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mihalych1998 [28]
3 years ago
11

Using a hammer to drive a nail, it is the reaction force that brings the motion of the hammer to a stop but the _______________

that drives the nail into the wood.
Physics
2 answers:
4vir4ik [10]3 years ago
8 0

The correct answer for the question that is being presented above is this one: "action force." Using a hammer to drive a nail, it is the reaction force that brings the motion of the hammer to a stop but the action force that drives the nail into the wood.

Explanation:

The reaction force is what presents you drive because it acts on you. Newton's Third Law of Motion describes that forces perpetually come in action-reaction pairs. The Third Law states that for every action force, there is an identical and opposite reaction force. The ball exerts an equal and opposite force on the bat.

statuscvo [17]3 years ago
6 0
The correct answer for the question that is being presented above is this one: "action force." Using a hammer to drive a nail, it is the reaction force that brings the motion of the hammer to a stop but the action force that drives the nail into the wood.
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What information about elements can be collected from the periodic table?
Dafna1 [17]

Answer:

Using the data in the table scientists, students, and others that are familiar with the periodic table can extract information concerning individual elements. For instance, a scientist can use carbon's atomic mass to determine how many carbon atoms there are in a 1 kilogram block of carbon.

Explanation:

HOPE THIS HELPS LIKE AN RATE PLZ

3 0
3 years ago
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A shield of material that reflects neutrons is placed around a radioactive sample. Which change in the nuclear reaction is most
Crazy boy [7]

Answer:

A chain reaction will be sustained in a sub-critical mass.

Explanation:

Hope this helps!

If not, I am sorry.

3 0
2 years ago
A 45-year-old man is meeting his doctor for his annual physical. He suffers from Type 2 diabetes and has a family history of cor
shutvik [7]

Answer:

The man is on the verge of having a heart attack or a stroke.

Explanation:

If he has a family history of coronary (heart) disease, it means it could normally affect. Normally here means without anything aggravating it. It's already in his lineage so he could have it.

Now, he's past middle age - he's 45. He's past the growing stages of life. His organs are fully developed herefore.

Now also, he suffers from Type 2 diabetes. Although this is sometimes milder than Type 1 diabetes, it increases the risk of having a heart disease or a stroke!

Soda, especially sweetened one, is not to be taken too often because it can cause Diabetes Mellitus. For a diabetes patient, this should be a "no-go-area". Taking this constantly (everyday at work) will now put this 45-year-old man in harm's way.

He is no more at risk of having complications but already on the path to a heart disease or a stroke.

4 0
3 years ago
The density for gold is 19.3 g/cm3. What would be the mass of a 45 cm3 piece of gold?
lyudmila [28]

Answer:

868.5 g

Explanation:

Mass= Density x Volume

Mass= 19.3 x 45

=868.5

6 0
3 years ago
Read 2 more answers
Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not
d1i1m1o1n [39]

Answer:

a) 3000 Hz;

b) 30 dB;

c) 1000 times.

Explanation:

a) From the human audiogram given on the figure below the black line represents the threshold for hearing the sound at each frequency. We see that the least intensity is necessary for the frequency of about 3000 Hz.

b) Using the same audiogram we see that we would need the sound of the intensity of about 30dB.

c) The least perceptible sound at 1000 Hz must be 0dB while at 100 Hz it is 30dB. These are logarithmic quantities. To transform them to the linear quantities we use the formula

I(\text{in dB})=10\log\frac{I}{I_0(\text{at }1000\text{ Hz})},

where  I_0(\text{at }1000\text{ Hz}) is the hearing threshold at 1000 Hz.

Therefore we have the following

0\text{ dB}=10\log\frac{I_1}{I_0(\text{at }1000\text{ Hz})}\quad 30\text{ dB}=10\log\frac{I_2}{I_0(\text{at }1000\text{ Hz})}

I_1 is the threshold at 1000Hz and I_2 is the threshold at 100Hz.

By exponentiating we have

10^0=\frac{I_1}{I_0(\text{at }1000\text{ Hz})},\quad 10^3=\frac{I_2}{I_0\text{at }1000\text{ Hz}}.

Now dividing these two equations we get

\frac{I_2}{I_1}=\frac{10^3}{10^0}=1000.

Therefore, the least perceptible sound at 100Hz is 1000 times more intense than the least perceptible sound at 1000Hz.

Note: I got these values unisng the audiogram that is attached here. The one that you have might be slightly different and might yield different answers.

7 0
3 years ago
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