Answer:
i) 21 cm
ii) At infinity behind the lens.
iii) A virtual, upright, enlarged image behind the object
Explanation:
First identify,
object distance (u) = 42 cm (distance between object and lens, 50 cm - 8 cm)
image distance (v) = 42 cm (distance between image and lens, 92 cm - 50 cm)
The lens formula,
Then applying the new Cartesian sign convention to it,
Where f is (-), u is (+) and v is (-) in all 3 cases. (If not values with signs have to considered, this method that need will not arise)
Substituting values you get,
i)
f = 21 cm
ii) u =21 cm, f = 21 cm v = ?
Substituting in same equation
v ⇒ ∞ and image will form behind the lens
iii) Now the object will be within the focal length of the lens. So like in the attachment, a virtual, upright, enlarged image behind the object.
Answer:
-40 kJ
80 kJ
Explanation:
Work is equal to the area under the pressure vs volume graph.
W = ∫ᵥ₁ᵛ² P dV
2.27) Pressure and volume are linearly related. When we graph P vs V, the area under the line is a trapezoid. So the work is:
W = ½ (P₁ + P₂) (V₂ − V₁)
W = ½ (100 kPa + 300 kPa) (0.1 m³ − 0.3 m³)
W = -40 kJ
2.29) Pressure and volume are inversely proportional:
pV = k
The initial pressure and volume are 500 kPa and 0.1 m³. So the constant is:
(500) (0.1) = k
k = 50
The final pressure is 100 kPa. So the final volume is:
(100) V = 50
V = 0.5
The work is therefore:
W = ∫ᵥ₁ᵛ² P dV
W = ∫₀₁⁰⁵ (50/V) dV
W = 50 ln(V) |₀₁⁰⁵
W = 50 (ln 0.5 − ln 0.1)
W ≈ 80 kJ
The north and the south of the globe are magnitized and the needles of a compass are magnetized. Since positives attract negatives. The negative end of the earth attarcts the positive point of the needle to point towards it. The same happens with positive end of the earth and the negative point of the needle.
Answer:
Explanation:
The speed of a wave in a string is given by:
where
T is the tension in the string
m is the mass of the string
L is the length
In this problem, the mass of the string is increased to 2m: m' = 2 m, while the length is not changed, L'=L. If the tension in the string is not changed, then the new speed of the wave in the string will be:
so, the speed of the wave decreases by a factor
Do both cars leave the starting line at the same time ? That's kind of important. I'll assume that they both start out at the same time, and now I'll proceed to answer the question that I have invented.
-- Car B moves 4 m/s faster than Car A .
-- So Car-A's lead shrinks by 4 meters every second after they start.
-- It takes (10/4) = 2.5 seconds for Car-A's lead to shrink to zero.
-- So Car-B overtakes Car-A <em>2.5 seconds</em> after they start.