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2 years ago

11

What graph shows the relationship between kinetic energy and mass

1 answer:

Mrrafil [7]2 years ago

5 0

It's definitely not one of the graphs you gave us to choose from.

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A high school has started a community service program. Under the program, students must participate in volunteer activities, suc

GrogVix [38]

I think the answer is reduction in tuition

3 0

2 years ago

Read 2 more answers

Billy picks up a 40 lb. dumbbell (mass = 18.14 kg). The center of his hand, where the dumbbell is held, is 56 cm (0.56 m) from t

umka21 [38]

Answer:

<h2>Force due to biceps is given as</h2><h2> $F = 1991.05 N$ </h2>

Explanation:

For balancing the force we know that

Torque due to weight hold on his hand = torque due to force applied by biceps

So here we will have

$mg \times L = F \times d$

so we have

$18.14 \times 9.8 \times 0.56 = F \times (0.05)$

$F = 1991.05 N$

8 0

2 years ago

Thanks + BRAINLIST <br><br> Please need correct answer asappp

nirvana33 [79]

Answer:

Standing waves can be thought of as a sin wave and a cos wave overlapping each other. They go in different direction hence C is correct
Wave interference can be thought of as the opposite of destructive ---> constructive anda hencd meet and interact on the same medium such that answer B is correct

6 0

2 years ago

A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be com

bixtya [17]

Answer:L=109.16 m

Explanation:

Given

initial temperature $=20^{\circ}C$

Final Temperature $=80^{\circ}C$

mass flow rate of cold fluid $\dot{m_c}=1.2 kg/s$

Initial Geothermal water temperature $T_h_i=160^{\circ}C$

Let final Temperature be T

mass flow rate of geothermal water $\dot{m_h}=2 kg/s$

diameter of inner wall $d_i=1.5 cm$

$U_{overall}=640 W/m^2K$

specific heat of water $c=4.18 kJ/kg-K$

balancing energy

Heat lost by hot fluid=heat gained by cold Fluid

$\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)$

$2\times (160-T)=1.2\times (80-20)$

$160-T=36$

$T=124^{\circ}C$

As heat exchanger is counter flow therefore

$\Delta T_1=160-80=80^{\circ}C$

$\Delta T_2=124-20=104^{\circ}C$

$LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}$

$LMTD=\frac{80-104}{\ln \frac{80}{104}}$

$LMTD=91.49^{\circ}C$

heat lost or gain by Fluid is equal to heat transfer in the heat exchanger

$\dot{m_c}c(80-20)=U\cdot A\cdot$ (LMTD)

$A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2$

$A=\pi DL=5.144$

$L=\frac{5.144}{\pi \times 0.015}$

$L=109.16 m$

6 0

2 years ago

Heat is transferred at a rate of 2 kW from a hot reservoir at 800 K to a cold reservoir at 300 K. Calculate the rate at which th

vekshin1

Answer:

0.00417 kW/K or 4.17 W/K

Second law is satisfied.

Explanation:

Parameters given:

Rate of heat transfer, Q = 2kW

Temperature of hot reservoir, Th = 800K

Temperature of cold reservoir, Tc = 300K

The rate of entropy change is given as:

ΔS = Q * [(1/Tc) - (1/Th)]

ΔS = 2 * (1/300 - 1/800)

ΔS = 2 * 0.002085

ΔS = 0.00417 kW/K or 4.17 W/K

Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.

6 0

3 years ago