.8 m H = 4.8 x 106-3 H

C = 330 nF = 330 x 10^-9 F

Vo = 6 V

(G)

f = 5000 Hz

Let the impedance is Z.

X_{L}= 2 \pi fL = 2 \times 3.14\times 5000\times 4.8\times 10^{-3}=150.72 ohm

X_{c}= \frac{1}{2 \pi fC}=\frac{1}{2\times 3.14\times 5000\times 330\times 10^{-9}}=96.51 ohm

Z=\sqrt{R^{2}+\left ( X_{L}-X_{c} \right )^{2}}

Z=\sqrt{52^{2}+\left (150.72-96.51)^{2}}=75.11 ohm

(H) Let Io be the peak current

I_{0}=\frac{V_{0}}{Z}=\frac{6}{75.11}=0.0798 A = 0.08 A

(I) Let Ф be the phase angle

tan\phi = \frac{X_{L}-X_{C}}{R}

tan\phi =\frac{150.72-96.51}}{52}=1.0425

Ф = 46.2 degree

4 0
3 years ago
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

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

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

7 0
3 years ago
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