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10 days ago

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A boat is headed with a velocity of 18 meters/second toward the last with respect to the water in the river. if the river is Flo

wing with the velocity of 2.5 meters/second in the same direction as the boat. what would the magnitude of the boats velocity be?

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3.113 A heat pump is under consideration for heating a research station located on Antarctica ice shelf. The interior of the sta

ValentinkaMS [3465]

Answer:

a. β = 8.23 K

b. β = 28.815 K

Explanation:

The performance of the heat pump can be calculated using the formula

β = TH / (TH - TC)

a.

TH = 15 ° C + 273.15 K = 288.15 K

TC = - 20 ° C + 273.15 K = 253.15 K

β = 288.15 K / (288.15 K - 253.15 K)

β = 8.23 K

b.

TH = 15 ° C + 273.15 K = 288.15 K

TC = 5 ° C + 273.15 K = 278.15 K

β = 288.15 K / (288.15 K - 278.15 K)

β = 28.815 K

6 0

2 months ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

serg [3582]

Answer:

$4.1\cdot 10^8 N$

Explanation:

To begin with, we must determine the pressure acting on the sphere, which is calculated using:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ denotes the atmospheric pressure

$\rho = 1000 kg/m^3$ represents the density of the water

$g=9.8 m/s^2$ signifies the acceleration due to gravity

$h=11,000 m$ indicates the depth

By substituting these values,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The sphere's radius is calculated as r = d/2 = 1.1 m/2 = 0.55 m

Thus, the sphere's total surface area can be expressed as

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

Consequently, the inward force acting on the sphere equals

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

2 months ago

Read 2 more answers

A student wishes to determine the heat capacity of a coffee-cup calorimeter. After she mixes 108.7 g of water at 60.2°C with 108

Ostrovityanka [3204]

Answer: The calorimeter's heat capacity is $6.72J/g^oC$

Explanation:

This scenario assumes the amount of heat lost by the hot object equals the amount of heat gained by the cold object.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat capacity of water = $4.184J/g^oC$

$c_2$ = specific heat capacity of calorimeter =?

$m_1$ = mass of water = 108.7 g

$m_2$ = mass of calorimeter = 108.7 g

$T_f$ = final temperature of the mixture = $35.0^oC$

$T_1$ = initial temperature of the water = $60.2^oC$

$T_2$ = initial temperature of calorimeter = $19.3^oC$

Now substituting all provided values into the formula, we obtain

$(108.7g)\times (4.184J/g^oC)\times (35.0-60.2)^oC=-(108.7g)\times c_2\times (35.0-19.3)^oC$

$c_2=6.72J/g^oC$

Hence, the heat capacity of the calorimeter is $6.72J/g^oC$

3 0

3 months ago