Answer:
16 cm
Explanation:
For protons:
Energy, E = 300 keV
radius of orbit, r1 = 16 cm
the relation for the energy and velocity is given by
So, 
.... (1)
Now,
Substitute the value of v from equation (1), we get
Let the radius of the alpha particle is r2.
For proton
So, 
... (2)
Where, m1 is the mass of proton, q1 is the charge of proton
For alpha particle
So, 
... (3)
Where, m2 is the mass of alpha particle, q2 is the charge of alpha particle
Divide equation (2) by equation (3), we get
q1 = q
q2 = 2q
m1 = m
m2 = 4m
By substituting the values
So, r2 = r1 = 16 cm
Thus, the radius of the alpha particle is 16 cm.
Hi there!
Voltage in a series can be expressed by the following:
In words, the total voltage is equal to the sum of the individual voltage drops in a SERIES circuit.
We can solve for the total voltage:
KE = (1/2) (mass) (speed)²
KE = (1/2) (20 kg) (40 m/s)²
KE = (1/2) (20 kg) (1,600 m²/s²)
KE = (10 kg) (1,600 m²/s²)
KE = 16,000 Joules
Answer:
Scandium with an ion charge of +3
Explanation:
The magnification of the ornament is 0.25
To calculate the magnification of the ornament, first, we need to find the image distance.
Formula:
- 1/f = u⁻¹+v⁻¹.................... Equation 1
Where:
- f = Focal length of the ornament
- u = image distance
- v = object distance.
make u the subject of the equation
- u = fv/(f+v)................ Equation 2
From the question,
Given:
Substitute these values into equation 2
- u = (12×4)/(12+4)
- u = 48/16
- u = 3 cm.
Finally, to get the magnification of the ornament, we use the formula below.
- M = u/v.................. Equation 3
Where
- M = magnification of the ornament.
Substitute these values above into equation 3
Hence, The magnification of the ornament is 0.25
