As we know that
here we know that
now from above equation we have
so image will form on left side of lens at a distance of 15 cm
This image will be magnified and virtual image
Ray diagram is attached below here
Answer:
B
Explanation:
Icicles melt from a rooftop
Answer:
Explanation:
i = Imax sin2πft
given i = 180 , Imax = 200 , f = 50 , t = ?
Put the give values in the equation above
180 = 200 sin 2πft
sin 2πft = .9
sin2π x 50t = .9
sin 360 x 50 t = sin ( 360n + 64 )
360 x 50 t = 360n + 64
360 x 50 t = 64 , ( putting n = 0 for least value of t )
18000 t = 64
t = 3.55 ms .
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
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
