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Travka [436]
3 years ago
11

What is the frequency of a wave if the speed is 4300 m/s and the wavelength is 32 m?

Physics
1 answer:
givi [52]3 years ago
8 0
4300 they didn’t look alike because DJ sinwosnbube is wow
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Easy question. Answer should include one line reasoning
KATRIN_1 [288]
C.figure 3 is the answer had the same and got is right
7 0
3 years ago
Read 2 more answers
Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo
Blizzard [7]

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

x=2.54\ cmx=2.54 cm

3 0
2 years ago
A pressure of 400 Pa is applied to an area of 2.5 m2.What force applies this pressure?
Irina-Kira [14]
F = 400 Pa x 2.5 m2
F = 1 kN
4 0
3 years ago
Yashoda prepares some lime juice on a hot day. She adds 80 g of ice at a temperature of 0°C to 0.32 kg of lime juice. The temper
Vikentia [17]

Answer:

Explanation:

a )

hear energy required to melt 1 g of ice = 340 J ,

hear energy required to melt 80 g of ice = 340 x 80  J = 27220 J .

b ) energy gained by the melted ice ( water at O°C ) = m ct

where m is mass of water , s is specific heat and t is rise in temperature

= 80 x 4.2 x ( 8°C - 0°C)

= 2688 J .

c )

energy lost by lime juice = energy gained by ice and water

= 27220 J + 2688 J .

= 29908 J .

d )

Let specific heat required be S

Heat lost by lime juice = M S T

M is mass of lime juice , S is specific heat , T is decrease in temperature

= 320 g x S x ( 29 - 8 )°C

= 6720 S

For equilibrium

Heat lost = heat gained

6720 S = 29908 J

S = 4.45 J /g °C .

4 0
3 years ago
PLEASE HELP ME WITH THIS ONE QUESTION
Vlad [161]

Answer: 7.95^{\circ}C

Explanation:

Given

Mass of water is m_1=1\ kg

mass of ice is m_2=0.1\ kg

Latent heat of fusion L=3.33\times 10^5\ kJ/kg

The heat capacity of water is c_{w}=4186\ J/kg^{\circ}C

Suppose water is at  T^{\circ} C  and it reaches to 0^{\circ}C to melt the ice

the heat released by water must be equivalent to heat absorbed by the ice

\therefore \quad m_1c_w(T-0)=m_2\times L\\\Rightarrow 1\times 4186\times T=0.1\times 3.33\times 10^5\\\Rightarrow T=7.95^{\circ}C

6 0
2 years ago
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