All categories

3 years ago

Which of the following is a potential explanation to resolve the Fermi paradox?

Physics

1 answer:

aksik [14]3 years ago

8 0

Haiagakavsusbwosvahaoabhsosvsnajsksbzjd sos Kansas

You might be interested in

A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field

Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

8 0

3 years ago

Bailey wants to find out which frozen solid melts the fastest: soda, ice, or orange juice. She pours each of the three liquids i

777dan777 [17]

Answer:

its 45 over 6

Explanation:the answer is in the question

8 0

3 years ago

Read 2 more answers

A spring has a spring constant of 105 N/m. If you compress the spring 0,1 m

Basile [38]

(105 N/m) x (0.1 m) = <em>10.5 Newtons</em>.

4 0

4 years ago

Guys please help me on the rest of the numbers

wel

Answer:

C. 100

D.3

E. 33.3

Explanation:

C. Mechanical Advantage=Load / Effort

= 200N

--------

100N

Therefore,. = 100

D. I. Velocity Ratio= distance moved by the effort / distance moved by load

= 30cm/10cm

= 3

II. Efficiency= M.A / V.R

= 100/3

= 33.33

8 0

3 years ago

Hi.

docker41 [41]

<h3>Answer :</h3>

Let the final temperature be "T".

For the piece of copper :

mass, $\sf{m_c=40\ g.}$

specific heat capacity, $\sf{c_c=0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_c=200^{\circ}C.}$

Then the heat of copper :

$\sf{\dashrightarrow Q_c=m_cc_c\,\Delta\!T_c}$

$\sf{\dashrightarrow Q_c =16(T-200)\ J}$

For copper calorimeter :

mass, $\sf{m_{cc} =60\ g.}$

specific heat capacity, $\sf{c_{cc} =0.4\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_{cc} =25^{\circ}C.}$

Then the heat of copper calorimeter :

$\sf{\dashrightarrow Q_{cc} =m_{cc}c_{cc}\,\Delta\!T_{cc}}$

$\sf{\dashrightarrow Q_{cc} =24(T-25)\ J}$

For water :

mass, $\sf{m_w=50\ g. }$

specific heat capacity, $\sf{c_w= 4.2\ J\,g^{-1}\,K^{-1}.}$

initial temperature, $\sf{T_w=25^{\circ}C.}$

Then heat of water :

$\sf{\dashrightarrow Q_w=m_wc_w\,\Delta\!T_w}$

$\sf{\dashrightarrow Q_w=210(T-25)\ J}$

By energy conservation, the sum of all these energies should be zero as there were no heat energy change before the process, i.e.,

$\sf{\dashrightarrow Q_c+Q_{cc}+Q_w=0}$

$\sf{\dashrightarrow16(T-200)+24(T-25)+210(T-25)=0}$

$\sf{\dashrightarrow 250T- 9050=0}$

$\sf{\dashrightarrow T=36.2^{\circ}C}$

$\large \underline{\underline{\boxed{\sf T=36.2^{\circ}C}}}$

<u>____________________________</u>

[Note: in case of considering temperature difference it's not required to convert the temperatures from $\sf{^{\circ}C}$ to K or K to $\sf{^{\circ}C}$ .]

7 0

3 years ago