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Schach [20]
3 years ago
10

Which statement should she place in the region marked X?

Physics
2 answers:
Sidana [21]3 years ago
6 0

Answer:

DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

Explanation:

Orlov [11]3 years ago
3 0
I think the answer is D
You might be interested in
Question 23 of 32
marshall27 [118]
Do you still need help?!?!
7 0
3 years ago
A 113 kg man sits on the stern of a 6.3 m long boat. The prow of the
Usimov [2.4K]

Answer:

m = 105.37 kg

Explanation:

We are given;

Mass of man; m = 113 kg

Length of boat = 6.3m

Now, The position of the center of mass will not change during the motion of the man.

Thus,

X_g,i = X_g,f

So,

[113(6.3) + ma]/(113 + m) = [113(3.26) + m(a +3.26)]/(113 + m)

113 + m will cancel on both sides to give;

113(6.3) + ma = [113(3.26) + m(a +3.26)]

711.9 + ma = 368.38 + ma + 3.26m

ma will cancel out to give;

711.9 - 368.38 = 3.26m

343.52/3.26 = m

m = 105.37 kg

3 0
3 years ago
What is the equivalent resistance of a circuit that contains three 10.0 12
lana66690 [7]

Answer:

Explanation:

The equivalent resistance for three resistors connected in parallel is given as

(1/R)=(1/R₁)+(1/R₂)+(1/R₃)

now we.need to.insert the value of 3 resistances but only 2 are given in the question.

3 0
3 years ago
Read 2 more answers
You drop a stone down a well that is 19.60 m deep. How long is it before you hear the splash? The speed of sound in air is 343 m
ki77a [65]

So, the time needed before you hear the splash is approximately <u>2.06 s</u>.

<h3>Introduction</h3>

Hi ! In this question, I will help you. This question uses two principles, namely the time for an object to fall freely and the time for sound to propagate through air. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

Meanwhile, for sound propagation (without sound reflection), time propagates is the same as the quotient of distance by time. Or it can be formulated by :

\boxed{\sf{\bold{t = \frac{s}{v}}}}

With the following condition :

  • t = interval of the time (s)
  • s = shift or displacement (m)
  • v = velocity (m/s)

<h3>Problem Solving</h3>

We know that :

  • h = height or any other displacement at vertical line = 19.6 m
  • g = acceleration of the gravity = 9.8 m/s²
  • v = velocity = 343 m/s

What was asked :

  • \sf{\sum t} = ... s

Step by step :

  • Find the time when the object falls freely until it hits the water. Save value as \sf{\bold{t_1}}

\sf{t_1 = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t_1 = \sqrt{\frac{2 \cdot \cancel{19.6} \:_2}{\cancel{9.8}}}}

\sf{t_1 = \sqrt{4}}

\sf{\bold{t_1 = 2 \: s}}

  • Find the time when the sound propagate through air. Save value as \sf{\bold{t_2}}

\sf{t_2 = \frac{h}{v}}

\sf{t_2 = \frac{19.6}{343}}

\sf{\bold{t_2 \approx 0.06 \: s}}

  • Find the total time \sf{\bold{\sum t}}

\sf{\sum t = t_1 + t_2}

\sf{\sum t \approx 2 + 0.06}

\boxed{\sf{\sum t \approx 2.06}}

<h3>Conclusion</h3>

So, the time needed before you hear the splash is approximately 2.06 s.

3 0
3 years ago
The pupil of an eagle’s eye has a diameter of 6.0 mm. Two field mice are separated by 0.010 m. From a distance of 185 m, the eag
max2010maxim [7]

Answer:

time is 5.973826 sec

Explanation:

Given data

diameter D =  6.0 mm  6× 10^{-3} m

separated d =  0.010 m

distance (dis) 185 m

speed s = 16 m/s

wavelength = 550 nm = 550  10^{-9} m

to find out

How much time passes

solution

we know that for resolution  we use Rayleigh's Criterion i.e

θ = 1.22 wavelength  / diameter  =  separated / distance 1

we calculate distance 1 by put value wavelength, diameter and   separated

distance 1  = diameter × separated / 1.22 wavelength

distance 1  = 6× 10^{-3} × 0.010   / 1.22 × 550 × 10^{-9}

distance 1 = 89.418778

so time will be i.e = distance (dis) - distance 1  / speed

time = ( 185 -  89.418778) / 16

time = 5.973826 sec

time is 5.973826 sec

4 0
3 years ago
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