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2 years ago

What is the probability that time headway is at most 6 sec? (round your answer to three decimal places.)

Physics

1 answer:

FromTheMoon [43]2 years ago

7 0

The probability of that time at which the headway is at most 6 sec is 0.585.

<h3>What is probability?</h3>

The area of mathematics known as probability deals with numerical representations of the likelihood that an event will occur or that a statement is true. An event's probability is a number between 0 and 1, where, roughly speaking, 0 denotes the event's impossibility and 1 denotes certainty. The likelihood that an event will occur increases with its probability. A straightforward illustration is tossing a fair (impartial) coin. Since there are no other conceivable outcomes and the coin is fair, the odds of both the possibilities, "heads" and "tails," are equally likely to occur. As a result, the probability of either occurrence is half.

P(X<6) =(1-exp(-0.16*(6-0.5)))

=0.585

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A tin can has a radius of 0.0410m and is 185 m high if the air is removed from inside the can how much forces will the outside a

grin007 [14]

The force exerted on the bottle is 532 N

Explanation:

When the air is removed from inside the bottle, vacuum is created, therefore the external pressure (the atmospheric pressure) is no longer balanced, and it creates a net force downward on the can.

The pressure is related to the force by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area of the can

Here we have:

$p=1.01\cdot 10^5 Pa$ is the atmospheric pressure

r = 0.0410 m is the radius of the can, so the area is

$A=\pi r^2 = \pi (0.0410)^2=5.27\cdot 10^{-3}m^2$

And solving for F, we can find the force on the can:

$F=pA=(1.01\cdot 10^5)(5.27\cdot 10^{-3})=532 N$

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3 0

4 years ago

A parallel combination of a 1.13-μF capacitor and a 2.85-μF one is connected in series to a 4.25-μF capacitor. This three-capaci

Nata [24]

Answer:

(a) Charge of 4.25 μF capacitor is 35.46 μC.

(b) Charge of 1.13 μF capacitor is 10.05 μC.

(c) Charge of 2.85 μF capacitor is 25.36 μC.

Explanation:

Let C₁ , C₂ and C₃ are the capacitor which are connected to the battery having voltage V. According to the problem, C₁ and C₂ are connected in parallel. There equivalent capacitance is:

C₄ = C₁ + C₂

Substitute 1.13 μF for C₁ and 2.85 μF for C₂ in the above equation.

C₄ = ( 1.13 + 2.85 ) μF = 3.98 μF

Since, C₄ and C₃ are connected in series, there equivalent capacitance is:

C₅ = $\frac{C_{3}C_{4} }{C_{3} + C_{4} }$

Substitute 4.25 μF for C₃ and 3.98 μF for C₄ in the above equation.

C₅ = $\frac{4.25\times3.98 }{4.25 + 3.98 }$

C₅ = 2.05 μF

The charge on the equivalent capacitance is determine by the relation :

Q = C₅ V

Substitute 2.05 μF for C₅ and 17.3 volts for V in the above equation.

Q = 2.05 μF x 17.3 = 35.46 μC

Since, the capacitors C₃ and C₄ are connected in series, so the charge on these capacitors are equal to the charge on the equivalent capacitor C₅.

Charge on the capacitor, C₃ = 35.46 μC

Charge on the capacitor, C₄ = 35.46 μC

Voltage on the capacitor C₄ = $\frac{Q}{C_{4} }$ = $\frac{35.46\times10^{-6} }{3.98\times10^{-6}}$ = 8.90 volts

Since, C₁ and C₂ are connected in parallel, the voltage drop on both the capacitors are same, that is equal to 8.90 volts.

Charge on the capacitor, C₁ = C₁ V = 1.13 μF x 8.90 = 10.05 μC

Charge on the capacitor, C₂ = C₂ V = 2.85 μF x 8.90 = 25.36 μC

5 0

3 years ago

How many miles will be completed by a runner with a constant speed of 5 mph in 30 minutes?

Mrac [35]

Distance=speed times time
speed=5mph=5 $\frac{mi}{hr}$
time=30mins=30/60hr=1/2hr

so
distance=speed times time
$distance=(5\frac{mi}{hr})(\frac{1}{2}hr)$
$distance=\frac{5mi \space\ hr}{2hr}$
$distance=2.5mi$

he will run 2.5 miles in 30 mins at the speed of 5mph

5 0

3 years ago

An electromagnetic wave of wavelength 435 nm is traveling in vacuum in the —z direction. The electric field has an amplitude of

joja [24]

Answer:

a) 6.9*10^14 Hz

b) 9*10^-12 T

Explanation:

Given that

The wavelength of the wave, λ = 435 nm

Amplitude of the electric field, E(max) = 2.7*10^-3 V/m

The frequency of the wave can be found by using the formula

c = fλ, where c = speed of light

f = c/λ

f = 3*10^8 / 435*10^-9

f = 6.90*10^14 Hz

E(max) = B(max) * c, magnetic field amplitude, B(max) =

B(max) = E(max)/c

B(max) = 2.7*10^-3 / 3*10^8

B(max) = 9*10^-12 T

1T = 1 (V.s/m^2)

8 0

3 years ago

Three identical capacitors are connected in parallel to a battery. if a total charge of q flows from the battery, how much charg

muminat

Each capacitor carry the same charge 'q'.

Discussion:
The voltage from the battery is distributed equally across all of the capacitors when they are linked in series. The three identical capacitors' combined voltage is computed as follows:

$V_{T}$ = V₁ +V₂ +V₃

This voltage may also be calculated using capacitance and charge;

V = Q/ C

$V_{T}$ = V₁ +V₂ +V₃

Provided that the total charge is 'q', hence the total voltage can be expressed as:

$V_{T}$ = (Q/C₁) + (Q/C₂) + (Q/C₃) = Q(1/C₁ +1/C₂ +1/C₃)

Therefore from the above explanation, it is concluded that each and every capacitor carry same charge 'q'.

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7 0

1 year ago