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eimsori [14]
3 years ago
11

Many machines-including inclined planes such as ramps- increase the strength of the force put into the machine but __________ th

e distance over which the force is applied.
A.) Increase

B.) Decrease

C.) Does not change

D.) None of the above
Physics
1 answer:
Leviafan [203]3 years ago
5 0
<h3><u>Answer;</u></h3>

B.) Decrease

<h3><u>Explanation;</u></h3>
  • Many machines including inclined planes such as ramps; increase the strength of the force put into the machine but <u>decrease </u>the distance over which the force is applied.
  • Other machines increase the distance over which the force is applied but decrease the strength of the force. Also other machines change the direction of the force, with or without also increasing its strength or distance.
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Which statements describe nuclear reactions? Check all that apply.
Alexandra [31]

Explanation:

Nuclear reactions are the reactions in which nucleus of an atom changes either by splitting or joining with the nucleus of another atom.

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  • Nuclear fission - In this process, large atomic nuclei splits into smaller nuclei.  
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Both nuclear fission and fusion processes involve nuclei of atoms.

For example, ^{0}n + ^{235}_{92}U \rightarrow ^{236}_{92}U \rightarrow ^{144}_{56}Ba + ^{89}_{36}Kr + 3n + 177 MeV

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7 0
3 years ago
Read 2 more answers
A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
Bingel [31]

Answer:

E_T= 28J

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2

Where:

\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

8 0
4 years ago
The magnitude​ R, measured on the Richter​ scale, of an earthquake of intensity I is defined as Requalslog StartFraction Upper I
lapo4ka [179]

Answer:

R = 6.8

Explanation:

Given data:

Richter scaleR = log(\frac{I}{I_o})

where R - magnitude of earthquake of Richter scale

I - quake's intensity =  10^{6.8} \times I_o

I_o - minimum intensity earthquake

Plugging all information in the equation to get Richter's scale

R = log(\frac{10^{6.8} \times I_o}{I_o})

R = log(10^{6.8})

R = 6.8

6 0
3 years ago
What is the energy released from nuclear fuels inside a nuclear reactor used for
belka [17]
Each time a uranium nucleus splits<span> up it releases energy and three neutrons. If all the neutrons are allowed to be absorbed by other uranium nuclei the chain reaction will spiral out of control causing an explosion. To control the energy released in the reactor moveable control rods are placed between the fuel rods.</span>
6 0
4 years ago
A particle of charge 3.0 x 10 C experiences an upward force of magnitude 4.8 x 10-6N when it is placed in a particular point in
MA_775_DIABLO [31]

Answer: a) 1.6 * 10 ^-7 N/C (upward) ; b) -2.5*10^-6N (downward)

Explanation: In order to solve this proble we have to use the Coulomb law given by:

F=q*E from this expression we have

E=F/q=4.8*10^-6/30 C= 1.6 * 10 ^-7 N/C

The force on the particle charge by -1.6 X10 C place intead of the initial charge we have

F=q*E= -16 C* 1.6 * 10 ^-7 N/C= -2.5*10^-6N

6 0
3 years ago
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