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ruslelena [56]
3 years ago
8

In Rutherford's gold foil experiment, the alpha particles that traveled through the gold foil hit the screen behind it in more t

han one spot. Explain the significance of this finding and the discovery it led to.
Physics
1 answer:
sattari [20]3 years ago
4 0

before Rutherford explanation of atomic structure we know that atom is made up of solid positive sphere in which equal negative charge is randomly distributed

So here we can say if atom is like above then on incident of alpha particles they should deflect from the atom and no alpha particle must have to detect behind the atom on screen.

But here we can see in Rutherford experiment we have more than one spot where alpha particles hit the screen

So as per above result we can further say that positive part of nuclei is very small and it will deflect the alpha particle on if alpha particle pass near it.

So here we got more alpha particle behind the atom and then the new model of Rutherford is proposed for an atom

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Please answer both questions and not just one. Thanks!
Margarita [4]
It would be d and c hoped i helped!
8 0
3 years ago
A tree loses water to the air by the process of transpiration at the rate of 110 g/h. This water is replaced by the upward flow
Vinvika [58]

Answer:

The answer is 1.87nm/s.

Explanation:

The 110g/hr  water loss must be replaced by 110g/hr of sap. 110g of sap corresponds to a volume of  

110g \div \dfrac{1040*10^3g}{1*10^6cm^3}  = 106cm^3

thus rate of sap replacement is

106cm^3/hr = 106*10^{-6}m^3/3600s  = 2.94*10^{-8}m^3/s

The volume of sap in the vessel of length x is

V = Ax,

where A is the cross sectional area of the vessel.

For 2000 such vessels, the volume is

V = 2000Ax

taking the derivative of both sides we get:

\dfrac{dV}{dt} = 2000A \dfrac{dx}{dt}

on the left-hand-side \dfrac{dx}{dt} is the velocity v of the sap, and on right-hand-side \dfrac{dV}{dt}  = 2.94*10^{-8}m^3/s; therefore,

2.94*10^{-8}m^3/s=2000Av

and since the cross-sectional area is

A = \pi (\dfrac{100*10^{-3}m}{2} )^2 = 7.85*10^{-3}m^2;

therefore,

2.94*10^{-8}m^3/s =2000(7.85*10^{-3}m^2)v

solving for v we get:

v = \dfrac{2.94*10^{-8}m^3/s}{2000(7.85*10^{-3}m^2)}

\boxed{v =1.875*10^{-9}m/s = 1.875nm/s}

which is the upward speed of the sap in each vessel.

4 0
4 years ago
What is the best definition of
Brut [27]

Answer:

Work is best defined as a force exerted on a body to cause it move over a certain distance.

work=force×displacement×cosθ

Explanation:

7 0
2 years ago
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
wolverine [178]

Answer:

0.304 m/s2

Explanation:

If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.

We can use Newton's 2nd law to calculate the net acceleration of the system

a = F/m = 7 / 23 = 0.304 m/s^2

5 0
3 years ago
What is the formula for calculating the net force of an object
stepan [7]

Answer:

Fg = gravitational force. When a force is applied on the body, not only the applied force is acting there are many other forces like gravitational force Fg, frictional force Ff and the normal force that balances the other force. Therefore, the net force formula is given by, FNet = Fa + Fg + Ff + FN.

8 0
3 years ago
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