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3 years ago

14

Uses and effects of UV rays?

1 answer:

anygoal [31]3 years ago

6 0

Answer:

The above pic may help you :)

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A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

Luden [163]

Answer:

$\rho _{liquid}=1995.07kg/m^{3}$

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have $T_{3}+B = W_{rock}$

$T_{3}+\rho _{Liquid}V_{rock}g$ = $W_{rock}.....(\alpha)$

When the rock is suspended in air for equilibrium we have

$T_{1}=W_{rock}....(\beta)$

When the rock is suspended in water for equilibrium we have

$T_{2}$ + $\rho _{water}V_{rock}g$ = $W_{rock}.....(\gamma)$

Using the given values of tension and solving α,β,γ simultaneously for $\rho _{Liquid}$ we get

$W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\$

Solving for density of liquid we get

$\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000$

$\rho _{liquid}=1995.07kg/m^{3}$

5 0

3 years ago

Which group within the Senate has the most power over legislation?

ozzi

<span>the majority party................</span>

5 0

4 years ago

White light (light containing all frequencies of light in the visible spectrum) passes through a diffraction grating with 1300 l

lara31 [8.8K]

Answer:

The distance between the 1st order blue fringe and the 1st order red fringe is 6.5 cm.

Explanation:

Given that,

Diffraction grating =1300 lines/cm

Distance of screen = 2 m

Order number = 1

We need to calculate the distance between the 1st order blue fringe and the 1st order red fringe

Using formula of distance

For first order red fringe

$d\sin\theta=m\lambda$

$d\times\dfrac{y}{l}=m\times\lambda$

$y_{r}=\dfrac{m\times\lambda l}{d}$

$y_{r}=\dfrac{1\times700\times10^{-9}\times2}{\dfrac{1}{1300}\times10^{-2}}$

$y_{r}=0.182\ m$

For first order blue fringe

$y_{b}=\dfrac{1\times450\times10^{-9}\times2}{\dfrac{1}{1300}\times10^{-2}}$

$y_{b}=0.117\ m$

We need to calculate the distance between blue fringe and red fringe

$\Delta y=y_{r}-y_{b}$

$\Delta y=0.182-0.117$

$\Delta y=0.065\ m=6.5 cm$

Hence, The distance between the 1st order blue fringe and the 1st order red fringe is 6.5 cm.

7 0

3 years ago

Samples of different materials, A and B, have the same mass, but the sample

NISA [10]

Answer: C. The particles that make up material A have more mass than the

particles that make up material B.

7 0

3 years ago

Read 2 more answers

Nina [5.8K]

Ur right 2.23848):$/)/8-&:

6 0

3 years ago