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hammer [34]
3 years ago
12

PLEASE HELP When astronauts eliminate waste in space, which is likely to be used?  [A] showers and faucets, spraying hot water..

... [B] vacuum to secure solid and liquid waste.... [C]wet towels that are later laundered.... [D] radiation to disinfect their bodies. WILL GIVE 9000000000000000000000 POINTS :) :) :)..................
Physics
1 answer:
lesya [120]3 years ago
3 0

Answer:

B. vacuum to secure solid and liquid waste

Explanation:

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You have 4.9 g of an unknown substance. To identify the substance, you decide to measure its specific heat and find that it requ
kifflom [539]

The unknown substance can be lithium, which has a specific heat capacity of approximately 3.5 J/gK.

Explanation:

When heat energy is supplied to a certain substance, the temperature of the substance increases according to the equation:

Q=mC_s \Delta T

where

Q is the amount of energy supplied

m is the mass of the sample

C_s is the specific heat capacity of the substance

\Delta T is the change in temperature

In this problem, we have

m = 4.9 g is the mass

Q = 668.85 J is the specific heat capacity

\Delta T = 39 K is the change in temperature

Solving for C_s, we find the specific heat capacity of the substance:

C_s = \frac{Q}{m\Delta T}=\frac{668.85}{(4.9)(39)}=3.5 J/gK

Looking at tables of specific heat capacity, we can see that the unknown substance can be lithium, which has a specific heat capacity of approximately 3.5 J/gK.

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0
3 years ago
The spherical annulus of cometary nuclei surrounding the solar system is called the ___.
Bingel [31]

Answer:

Oort Cloud

Explanation:

  • Hundreds of millions of comet nuclei forming a spherical region that surrounds the solar system is called Oort cloud.
  • The Oort cloud revolves at tens of thousands of the distance of the earth from the sun.
  • Comets are the condensed water and dust particles that orbit around the sun with inward helix leaving behind a tail of vapours and vaporize completely as they get closer to the sun during their revolution around it.
  • They have a nebulous appearance and extremely elongated orbits.

3 0
3 years ago
A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet
lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

  • <em>Current flowing in the wire, I = 4.00 mA</em>
  • <em>Initial diameter of the wire, d₁ = 4 mm = 0.004 m</em>
  • <em>Final diameter of the wire, d₂ = 1 mm = 0.001 m</em>
  • <em>Length of wire, L = 2.00 m</em>
  • <em>Density of electron in the copper, n = 8.5 x 10²⁸ /m³</em>

<em />

The initial area of the copper wire;

A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2

The final area of the copper wire;

A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2

The initial drift velocity of the electrons is calculated as;

v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s

The final drift velocity of the electrons is calculated as;

v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7}  \ m/s

The change in the mean drift velocity is calculated as;

\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s

The time of motion of electrons for the initial wire diameter is calculated as;

t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s

The time of motion of electrons for the final wire diameter is calculated as;

t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s

The average acceleration of the electrons is calculated as;

a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0
2 years ago
Monarch butterflies can sustain air speeds of 4.8 m/s while migrating. A certain monarch wants to fly to a location due east 4.7
AnnyKZ [126]
It will take the butterfly 2.3 km to reach the destination
6 0
3 years ago
Read 2 more answers
Olivia wants to find out whether a substance will fluoresce. She says she should put it in a microwave oven. Do you agree with h
andreyandreev [35.5K]
Disagree.
Fluoresce objects will only glow when put under actual Ultraviolet light. This is due to the molecules becoming excited by the ultraviolet radiation.


Microwaves give micro-waves that are present in another spectrum of wave length and will not be able to fluoresce the molecules. If it’s not “ultra violet “.... it’s not going to glow.
4 0
2 years ago
Read 2 more answers
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