Answer:
The angular magnification is 
Explanation:
From the question we are told
The focal length is 
The near point is 
The angular magnification is mathematically represented as

Substituting values

Any kind of frequency, including the angular kind, is closely involved with
time. Still, for some unknown reason,you've given us no time information
whatsoever ... a peculiar decision on your part, since we can be sure that
it's right there, inexorably intertwined with the part of the question that you
DID copy and share with us.
Furthermore and moreover, for one with no prior experience with simple
harmonic motion, the many symbols in this question such as ' d ', ' a ',
' << ', ' d₂ ', and ' a₂ ' would be of no help at all to guide him toward a
solution. On the contrary, he would conclude that the question itself
had been posted by some alien life form.
To sum up: Come back and post the drawing that goes along with the
question, make sure you have presented all of the information that the
question includes, and then we'll talk.
Answer:
Answer:
15.67 seconds
Explanation:
Using first equation of Motion
Final Velocity= Initial Velocity + (Acceleration * Time)
v= u + at
v=3
u=50
a= - 4 (negative acceleration or deceleration)
3= 50 +( -4 * t)
-47/-4 = t
Time = 15.67 seconds
This question is incomplete, the complete question is;
The Figure shows a container that is sealed at the top by a moveable piston, Inside the container is an ideal gas at 1.00 atm. 20.0°C and 1.00 L.
"What will the pressure inside the container become if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant?"
Answer:
the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Explanation:
Given that;
P₁ = 1.00 atm
P₂ = ?
V₁ = 1 L
V₂ = 1.60 L
the temperature of the gas is kept constant
we know that;
P₁V₁ = P₂V₂
so we substitute
1 × 1 = P₂ × 1.60
P₂ = 1 / 1.60
P₂ = 0.625 atm
Therefore the pressure inside the container become 0.625 atm if the piston is moved to the 1.60 L mark while the temperature of the gas is kept constant
Answer:
40 MJ (D)
Explanation:
Quantity of heat (Qh) = 100 MJ
temperature of steam (Th) = 450°c = 450 + 273 = 723 K
emperature of water (TI) = 20 °c = 20 + 273 = 293 k
efficiency = (Qh-Qi)/Qh = (Th-Ti)/Th

- Qi= 0.5947 x 
- (0.5947 x
) = Qi
Qi = 40.5 MJ equivalent to 40 MJ (D)