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3 years ago

Describe two ways unbalanced forces help you in your day to day life.

Physics

1 answer:

Alexxandr [17]3 years ago

6 0

Answer:

we need unbalance force to lift objects
we need unbalance force to drag objects

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How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)

Elan Coil [88]

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

$\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c$

Option (D) is correct.

4 0

3 years ago

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

kogti [31]

Answer:

a) $y_{first}=5.3mm$

b) $y_{second}=10.6-5.3 =5.3 mm$

Explanation:

The width of the central bright in this diffraction pattern is given by:

$y=\frac{m\lambda D}{a}$ when m is a natural number.

here:

m is 1 (to find the central bright fringe)
D is the distance from the slit to the screen
a is the slit wide
λ is the wavelength

So we have:

$y_{first}=\frac{633*10^{9}*3.35}{0.0004}$

$y_{first}=5.3mm$

Now, if we do m=2 we can find the distance to the second minima.

$y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}$

$y_{2}=10.6 mm$

Now we need to subtract these distance, to get the width of the first bright fringe :

$y_{second}=10.6-5.3 =5.3 mm$

I hope it heps you!

4 0

3 years ago

Solve the below problems being sure to provide the correct significant figures.

morpeh [17]

Sum of the first ten digits is:
7+3+4+5+4+1+7+8+0 = 39

39, when divided by 9 give you the remainder of 3

9 x4, is 36
36 +3 equals 39
(39 = 9 x 4 + 3)

So X= 0

5 0

4 years ago

I need help with question 26<br>this is from my physics textbook

Neko [114]

Suppose $a,b,c$ are measured in meters. Then the semiperimeter $s=\dfrac{a+b+c}2$ is also given in meters, so $s-a,s-b,s-c$ are all also given in meters. When multiplied together, $(s-a)(s-b)(s-c)$ is measured in cubic meters. Divide this by $s$ , measured in meters, and this reduces to square meters. Take the square root and you end up back with meters once again. So the formula is indeed dimensionally consistent.

6 0

4 years ago

Water drips from the nozzle of a shower onto the floor 189 cm below. The drops fall at regular (equal) intervals of time, the fi

laiz [17]

Answer:

0.83999 m

0.20999 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s² = a

s = 189 cm

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.89=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.89\times 2}{9.81}}\\\Rightarrow t=0.62074\ s$

When the time intervals are equal, if four drops are falling then we have 3 time intervals.

So, the time interval is

$t'=\dfrac{t}{3}\\\Rightarrow t'=\dfrac{0.62074}{3}\\\Rightarrow t'=0.206913\ s$

For second drop time is given by

$t''=2t'\\\Rightarrow t''=2\times 0.2069133\\\Rightarrow t''=0.4138266\ s$

Distance from second drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut''+\dfrac{1}{2}at''^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.4138266^2\\\Rightarrow s=0.839993\ m$

Distance from second drop is 0.83999 m

Distance from third drop

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow y''=ut'+\dfrac{1}{2}at'^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 9.81\times 0.206913^2\\\Rightarrow s=0.20999\ m$

Distance from third drop is 0.20999 m

6 0

4 years ago