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olasank [31]
3 years ago
7

An airplane accelerates down a runway at 3 m/s2 for 32 s until is finally lifts off the ground. Determine the distance traveled

before takeoff.
Physics
1 answer:
Alex Ar [27]3 years ago
4 0

Answer:

12 or 24

Explanation:

i think it is i hope it is right

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An air bubble has a volume of 1.3 cm3 when it is released by a submarine 160 m below the surface of a freshwater lake. What is t
murzikaleks [220]

Answer:

V2 = 21.44cm^3

Explanation:

Given that: the initial volume of the bubble = 1.3 cm^3

Depth = h = 160m

Where P2 is the atmospheric pressure = Patm

P1 is the pressure at depth 'h'

Density of water = ρ = 10^3kg/m^3

Patm = 1.013×10^5 Pa.

Patm = 101300Pa

g = 9.81m/s^2

P1 = P2+ρgh

P1 = Patm +ρgh

P1 = 1.013×10^5+10^3×9.81×160.

P1 = 101300+1569600

P1 = 1670900 Pa

For an ideal gas law

PV =nRT

P1V1/P2V2 = 1

V2 = ( P1/P2)V1

V2 = (P1/Patm)V1

V2 = ( 1670900 /101300 Pa) × 1.3

V2 = 1670900/101300

V2 = 16.494×1.3

V2 = 21.44cm^3

4 0
3 years ago
Why does a moving object come to a stop on a frictional surface?
ohaa [14]
It stops cause of gravitaional pull 
3 0
3 years ago
two coils close to each other have a mutual inductance of 32 mh. if the current in one coil decays according to , where i0
Harlamova29_29 [7]

The EMF induced in the second coil is 43 Volts.

Michael Faraday was the first to discover electromagnetic induction back in the 1830s. Faraday discovered that moving a permanent magnet in and out of a coil or a single loop of wire caused an electromotive force, or EMF—otherwise known as a voltage—to be produced.

Changing magnetic flux results in varied currents flowing through the coil, which in turn generates its own magnetic field. This self-induced EMF opposes the change that is creating it, and the stronger the opposing EMF is, the faster the rate at which the current is changing. According to Lenz's law, this self-induced EMF will oppose the change in current in the coil, and because of its orientation, it is typically referred to as a back-EMF.

To learn more about EMF please visit-

brainly.com/question/15121836

#SPJ4

3 0
2 years ago
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc
4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: t=\frac{D_{A}}{v_{B}-v_{A}}

              Part B: x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

where

x_{0} is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is D_{A}.

The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

5 0
3 years ago
A lamp is 10% efficient. How much electrical energy must be supplied to the lamp each second if it produces 20 J of light energy
Leto [7]

Answer:

200J

Explanation:

200J is 100%, 10% of 200J would be 20J

3 0
3 years ago
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