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Vaselesa [24]
3 years ago
10

The water temperature in the ocean varies inversely with the depth of the water. The deeper a person dives, the colder the water

becomes. At a depth of 700 meters, the water temperature is 3º Celsius. What is the water temperature at a depth of 300 meters?
Physics
1 answer:
VMariaS [17]3 years ago
6 0

Answer:

7°C

Explanation:

Given that ;

The water temperature in the ocean varies inversely with the depth of the water.

Let the water temperature be =   T_w

Let the depth of the ocean = d

So: T_w \alpha  \frac{1}{d}

T_w=\frac{1}{d}    ---- Equation 1

At a depth of 700 meters, the water temperature is 3º

3= \frac{1}{700}  

3 × 700 = 2100  ----- Equation 2

What is the water temperature at a depth of 300 meters?

T_2=\frac{1}{300}

300 T_2            ------- Equation (3)

Equating equation 2 and 3 together; we have:

2100 = 300 T_2

T_2 =  \frac{2100}{300}

T_2 = 7°C

∴ the water temperature is 7°C at a depth of 300 meters

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To win a prize at the county fair, you're trying to knock down a heavy bowling pin by hitting it with a thrown object. Should yo
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Answer:

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For Rubber Ball

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For Beanbag

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Explanation:

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3 years ago
about how much more energy is released in a 6.5 richter magnitude earthquake than in one with magnitude 5.5?
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2 years ago
Determine the thrust that a boat with a volume of 1.2m³ receives when it is stranded at sea. The density of seawater is 1020kg /
puteri [66]

Answer:

The maximum possible up-thrust on the boat is 11,995.2 N

Explanation:

According to Archimedes' principle, the thrust received by an object immersed a fluid is equal to the weight of the fluid displaced;

The given parameter of the boat in sea water are;

The volume of the boat = 1.2 m³

The density of seawater = 1020 kg/m³

Density = Mass/Volume

Therefore, Mass = Density × Volume

The maximum volume of water that the boat displaces = 1.2 m³

The mass of the water displaced by the boat = (Density of seawater) × (Volume of seawater displaced)

∴ The maximum possible mass of the water displaced by the boat = 1.2 m³ × 1020 kg/m³ = 1224 kg

The maximum possible mass of the water displaced by the boat, m = 1224 kg

Weight = Mass, m × g

Where;

g = The acceleration due to gravity = 9.8 m/s²

The up-thrust on the boat = The weight of the seawater displaced

∴ The maximum possible up-thrust on the boat = m × g = 1224 kg × 9.8 m/s² = 11,995.2 N

The maximum possible up-thrust on the boat = 11,995.2 N.

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3 years ago
1.A student attaches a string to a puck on a frictionless air table, and pulls with a constant force on the puck. a. Draw the fo
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a)

for the puck :

F = force applied in the direction of pull

N = normal force on the puck in upward direction by the surface of table

W = weight of the puck in down direction due to force of gravity


b)

along the vertical direction , normal force balance the weight of the puck , hence the net force is same as the force of pull F .

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Fnet = F


c)

since the net force acts in the direction of force of pull F , hence the puck accelerates in the same direction .

6 0
3 years ago
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