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3 years ago

When a puddle dries up what are the particles really doing

Physics

1 answer:

Alexandra [31]3 years ago

7 0

The particles are either being absorbed or evaporating

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How many turns are needed in a solenoid of radius 10 cm and length 20 cm for its self-inductance to be 6.0 H?

MA_775_DIABLO [31]

Answer:

Explanation:

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3 years ago

if the car went 30 km west in 25 min. and then 40 km south in 35 min., what would be its average speed?

igomit [66]

Average speed = (distance covered) / (time to cover the distance)

   = (30km + 40 km) / (25min + 35min)

   =    (70 km) / (60 min)

   = 70 km/hour

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3 years ago

Read 2 more answers

How does a laser work?

a_sh-v [17]

Answer:

A laser is created when the electrons in atoms in special glasses, crystals, or gases absorb energy from an electrical current or another laser and become “excited.” The excited electrons move from a lower-energy orbit to a higher-energy orbit around the atom's nucleus.

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4 years ago

A circuit contains an EMF source, a resistor R, a capacitor C, and an open switch in series. The capacitor initially carries zer

Flura [38]

Answer:

t = 1.098*RC

Explanation:

In order to calculate the time that the capacitor takes to reach 2/3 of its maximum charge, you use the following formula for the charge of the capacitor:

$Q=Q_{max}[1-e^{-\frac{t}{RC}}]$ (1)

Qmax: maximum charge capacity of the capacitor

t: time

R: resistor of the circuit

C: capacitance of the circuit

When the capacitor has 2/3 of its maximum charge, you have that

Q=(2/3)Qmax

You replace the previous expression for Q in the equation (1), and use properties of logarithms to solve for t:

$Q=\frac{2}{3}Q_{max}=Q_{max}[1-e^{-\frac{t}{RC}}]\\\\\frac{2}{3}=1-e^{-\frac{t}{RC}}\\\\e^{-\frac{t}{RC}}=\frac{1}{3}\\\\-\frac{t}{RC}=ln(\frac{1}{3})\\\\t=-RCln(\frac{1}{3})=1.098RC$

The charge in the capacitor reaches 2/3 of its maximum charge in a time equal to 1.098RC

6 0

4 years ago

As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th

NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

$v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek$

Conservation of Momentum:

$P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1$

Energy Balance:

$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

$0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}$

Using Eq 1

$0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m$

7 0

3 years ago