A parallel-plate air capacitor is made from two plates 0.200 m square, spaced 0.900 cm apart. It is connected to a 140 V battery.
A. What is the capacitance?
B. What is the charge on each plate?
C. What is the electric field between the plates?
D. What is the energy stored in the capacitor?
E. If the battery is disconnected and then the plates are pulled apart to a separation of 1.40 cm , what are the answers to parts A, B, C, and D?
Enter your answer as four numbers corresponding to C, Q, E, U. Please enter the answer in the given order and in the same units as in parts A, B, C, and D.

Answer:

5000N

Explanation:

According to Newton's second law of motion, the net force (∑F) acting on a body is the product of the mass (m) of the body and the acceleration (a) of the body caused by the force. i.e

∑F = m x a             -------------(i)

From the question, the net force is the combined effect of the thrust (F) and the friction force (Fₓ). i.e

∑F = F + Fₓ             -------------(ii)

Where;

Fₓ = -200N       [negative sign because the friction force opposes motion]

Combine equations(i) and (ii) together to get;

F + Fₓ = m x a

F = ma - Fₓ         -------------(iii)

Where;

m = mass of car = 1200kg

a = acceleration of the car = 4m/s²

Now substitute the values of m, a and Fₓ into equation (iii) as follows;

F = (1200 x 4) - (-200)

F = 4800 + 200

F = 5000N

Therefore, the force the thrust is providing is 5000N

Answer:

Explanation:

When a body is launched in air and allowed to fall freely under the influence of gravity, the motion experienced by the body is known as a projectile motion. The body is launched at a particular velocity and at an angle theta to the horizontal. The velocity of the body ca be resolved towards the horizontal component and the vertical component.

Along the horizontal Ux = Ucos(theta)

Along the vertical Uy = Ucos(theta)

Ux and Uy are the velocities of the body along the horizontal and vertical components respectively.

This means that the only factor connecting horizontal and vertical components of projectile motion is its velocity since we are able to calculate the velocity of the body along both components irrespective of its initial velocity.

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

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]

The correct answer is B. thrill and adventure seeking, experience seeking, disinhibition, and susceptibility to boredom

Explanation:

Marvin Zuckerman was an important American Psychologists mainly known for his research about personality and the creation of a model to study this aspect of human psychology. This model purposes five factors define personality, these are the thrill and adventure-seeking that involves seeking for adventures and danger; experience seeking that implies a strong interest in participating in new activities; disinhibition that implies being open and extrovert; and susceptibility to boredom that implies avoiding boredom or repetition. Thus, option B correctly describes the characteristics used in Zuckerman's test.

Answer:

HSBC keen vs kg get it yyyyyuuy

Explanation:

hgccccxfcffgbbbbbbbbbbghhyhhhgdghcjyddhhyfdghhhfdgbxbbndgnncvbhcxgnjffccggshgdggjhddh

nnnbvvvvvggfxrugdfutdfjhyfggigftffghhjjhhjyhrdffddfvvvvvvvvvvvbbbbbbbbbvvcxccghhyhhhjjjhjnnnnnnnnnnnnnbhbfgjgfhhccccccvvjjfdbngxvncnccbnxcvbchvxxghfdgvvhhihbvhbbhhvxcgbbbcxzxvbjhcxvvbnnxvnn

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:

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:

20,000 kg m/s

Explanation:

Given:

v₀ = 50 m/s

a = -5 m/s²

t = 2 s

Find: v

v = at + v₀

v = (-5 m/s²) (2 s) + (50 m/s)

v = 40 m/s

p = mv

p = (500 kg) (40 m/s)

p = 20,000 kg m/s

Answer:

The two small charged spheres are now 4.382 cm apart

Explanation:

Given;

distance between the two small charged sphere, r = 7.59 cm

The force on each of the charged sphere can be calculated by applying Coulomb's law;

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where;

F is the force on each sphere

q₁ and q₂ are the charges of the spheres

r is the distance between the spheres

[tex]F = \frac{kq_1q_2}{r^2} \\\\kq_1q_2 = Fr^2 \ \ (keep \ kq_1q_2 \ constant)\\\\F_1r_1^2 = F_2r_2^2\\\\r_2^2 = \frac{F_1r_1^2}{F_2} \\\\r_2 = \sqrt{\frac{F_1r_1^2}{F_2}} \\\\r_2 = r_1\sqrt{\frac{F_1}{F_2}}\\\\(r_1 = 7.59 \ cm, \ F_2 = 3F_1)\\\\r_2 = 7.59cm\sqrt{\frac{F_1}{3F_1}}\\\\r_2 = 7.59cm\sqrt{\frac{1}{3}}\\\\r_2 = 7.59cm *0.5773\\\\r_2 = 4.382 \ cm[/tex]

Therefore, the two small charged spheres are now 4.382 cm apart.

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:

F = umg where u is coefficient of dynamic friction

Explanation:

F = 0.4 x 80 x 9.81 = 313.92 N

Answer:

2000 N

Explanation:

20 km/h = 5.56 m/s

100 km/h = 27.78 m/s

F = ma

F = m Δv/Δt

F = (200 kg + 80 kg) (27.78 m/s − 5.56 m/s) / (3 s)

F = 2074 N

Rounded to one significant figure, the force is 2000 N.

Answer:

8 ft/s

Explanation:

This is a straight forward question without much ado.

It is given from the question that she walks with a speed of 8 ft/s

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:

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:

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:

1 yr = 24 * 3600 * 365 = 3.2 * 10E7 sec

C = 6 R = 1.5 * 10E8 * 6 = 9 * 10E8 km     circumference of orbit

v = C / t = 9 * 10E8 km / 3 * 10E7  sec = 30 km / sec = 18 mi/sec

Answer:

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

Explanation:

Let suppose that lion and Thomson's gazelle are running at constant speed before and after collision and that collision is entirely inelastic. Given the absence of external force, the Principle of Momentum Conservation is applied such that:

[tex]\vec p_{L} + \vec p_{G} = \vec p_{F}[/tex]

Where:

[tex]\vec p_{L}[/tex] - Linear momentum of the lion, measured in kilograms-meters per second.

[tex]\vec p_{G}[/tex] - Linear momentum of the Thomson's gazelle, measured in kilograms-meters per second.

[tex]\vec p_{F}[/tex] - Linear momentum of the lion-Thomson's gazelle, measured in kilograms-meters per second.

After using the definition of momentum, the system is expanded:

[tex]m_{L}\cdot \vec v_{L} + m_{G}\cdot \vec v_{G} = (m_{L} + m_{G})\cdot \vec v_{F}[/tex]

Vectorially speaking, the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{m_{L}}{m_{L}+m_{G}}\cdot \vec v_{L} + \frac{m_{G}}{m_{L}+m_{G}}\cdot \vec v_{G}[/tex]

Where:

[tex]m_{L}[/tex], [tex]m_{G}[/tex] - Masses of the lion and the Thomson's gazelle, respectively. Measured in kilograms.

[tex]\vec v_{L}[/tex], [tex]\vec v_{G}[/tex], [tex]\vec v_{F}[/tex] - Velocities of the lion, Thomson's gazelle and the lion-gazelle system. respectively. Measured in meters per second.

If [tex]m_{L} = 177\,kg[/tex], [tex]m_{G} = 32\,kg[/tex], [tex]\vec v_{L} = 81.8\cdot j\,\left[\frac{km}{h} \right][/tex] and [tex]\vec v_{G} = 59.0\cdot i\,\left[\frac{km}{h} \right][/tex], the final velocity of the lion-gazelle system is:

[tex]\vec v_{F} = \frac{177\,kg}{177\,kg+32\,kg}\cdot \left(81.8\cdot j\right)\,\left[\frac{km}{h} \right] + \frac{32\,kg}{177\,kg+32\,kg}\cdot \left(59.0\cdot i\right)\,\left[\frac{km}{h} \right][/tex]

[tex]\vec v_{F} = 9.033\cdot i + 69.276\cdot j\,\left[\frac{km}{h} \right][/tex]

The speed of the system is the magnitude of the velocity vector, which can be found by means of the Pythagorean theorem:

[tex]\|\vec v_{F}\| = \sqrt{\left(9.033\frac{km}{h} \right)^{2}+\left(69.276\frac{km}{h} \right)^{2}}[/tex]

[tex]\|\vec v_{F}\| \approx 69.862\,\frac{km}{h}[/tex]

The final speed of the lion-gazelle system immediately after the attack is 69.862 kilometers per hour.

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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Complete Question

For a human body falling through air  in  a spread edge position , the numerical value of the constant D is about [tex]D = 0.2500 kg/m[/tex]

What value of D is required to make vt = 42.7 m/s the terminal velocity of a skydiver of mass 85.0 kg . Express your answer using two significant figures?

Answer:

The value of D is   [tex]D = 0.457 \ kg/m[/tex]

Explanation:

From the question we are told that

     The terminal velocity is  [tex]v_t = 42.7 \ m/s[/tex]

     The mass of the skydiver is  [tex]m = 85.0 \ kg[/tex]

      The numerical value of  D  is  [tex]D = 0.2500 kg/m[/tex]

From the unit of D  in the question we can evaluate D as  

       [tex]D = \frac{m * g }{v^2}[/tex]

substituting values  

        [tex]D = \frac{85 * 9.8 }{(42.7)^2}[/tex]

         [tex]D = 0.457 \ kg/m[/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:

E = 2750 J at h = 5 m

Explanation:

The gravitational potential energy is given by :

[tex]E=mgh[/tex]

In this case, m is the mass of swimmer is constant at every heights. So,

At h = 1 m, E = 550 J

[tex]550=m\times 10\times 1\\\\m=55\ kg[/tex]

So, at h = 5 m, gravitational potential energy is given by :

[tex]E=55\times 10\times 5\\\\E=2750\ J[/tex]

So, the correct option is (B).

Answer:

Explanation:

Mass ( m ) = 2 kg

Speed of the object (v) = 6 metre per second

Kinetic energy =?

Now,

We have,

Kinetic Energy = [tex] \frac{1}{2} \times m \times {v}^{2} [/tex]

Plugging the values,

[tex] = \frac{1}{2} \times 2 \times {(6)}^{2} [/tex]

Reduce the numbers with Greatest Common Factor 2

[tex] = {(6)}^{2} [/tex]

Calculate

[tex] = 36 \: joule[/tex]

Hope this helps...

Good luck on your assignment...

The Kinetic energy of the object will be "36 joules".

The excess energy of moving can be observed as that of the movement of an object, component, as well as the group of components. There would never be a negative (-) amount of kinetic energy.

According to the question,

Mass of object, m = 2 kg

Speed of object, v = 6 m/s

As we know the formula,

→ Kinetic energy (K.E),

= [tex]\frac{1}{2}[/tex] × m × v²

By substituting the values, we get

= [tex]\frac{1}{2}[/tex] × 2 × (6)²

=  [tex]\frac{1}{2}[/tex] × 2 × 36

= 36 joule

Thus the above answer is appropriate.

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A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

Answer:

Q = 20.22 x 10³ W = 20.22 KW

Explanation:

First we need to find the volume of water dropped.

Volume = V = πr²h

where,

r = radius of pan = 30.2 cm/2 = 15.1 cm = 0.151 m

h = height drop = 1.45 cm = 0.0145 m

Therefore,

V = π(0.151 m)²(0.0145 m)

V = 1.038 x 10⁻³ m³

Now, we find the mass of the water that is vaporized.

m = ρV

where,

m = mass = ?

ρ = density of water = 1000 kg/m³

Therefore,

m = (1000 kg/m³)(1.038 x 10⁻³ m³)

m = 1.038 kg

Now, we calculate the heat required to vaporize this amount of water.

q = mH

where,

H = Heat of vaporization of water = 22.6 x 10⁵ J/kg

Therefore,

q = (1.038 kg)(22.6 x 10⁵ J/kg)

q = 23.46 x 10⁵ J

Now, for the rate of heat transfer:

Rate of Heat Transfer = Q = q/t

where,

t = time = (18.6 min)(60 s/1 min) = 1116 s

Therefore,

Q = (23.46 x 10⁵ J)/1116 s

Q = 20.22 x 10³ W = 20.22 KW

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:

the result is the quantization of __Energy__ of the particle

Explanation:

Answer:

Explanation:

We shall apply conservation of mechanical energy law to solve the problem .

loss of height = L ( 1 -  cos 37 ) where L is length of rope

loss of potential energy at the bottom = gain of kinetic energy .

mg L ( 1 - cos 37 ) = 1/2 m v² where v is velocity at the bottom

v² = 2 L g ( 1 - cos 37 )

= 2 x 10 x 9.8 ( 1 - cos 37 )

= 39.46

v = 6.28 m /s

Answer:

0.79 cm

Explanation:

The computation is shown below:-

Particle acceleration is

[tex]a = \frac{qE}{m}[/tex]

We will take d which indicates distance as from the negative plate, so the travel by proton is 0.800 cm - d at the same time

[tex]d = \frac{1}{2} a_et^2\\\\0.800 cm - d = \frac{1}{2} a_pt^2\\\\\frac{d}{0.800 cm - d} = \frac{a_e}{a_p} \\\\\frac{d}{0.800 cm - d} = \frac{m_p}{m_e} \\\\\frac{d}{0.800 cm - d} = \frac{1836m_e}{m_e}[/tex]

After solving the equation we will get 0.79 cm from the negative plate.

Therefore it is 0.79 cm far from the negative pate i.e the point at which the electron and proton pass each other

The point at which the electron and proton pass each other will be 0.79 cm.

When the matter is put in an electromagnetic field, it has an electric charge, which causes it to experience a force. A positive or negative electric charge can exist.

The given data in the problem is;

d' is the distance between the two parallel plates= 0.800 cm

The acceleration is given as;

[tex]\rm a= \frac{qE}{m} \\\\[/tex]

The distance from Newton's law is found as;

[tex]d = ut+\frac{1}{2} at^2 \\\\ u=0 \\\\ d= \frac{1}{2} at^2 \\\\ d-d' = \frac{1}{2} a_pt^2 \\\\ 0.800-d= \frac{1}{2} a_pt^2 \\\\\ \frac{d}{0.800-d} =\frac{a}{a_p} \\\\ \frac{d}{0.800-d} =\frac{m_p}{m} \\\\ \frac{d}{0.800-d} =\frac{1836m_e}{m_e} \\\\ d=0.79 \ cm[/tex]

Hence the point at which the electron and proton pass each other will be 0.79 cm.

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