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3 years ago

The Rite of Spring was composed by

Physics

1 answer:

7nadin3 [17]3 years ago

7 0

The 'Rite of Spring' was composed by Igor Stravinsky.

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3. The pressure at the bottom of the ocean is great enough to crush submarines with steel walls that are 10 centimeters thick. S

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2 years ago

Read 2 more answers

You can also describe position by using coordinates of _____and _______

SOVA2 [1]

Answer:

x and y

Explanation:

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2 years ago

2. Light waves of the wavelength of 650 nm and 500 nm produce interference fringes on a screen at a distance of 1m from a double

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Answer:

least distance= 13mm

ratio of the lattice = 1 : 0.71 : 0.58

Explanation:

given λ₁ = 650nm = 650×10⁻⁹m, λ₂ = 500nm = 500×10⁻⁹m

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3 years ago

What would happen if the moon’s horizontal motion sped up too much

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3 years ago

an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

2 years ago