How is energy transferred when
hitting a nail?​

Answer:

a) A:170572.5 J

   C: 55794.9J

b) 170572.5 J

c) 41.4413265306m

d) 2.7455874717m/s

Explanation:

a) Kinetic energy = 0.5*m*v²

KE at A = 0.5*420*28.5² = 170572.5 J

KE at C = 0.5*420*16.3² = 55794.9 J

b) Mechanical energy is the total kinetic energy plus potential energy at a point on the system. There is no potential energy at A.

ANSWER: 170572.5 J

c) v²=u²+2as

28.5²=2(9.8)s

812.25/19.6=s

s=41.4413265306m

d) h=height from part c, r=radius of loop

v²=u²+2as

v²=gr or a=v²/r

Ei=Ef

mgh=0.5mv²+mg(2r)

gh=0.5v²+2gr

h=0.5r+2r

h=5/2r

r=2/5h=(2/5)(41.4413265306)=16.5765306122

F=ma

mg=m(v²/r)

g=v²/r

v²=(9.8)(16.5765306122)

v=√162.45

=12.7455874717m/s

Answer:

5766.7 K

Explanation:

We are given that

Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]

Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]

Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]

We have to find the temperature at the surface of the Sun.

We know that

Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]

Where [tex]K_{sc}=1350 W/m^2[/tex]

[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]

Using the formula

[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]

T=5766.7 K

Hence, the temperature at the surface of the sun=5766.7 K

Answer:

- 21⁰C .

Explanation:

Speed of jet = 2.05 x 10³ km /h

= 2050 x 1000 / (60 x 60 ) m /s

= 569.44 m / s

Mach no represents times of speed of sound , the speed of jet

1.79 x speed of sound = 569.44

speed of sound = 318.12 m /s

speed of sound at 20⁰C = 343 m /s

Difference = 343 - 318.12 = 24.88⁰C

We know that 1 ⁰C change in temperature changes speed of sound

by .61 m /s

So a change in speed of 24.88 will be produced by a change in temperature of

24.88 / .61

= 41⁰C  

temperature = 20 - 41 = - 21⁰C .  

Answer:

240 meters

Explanation:

The distance traveled by the vehicle can be calculated using the following equation:

[tex] v_{f}^{2} = v_{0}^{2} + 2ax [/tex]   (1)

Where:

x: is the displacement

[tex]v_{f}[/tex]: is the final speed = 0 (reduces its velocity back to zero)                    

[tex]v_{0}[/tex]: is the initial speed = 60 m/s

a: is the acceleration = -7.5 m/s²

By solving equation (1) for x we have:

[tex] x = \frac{v_{f}^{2} - v_{0}^{2}}{2a} = \frac{0 - (60 m/s)^{2}}{2*(-7.5 m/s^{2})} = 240 m [/tex]

Therefore, the vehicle undergoes 240 meters of displacement during the deceleration period.

I hope it helps you!

Answer:

the diameter of the bright circular spot formed is 0.787 m  

Explanation:

Given that;

Radius of the flat circular mirror = 0.100 m

height of small ight source = 0.920 m

high ceiling = 2.70 m  

now;

Diameter(mirror) = 2×r = 2 × 0.100 = 0.2 m

D(spot) = [Diameter(mirror) × ( 2.70m + 0.920 m)] /  0.920 m

so

D(spot) = 0.2m × 3.62m /  0.920 m

D(spot) = 0.724 m / 0.920 m

D(spot) = 0.787 m  

Therefore, the diameter of the bright circular spot formed is 0.787 m  

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Data Given:

Electric Field between two parallel plates = 628 N/C

Separation = 4.22 cm

a) In this part, we are asked to calculate the distance from positive plate at which the electron and proton pass each other.

Solution:

First of all:

Force on proton due to the Electric field between the plates is:

[tex]F_{p}[/tex] = [tex]q_{p}[/tex]E

and, we know that, F = ma

So,

[tex]m_{p}[/tex]a = [tex]q_{p}[/tex]E

a = [tex]\frac{q_{p}.E }{m_{p} }[/tex]      Equation 1

So,

The distance covered by the electron is:

S = ut + 1/2[tex]at^{2}[/tex]

Here, u = 0.

S = 1/2[tex]at^{2}[/tex]

Put equation 1 into the above equation:

S = 1/2 x ([tex]\frac{q_{p}.E }{m_{p} }[/tex]  )[tex]t^{2}[/tex]      Equation 2

So,  

Similarly, the distance covered by electron will be:

(D-S) = 1/2 x ([tex]\frac{q_{e}.E }{m_{e} }[/tex]  )[tex]t^{2}[/tex]    Equation 3

We know that the charge of electron is equal to the charge of proton so,

[tex]q_{p}[/tex] = [tex]q_{e}[/tex] = q

By dividing the equation 2 by equation 3, we get:

[tex]\frac{S}{D-S}[/tex] = [tex]\frac{m_{e} }{m_{p} }[/tex]

Solve the above equation for S,

S[tex]m_{p}[/tex] = [tex]m_{e}[/tex]D - [tex]m_{e}[/tex]S

So,

S = [tex]\frac{m_{e}.D }{(m_{e} + m_{p}) }[/tex]

Plugging in the values,

As we know the mass of electron is 9.1 x [tex]10^{-31}[/tex] and the mass of proton is 1.67 x [tex]10^{-27}[/tex]

S = [tex]\frac{9.1 . 10^{-31} . 4.22 }{(9.1 . 10^{-31} + 1.67 . 10^{-27} }[/tex]

S = 0.002298 cm (Distance from the positive plate at which the two pass each other)

b) In this part, we to calculate distance for Sodium ion and chloride ion as above.

So,

we already have the equation, we need to put the values in it.

So,

S = [tex]\frac{m_{Cl}.D }{(m_{Cl} + m_{Na}) }[/tex]

As we know the mass of chlorine is 35.5 and of sodium is 23

S = [tex]\frac{35.5 . 4.22}{(35.5 + 23)}[/tex]

S = 2.56 cm

Answer:

Thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Explanation:

From the concept of Escape Velocity, derived from Newton's Law of Gravitation, definition of Work, Work-Energy Theorem and Principle of Energy Conservation, which is the minimum speed such that rocket can overcome gravitational forces exerted by the Earth, and according to the Tsiolkovski's Rocket Equation, which states that thrust done by the rocket is equal to the change in linear momentum of the rocket itself, we conclude that thrust due to fuel consumption must overcome gravitational force from the Earth to send the rocket up into space.

Answer:

La aceleración del autobús es -5.80 m/s².

Explanation:

Podemos encontrar la aceleración del autobús usando la siguiente ecuación:

[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]

Where:

[tex]v_{f}[/tex]: es la velocidad final = 0 (se detiene al final)

[tex]v_{0}[/tex]: es la velocidad inicial = 95 km/h

d: es la distancia recorrida = 60 m

Por lo tanto, la aceleración es:

[tex] a = \frac{v_{f}^{2} - v_{0}^{2}}{2d} = \frac{0 - (95 \frac{km}{h}*\frac{1000 m}{1 km}*\frac{1 h}{3600 s})^{2}}{2*60 m} = -5.80 m/s^{2} [/tex]

El signo negativo se debe a que el autobús está desacelerando (hasta que se detiene).

Entonces, la aceleración del autobús es -5.80 m/s².

Espero que te sea de utilidad!                      

Answer:

A)  q = -8.488 cm ,  B)  m = 0.29

Explanation:

A) For this exercise in geometric optics, we will use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where p and q are the distance to the object and image, respectively and f is the focal length

in our case the distance the object is p = 29 cm the focal length of a diverging lens is negative and indicates that it is f = - 12 cm

         [tex]\frac{1}{q} = \frac{1}{f} - \frac{1}{p}[/tex]

we calculate

          [tex]\frac{1}{q} = - \frac{1}{12} - \frac{1}{29}[/tex]

          [tex]\frac{1}{q}[/tex] = - 0.1178

          q = -8.488 cm

the negative sign indicates that the image is virtual

B) the magnification is given

          [tex]m = \frac{h'}{h} = - \frac{q}{p}[/tex]

we substitute

          m = [tex]- \frac{-8.488}{29}[/tex]

          m = 0.29

the positive sign indicates that the image is right

Answer:

it will option option A hope it helps

Answer:

Explanation:

Sandy hears 8 such clicks every 3 seconds and a small twig, caught in the spokes, causes the tire to click once each revolution that means the wheel of the cycle is rotating at 8 rotations every 3 seconds or 8/3 rotation per second . In each rotation , it moves distance equal to its circumference .

circumference = 2π r = 2 x 3.14 x .65 / 2 m

= 2.041 m

In 8/3 rotation , distance covered = 8/3 x 2.041 = 5.44 m

So speed of cycle is 5.44 m per second

5.44 x 60 x 60 m per hour

19584 m per hour

= 19.584 km per hour .

= 20 km per hour approx.

Answer:

E = 1,720,779.221 or 1.720779221 * 10^ 6V/m

Explanation:

The electric field between the parallel conducting plates is given by

E =V / d

where V is the potential difference and d is the distance between the plates.

E = 26.5 kV/ 1.54 cm

Now we have to convert into proper units

26.5 kv= 26.5 * 1000 v=  26500 volts

1  kv= 1000 volts

1.54 cm = 1.54/ 100 m= 0.0154m

1m = 100cm

Now putting the values

E= 26500/0.0154 = 1,720,779.221 V/m

The Electric field is equal to E= 1,720,799.221 or 1.7220799221 * 10 ^6 Volts per meter.

In scientific notation this can be written as 1.7220799221 *10^6 V/m

Answer:

2.59m/s

Explanation:

Using the equation of motion

v = u+at

v is the final velocity = 35ms

u is the initially velocity = 9m/s

t is the time = 13.5s

a is the acceleration

Substitute into the formula

35 = 0+13.5a

a = 35/13.5

a = 2.59m/s²

Hence the acceleration of the motorcycle is 2.59m/s

Answer:

P.E = 98 Joules

Explanation:

Given the following data;

Mass = 5kg

Speed = 4m/s

Height = 2m

We know that acceleration due to gravity is equal to 9.8m/s²

To find the potential energy;

Potential energy can be defined as an energy possessed by an object or body due to its position.

Mathematically, potential energy is given by the formula;

[tex] P.E = mgh[/tex]

Where, P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the equation, we have;

[tex] P.E = 5*9.8*2[/tex]

P.E = 98 Joules

Answer:

c

Explanation:

a vector quantity has both magnitude and direction

The value of  20m/s, east is a vector quantity is Hence, option (C) is correct.

A physical quantity that has both directions and magnitude is referred to as a vector quantity.

A lowercase letter with a "hat" circumflex, such as "û," is used to denote a vector with a magnitude equal to one. This type of vector is known as a unit vector.

Given values  200V, 100kg/m, 50J/(kg°C) are denoting magnitude of different physical quantity. Hence, they and scalar quantity ( Physical quantities with merely magnitude and no direction are referred to as scalar quantities. These physical quantities can be explained just by their numerical value without any further guidance.).

But The value of  20m/s, east has a magnitude of 20 m/s and a direction along east. Hence, 20m/s, east denotes a  vector quantity is Hence, option (C) is correct.

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Answer:

average speed is 1 m/s and average velocity is 0 m/s.

Explanation:

Given that :

Length of round trip = 50 m

Time taken = 100 seconds

The average speed :

Total distance / total time taken

Length of complete round trip :

(50 + 50) m, total. Distance = 100 m

100 / 100 = 1m/s

The average velocity :

Total Displacement / total time taken

Total Displacement of round trip = end point - start point = 0

0 / 100 = 0

Average speed is 1 m/s and average velocity is 0 m/s.

The average speed is defined as the ratio of distance to time. Speed is a scalar quantity hence it does not take direction into account while velocity is a vector quantity hence it takes direction into account.

The speed is obtained from;

Speed = Distance/time = 2(50 m)/100 s = 1 m/s.

The velocity is 0 m/s since it is complete round-trip lap.

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Answer:

3kg

Explanation:

Given parameters:

Force  = 9N

Acceleration  = 3m/s²

Unknown:

Mass = ?

Solution:

From Newton's second law of motion:

        Force  = mass x acceleration

So;

             9  = mass x 3

             mass  = 3kg

Answer:

2.88m/s

Explanation:

Given parameters:

Displacement  = 7.2m

Time taken  = 2.5s

Unknown:

Velocity of the plane  = ?

Solution:

Velocity is the displacement divided by the time taken.

  Velocity  = [tex]\frac{displacement}{time taken}[/tex]  

 So;

   Velocity  = [tex]\frac{7.2}{2.5}[/tex]    = 2.88m/s

Answer:

The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Explanation:

When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Answer:

30J

Explanation:

Given parameters:

Mass of hamster  = 0.104kg

Velocity  = 24m/s

Unknown:

Kinetic energy  = ?

Solution:

Kinetic energy is the energy due to the motion of a body. It is mathematically derived by;

  Kinetic energy  = [tex]\frac{1}{2}[/tex] m v²  

m is the mass

v is the velocity

  Kinetic energy  = [tex]\frac{1}{2}[/tex] x 0.104 x 24²   = 30J

Answer:

Explanation:

a

Answer:

hello your question has a missing information

The other is Δ-connected with an impedance of (60 - j45) ohm per phase.

answer : A) 5A ∠0° ,

               p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

 B) 193.64 v

C) current at load 1 = 2.236 A , current at load 2 = 4.472 A

 D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )

      Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )

Explanation:

First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution

A) determine the current, real power and reactive power delivered by the sending-end source

current power delivered (Is)  =  5A ∠0°

complex power delivered ( s ) = 3vs Is  

                                                  = 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )

also s = p + jQ  ------ ( 2 )

comparing equation 1 and 2

p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

B) determine Line-to-line voltage at the load

Vload = √3 * 111.8

           = 193.64 v

c) Determine current per phase in each load

[tex]I_{l1} = Vl1 / Zl1[/tex]

     = [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A   hence current at load 1 = 2.236 A

[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]  

     = [tex]\frac{111.8<-10.3}{25<-36.87}[/tex]  = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A

D) Determine the Total three-phase real and reactive powers absorbed by each load

For load 1

3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR

for load 2

3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex]  = 1200 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR

The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.

(a) The impedance per phase of the equivalent Y,

[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]

The phase voltage,

[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]

Total impedance from the input terminals,

[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]

The three-phase complex power supplied  [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]  

P =1800 W and Q = 0 VAR delivered by the sending-end source.

(b) Phase voltage at load terminals will be,  

[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]  

The line voltage magnitude at the load terminal,  

[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]    

(c) The current per phase in the Y-connected load,  

[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} ​}[/tex]

The phase current magnitude,  

[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]

(d) The three-phase complex power absorbed by each load,

[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]

The three-phase complex power absorbed by the line is  

[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]

Since, the sum of load powers and line losses,  

[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]

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Answer:

Macroscopic energy is energy at a level of system while microscopic energy is energy at the level of atoms and molecules

Explanation:

1. Macroscopic energy is possessed by a system as whole while microscopic energy is possessed by its constituents’ atoms or molecules.  

2. The common form of macroscopic energy is Kinetic and potential energy while the microscopic form of energy are atomic forces due its random, disordered motion and due to intermolecular forces

3. At microscopic level we consider behaviour of every molecule and in macroscopic approach we consider gross or average effects of various molecular infractions

Answer:

a) k= 3232.30 N / m,  b)  f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

          w = √k/m

the angular velocity is related to the period

          w = 2π / T

we substitute

          T = 2[tex]\pi[/tex]  √m/ k

a) empty car

           k = 4π² m / T²

           k = 4 π² 1310/2 2

           k = 12929.18 N / m

This is the equivalent constant of the short springs

           F1 + F2 + F3 + F4 = k_eq x

           k x + kx + kx + kx = k_eq x

           k_eq = 4 k

           k = k_eq / 4

           k = 12 929.18 / 4

            k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

            m_total = 1360 + 4 70

            m_total = 1640 kg

angular velocity and frequency are related

              w = 2pi f

we substitute

             2 pi f = Ra K / m

in this case the spring constant changes us

             k_eq = 12929.18 N / m

             f = 1 / 2π √ 12929.18 / 1640

             f = π / 2 2.80778

             f = 4,410 Hz

Answer:

v = 8.1 m/s

θ = -36.4º (36.4º South of East).

Explanation:

        [tex]p_{ox} = p_{fx} (1)[/tex]

         ⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]

       [tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]

        [tex]p_{oy} = p_{fy} (4)[/tex]

        ⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]

       [tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]

       [tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]

       [tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]

      ⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)

      ⇒ θ = tg⁻¹ (-0.738) = -36.4º

The question is incomplete, the complete question is;

When an object with an electric charge of −7.0μC is 5.0cm from an object with an electric charge of 4.0μC, the force between them has a strength of 100.7N. Calculate the strength of the force between the two objects if they are 1.7cm apart. Round your answer to 2 significant digits

Answer:

865.1 N

Explanation:

F1 = Kq1q2/r1^2 ---------1

F2 = Kq1q2/r2^2 -------2

We have that;

r1 = 5cm

r2 =1.7 cm

F1 = 100.7 N

Comparing equations 1 and 2

F2 = F1r1^2/r2^2

F2 = 100.7N[(5cm)^2/(1.7cm)^2]

F2= 865.1 N

I believe the answer would be protentional because they have the potential energy in them to lift the hammer.

Answer:

Explanation:

For resistance of a wire , the formula is as follows .

R = ρ L/S

where ρ is specific resistance , L is length and S is cross sectional area of wire .

for first wire resistance

R₁ =  ρ 3L/3a = ρ L/a

for second wire , resistance

R₂ = ρ 3L/6a

= .5 ρ L/a

For 3 rd wire resistance

R₃ = ρ 6L/3a

= 2ρ L/a

For fourth wire , resistance

R₄ = ρ 6L/6a

=  ρ L/a

So the smallest resistance is of second wire .

Its resistance is .5 ρ L/a