A car travels 120 mi averaging a certain speed. If the car had gone 10 mph​ faster, the trip would have taken 1 hr less. Find the​ car's average speed.

Answer:

Moment of inertia of the flywheel is equal to 10.19 kg-m^2

Explanation:

The maximum rotational energy to be stored by the flywheel [tex]E_{r}[/tex] = 2.00 x 10^6 J

Angular speed with which to store this energy ω =  443 rad/s

moment of inertia of the flywheel [tex]I[/tex] = ?

Recall that the energy of a rotating body is gotten from the equation

[tex]E_{r} = Iw^{2}[/tex]

Where [tex]E_{r}[/tex] is the rotational energy of the rotating body

[tex]I[/tex] = moment of inertia of the body

ω = angular speed of the rotating body

imputing the values into the equation, we'll have

2.00 x 10^6 = [tex]I[/tex] x [tex]443^{2}[/tex]

2.00 x 10^6 =  [tex]I[/tex] x 196249

[tex]I[/tex] = (2.00 x 10^6) ÷ 196249 = 10.19 kg-m^2

Answer:

Explanation:

check it out and rate me

Explanation:

Mass and energy are closely related. Due to mass–energy equivalence, any object that has mass when stationary (called rest mass) also has an equivalent amount of energy whose form is called rest energy, and any additional energy (of any form) acquired by the object above that rest energy will increase the object's total mass just as it increases its total energy. For example, after heating an object, its increase in energy could be measured as a small increase in mass, with a sensitive enough scale.

Answer:

Option 5:

The 60W bulb has a greater resistance and a lower current than the 100 W bulb.

Explanation:

We have to compare the resistance and current of both bulbs.

Bulb A

Power = 60 W

Voltage = 110 V

Power is given as:

[tex]P = V^2 /R[/tex]

where V= voltage and R = resistance

[tex]=> 60 = 110^2 / R\\\\R = 201.6 \Omega[/tex]

Power is also given as:

P = IV

where I = current

=> 60 = I * 110

I = 60/110 = 0.54 A

Bulb B

Power = 100 W

Voltage = 110 V

To get resistance:

[tex]100 = 110^2 / R\\\\R = 121 \Omega[/tex]

To get current:

100 = I * 110

I = 100 / 110

I = 0.91 A

Therefore, by comparison, the 60W bulb has a greater resistance and a lower current.

Answer:

λ = 1.4 × 10^(-7) m

Explanation:

We are given;

distance of eye piece from the source;D = 1.5 m

distance between the virtual sources;d = 7.5 × 10^(-4) m

To find the wavelength, we will use the formula for fringe width;

X = λD/d

Where X is fringe width, λ is wavelength, while d and D remain as before.

Now, fringe width = eye-piece distance moved transversely/number of fringes

Eye piece distance moved transversely = 1.88 cm = 1.88 × 10^(-2) m

Thus,

Fringe width = (1.88 × 10^(-2))/10 = 1.88 × 10^(-3) m

Thus;

1.88 × 10^(-3) = λ(1.5)/(7.5 × 10^(-4))

λ = [1.88 × 10^(-3) × (7.5 × 10^(-4))]/1.5

λ = 1.4 × 10^(-7) m

Answer:

The  corresponding  magnetic field is  

Explanation:

From the question we are told that

    The electric field amplitude is  [tex]E_o = 611\ V/m[/tex]

Generally the  magnetic  field amplitude is  mathematically represented as

              [tex]B_o = \frac{E_o }{c }[/tex]

Where c is the speed of light with a constant value

         [tex]c = 3.0 *0^{8} \ m/s[/tex]

So  

        [tex]B_o = \frac{611 }{3.0*10^{8}}[/tex]

         [tex]B_o = 2.0 4 *10^{-6} \ Vm^{-2} s[/tex]

Since 1  T  is  equivalent to  [tex]V m^{-2} \cdot s[/tex]

         [tex]B_o = 2.0 4 *10^{-6} \ T[/tex]

Answer:

T= 2p√m/k

Explanation:

This is because the period of oscillation of the mass of spring system is directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

The period of a mass on a spring is given by the equation

T=2π√m/k.

Where T is the period,

M is mass

K is spring constant.

An increase in mass in a spring increases the period of oscillation and decrease in mass decrease period of oscillation.

When there is the relationship between the Period (T) caused by the oscillation of the mass should be considered as the T= 2p√m/k.

The mass of the spring system with respect to period of oscillation should be directly proportional to the square root of the mass and it is inversely proportional to the square root of the spring constant.

So the following equation should be considered

T=2π√m/k.

Here,

T is the period,

M is mass

K is spring constant.

An increase in mass in a spring rises the period of oscillation and reduce in mass decrease period of oscillation.

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Answer:

(a) force is 1066.56N

Explanation:

(a) MV²/R

Answer:

U =-2.39*10^-18 J

Explanation:

In order to calculate the electric potential energy of the electron you use the following formula:

[tex]U=k\frac{q_1q_2}{r}[/tex]           (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between charges

In this case the electron is at point midway between two charges, then the electric potential energy is the sum of two contributions:

[tex]U=U_1+U_2=k\frac{eq_1}{r}+k\frac{eq_2}{r}=\frac{ke}{r}[q_1+q_2][/tex]

e: charge of the electron = 1.6*10^-19C

q1: charge 1 = 3.00nC = 3.00*10^-9C

q2: charge 2 = 2.00nC = 3.00*10^-9C

r: distance to each charge = 60.0cm/2 = 30.0cm = 0.3m

If you consider that the electron is at the origin of coordinates, with the first charge in the negative x axis, and the other one in the positive x axis, you have:

[tex]U=\frac{(8.98*10^9Nm^2/C^2)(1.6*10^{-19}C)}{0.6m}[-3.0*10^{-9}C+2.0*10^{-9}C]\\\\U=-2.39*10^{-18}J[/tex]

The electric potential energy of the electron is -2.39*10^-18 J

Answer:

Explanation:

this is the answer to your question

The electron is going with a velocity of 1.25 × 10⁷ m/s when it hits the positive plate.

According to the law of the conservation of mechanical energy, the total mechanical energy is always conserved by an electron. We can say that the sum of potential energy (U) and kinetic energy (K) is always constant.

K + U = E

Given, the distance between the two parallel plates = 1.5 mm

The potential difference between the plates, V = 450V

The charge on an electron, q = [tex]-1.6\times 10^{-19} C[/tex]

The mass of an electron, m = 9.1× 10⁻³¹ Kg

The change in the potential energy of the charge moving through the potential difference of 450V.

ΔU = qΔV = (-1.6× 10⁻¹⁹)(450)  = -7.2 × 10⁻¹⁷J

From the law of the conservation of mechanical energy, we can write:

K + U = E

ΔK + ΔU = 0

ΔK = -ΔU

1/2mv² = -ΔU

v² = -2ΔU/m

[tex]v^2 =\frac{-2\times (-7.2\times 10{-17})}{9.1\times 10^{-31}}[/tex]

[tex]v=\sqrt{1.58\times 10^{14}}[/tex]

v = 1.25 × 10⁷ m/s

Therefore, the electron is going with the speed of 1.25 × 10⁷ m/s  when it hits the positive plate.

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Answer:

The atomic number

Explanation:

Answer:

power dissipated in the target is 100 W

Explanation:

given data

electrons = 1 mev = [tex]10^{6}[/tex] eV

1 eV = 1.6 × [tex]10^{-19}[/tex] J

current =  100 microampere = 100 × [tex]10^{-6}[/tex] A

solution

when energy of beam strike with 1 MeV so energy of electron is

E = e × v   ...................1

e is charge of electron and v is voltage

so put here value and we get voltage

v = 1 ÷ 1.6 × [tex]10^{-19}[/tex]

v =  [tex]10^{6}[/tex] volt

so power dissipated in target

P = voltage × current   ..............2

put here value

P =  [tex]10^{6}[/tex]  × 100 × [tex]10^{-6}[/tex]

P = 100 W

so power dissipated in the target is 100 W

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

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nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn

Answer:

See detailed solution below

Explanation:

a) From C= εoεrA/d

Where;

C= capacitance of the capacitor

εo= permittivity of free space

εr= relative permittivity

A= cross sectional area

d= distance between the plates

Since the relative permittivity of air=1 and permittivity of free space = 8.85 × 10^−12 Fm−1

Then;

C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.009 m

C= 196.67 × 10^-12 F or 1.967 ×10^-10 F

b) Q= CV = 1.967 ×10^-10 F × 140 V = 2.75 × 10^-8 C

c) E= V/d = 140 V/0.009m = 15.56 Vm-1

d) W= 1/2 CV^2 = 1/2 × 1.967 ×10^-10 F × (140)^2 =1.93×10^-6J

Part II

When the distance is now 0.014 m

a) C= 8.85 × 10^−12 Fm−1 × 0.2m^2/0.014 m = 1.26×10^-10 F

b) W= 1/2 Q^2/C = 1/2 × ( 2.75 × 10^-8 C)^2 / 1.26×10^-10= 3×10^-6 J

Note that the voltage changes when the distance is changed but the charge remains the same

Answer:

C. R^2

Explanation:

A cyclotron is a particle accelerator which employs the use of electric and magnetic fields for its functioning. It consists of two D shaped region called dees and the magnetic field present in the dee is responsible for making sure the charges follow the half-circle and then to a gap in between the dees.

R is denoted as the radius of the final orbit then the final particle energy is proportional to the radius of the two dees. This however translates to the energy being proportional to R^2.

Answer:

A.  Z = 185.87Ω

B.  I  =  0.16A

C.  V = 1mV

D.  VL = 68.8V

E.  Ф = 30.59°

Explanation:

A. The impedance of a RL circuit is given by the following formula:

[tex]Z=\sqrt{R^2+\omega^2L^2}[/tex]       (1)

R: resistance of the circuit = 160-Ω

w: angular frequency = 220 rad/s

L: inductance of the circuit = 0.430H

You replace in the equation (1):

[tex]Z=\sqrt{(160\Omega)^2+(220rad/s)^2(0.430H)^2}=185.87\Omega[/tex]

The impedance of the circuit is 185.87Ω

B. The current amplitude is:

[tex]I=\frac{V}{Z}[/tex]                     (2)

V: voltage amplitude = 30.0V

[tex]I=\frac{30.0V}{185.87\Omega}=0.16A[/tex]

The current amplitude is 0.16A

C. The current I is the same for each component of the circuit. Then, the voltage in the resistor is:

[tex]V=\frac{I}{R}=\frac{0.16A}{160\Omega}=1*10^{-3}V=1mV[/tex]            (3)

D. The voltage across the inductor is:

[tex]V_L=L\frac{dI}{dt}=L\frac{d(Icos(\omega t))}{dt}=-LIsin(\omega t)\\\\V_L=-(0.430H)(160\Omega)sin(220 t)=68.8sin(220t)\\\\V_L_{max}=68.8V[/tex]

E. The phase difference is given by:

[tex]\phi=tan^{-1}(\frac{\omega L}{R})=tan^{-1}(\frac{(220rad/s)(0.430H)}{160\Omega})\\\\\phi=30.59\°[/tex]

Answer:

did you mean released?

Explanation:

If so the process is called respiration

Answer:

a) 43.98 V

b) E = 21.37 MJ

Explanation:

Parameters given:

Length of cable = 180 km = 180000 m

Diameter of cable = 11 cm = 0.11 m

Radius = 0.11 / 2 = 0.055 m

Current, I = 135 A

a) To find the potential drop, we have to find the voltage across the wire:

V = IR

=> V = IρL / A

where R = resistance

L = length of cable

A = cross-sectional area

ρ = resistivity of the copper wire = 1.72 * 10^(-8) Ωm

Therefore:

V = (135 * 1.72 * 10^(-8) * 180000) / (π * 0.055^2)

V = 43.98 V

The potential drop across the cable is 43.98 V

b) Electrical energy is given as:

E = IVt

where t = time taken = 1 hour = 3600 s

Therefore, the energy dissipated per hour is:

E = 135 * 43.98 * 3600

E = 21.37 MJ (mega joules, 10^6)

Answer:

The spring constant is 60,000 N

The total work done on it during the compression is 3 J

Explanation:

Given;

weight of the girl, W = 600 N

compression of the spring, x = 1 cm = 0.01 m

To determine the spring constant, we apply hook's law;

F = kx

where;

F is applied force or weight on the spring

k is the spring constant

x is the compression of the spring

k = F / x

k = 600 / 0.01

k = 60,000 N

The total work done on the spring = elastic potential energy of the spring, U;

U = ¹/₂kx²

U = ¹/₂(60000)(0.01)²

U = 3 J

Thus, the total work done on it during the compression is 3 J

Answer:

A. 1.6 N/cm

Explanation:

spring constant = 21/13 = 1.6 N/cm

Answer:

The  length is  [tex]D = 5 \ cm[/tex]

Explanation:

From the question we are told  that

     The  length of the  hand is  [tex]l = 10.0 \ cm[/tex]

      The  angle at the hand is  held is  [tex]\theta = 30 ^o[/tex]

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

             [tex]D = l * sin(\theta )[/tex]

substituting values

             [tex]D = 10 * sin (30)[/tex]

             [tex]D = 5 \ cm[/tex]

Answer:

 e = 50.27 give / s

Explanation:

The expression for simple harmonic motion is

    x = A cos (wt + Ф)

in this case they give us the amplitude A = 3.9 cm and frequency f = 8.0 Hz

The angular and linФear variables are related

      e = 2π d

      e = 2π 8

      e = 50.27 give / s

let's look for the constant fi

       so let's find the time to have the maximum displacement

       v = dx / dt

       v = -A w sin (wt +Ф)

for the point of maximum displacement the speed is I think

        0 = - sin (0 + Ф)

therefore fi = 0

Let's put together the equation of motion

          x = 0.039 sin (50.27 t)

          v = 0.039 50.27 sin (50.27 3)

           v = 1.96 50 0.01355

            v = 0.0266 m / s

Answer:

Approximately [tex]0.45\; \rm m \cdot s^{-1}[/tex].

Explanation:

The mechanical energy of an object is the sum of its potential energy and kinetic energy. Consider this question from the energy point of view:

Mechanical energy of the block [tex]0.04\; \rm m[/tex] away from the equilibrium position:

While the block moves back to the equilibrium position, it keeps losing (mechanical) energy due to friction:

[tex]\begin{aligned}& \text{Work done by friction} = (-0.3\; \rm N) \times (0.04 \; \rm m) = -0.012\; \rm J\end{aligned}[/tex].

The opposite ([tex]0.012\; \rm N[/tex]) of that value would be the amount of energy lost to friction. Since there's no other form of energy loss, the mechanical energy of the block at the equilibrium position would be [tex]0.032\; \rm N - 0.012\; \rm N = 0.020\; \rm N[/tex].

The elastic potential energy of the block at the equilibrium position is zero. As a result, all that [tex]0.020\; \rm N[/tex] of mechanical energy would all be in the form of the kinetic energy of that block.

Given that the mass of this block is [tex]0.020\; \rm kg[/tex], calculate its speed:

[tex]\begin{aligned}v &= \sqrt{\frac{2\, \mathrm{KE}}{m}} \\ &= \sqrt{\frac{2 \times 0.020\; \rm J}{0.20\; \rm kg}} \approx 0.45\; \rm m\cdot s^{-1}\end{aligned}[/tex].

Answer:

(a) V = 65.625 Volts

(b) V = 131.25 Volts

(c) V = 131.25 Volts

Explanation:

Recall that:

1) in a metal sphere the charges distribute uniformly around the surface, and the electric field inside the sphere is zero, and the potential is constant equal to:

[tex]V=k\frac{Q}{R}[/tex]

2) the electric potential outside of a charged metal sphere is the same as that of a charge of the same value located at the sphere's center:

[tex]V=k\frac{Q}{r}[/tex]

where k is the Coulomb constant ( [tex]9\,\,10^9\,\,\frac{N\,m^2}{C^2}[/tex] ), Q is the total charge of the sphere, R is the sphere's radius (0.24 m), and r is the distance at which the potential is calculated measured from the sphere's center.

Then, at a distance of:

(a) 48 cm = 0.48 m, the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.48} =65.625\,\,V[/tex]

(b) 24 cm = 0.24 m, - notice we are exactly at the sphere's surface - the electric potential is:

[tex]V=k\frac{Q}{r}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

(c) 12 cm (notice we are inside the sphere, and therefore the potential is constant and the same as we calculated for the sphere's surface:

[tex]V=k\frac{Q}{R}=9\,\,10^9 \,\frac{3.5\,\,10^{-9}}{0.24} =131.25\,\,V[/tex]

Answer:

c) a difference in electric potential

Explanation:

my insta: priscillamarquezz

Answer:

F = umg where u is coefficient of dynamic friction

Explanation:

F = 0.4 x 80 x 9.81 = 313.92 N

Answer:

Wavelength is 0.359 m

Explanation:

Given that,

Magnetic field, B = 0.547 T

We need to find the wavelength of radiation produced by a proton in a cyclotron with a magnetic field of 0.547 T.

The frequency of revolution of proton in the cyclotron is given by :

[tex]f=\dfrac{qB}{2\pi m}[/tex]

m is mass of proton

q is charge on proton

So,

[tex]f=\dfrac{1.6\times 10^{-19}\times 0.547}{2\pi \times 1.67\times 10^{-27}}\\\\f=8.34\times 10^6\ Hz[/tex]

We know that,

Speed of light, [tex]c=f\lambda[/tex]

[tex]\lambda[/tex] = wavelength

[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{834\times 10^6}\\\\\lambda=0.359\ m[/tex]

So, the wavelength of the radiation produced by a proton is 0.359 m.

Answer: A. Strong nuclear

The max effective range of strong nuclear force is about 1.2 femtometers ( which is 1.2*10^(-15) meters). This is well below 1 meter. Strong nuclear forces are the forces that hold together a nucleus. Specifically it holds together the protons that would otherwise repel one another due to similar charge.

Answer:

The horizontal force acting on m2 is F + 9.8m1

Explanation:

Given;

Block m1 on left of block m2

Make a sketch of this problem;

                         F →→→→→→→→→→→-------m1--------m2

Apply Newton's second law of motion;

F = ma

where;

m is the total mass of the body

a is the acceleration of the body

The horizontal force acting on block m2 is the force applied to block m1 and force due to weight of block m1

F₂ = F + W1

F₂ = F + m1g

F₂ = F + 9.8m1

Therefore, the horizontal force acting on m2 is F + 9.8m1

The force acting on the block of mass m₂ is  [tex]\frac{m_2F}{m_1+m_2}[/tex]

Given that there are two blocks of mass m₁ and m₂.

m₁ is on the left of block m₂. They are in contact with each other.

A force F is applied on m₁ to the right.

According to Newton's laws of motion:

The equation of motion of the blocks can be written as:

F = (m₁ + m₂)a

here, a is the acceleration.

so, acceleration:

a = F / (m₁ + m₂)

Now, the force acting on the block of mass m₂ is:

f = m₂a

[tex]f = \frac{m_2F}{m_1+m_2}[/tex]

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Answer:

a) ±7.08×10⁻⁷ C/m²

b) 1.77×10⁻⁷ C

Explanation:

For a conductor,

σ = ±Eε₀,

where σ is the charge density,

E is the electric field,

and ε₀ is the permittivity of space.

a)

σ = ±Eε₀

σ = ±(8×10⁴ N/C) (8.85×10⁻¹² F/m)

σ = ±7.08×10⁻⁷ C/m²

b)

σ = q/A

7.08×10⁻⁷ C/m² = q / (0.5 m)²

q = 1.77×10⁻⁷ C

Answer:

u = 10.02m/s

Explanation:

a = f/m

a = 20/12 = 1.67m/s²

U =2aS

u = 2 x 1.67 x 3

U = 10.02m/s