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3 years ago

A stone is dropped from a bridge abd it hits the water 2.2 seconds later how high is the bridge above the water

Physics

1 answer:

Bess [88]3 years ago

8 0

Answer:

h = 23.716 m

Explanation:

Given that,

The time taken by the stone to hit the water is, t = 2.2 s

Height of the bridge above the ground, h = ?

The distance that the body will fall through the time is given by the formula

S = 1/2 gt² m

Where,

g - acceleration due to gravity

Substituting the values in the above equation

S = 1/2 x 9.8 m/s² x (2.2 s)²

= 23.716 m

Therefore, the height of the bridge from the surface of the water is h = 23.716 m

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In young’s double slit experiment, the measured fringe width is 0.5 mm for a Sodium light of 589 nm at a distance of 1.5 m. A br

Levart [38]

Answer:

(A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

Explanation:

Given that,

Fringe width d = 0.5 mm

Wavelength = 589 nm

Distance of screen and slit D = 1.5 m

Distance of bright fringe y = 1 cm

(A) We need to calculate the order of the bright fringe

Using formula of wavelength

$\lambda=\dfrac{dy}{mD}$

$m=\dfrac{d y}{\lambda D}$

Put the value into the formula

$m=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{589\times10^{-9}\times1.5}$

$m=5.65 = 6$

(B). We need to calculate the width of the bright fringe

Using formula of width of fringe

$\beta=\dfrac{yd}{D}$

Put the value in to the formula

$\beta=\dfrac{1\times10^{-2}\times0.5\times10^{-3}}{1.5}$

$\beta=3.33\times10^{-6}\ m$

$\beta=3.33\ \mu m$

Hence, (A). The order of the bright fringe is 6.

(B). The width of the bright fringe is 3.33 μm.

3 0

3 years ago

Which temperature is the hottest? 98 F or 39 C or 303K?<br> F= 1.8C + 32<br> C= (F-32)/1.8

sergejj [24]

Answer:

The hottest temperature is $T_2 = 39^o C$

Explanation:

From the question we are given

$T_1 = 98 F$

$T_2 = 39^oC$

$T_3 = 303 \ K$

Generally converting $T_3$ to Fahrenheit

$T_3' = (T_3 -273 ) * \frac{9}{5} + 32$

=> $T_3' = (303 -273 ) * \frac{9}{5} + 32$

=> $T_3' = 86 F$

Converting $T_2$ to Fahrenheit

$T_2' = T_2 * \frac{9}{5} + 32$

=> $T_2' = 39 * \frac{9}{5} + 32$

=> $T_2' =102.2 F$

Now comparing the temperature in Fahrenheit we see that $T_2$ is the hottest

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Answer:

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Explanation:

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