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allochka39001 [22]

3 years ago

7

A 50 kg object traveling at 100 m/s collides (perfectly elastic) with a 50 kg object initially at rest.

1 answer:

tino4ka555 [31]3 years ago

8 0

Answer:

a

Explanation:

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Identify two factors that determine the state of matter

vova2212 [387]

Two factors determine whether a substance is a solid, a liquid, or a gas: The kinetic energies of the particles (atoms, molecules, or ions) that make up a substance. Kinetic energy tends to keep the particles moving apart. The attractive intermolecular forces between particles that tend to draw the particles together.

4 0

3 years ago

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If Scoobie could drive a Jetson's flying car at a constant speed of 450.0 km/hr across oceans and space, approximately how long

vladimir2022 [97]

Answer:

The value is $t = 3.6 \ days$

Explanation:

From the question we are told that

The speed is $v = 450.0 km/h$

The radius of the earth is $R = 6200 \ km$

Generally the circumfernce of the earth is mathematically evaluated as

$C = 2\pi R$

=> $C = 2 * 3.142 * 6200$

=> $C = 38960.8 \ km$

Generally the time taken is mathematically represented as

$t = \frac{38960.8}{450.0}$

$t = 86.6 \ hr$

Converting to days

$t = \frac{86.6}{24}$

=> $t = 3.6 \ days$

7 0

3 years ago

What happens to light when it changes speed?

Brut [27]

Hey there!

When light changes speed, it REFRACTS.
Your answer is going to be option C.

Hope this helps you.
Have a great day!

5 0

3 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

3 years ago

Force F=2.0N i - 3.0N k acts on a pebble with position vectorr=0.50m j - 2.0m k relative to the origin. In unit vector notation,

klasskru [66]

Answer with Explanation:

We are given that

Force acts on a pebble= $2\hat{i}-3\hat{k}$ N

Position vector= $r=0.5\hat{j}-2\hat{k}$ m

a.We have to find the resulting torque on the pebble about origin.

Torque= $r\times F$

Substitute the values then we get

$Torque= (0.5j-2k)\times (2i-3k)$

Torque= $-k-1.5i-4j$ N-m

By using $i\times j=k,j\times k=i,k\times i=j,j\times i=-k,k\times j=-i,i\times k=-j,i\times i=j\times j=k\times k=0$

b. $r=2i-3k$

$r-r_1=(0.5j-2k)-(2i-3k)=-2i+0.5j+k$

Torque about point (2,0,-3)

$\tau=(-2i+0.5j+k)\times (2i-3k)$

$\tau=-6j-k-1.5i+2j=-1.5i-4j-k$ N-m

6 0

3 years ago