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const2013 [10]
4 years ago
15

Explain what the interviewer meant when he said that genetic mutations are like misprints in a book.

Physics
1 answer:
asambeis [7]4 years ago
7 0

Answer:

A publisher can misprint a book in several different ways. The publisher can miss words, make spelling errors, mix up paragraphs, or delete some text entirely. With such errors, the text will be inaccurate, and people reading that part will get confused. Mutations in genes happen similarly. At times, parts of the gene may change entirely, leading to altered products. There are many types of gene mutations, but all types of mutations can potentially affect the cell and its function.

Explanation:

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all of the following answer choices are ideas from daltons atomic theory. which one of them do we now know is not true?
dem82 [27]
Did you forget to include the options?
5 0
3 years ago
What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h
Citrus2011 [14]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
C_s is its specific heat capacity
\Delta T is the increase of temperature

The sample of silver of our problem has a mass of m=24.7 kg. Its specific heat capacity is C_s = 236 J/g^{\circ}C and the increase in temperature is
\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C

Therefore, the amount of heat needed is
Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J
8 0
4 years ago
An object weighs 60.0 kg on the surface of the earth. How much does it weigh 4R from the surface? (5R from the center)
Alecsey [184]
"60 kg" is not a weight.  It's a mass, and it's always the same
no matter where the object goes.

The weight of the object is   

                                 (mass) x (gravity in the place where the object is) .

On the surface of the Earth,

                   Weight = (60 kg) x (9.8 m/s²)

                                =      588 Newtons.

Now, the force of gravity varies as the inverse of the square of the distance from the center of the Earth.
On the surface, the distance from the center of the Earth is 1R.
So if you move out to  5R  from the center, the gravity out there is

                    (1R/5R)²  =  (1/5)²  =  1/25  =  0.04 of its value on the surface.

The object's weight would also be 0.04 of its weight on the surface.

                 (0.04) x (588 Newtons)  =  23.52 Newtons.

Again, the object's mass is still 60 kg out there.
___________________________________________

If you have a textbook, or handout material, or a lesson DVD,
or a teacher, or an on-line unit, that says the object "weighs"
60 kilograms, then you should be raising a holy stink. 
You are being planted with sloppy, inaccurate, misleading
information, and it's going to be YOUR problem to UN-learn it later.
They owe you better material.
6 0
3 years ago
Henry ran a 100 m race. Use the graph to answer the following:
horrorfan [7]

Using the graph, which describes how Henry ran the 100m race;

a) It takes Henry 20seconds to run 100m

b) Henry's average speed over the race is; 5m/s.

According to the linear graph which describes the distance ran by Henry during the 100m race as a function of time.

a) Since the distance from start ran by Henry is plotted on the vertical axis, and the time is plotted on the horizontal axis;

To determine how long it took Henry to run 100m; The point corresponding to 100m is traced downward from the line of the graph and we find out that;

It takes Henry 20seconds to run 100m

b) Henry's average speed over the race is simply;

The slope of the distance-time graph.

Therefore,

  • Average speed = (100-0)/(20-0)

  • Average speed = 100/20

  • Average speed = 5m/s.

Therefore, Henry's average speed over the race is; 5m/s.

Read more:

brainly.com/question/22125199

6 0
2 years ago
The construction of a flat rectangular roof (6.17 m × 5.92 m) allows it to withstand a maximum net outward force of 2.00 × 104 N
Blababa [14]

Answer:

v_2=29.13\ m/s

Explanation:

It is given that,

Dimension of the rectangular roof, (6.17 m × 5.92 m)

The maximum net outward force, F=2\times 10^4\ N

The density of air, \rho=1.29\ kg/m^3

The Bernoulli equation is used to find wind speed of this roof blow outward. It is given by :

P_1+\dfrac{1}{2}\rho v_1^2=P_2+\dfrac{1}{2}\rho v_2^2

Here, v_1=0 (since air inside the roof is not moving)

v_2=\sqrt{\dfrac{2(P_1-P_2)}{\rho}}

Since, F=(P_1-P_2)A

v_2=\sqrt{\dfrac{2F}{\rho A}}

v_2=\sqrt{\dfrac{2\times 2\times 10^4}{1.29\times 6.17\times 5.92 }}

v_2=29.13\ m/s

So, the wind speed of  this roof blow outward is 29.13 m/s. Hence, this is the required solution.

5 0
4 years ago
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