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
Look at the first person’s answer. Cause I know I’m wrong
Answer:
a. 1.027 x 10^7 m/s b. 3600 V c. 0 V and d. 1.08 MeV
Explanation:
a. KE =1/2 (MV^2) where the M is mass of electron
b. E = V/d
c. V= 0 V (momentarily the pd changes to zero)
d KE= 300*3600 v = 1.08 MeV
Answer:
(A) Consists of a small number of tiny particles that are far apart- relative in their size.
Explanation:
An <em>ideal gas</em> is defined as a simplification of a real gas, with punctual particles, in which all collisions are elastic, with random displacements and with no attractive force between them.
The assumption of the particles being punctual make clear that they do not have size at all. So if they were far apart-relative in their size, they can not collide each other, that is why assumption (B) can not be possible (<u><em>for that particular case</em></u>).
It is clear that (A) is not an assumption for an ideal gas, because do not fit in any of its properties.
Elastic collision: It is a case in which the energy is conserved (Kinetic Energy).
Kinetic Energy: It is the energy that will have an object as a consequence of its movement.
Answer:
0.1 s
Explanation:
The net force on the log is F - f = ma where F = force due to winch = 2850 N, f = kinetic frictional force = μmg where μ = coefficient of kinetic friction between log and ground = 0.45, m = mass of log = 300 kg and g = acceleration due to gravity = 9.8 m/s² and a = acceleration of log
So F - f = ma
F - μmg = ma
F/m - μg = a
So, substituting the values of the variables into the equation, we have
a = F/m - μg
a = 2850 N/300 kg - 0.45 × 9.8 m/s²
a = 9.5 m/s² - 4.41 m/s²
a = 5.09 m/s²
Since acceleration, a = (v - u)/t where u = initial velocity of log = 0 m/s (since it was a rest before being pulled out of the ditch), v = final velocity of log = 0.5 m/s and t = time taken for the log to reach a speed of 0.5 m/s.
So, making t subject of the formula, we have
t = (v - u)/a
substituting the values of the variables into the equation, we have
t = (v - u)/a
t = (0.5 m/s - 0 m/s)/5.09 m/s²
t = 0.5 m/s ÷ 5.09 m/s²
t = 0.098 s
t ≅ 0.1 s
