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Master Mind

रायफल का द्रव्यमान M = 4 kg

गोली का द्रव्यमान m = 50 g = 0.050 kg

गोली का प्रारंभिक वेग u = 0 ms-1

गोली की अंतिम वेग v = 35 ms-1

राइफल का प्रारंभिक वेग U = 0 ms-1

राइफल का प्रतिक्षेपित वेग V = ?

गोली छूटने के पहले का संवेग = MU + mu

= 4 × 0  + 0.050 × 0

= 0 kg ms-1

गोली छूटने के बाद का संवेग = MV + mv

= 4 × V + 0.050 × 35

= 4V + 1.750  kg ms-1

संवेग संरक्षण के नियम से

गोली छूटने के बाद का संवेग  = गोली छूटने के पहले का संवेग

4V + 1.750 = 0

4V = - 1.750

V = - 1.7504 = - 0.4375 ms-1

-Answered by Master Mind On 30 September 2020:03:59:53 AM

Question 2. Asked on :05 October 2019:10:39:04 AM

#### Write the First Law of motion.

-Added by Himanshi Verma Science » Force And Laws Of Motion

Nishant Verma

An object will be in rest or in motion until unbalanced force applied on it.

An object continues to be in a state of rest or of uniform motion along a straight line unless acted upon by an unbalanced force.

-Answered by Nishant Verma On 06 October 2019:09:09:08 AM

Question 3. Asked on :15 January 2019:05:44:19 PM

#### A bullet is fired from a rifle the kinetic energy of the rifle in comparison to that of the bullet is

-Added by Sumitra Mahajan Science » Force And Laws Of Motion

Master Purushottam

a) more than one b) less than one c) equal to one d) zero

Let,
$M$ -mass of life
$v$- velocity of rifle
$m$ -mass of bullet
$u$ -velocity of bullet
$Mv=mu$
$\frac{{K}_{r}}{{K}_{b}}=\frac{\frac{1}{2}M{v}^{2}}{\frac{1}{2}m{v}^{2}}$
$\frac{\left(Mv{\right)}^{2}}{\left(Mu{\right)}^{2}}$$×\frac{m}{M}=\frac{m}{M}$
Since $m$ < $M$
${K}_{b}\phantom{\rule{thickmathspace}{0ex}}>\phantom{\rule{thickmathspace}{0ex}}{K}_{r}$
Therefore $\frac{Kinetic\phantom{\rule{thickmathspace}{0ex}}energy\phantom{\rule{thickmathspace}{0ex}}of\phantom{\rule{thickmathspace}{0ex}}rifle}{Kinetic\phantom{\rule{thickmathspace}{0ex}}energy\phantom{\rule{thickmathspace}{0ex}}of\phantom{\rule{thickmathspace}{0ex}}bullet}\phantom{\rule{thickmathspace}{0ex}}<\phantom{\rule{thickmathspace}{0ex}}1$
Hence b is the correct answer. Hence b is the correct answer.

-Answered by Master Purushottam On 24 August 2019:12:12:30 AM

Question 4. Asked on :13 January 2019:04:49:55 PM

#### A tortoise moves a distance of 100m in 15min. What is its average speed in km/hr?

-Added by Himanshi Verma Science » Force And Laws Of Motion

Master Purushottam

Given:  Tortoise moves 100 meters in 15 minutes.

Therefore, Average Speed = 100 meters /15 minutes

(as Speed = Distance/time)

= 0.10 Km/ 0.25 hours

(as 1 Km

= 1000 m and 1 hour = 60 min)

= 0.4 Km/hour

Answer – The average speed of tortoise is 0.4 Km/hour

-Answered by Master Purushottam On 23 August 2019:11:38:55 PM

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