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LenKa [72]
3 years ago
8

A jet plane travels at 40.8 km at an average speed of 340 m/s calculate how long this journey took- giveyour answer to the neare

st minute
Physics
1 answer:
Tju [1.3M]3 years ago
6 0

Answer:

2\ \text{minutes}

Explanation:

d = Distance traveled = 40.8 km

s = Speed of jet = 340 m/s

Time is given by

t=\dfrac{d}{s}

\Rightarrow t=\dfrac{40.8\times 10^3}{340}

\Rightarrow t=120\ \text{s}

\Rightarrow t=\dfrac{120}{60}=2\ \text{minutes}

The time taken to complete the journey is 2\ \text{minutes}.

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The answer is the first option. The one about doing your best. trust me I read the book today!
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3 years ago
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A linear accelerator produces a pulsed beam of electrons. The pulse current is 0.50 A, and the pulse duration is 0.10 μs. (a) Ho
Crank

Answer:

a)N = 3.125 * 10¹¹

b) I(avg)  = 2.5 × 10⁻⁵A

c)P(avg) = 1250W

d)P = 2.5 × 10⁷W

Explanation:

Given that,

pulse current is 0.50 A

duration of pulse Δt = 0.1 × 10⁻⁶s

a) The number of particles equal to the amount of charge in a single pulse divided by the charge of a single particles

N = Δq/e

charge is given by Δq = IΔt

so,

N = IΔt / e

N = \frac{(0.5)(0.1 * 10^-^6)}{(1.6 * 10^-^1^9)} \\= 3.125 * 10^1^1

N = 3.125 * 10¹¹

b) Q = nqt

where q is the charge of 1puse

n = number of pulse

the average current is given as I(avg) = Q/t

I(avg) = nq

I(avg) = nIΔt

         = (500)(0.5)(0.1 × 10⁻⁶)

         = 2.5 × 10⁻⁵A

C)  If the electrons are accelerated to an energy of 50 MeV, the acceleration voltage must,

eV = K

V = K/e

the power is given by

P = IV

P(avg) = I(avg)K / e

P(avg) = \frac{(2.5 * 10^-^5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}

= 1250W

d) Final peak=

P= Ik/e

= = P(avg) = \frac{(0.5)(50 * 10^6 . 1.6 * 10^-^1^9)}{1.6 * 10^-^1^9}\\2.5 * 10^7W

P = 2.5 × 10⁷W

5 0
3 years ago
Read 2 more answers
two-point charges are 10.0 cm apart and have charges of 2.0 uc and -2.0uc respectively What is the magnitude of the electrical f
Scorpion4ik [409]

Answer:

Electric field due to two charges is given as

E = 1.44 \times 10^7 N/C

Explanation:

As we know that two charges are opposite in nature

So the electric field at the mid point of two charges will add together

so the net field is given as

E = 2\frac{kq}{r^2}

now we have

q = 2\times 10^{-6} C

r = 5 cm = 0.05 m

now we have

E = 2\frac{(9\times 10^9)(2\times 10^{-6})}{0.05^2}

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6 0
3 years ago
2. A solid plastic cube of side 0.2 m is submerged in a liquid of density 0.8 hgm calculate the
kotegsom [21]

Answer:

vpg = 0.064 N

Explanation:

Upthrust = Volume of fluid displaced

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vpg = 0.064 N

hope it helps.

3 0
3 years ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

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Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

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