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4 years ago

Help with this please.

Physics

1 answer:

gregori [183]4 years ago

8 0

Rutherford's experiment utilized positively charged alpha particles (He with a +2 charge) which were deflected by the dense inner mass (nucleus). The conclusion that could be formed from this result was that atoms had an inner core which contained most of the mass of an atom and was positively charged.

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A woman standing before a cliff claps her hands, and 2.8s later she hears the echo. How far away is the cliff? The speed of soun

ella [17]

Answer:

480.2 m

Explanation:

The following data were obtained from the question:

Speed of sound (v) = 343 m/s.

Time (t) = 2.8 s

Distance (x) of the cliff =?

The distance of the cliff from the woman can be obtained as follow:

v = 2x /t

343 = 2x /2.8

Cross multiply

2x = 343 × 2.8

2x = 960.4

Divide both side by the coefficient of x i.e 2

x = 960.4/2

x = 480.2 m

Therefore, the cliff is 480.2 m away from the woman.

4 0

3 years ago

A satellite has a mass of 5850 kg and is in a circular orbit 4.1 x10 to the 5th power m above the surface of a planet. The perio

koban [17]

Answer:

W = 24.28 kN

Explanation:

given,

Mass of satellite = 5850 Kg

height , h = 4.1 x 10⁵ m

Radius of planet = 4.15 x 10⁶ m

Time period = 2 h

= 2 x 3600 = 7200 s

Time period of satellite

$T = \dfrac{2\pi}{R}\sqrt{\dfrac{(R+h)^3}{g}}$

R is the radius of planet

h is the height of satellite

$T^2 = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{g}}$

now calculation of acceleration due to gravity

$g = \dfrac{4\pi^2}{R^2}\ {\dfrac{(R+h)^3}{T^2}}$

$g = \dfrac{4\pi^2}{(4.15\times 10^6)^2}\ {\dfrac{(4.15\times 10^6+4.1\times 10^5)^3}{(7200)^2}}$

g = 4.15 m/s²

True weight of satellite

W = m g

W = 5850 x 4.15

W = 24277.5 N

W = 24.28 kN

True weight of the satellite is W = 24.28 kN

8 0

3 years ago

1 Point

prohojiy [21]

Answer:

Explanation:

because it taken the rock more time to hit the water

8 0

4 years ago

A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so

joja [24]

ANSWER:100dB

from the sound intensity level,the sound intensity is calculated as:

$\beta$ ₁= $\beta$ =(10dB)㏒₁₀(l₁/l₀)

inserting numbers:

120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²

Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which

can be used to solve for l₁ and get

l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²

since the sound intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in

ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:

l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²

since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:

$\beta$ ₂= $\beta$ =(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)

$\beta$ ₂=(10dB)㏒₁₀(2.04×10¹⁰)

$\beta$ =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB

3 0

4 years ago

A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic

Mazyrski [523]

Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, $R_{Total}$