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Natasha_Volkova [10]
3 years ago
11

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles trave

led in a straight line and some were deflected at different angles.
Physics
1 answer:
jeyben [28]3 years ago
4 0

The question is incomplete, the complete question is;

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles traveled in a straight line and some were deflected at different angles. Which statement best describes what Rutherford concluded from the motion of the particles? Some particles traveled through empty spaces between atoms and some particles were deflected by electrons. Some particles traveled through empty parts of the atom and some particles were deflected by electrons. Some particles traveled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms. Some particles traveled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Answer:

Some particles traveled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.

Explanation:

Rutherford first proposed the nuclear model of the atom after his landmark experiment.

In this experiment, alpha particles from a source was focused on a thin gold foil. Some of the particles passed through empty spaces within the atom but were deflected at different angles by a small area of high-density positive charge within an atom which Rutherford later called the atomic nucleus.

Hence the answer above.

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The applied force required to push something across a surface as friction increases is what?
katovenus [111]
Is proportional to the value of the normal force acting on the object.
7 0
3 years ago
A flat coil of wire has an area A, N turns, and a resistance R. It is situated in a magnetic field, such that the normal to the
Phantasy [73]

Answer:

3.4 x 10^-4 T

Explanation:

A = 1.5 x 10^-3 m^2

N = 50

R = 180 ohm

q = 9.3 x 106-5 c

Let B be the magnetic field.

Initially the normal of coil is parallel to the magnetic field so the magnetic flux is maximum and then it is rotated by 90 degree, it means the normal of the coil makes an angle 90 degree with the magnetic field so the flux is zero .

Let e be the induced emf and i be the induced current

e = rate of change of magnetic flux

e = dФ / dt

i / R = B x A / t

i x t / ( A x R) = B

B = q / ( A x R)

B = (9.3 x 10^-5) / (1.5 x 10^-3 x 180) = 3.4 x 10^-4 T

6 0
4 years ago
X rays of wavelength 0.0100 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
Fofino [41]

Answer:

a)  =4.84*10^{-12}

b)= -2.76*10^{-14} J

c)i.e -2.76*10^{-14} J

d)= 0 and the direction of motion is equal to zero

Explanation:

a) compton shift

\Delta\lambda = \frac{h}{mc} (1-cos\theta)

\Delta\lambda = \frac{6.626*10^{-34}}{9.11*10^{-11}3*10^8} (1-cos180)

                        =4.84*10^{-12}

b) the new wavelength

\lambda' = 10.0*10^{-12} +4.84^10^{-12}

               =14.84*10^{-12} m

\Delta E = E' - E

              =hc[\frac{1}{\lambda'}-\frac{1}{\lambda}]

\Delta E = 6.626*10^{-34}*(3*10^8)[\frac{1}{14.84*10^{-12}}-\frac{1}{4.8*10^{-12}}]

= -2.76*10^{-14} J

C)By conservation of energy, the kinetic energy of recoiling electron is equal to the magnitude of energy between the photon energy

i.e -2.76*10^{-14} J

d) the angle between the positive direction of motion

sin\phy = \frac{\lambda_t sin\theta}{\lambda'}

            =\frac{2.43*10^{-12}sin180}{14.84*10^{-12}}  

             = 0

the direction of motion is equal to zero.

4 0
4 years ago
What atom has 5 neutrons and an atomic number of 5?!
meriva

Answer:

boron has an atomic number of 5

but it has 6 neutron

Explanation:

there is no atom which has 5 neutrons and an atomic number of 5.

8 0
3 years ago
Create a file "parts_inv.dat" that stores on each line a part number, cost, and quantity in inventory, in the following format:1
andriy [413]

Answer:

%Open the file.

fID = fopen('parts_inv.dat');

%Read from the file.

data = fscanf(fID,'%d\t%f\t%d',[3,inf]);

%Close

fclose(fID);

%Restore the data.

data = data';

%Get the size

[rs, cs] = size(data);

%Set value.

invCost = 0;

%Loop

for rw = 1 : rs

%Find cost

invCost = invCost + (data(rw, 2) * data(rw, 3));

%Loop end

end

%Display the cost.

fprintf('Total cost: %4.2f\n\n', invCost);

Explanation:

7 0
4 years ago
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