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2 years ago

Help me pls I WILL GIVE BRAINIEST I DONT CARE PLS HELP

Physics

2 answers:

borishaifa [10]2 years ago

7 0

Answer:

Explanation:

PilotLPTM [1.2K]2 years ago

3 0

Answer:

A) was reusable

Explanation:

Check this website out for more information about the space shuttle: https://www.nasa.gov/audience/forstudents/k-4/stories/nasa-knows/what-is-the-space-shuttle-k4.html

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During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Luden [163]

10.8 seconds is the correct answ

3 0

3 years ago

What is the lupac name H3C – CHT the H₂ - CH - CH2 CH₂ CH₂ CH3

dimulka [17.4K]

Answer:??

Explanation:

5 0

3 years ago

A 2.0-kg object moving with a velocity of 5.0 m/s in the positive x direction strikes and sticks to a 3.0-kg object moving with

Andrej [43]

Answer:

5.4 J.

Explanation:

Given,

mass of the object, m = 2 Kg

initial speed, u = 5 m/s

mass of another object,m' = 3 kg

initial speed of another orbit,u' = 2 m/s

KE lost after collusion = ?

Final velocity of the system

Using conservation of momentum

m u + m'u' = (m + m') V

2 x 5 + 3 x 2 = ( 2 + 3 )V

16 = 5 V

V = 3.2 m/s

Initial KE = $\dfrac{1}{2}mu^2 + \dfrac{1}{2}m'u'^2$

= $\dfrac{1}{2}\times 2\times 5^2 + \dfrac{1}{2}\times 3 \times 2^2$

= 31 J

Final KE = $\dfrac{1}{2} (m+m')V^2 = \dfrac{1}{2}\times 5 \times 3.2^2 = 25.6 J$

Loss in KE = 31 J - 25.6 J = 5.4 J.

4 0

3 years ago

Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur

Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

$\mu_{J,T}=(\frac{dT}{dP})_H$

or,

$\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}$

As per question the formula will be:

$\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}$ .........(1)

where,

$\mu_{J,T}$ = Joule-Thomson coefficient of the gas = $0.13K/atm$

$T_1$ = initial temperature = $19.0^oC=273+19.0=292.0K$

$T_2$ = final temperature = ?

$P_1$ = initial pressure = 200.0 atm

$P_2$ = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

$0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}$

$T_2=266.12K$

Therefore, the final temperature of gas is 266.12 K

5 0

3 years ago

Answer True or Flase1-Electric potential due to a uniform E field doesn’t change with location.2-The equipotential surfaces asso

TEA [102]

Answer:

1. False

2. True

3. True

Explanation:

1- False —> The relation between electric potential and electric field is given such that

$-\int\limits^a_b \vec{E}d\vec{l} = V_{ab}$

Therefore, for a uniform E field, electric potential is linearly proportional to the distance.

2- True —> The electric field lines always cross the equipotential lines perpendicularly.

3- True —> In order to be a potential difference, one source of electric field is enough. The electric potential will decrease radially according to the following formula:

$V = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$

There is no test charge in the formula, only the source charge. Even when there is no test charge, the potential difference between points in space can exist.

3 0

3 years ago