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3 years ago

A boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s .

Physics

1 answer:

zimovet [89]3 years ago

6 0

Answer:

Force exerted, F = 1.5 N

Explanation:

It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.

i.e. u = 0

v = 30 m/s

Time taken, t = 0.06 s

Mass of the paper, m = 0.003 kg

We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :

$F=m\times a$

$F=m\times (\dfrac{v-u}{t})$

$F=0.003\times (\dfrac{30}{0.06})$

F = 1.5 N

So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.

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Ethan made a diagram to compare examples of the first and second laws of thermodynamics. What belongs in the areas marked X and

Thepotemich [5.8K]

First law of thermodynamics:

It states that "Energy neither be created nor it can be destroyed". simply it converts one form of energy into another form.

It is also known as "law of conservation of energy"

Limitations of First law

It doesn't provide a clear idea about the direction of transfer of heat.
It doesn't provide the information that how much heat energy converted inti work.
Its not given any practical applications.

II law of thermodynamics:

It states that "the total entropy of the system can never decrease over time"

It is strongly proved by two laws, they are

1. Kelvin-plank statement:

He stated that "any engine does not give 100% efficiency". It violates the Perpetual motion of machine II kind(PMM-II).

2. Classius statement: 

  It states that "Heat always flows from high temperature body to low temperature body, without aid of external energy".

Also it stated that " Heat can also be transferred from low temperature body to high temperature body, by the aid of an external energy".

Applications of II law: 

Refrigeration &Air conditioning, Heat transfer, I.C. engines, etc.

3 0

3 years ago

Read 2 more answers

A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic

Mazyrski [523]

Answer:

The total resistance of the wire is = $1.917\times10^{-3}$

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, $R_{Total}$

$\frac{1}{R_{Total}} =\frac{1}{R_{cu} }+\frac{1}{R_{al}}$

Parameters given:

Length of wire = 1 m

Cross sectional area of copper $A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3} )^{2} =3.142\times10^{-6} m^{2}$

Cross sectional area of aluminium wire

$A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3} )^{2}-(1\times 10^{-3} )^{2}] =9.42\times10^{-6} m^{2}\\$

Resistivity of copper $\rho _{cu}=1.7\times 10^{-8} \Omega .m$

Resistivity of Aluminium $\rho _{al}=2.8\times 10^{-8} \Omega .m$

Resistance of copper $R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} } =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega$

Resistance of aluminium $R_{al}= \frac{\rho_{al} \times l}{A_{al} } =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega$

The total resistance of the wire can be obtained as follows;

$\frac{1}{R_{Total}} =\frac{1}{5.41\times10^{-3} }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}$

$R_{Total}= 1.917\times 10^{-3}\Omega$

∴ The total resistance of the wire = $1.917\times 10^{-3}\Omega$

4 0

3 years ago

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Using the equation for decay, calculate the amount left of a radioactive sample amount N0 if the decay constant is 0.00125 secon

makkiz [27]

The amount left of a radioactive sample amount N0 if the decay constant is 0.00125 seconds and the time is 180 seconds is 0.7999 N.

The time it takes for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life T1/2 and the decay constant is given by T1/2 = 0.693/λ.

N=N0e−λt
given λ = 0.00125 seconds
t = 180 seconds
Now putting values.
N=N0e−λt = 0.799
N= 0.7999.