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kiruha [24]
2 years ago
11

Find the mean of set values 12g 9g 13g 12g 20g 17g 15g

Physics
1 answer:
nalin [4]2 years ago
5 0

{\huge{\boxed{\mathcal{\green{Answer}}}}} \\ \frac{12 + 9 + 13 + 12 + 20 + 17 + 15}{7}  \\  =  \frac{98}{7}  \\  = 14 \\ {\huge{\boxed{\mathcal{\green{Hope \: it \: Helps}}}}}

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Coherent light of frequency 6.37×1014 Hz passes through two thin slits and falls on a screen 88.0 cm away. You observe that the
IgorC [24]

Answer:

The distance between the two slits is 40.11 μm.

Explanation:

Given that,

Frequency f= 6.37\times10^{14}\ Hz

Distance of the screen l = 88.0 cm

Position of the third order y =3.10 cm

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{c}{f}

where, c = speed of light

f = frequency

Put the value into the formula

\lambda=\dfrac{3\times10^{8}}{6.37\times10^{14}}

\lambda=471\ nm

We need to calculate the distance between the two slits

m\times \lambda=d\sin\theta

d =\dfrac{m\times\lambda}{\sin\theta}

Where, m = number of fringe

d = distance between the two slits

Here, \sin\theta =\dfrac{y}{l}

Put the value into the formula

d=\dfrac{3\times471\times10^{-9}\times88.0\times10^{-2}}{3.10\times10^{-2}}

d=40.11\times10^{-6}\ m

d = 40.11\ \mu m

Hence, The distance between the two slits is 40.11 μm.

7 0
3 years ago
You toss a walnut at a speed of 15.0 m/s at an angle of 50.0∘ above the horizontal. The launch point is on the roof of a buildin
kaheart [24]

Answer:

a.3.51s

b.33.8m

c. 9.64,-22.9

Explanation:

8 0
2 years ago
You are standing on a street corner and you hear an ambulance siren. The pitch of the siren seems to be getting higher. What do
Gre4nikov [31]

B.The siren will get louder and higher as the ambulance moves towards you


Hope this helped !

4 0
3 years ago
Read 2 more answers
5. Harrison claims that waves travel faster when the particles within the medium are farther apart. Is Harrison correct?
andrew11 [14]

Answer:

सं देश भर भर में नौ महिना अघिदेखि महिना अघिदेखि नै जातीय आधारमा प्रान्त पुगें र नेपाली जनताको लागि फरक छ तर त्यो पनि पाँचै वर्षमा धरासायी भयो पाँचै वर्षमा धरासायी भयो भने हेर्न त्यो पनि हो र स्वार्थी हुन्छ भन्थे यो मेरो पहिलो मृत्यु भएको हो। छ। यो म भने त्यो प्रेम कति

5 0
3 years ago
A man pulled a 13.0 kg object 11.8 cm vertically with his teeth. (a) How much work (in J) was done on the object by the man in t
jonny [76]

Answer:

(a)The work done by the man is -15.03J.

(b)The force exerted on the object is 127.4N.

Explanation:

Mass of the object pulled by the man is -13kg

Object is lifted 11.8 cm vertical with his teeth it means (displacement = +11.8cm = +0.118m)

Acceleration due to gravity is 9.8 \mathrm{m} / \mathrm{s}^{2}

(a) <u>Calculating the work done</u>:

Work done = mgh

Where "m" is mass of an object, "g" is acceleration due to gravity and "h" is the displacement.

\text { Work }=-13 \times 9.8 \times(+0.118 \mathrm{m})

\text { Work }=-15.03 \mathrm{J}

The work done by the man is -15.03J.

(b) <u>Calculating the force</u>:

Probably the man and the object are close to the exterior of the earth. If the rigidity required to maintained the object of consistent velocity interior the gravitational field of the earth is \mathrm{g}=9.8 \mathrm{m} / \mathrm{s}^{2}

Thus the weight of the object is balanced by the force of the man's teeth on the object. That is

F = mg

\mathrm{F}=13 \times 9.8

F = 127.4N

The force exerted on the object is 127.4N.

4 0
2 years ago
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