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Kay [80]
3 years ago
13

Chromosomes usually appear in pairs. If there are three chromosomes where there should be two, what kind of mutation is this?

Physics
2 answers:
WITCHER [35]3 years ago
8 0
The answer is genetic.
Dvinal [7]3 years ago
8 0
The answer here is sex cells
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Paleobotany is the study of fossil insects. True False
alexira [117]

Answer:

False

Explanation:

Paleobotany is the study of fossil plants.

6 0
3 years ago
Read 2 more answers
At an amusement park there is a ride in which cylindrically shaped chambers spin around a central axis. People sit in seats faci
ozzi

Answer:

r=2.4m

Explanation:

We have to use the centripetal force  equation

Fc=\frac{mv^{2} }{r}

we  need the radious so we have to isolate "r" and we get

r=\frac{mv^{2} }{Fc}

replacing m=65 kg, v= 4.1 m/s and Fc=455N we get

r=\frac{65*4.1^{2} }{455}

r=2.4m

The radius of the amusement park chamber is 2.4m

4 0
3 years ago
What is the most appropriate tool to use when collecting data from the investigation? List, table, graph or chart?
forsale [732]
A table would be the most appropriate because that way you can compare the data.
6 0
3 years ago
Two teams of nine members each engage in tug-of-war. Each of the first team's members has an average mass of 68 kg and exerts an
diamong [38]

Answer:

(a) Acceleration  = 0.1063 m/s^2      (Second team wins)

(b) Tension in rope = 65.106 N

Explanation:

Total mass of first team = 68 * 9 = 612 kg

Total force of first team = 1350 * 9 = 12150 N

Total mass of second team = 73 * 9 = 657 kg

Total force of seconds team = 1365 * 9 = 12285 N

Difference in force = 12285 - 12150 = 135 N   (towards the second team as it has more force)

(a) For acceleration we get:

F = m * a

135 = (mass of both teams) * a

a = 135 / (612 + 657)

acceleration  = 0.1063 m/s^2      (Second team wins)

(b) Since we know the acceleration of the first team (pulling being pulled towards the second team at an acceleration of 0.1063 m/s^2) , we can find out the force required to move them:

Force required for first team = mass of first team * acceleration

Force required = 612 * 0.1063

Force required = 65.106 N

This is the force exerted on the first team through the rope, so the tension in the rope will also be 65.106 N.

7 0
4 years ago
A mixture of nitrogen and xenon gases, at a total pressure of 836 mm Hg, contains 2.80 grams of nitrogen and 24.9 grams of xenon
larisa86 [58]

Answer: Partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

Explanation:

The partial pressure of a gas is given by Raoult's law, which is:

p_A=p_T\times \chi_A

where,

p_A = partial pressure of substance A

p_T = total pressure

\chi_A = mole fraction of substance A

We are given:

m_{N_2}=2.80g

m_{Xe}=24.9g

Mole fraction of a substance is given by:

\chi_A=\frac{n_A}{n_A+n_B}

And,

n_A=\frac{m_A}{M_A}

Mole fraction of nitrogen is given as:

\chi_{N_2}=\frac{\frac{m_{N_2}}{M_{N_2}}}{(\frac{m_{N_2}}{M_{N_2}}+\frac{m_{Xe}}{M_{Xe}})}

Molar mass of N_2 = 28 g/mol

Molar mass of Xe =  g/mol

Putting values in above equation, we get:

\chi_{N_2}=\frac{\frac{2.80}{28}}{\frac{2.80}{28}+\frac{24.9}{131}}

\chi_{N_2}=\frac{0.100}{0.100+0.190}=0.345

To calculate the mole fraction of xenon, we use the equation:

\chi_{Xe}+\chi_{N_2}=1\\\\\chi_{Xe}=1-0.345=0.655

p_{N_2}=836mmHg\times 0.345=288mmHg

p_{Xe}=836mmHg\times 0.655=548mmHg

Thus partial pressure of nitrogen and xenon are 288mmHg and 548 mmHg respectively.

6 0
4 years ago
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