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muminat
3 years ago
14

The runner reaches the first camera, at the 500 m mark, at 9:03:05 a.m. and the second camera, at the 600 m mark, at 9:03:25 a.m

. What is her average speed over that time interval
Physics
1 answer:
SVEN [57.7K]3 years ago
7 0

Answer:

Explanation:

Distance travelled = 600 - 500 = 100 m

Time taken = 9:03:25 a.m.-  9:03 :05 a.m.

= 20 s

average speed = distance / time taken

= 100 / 20 m /s

= 5 m /s

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givi [52]

Answer:

decreases.

Explanation:

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as we know that

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here u is  Local flow velocity with respect to the boundarie and v is the speed of sound in the medium

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7 0
3 years ago
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
3 0
3 years ago
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A certain type of laser emits light that has a frequency of 4.2 × 1014 Hz. The light, however, occurs as a series of short pulse
bogdanovich [222]

Explanation:

It is given that,

Frequency of the laser light, f=4.2\times 10^{14}\ Hz

Time, t=3.2\times 10^{-11}\ s

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\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{4.2\times 10^{14}}

\lambda=7.14\times 10^{-7}\ m

or

\lambda=714\ nm

(b) Let n is the number of the wavelengths in one pulse. It can be calculated as :

n=f\times t

n=4.2\times 10^{14}\times 3.2\times 10^{-11}

n = 13440

Hence, this is the required solution.

8 0
3 years ago
What is the total resistance in this circuit?
saul85 [17]

Answer:B

Explanation:

6 0
3 years ago
How should you draw the field lines for Earth's magnetic field?
yuradex [85]
A. Coming out near the South Pole and going in near the North Pole
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