Which two measures are needed to calculate the distance between the seafloor and the ship?
O A. The number of sound waves emitted from the ship
B. The size of the ship
C. The speed of sound waves
D. The amount of time it takes sound waves to travel to the seafloor
and back

The two measures required are:

C. The speed of sound waves.

D. The amount of time it takes sound waves to travel to the seafloor and back.

Explanation:

To measure the distance between the seafloor and the ship, it's very important to know:

i. the speed of the waves

ii. the amount of time it takes the sound to travel and back.

When estimating water depth, the speed of sound through the water is used. Then the formula for the simple calculation is given as:

Distance = speed x time.

Where speed is the speed of the sound waves and time is the amount of time it takes the sound to travel and back.

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explanation:

i need some with some bio stuff see if anyone will

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol