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3 years ago

3

Five workers from the same factory developed angiosarcoma, a rare cancer of the veins and arteries. Which part of the scientific

method is this? Group of answer choices conclusion experiment observation none of these hypothesis

1 answer:

Vinil7 [7]3 years ago

4 0

Answer:

a

Explanation:

i took the test

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Which of the following statements are true? Positively charged objects attract other positively charged objects. Negatively char

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The statement that is true is that positively charged objects attract negatively charged objects. This is due to a law that states 'like forces attract while unlike forces repel. This same concept applies to magnetism. If you put two similar poles together, for example; if you place two south poles together. You feel a separating force between the two poles. But if you place two opposite poles together they attract each other. Hope i helped. <span />

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3 years ago

Suppose that 4 charged particles, all having a positive charge Q, are placed and fixed at the vertices of a square, where every

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3 years ago

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point

Salsk061 [2.6K]

Answer:

The fluids speed at a) $0.105\,m^{2}$ and b) $0.047\,m^{2}$ are $2.33\,\frac{m}{s^{2}}$ and $5.21\,\frac{m}{s^{2}}$ respectively

c) Th volume of water the pipe discharges is: $882\,m^{3}$

Explanation:

To solve a) and b) we should use flow continuity for ideal fluids:

$\Delta Q=0$ (1)

With Q the flux of water, but Q is $Av$ using this on (1) we have:

$A_{2}v_{2}-A_{1}v_{1}=0$ (2)

With A the cross sectional areas and v the velocities of the fluid.

a) Here, we use that point 2 has a cross-sectional area equal to $A_{2}=0.105\,m^{2}$ , so now we can solve (2) for $v_{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.105}\approx2.33\,\frac{m}{s^{2}}$

b) Here we use point 2 as $A_{2}=0.047\,m^{2}$ :

$v_{2}=\frac{A_{1}v_{1}}{A_{2}}=\frac{(0.070)(3.5)}{0.047}\approx5.21\,\frac{m}{s^{2}}$

c) Here we need to know that in this case the flow is the volume of water that passes a cross-sectional area per unit time, this is $Q=\frac{V}{t}$ , so we can write:

$A_{1}v_{1}=\frac{V}{t}$ , solving for V:

$V=A_{1}v_{1}t=(0.070m^{2})(3.5\frac{m}{s})(3600s)=882\,m^{3}$

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4 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:

A)

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)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s _____ eqn (2)

<u>FOR OXYGEN:</u>

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)

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

B)

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

<u>K.E of Oxygen = 4000 J</u>

C)

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)

<u>v = 2149.24 m/s</u>

<u></u>

8 0

4 years ago