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3 years ago

Crews at the International Space Station are researching the effects of the weightlessness of space on ________.

Physics

2 answers:

Natalija [7]3 years ago

8 0

The answer is A. The human body

andrezito [222]3 years ago

6 0

A: Human Body

C is wrong because they don’t have the tools to test it on another planet

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which of the following statements best explains how consumers determine growth in technological areas

Mrac [35]

If these were the missing choices:

a) Consumers fill out questionnaires concerning their need for new products.

b) Consumers vote for politicians who decide which kind of research to support

c) Consumers decide what to buy and what not to buy

d) Consumers influence the decisions of private foundations by deciding where to donate money.

My answer would be: c) Consumers decide what to buy and what not to buy

Every growth is based on the demand of the people. If a good or service is needed then its demand will increase. If a good or service is not needed then its demand will decrease until such time that said good or service will be eliminated.

6 0

3 years ago

1.which of the following are true.

Ksivusya [100]

B is the correct answer

3 0

3 years ago

Read 2 more answers

two forces whose magnitude are in ratio of 3:5 gives a resultant of 35N.if the angle of inclination is 60degree.calculate the ma

nadya68 [22]

Answer:

the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

Explanation:

The computation of the magnitude of the each force is shown below:

Provided that

Ratio of forces = 3: 5

Let us assume the common factor is x

Now

first force = 3x

And, the second force = 5x

Resultant force = 35 N

The Angle between the forces = 60 degrees

Based on the above information

Resultant force i.e. F = √ F_1^2 +F_2^2 + 2 F_1F_2cos $\theta$

35 = √[(3x)²+ (5x)²+ 2 (3x)(5x) cos 60°]

35 =√ 9x² + 25x² + 15x² (cos 60° = 0.5)

35 = √49 x²

x = 5

So, the magnitude of first force = 3 × 5= 15 N

ANd, the magnitude of second force = 5 × 5 = 25 N

7 0

3 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

3 years ago

A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat

AleksandrR [38]

Answer:

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

$L_i = L_f$

now we have

$L_i = (\frac{1}{2}MR^2)\omega_o$

also we have

$L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now from above equation we have

$(\frac{1}{2}MR^2)\omega_o = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega$

now we have

$\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}$

$\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}$

6 0

3 years ago