The Silence of the Lambs ends when Hannibal Lecter, from a payphone in the tropics, congratulates FBI Academy graduate Clarice Starling and gently warns her not to hunt him, ending the call by saying he had to go because he was having a friend for dinner, as he watched his hospital tormenter, Dr. Chilton, disembark from a plane. While that nervous laugh allowed movie goers to summon the courage to leave the theater and run to their cars, the original ending scripted by Tally gave no such quarter. When Lecter speaks to Starling, he compliments her outfit, which makes her realize he had watched from a distance. In the original ending, Lecter is cutting orange segments with a small paring knife, while he speaks to Clarice. As he hangs up the phone, the camera shot widens. We discover that he”s at a desk in a book lined office. There is the body of a bodyguard on the floor, and then we see Lecter is not alone. Chilton is trussed up in a chair across from him, the same method of restraints the doctor used on Lecter earlier in the movie. Lecter rises, slowly, a dreamy gleam in his eye, as he approaches his terrified victim, paring knife in hand. “Shall we begin?”
Answer: The atomic mass of a Europium atom is 151.96445 amu.
From the given information:
Percent intensity is 91.61% of Europium atom of molecular weight 150.91986 amu.
Percent intensity is 100.00% of Europium atom of molecular weight 152.92138 amu.
Abundance of Eu-151 atom:
Abundance of Eu-153 atom:
Atomic mass of Europium atom:
Therefore, the atomic mass of a Europium atom is 151.96445 amu.
Answer:
Explanation:
Given that,
Mass attached m = 0.95kg
Spring constant k = 16N/m
Instantaneous speed v = 36cm/s = 0.36m/s
Amplitude A=?
When x = 0.7A
Using conservation of energy
∆K.E + ∆P.E = 0
K.E(final) — K.E(initial) + P.E(final) — P.E(initial) = 0
At the beginning immediately the hammer hits the mass, the potential energy is 0J, Therefore, P.E(initial) = 0J, so the speed is maximum.
Also, at the end, at maximum displacement, the speed is zero, therefore, K.E(final) = 0
So, the equation becomes
— K.E(initial) + P.E(final) = 0
K.E(initial) = P.E(final)
½mv² = ½kA²
mv² = kA²
0.95 × 0.36² = 16×A²
0.12312 = 16•A²
A² = 0.12312/16
A² = 0.007695
A = √0.007695
A = 0.088 m
A = 8.8cm
B. Speed at x = 0.7A
Using the same principle above
K.E(initial) = P.E(final)
½mv² = ½kA²
Where A = 0.7A = 0.7 × 0.088 = 0.0614m
Then,
½× 0.95 × v² = ½ × 16 × 0.0614²
0.475v² = 0.0310644
v² = 0.0310644/0.475
v² = 0.0635
v = √0.0635
v = 0.252 m/s
v = 25.2 cm/s
Answer:
I = 0.2 A
Explanation:
Lamp is rated at 300 mA
I_lamp = 0.3 A
Voltage is; V = 3V
Thus; Resistance is given by;
R = V/I
R = 3/0.3
R = 10 ohms
Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;
R_eq = 10 + 5
R_eq = 15 ohms
Ammeter current will be;
I = V/R_eq
I = 3/15
I = 0.2 A
