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2 years ago

Which type of evidence was used to tie Ted Bundy to the murder of Lisa Levy?

Physics

1 answer:

Elodia [21]2 years ago

6 0

Explanation:

identified by one of the Chi Omega survivors as the assailant, and hair and fiber evidence tied him to the Leach murder. His trial for the Florida attacks in June 1979 was the first-ever nationally televised trial in the United States.

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What are the six different character traits scientists should have?

frozen [14]

careful observation, curiosity, logic, creativity, skepticism, and objectivity

3 0

3 years ago

Read 2 more answers

Suppose you had the same laser and diffraction grating from the previous question but now you had a flat detection screen. You w

kolezko [41]

Answer:

measuring the zero intensity point, we can deduce the movement of the screen.

The distance from the center of the pattern to the first zero is proportional to the distance to the screen,

Explanation:

The expression for the diffraction phenomenon is

a sin θ = m λ

for the case of destructive interference. In general the detection screen is quite far from the grid, let's use trigonometry to find the angles

tan θ = y / L

in these experiments the angles are small

tan θ = sin θ / cos θ = sin θ

sunt θ = y / L

we substitute

a $\frac{y}{L}$ = m λ

y = m L λ / a

therefore, by carefully measuring the zero intensity point, we can deduce the movement of the screen.

The distance from the center of the pattern to the first zero is proportional to the distance to the screen, so you can know where the displacement occurs, it should be clarified that these displacements are very small so the measurement system must be capable To measure quantities on the order of hundredths of a millimeter, a micrometer screw could be used.

4 0

2 years ago

Which description does the phrase "mechanical advantage" most closely relate to?

yKpoI14uk [10]

<span>the ratio of the force produced by a machine to the force applied to it, used in assessing the performance of a machine. I would say the answer is D, but i'm not sure. :)</span>

4 0

3 years ago

The voltage across the terminals of a 250nF capacitor is푣푣=�50푉푉, 푡푡≤0(푚푚1푒푒−4000푡푡+푚푚2푡푡푒푒−4000푡푡)푉푉, 푡푡 ≥0The initial current

olga2289 [7]

The first part of the question is not complete and it is;

The voltage across the terminals of a 250 nF capacitor is 50 V, A1e^(-4000t) + (A2)te^(-4000t) V, t0, What is the initial energy stored in the capacitor? Express your answer to three significant figures and include the appropriate units. t

Answer:

A) initial energy = 0.3125 mJ

B) A1 = 50 and A2 = 1,800,000

C) Capacitor Current is given by the expression;

I = e^(-4000t)[0.95 - 1800t]

Explanation:

A) In capacitors, Energy stored is given as;

U = (1/2)Cv²

Where C is capacitance and v is voltage.

So initial kinetic energy;

U(0) = (1/2)C(vo)²

From the question, C = 250 nF and v = 50V

So, U(0) = (1/2)(250 x 10^(-9))(50²) = 0.3125 x 10^(-3)J = 0.3125 mJ

B) from the question, we know that;

A1e^(-4000t) + (A2)te^(-4000t)

So, v(0) = A1e^(0) + A2(0)e^(0)

v(0) = 50

Thus;

50 = A1

Now for A2; let's differentiate the equation A1e^(-4000t) + (A2)te^(-4000t) ;

And so;

dv/dt = -4000A1e^(-4000t) + A2[e^(-4000t) - 4000e^(-4000t)

Simplifying this, we obtain;

dv/dt = e^(-4000t)[-4000A1 + A2 - 4000A2]

Current (I) = C(dv/dt)

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

Thus, Initial current (Io) is;

Io = (250 x 10^(-9))[e^(0)[-4000A1 + A2]]

We know that Io = 400mA from the question or 0.4 A

Thus;

0.4 = (250 x 10^(-9))[-4000A1 + A2]

0.4 = 0.001A1 - (250 x 10^(-9)A2)

Substituting the value of A1 = 50V;

0.4 = 0.001(50) - (250 x 10^(-9)A2)

0.4 = 0.05 - (250 x 10^(-9)A2)

Thus, making A2 the subject, we obtain;

(0.4 + 0.05)/(250 x 10^(-9))= A2

A2 = 1,800,000

C) We have derived that ;

I = (250 x 10^(-9))e^(-4000t)[-4000A1 + A2 - 4000tA2]

So putting values of A1 = 50 and A2 = 1,800,000 we obtain;

I = (250 x 10^(-9))e^(-4000t)[(-4000 x 50) + 1,800,000 - 4000(1,800,000)t]

I = e^(-4000t)[0.05 + 0.45 - 1800t]

I = e^(-4000t)[0.95 - 1800t]

5 0

3 years ago

A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s

belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, $\omega_i=114\ rev/min = 11.93\ rad/s$

Final angular speed of the wheel, $\omega_f=0$

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

$I=mr^2$

$I=49\times (0.73)^2$

$I=26.11\ kg-m^2$

We know that the work done is equal to change in kinetic energy.

$W=\Delta E$

$W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)$

$W=-\dfrac{1}{2}\times 26.11\times (11.93^2)$

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

$P=\dfrac{W}{t}$

$P=\dfrac{1858.05\ J}{22\ s}$

P =84.45 watts

Hence, this is the required solution.

4 0

3 years ago