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Vikki [24]
3 years ago
8

What scientist shot alpha particles into gold

Physics
2 answers:
I am Lyosha [343]3 years ago
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]3 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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Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat
tigry1 [53]

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, T_i=4^{\circ} C

Final temperature of water, T_f=25^{\circ} C

The specific heat capacity of liquid water is, c=4.184\ J/g\ ^oC

Heat transferred is given by :

Q=mc(T_f-T_i)

Q=710\times 4.184\times (25-4)

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

3 0
3 years ago
Name each type of symbiosis and explain how the two species are affected
disa [49]
Frogs
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7 0
3 years ago
ASAP
pochemuha

Answer:

Explanation:

The fish falls from vertical rest in a time of

t = √(2h/g) = √(2(2.27)/9.81) = 0.68 s

v = d/t = 5.6 / 0.68 = 8.2 m/s

4 0
2 years ago
Why is the majority of Earth's freshwater not readily available for our use?
lara [203]
The answer is B. It is locked up in glaciers and ice caps.

Hope this helped. Good luck! Please give me a thanks and Brainliest.
8 0
3 years ago
Read 2 more answers
Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.
rjkz [21]

Answer:

Momentum of block B after collision =-50\ kg\ ms^{-1}

Explanation:

Given

Before collision:

Momentum of block A = p_{A1}= -100\ kg\ ms^{-1}

Momentum of block B = p_{B1}= -150\ kg\ ms^{-1}

After collision:

Momentum of block A = p_{A2}= -200\ kg\ ms^{-1}

Applying law of conservation of momentum to find momentum of block B after collision p_{B2}.

p_{A1}+p_{B1}=p_{A2}+p_{B2}

Plugging in the given values and simplifying.

-100-150=-200+p_{B2}

-250=-200+p_{B2}

Adding 200 to both sides.

200-250=-200+p_{B2}+200

-50=p_{B2}

∴ p_{B2}=-50\ kg\ ms^{-1}

Momentum of block B after collision =-50\ kg\ ms^{-1}

6 0
2 years ago
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