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2 years ago

What scientist shot alpha particles into gold

2 answers:

8 0

The scientist that shot alpha particles into gold was Earnest Rutherford. He discovered about electrons. Then, later, he tried to prove that atoms had neutrons. This was done with his gold foil experiment. When the atoms were shot and hit the nucleus, they bounced back... that is how he found out that the atoms had a nucleus, also where our nuclear structure of an atom comes from...

Hope this helps;)

laila [671]2 years ago

5 0

The Geiger–Marsden experiment(s) (also called the Rutherford gold foil experiment) were a landmark series of experiments by which scientists discovered that every atom contains a nucleus where its positive charge and most of its mass are concentrated

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what is cut off from the open sea coral reefs or sandbars A watershed B lagoon C reef or D atoll . also this is a science questi

Klio2033 [76]

Hello there.

Its D.

HOPE I HELPED

7 0

2 years ago

Read 2 more answers

Two astronauts, each with a mass of 50 kg, are connected by a 7 m massless rope. Initially they are rotating around their center

kiruha [24]

Answer:

The angular velocity is $w_f = 1.531 \ rad/ s$

Explanation:

From the question we are told that

The mass of each astronauts is $m = 50 \ kg$

The initial distance between the two astronauts $d_i = 7 \ m$

Generally the radius is mathematically represented as $r_i = \frac{d_i}{2} = \frac{7}{2} = 3.5 \ m$

The initial angular velocity is $w_1 = 0.5 \ rad /s$

The distance between the two astronauts after the rope is pulled is $d_f = 4 \ m$

Generally the radius is mathematically represented as $r_f = \frac{d_f}{2} = \frac{4}{2} = 2\ m$

Generally from the law of angular momentum conservation we have that

$I_{k_1} w_{k_1}+ I_{p_1} w_{p_1} = I_{k_2} w_{k_2}+ I_{p_2} w_{p_2}$

Here $I_{k_1 }$ is the initial moment of inertia of the first astronauts which is equal to $I_{p_1}$ the initial moment of inertia of the second astronauts So

$I_{k_1} = I_{p_1 } = m * r_i^2$

Also $w_{k_1 }$ is the initial angular velocity of the first astronauts which is equal to $w_{p_1}$ the initial angular velocity of the second astronauts So

$w_{k_1} =w_{p_1 } = w_1$

Here $I_{k_2 }$ is the final moment of inertia of the first astronauts which is equal to $I_{p_2}$ the final moment of inertia of the second astronauts So

$I_{k_2} = I_{p_2} = m * r_f^2$

Also $w_{k_2 }$ is the final angular velocity of the first astronauts which is equal to $w_{p_2}$ the final angular velocity of the second astronauts So

$w_{k_2} =w_{p_2 } = w_2$

$mr_i^2 w_1 + mr_i^2 w_1 = mr_f^2 w_2 + mr_f^2 w_2$

=> $2 mr_i^2 w_1 = 2 mr_f^2 w_2$

=> $w_f = \frac{2 * m * r_i^2 w_1}{2 * m * r_f^2 }$

=> $w_f = \frac{3.5^2 * 0.5}{ 2^2 }$

=> $w_f = 1.531 \ rad/ s$

3 0

2 years ago

You are listening to the radio when one of your favorite songs comes on, so you turn up the volume. If you managed to increase t

andrew-mc [135]

To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as

$\beta_{dB} = 10log_{10} \frac{I}{I_0}$

Where,

I = Acoustic intensity in linear scale

$I_0$ = Hearing threshold

The value in decibels is 17dB, then

$17dB = 10log_{10} \frac{I}{I_0}$

Using properties of logarithms we have,

$\frac{17}{10} = log_{10} \frac{I}{I_0}$

$log_{10} \frac{I}{I_0} = 1.7$

$\frac{I}{I_0} = 10^{1.7}$

$\frac{I}{I_0} = 50.12 W/m^2$

Therefore the factor that the intensity of the sound was $50.12W/m^2$

5 0

2 years ago

Two of the three tuning forks have known frequencies. When a 610 Hz fork and the unknown fork are struck together, four beats pe

RUDIKE [14]

-- In combination with 610 Hz, the beat frequency is 4 Hz.

So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.

In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.

-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.

So the unknown is (605+9) = 614 Hz.

6 0

3 years ago

A 5.0-ohm resistor, a 10.0-ohm resistor, and a 15.0-ohm resistor are connected in parallel with a battery. The current through t

Kaylis [27]

Answer:

E = 3456 J

Explanation:

The electrical energy expended in a resistor can be easily calculated by using the following formula:

$E = Pt$

where,

E = Energy Expended = ?

I = current through 5 ohm resistor = 2.4 A

R = Resistance = 5 ohms

P = Electrical Power = VI

Since,

V = IR (Ohm's Law)

Therefore,

P = (IR)(I) = I²R = (2.4 A)²(5 ohms) = 28.8 Watt

t = time taken = (2 min)(60 s/1 min) = 120 s

Therefore,

E = (28.8 Watt)(120 s)

8 0

2 years ago