To test the resiliency of its bumper during low-speed collisions, a 2 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 4.00 106 N/m and compresses 3.18 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall?

Answer:

a) before immersion

C = εA/d = (8.85e-12)(25e-4)/(1.31e-2) = 1.68e-12 F

q = CV = (1.68e-12)(255) = 4.28e-10 C

b) after immersion

q = 4.28e-10 C

Because the capacitor was disconnected before it was immersed, the charge remains the same.

c)*at 20° C

C = κεA/d = (80.4*)(8.85e-12)(25e-4)/(1.31e-2) = 5.62e-10 F

V = q/C = 4.28e-10 C/5.62e-10 C = 0.76 V

e)

U(i) = (1/2)CV^2 = (1/2)(1.68e-12)(255)^2 = 5.46e-8 J

U(f) = (1/2)(5.62e-10)(0.76)^2 = 1.62e-10 J

ΔU = 1.62e-10 J - 5.46e-8 J = -3.84e-8 J

Answer:

The magnitude of the induced Emf is [tex]0.003371V[/tex]

Explanation:

The width of the truck is given as 79inch but we need to convert to meter for consistency, then

The width= 79inch × 0.0254=2.0066 metres.

Now we can calculate the induced Emf using expresion below;

Then the [tex]induced EMF= B L v[/tex]

Where B= magnetic field component

L= width

V= velocity

=(40*10^-6) × (42) × (2.0066)

=0.003371V

Therefore, the magnitude emf that is induced between the driver and passenger sides of the truck is 0.003371V

Answer:

metre per seconds

Explanation:

because velocity = distance ÷ time

Answer:

Amplitude, A = 5 m

Explanation:

Let a progressive wave is given by equation :

[tex]y=5\sin (100\pi t-0.4\pi x)[/tex] .....(1)

The general equation of a progressive wave is given by :

[tex]y=A\sin (\omega t-kx)[/tex] ....(2)

Here,

A is the amplitude of the wave

[tex]\omega[/tex] is the angular frequency

k is propagation constant

We need to find the amplitude of the wave.

If we compare equations (1) and (2), we find that the amplitude of the given plane progressive wave is 5 m.

Answer:

2.78 cm³/s

Explanation:

From the question,

Q = v/A'.................... Equation 1

Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.

Given: v = 100 km/h, A' = 10 km/L

Substitute this value into equation 1

Q = 100/10

Q = 10 L/h.

Now, we convert L/h to cm³/s.

Since,

1 L = 1000 cm³, and

1 h = 3600 s

Therefore,

Q = 10(1000/3600) cm³/s

Q = 2.78 cm³/s

Answer:

a)v = 476.28 m / s , b) T = 6.69 10⁵ N , c)  λ = 0.486 m , d)     λ = 0.35 m

Explanation:

a) The speed of a wave on a string is

          v = √T /μ

also all the waves fulfill the relationship

          v = λ f

they indicate that the fundamental frequency is f = 980 Hz.

The wavelength that is fixed at its ends and has a maximum in the center

          L = λ / 2

          λ = 2L

we substitute

           v = 2 L f

let's calculate

           v = 2  0.243  980

           v = 476.28 m / s

b) The tension of the rope

             T = v² μ

the density of the string is

            μ = m / L

            T = v² m / L

            T = 476.28²   0.717 / 0.243

            T = 6.69 10⁵ N

c)          λ = 2L

            λ = 2  0.243

            λ = 0.486 m

d) The violin has a resonance process with the air therefore the frequency of the wave in the air is the same as the wave in the string. Let's find the wavelength in the air

          v = λ f

          λ= v / f

          λ = 343/980

          λ = 0.35 m

Answer:

7.68×10^25kg

Explanation:

The formula for energy used per year is calculated as

Energy used per year =12 x Energy used per month

By substituting Energy used per month in the above formula, we get

Energy used per year =12 x 1600kWh

= 19200kWh

Conversion:

From kWh to J:

1 kWh=3.6 x 10^6 J

Therefore, it is converted to J as

19200 kWh =19200 x 3.6 x 10^6 J

= 6.912×10^10 J

Hence, energy used per year is 6.912×10^10 J

To find the mass that is converted to energy per year.

E = MC^2 ............1

E is the energy used per year

C is the speed of light = 3.0× 10^8m/s

Where E= 6.912×10^10 J

Substituting the values into equation 1

6.912×10^10 J = M × 3.0× 10^8m/s

M = 6.912×10^10 J / (3.0× 10^8m/s)^2

M = 6.912×10^10 J/9×10^16

M = 7.68×10^25kg

Hence the mass to be converted is

7.68×10^25kg

Answer:

The current in the wire is 31.96 A.

Explanation:

The current in the wire can be calculated as follows:

[tex] I = \frac{q}{t} [/tex]

Where:

q: is the electric charge transferred through the surface

t: is the time      

The charge, q, is:

[tex] q = n*e [/tex]

Where:

n: is the number of electrons = 7.93x10²⁰

e: is the electron's charge = 1.6x10⁻¹⁹ C

[tex] q = n*e = 7.93 \cdot 10^{20}*1.6 \cdot 10^{-19} C = 126.88 C [/tex]

Hence, the current in the wire is:

[tex] I = \frac{126.88 C}{3.97 s} = 31.96 A [/tex]

Therefore, the current in the wire is 31.96 A.

I hope it helps you!

Answer:

(a) The speed of the block after the collision is 0.0476v0.

(b) The percentage of the ball's initial energy lost, is 0 % (energy is conserved)

Explanation:

Given;

mass of ball of clay, m₁ = 50 g = 0.05 kg

mass of brick, m₂ = 1 kg

initial velocity of the ball of clay, u₁ = v0

initial velocity of the brick, u₂ = 0

Since the clay ball sticks with the brick after collision, it is inelastic collision.

Therefore, let their final velocity = v

(a) What is the speed of the block after the collision?

Apply the principle of conservation linear momentum

m₁u₁ + m₂u₂ = v (m₁ + m₂)

0.05v₀ + 1(0) = v( 0.05 + 1)

0.05v₀ = 1.05v

v = 0.05v₀ / 1.05

v = 0.0476v₀

Thus, the speed of the block after the collision is 0.0476 of its initial velocity.

(b). What percentage of the ball's initial energy is "lost"?

Initial kinetic energy of the ball = ¹/₂mv₀²

                                                    = ¹/₂ x 0.05 x v₀²

                                                    = 0.025v₀²

Final kinetic energy of the ball, = ¹/₂(m₁ + m₂)v²

                                                    = ¹/₂ x 1.05 x 0.0476v₀²

                                                    = 0.025v₀²

Change in kinetic energy = 0.025v₀² - 0.025v₀²

                                           = 0

percentage change in the initial kinetic energy of the ball;

= (0 / 0.025v₀²) x 100%

= 0 x 100%

= 0 %

Therefore, the percentage of the ball's initial energy lost, is 0 % (energy is conserved)

Answer:

a)   f₁ = 3.50 m ,  b)     f₂ = 0.84 m  

Explanation:

For this exercise we must use the constructor equation

          1 / f = 1 / p + 1 / q

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

a) the distance where the image should be placed is q = 3.50 m and the object is located at infinity p = ∞

           1 / f₁ = 1 /∞ + 1 / 3.50

           f₁ = 3.50 m

b) in this case the image is at q = -0.600 m and the object p = 0.350 m

           1 / f₂ = 1 / 0.350 -1 / 0.600

the negative sign, is because the image is in front of the object

           1 / f₂ = 1,1905

            f₂ = 1 / 1,1905

            f₂ = 0.84 m

Answer:

F= 3.56e22N

Explanation:

Using the force of radiation acting on the earth which is

force = radiation pressure x area = (intensity/c)xpi R^2

force = 1370W/m^2 x pi x( 6.37x10^6m)^2/3x10^8m/s

force = 5.82x10^8 N

But the sun's gravitational attraction means the magnitude of the solar gravitational force on earth: If that's the case, the answer is approx 10^22 N:

F=GMm/r^2

G=6.67x10^(-11)=6.67e-11

M=mass sun = 2x10^30kg=2e30

m=mass earth = 6x10^24kg

r=earth sun distance = 1.5x10^11m

F=(6.6e-11)(2e30)(6e24)/(1.5e11)^2 =

F= 3.56e22N

Answer:

5.4x10^4km

Explanation:

See attached file

Answer:

This is because motion is intended to occur but at zero acceleration. It means at a constant velocity, henceFor that to happen the pulling force F must exactly equal the frictional force Fk .

Answer:

If they are metallic spheres  they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

Answer:

Explanation:

Let the weight of the truck be W . reaction force R = W

Maximum frictional force = μ R

= .8 x W

So for movement of truck

Pulling force = frictional force

10560 = .8W

W = 13200 N

weight of heaviest truck required = 13200 N .

Answer:

The object with the twice the area of the other object, will have the larger drag coefficient.

Explanation:

The equation for drag force is given as

[tex]F_{D} = \frac{1}{2}pu^{2} C_{D} A[/tex]

where [tex]F_{D}[/tex] IS the drag force on the object

p = density of the fluid through which the object moves

u = relative velocity of the object through the fluid

p = density of the fluid

[tex]C_{D}[/tex] = coefficient of drag

A = area of the object

Note that [tex]C_{D}[/tex] is a dimensionless coefficient related to the object's geometry and taking into account both skin friction and form drag. The most interesting things is that it is dependent on the linear dimension, which means that it will vary directly with the change in diameter of the fluid

The above equation can also be broken down as

[tex]F_{D}[/tex] ∝ [tex]P_{D}[/tex] A

where [tex]P_{D}[/tex] is the pressure exerted by the fluid on the area A

Also note that [tex]P_{D}[/tex] = [tex]\frac{1}{2}pu^{2}[/tex]

which also clarifies that the drag force is approximately proportional to the abject's area.

In this case, the object with the twice the area of the other object, will have the larger drag coefficient.

Answer:

Explanation:

Before we determine the amount of current in each bulb, we must first know that the same current flows in a series connected resistors. Since the Two 15-Ω and three 25-Ω light bulbs are connected in series, same current will flow in all of them.

According to Ohm's law, E = IRt where;

E is the supply voltage = 24V

I is the total current flowing in the circuit

Rt is the total equivalent resistance.

First, we need to calculate Rt.

Rt = 15Ω+15Ω+25Ω+25Ω+25Ω (Two 15-Ω and three 25-Ω light bulb in series)

Rt = 105Ω

From ohms law formula, I = E/Rt

I = 24/105

I = 0.229Amp

Since the total current in the circuit is 0.229A, therefore the amount of current that passes through each bulb is the same as the total current i.e  0.229A

The current that passes via each bulb is 0.229A

According to the above law,

E = IRt

Here

E should be the supply voltage = 24V

I should be the total current flowing in the circuit

Rt should be the total equivalent resistance.

Now Rt should be

Rt = 15Ω+15Ω+25Ω+25Ω+25Ω

Rt = 105Ω

Now the current is

I = E/Rt

I = 24/105

I = 0.229Amp

Therefore, The current that passes via each bulb is 0.229A.

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Answer:

 y <8 10⁻⁶ m

Explanation:

For this exercise, they indicate that we use the Raleigh criterion that establishes that two luminous objects are separated when the maximum diffraction of one of them coincides with the first minimum of the other.

 Therefore the diffraction equation for slits with m = 1 remains

             a sin θ = λ

in general these experiments occur for oblique angles so

             sin θ = θ

             θ = λ / a

in the case of circular openings we must use polar coordinates to solve the problem, the solution includes a numerical constant

           θ = 1.22 λ / a

The angles in these measurements are taken in radians, therefore

          θ = s / R

as the angle is small the arc approaches the distance s = y

          y / R = 1.22 λ / s

          y = 1.22 λ R / a

let's calculate

            y = 1.22 500 10⁻⁹ 0.42 / 0.032

            y = 8 10⁻⁶ m

with this separation the points are resolved according to the Raleigh criterion, so that it is not resolved (separated)

                 y <8 10⁻⁶ m

Answer:

The time interval required to lift the spacecraft to this specified height is 123.94 seconds

Explanation:

Height through which the spacecraft is to be lifted = 32.0 m

Mass of the spacecraft = 260.0 kg

Four crew member each pull with a power of 135 W

18.0% of the mechanical energy is lost to friction.

work done in this situation is proportional to the mechanical energy used to move the spacecraft up

work done = (weight of spacecraft) x (the height through which it is lifted)

but the weight of spacecraft = mg

where m is the mass,

and g is acceleration due to gravity 9.81 m/s

weight of spacecraft = 260 x 9.81 = 2550.6 N

work done on the space craft = weight x height

==> work = 2550.6 x 32 = 81619.2 J

this is equal to the mechanical energy delivered to the system

18.0% of this mechanical energy delivered to the pulley is lost to friction.

this means that

0.18 x 81619.2  = 14691.456 J   is lost to friction.

Total useful mechanical energy =  81619.2 J - 14691.456 J = 66927.74 J

Total power delivered by the crew to do this work = 135 x 4 = 540 W

But we know tat power is the rate at which work is done i.e

[tex]P = \frac{w}{t}[/tex]

where p is the power

where w is the useful work done

t is the time taken to do this work

imputing values, we'll have

540 = 66927.74/t

t = 66927.74/540

time taken t = 123.94 seconds

Answer:

E = 17.64 x 10⁶ N/C = 17.64 MN/C

Explanation:

The electric field is given by the following formula:

E = F/q

E= W/q

E = mg/q

where,

E = magnitude of electric field = ?

m = mass of plastic sphere = 5.4 g = 5.4 x 10⁻³ kg

g = acceleration due to gravity = 9.8 m/s²

= charge = 3 nC = 3 x 10⁻⁹ C

Therefore,

E = (5.4 x 10⁻³ kg)(9.8 m/s²)/(3 x 10⁻⁹ C)

E = 17.64 x 10⁶ N/C = 17.64 MN/C

Answer:

D) VL(t) leads IL(t) and VC(t) lags IC(t)

Explanation:

This is because The phase angle between voltage and current for inductors and capacitors is 90 degrees, or radians, also, this means that no power is dissipated in either the inductor or the capacitor, since the time average of current times voltage,

( I(t), V(t)), is zero.

Answer:

C) the rope tension is less than the objects weight

Explanation:

According to Newton's Second Law, when an unbalanced or net force is applied to a body, it produces an acceleration in the body in the direction of the net force itself.

In this scenario, we have two forces acting on the object. First is the weight of object acting downward. Second is the tension in the rope acting upwards.

Since, the object is being lowered in the direction of weight. Therefore, weight of the object must be greater than the tension in the rope. So, the net force has the downward direction and the object is lowered. Hence, the correct option is:

C) the rope tension is less than the objects weight

Answer:

3.867 μm

Explanation:

The index of refraction, μ = 1.6

Wavelength of the light, λ = 580 nm

N2 - N1 = (2L / λ) (n2 - n1), Making L subject of formula, we have

(N2 - N1) λ = 2L (n2 - n1)

L = [(N2 - N1) * λ] / 2(n2 - n1)

L = (8 * 580) / 2(1.6 - 1.0)

L = 4640 nm / 1.2

L = 3867 nm or 3.867 μm

Therefore we can come to the conclusion that the thickness of the film is 3.867 nm

Answer:

Explanation:

The formula for hydrogen atomic  spectrum is as follows

energy of photon due to transition from higher orbit n₂ to n₁

[tex]E=13.6 (\frac{1}{n_1^2 } - \frac{1}{n_2^2})eV[/tex]

For layman series n₁ = 1 and n₂ = 2 , 3 , 4 ,   ...   etc

energy of first line

[tex]E_1=13.6 (\frac{1}{1^2 } - \frac{1}{2 ^2})[/tex]

10.2 eV

wavelength of photon = 12375 / 10.2 = 1213.2 A

energy of 2 nd line

[tex]E_2=13.6 (\frac{1}{1^2 } - \frac{1}{3 ^2})[/tex]

= 12.08 eV

wavelength of photon = 12375 / 12.08 = 1024.4 A

energy of third line

[tex]E_3=13.6 (\frac{1}{1^2 } - \frac{1}{4 ^2})[/tex]

12.75 e V

wavelength of photon = 12375 / 12.75 = 970.6 A

energy of fourth line

[tex]E_4=13.6 (\frac{1}{1^2 } - \frac{1}{5 ^2})[/tex]

= 13.056 eV

wavelength of photon = 12375 / 13.05 = 948.3 A

energy of fifth line

[tex]E_5=13.6 (\frac{1}{1^2 } - \frac{1}{6 ^2})[/tex]

13.22 eV

wavelength of photon = 12375 / 13.22 = 936.1 A

Answer:

a. The primary turns is 60 turns

b. The secondary voltage will be 360 volts.

Explanation:

Given data

secondary turns N2= 40 turns

primary turns N1= ?

primary voltage V1= 120 volts

secondary voltage V2= 8 volts

Applying the transformer formula which is

[tex]\frac{N1}{N2} =\frac{V1}{V2}[/tex]

we can solve for N1 by substituting into the equation above

[tex]\frac{N1}{40} =\frac{120}{8} \\\ N1= \frac{40*120}{8} \\\ N1= \frac{4800}{8} \\\ N1= 60[/tex]

the primary turns is 60 turns

If the primary voltage is V1 240 volts hence the secondary voltage V2 will be (to get the voltage of the secondary coil using emf substitute the values of the previously gotten N1 and N2 using V1 as 240 volts)

[tex]\frac{40}{60} =\frac{240}{V2}\\\\V2= \frac{60*240}{40} \\\\V2=\frac{ 14400}{40} \\\\V2= 360[/tex]

the secondary voltage will be 360 volts.

(a) In the primary coil, you have "60 turns".

(b) The emf across the secondary coil would be "360 volts".

According to the question,

Primary voltage, V₁ = 120 volts

Secondary voltage, V₂ = 8 volts

Secondary turns, N₂ = 40 turns

(a) By applying transformer formula,

→ [tex]\frac{N_1}{N_2} = \frac{V_1}{V_2}[/tex]

or,

   N₁ = [tex]\frac{N_2\times V_1}{V_2}[/tex]

By substituting the values,

        = [tex]\frac{40\times 120}{8}[/tex]

        = [tex]\frac{4800}{8}[/tex]

        = 60

(2) Again by using the above formula,

→ V₂ = [tex]\frac{60\times 240}{40}[/tex]

       = [tex]\frac{14400}{40}[/tex]

       = 360 volts.

Thus the above approach is correct.  

Find out more information about voltage here:

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Answer:

θ = 26.19 x 10³ radians

Explanation:

FOR ACCELERATED MOTION:

we use 2nd equation of motion for accelerated motion:

θ₁ = ωi t + (1/2)αt²

Where,

θ₁ = Angular Displacement covered during accelerated motion = ?

ωi = Initial Angular Speed = 0 rad/s

t = Time Taken = 2.8 x 10³ s

α = Angular Acceleration = 2.24 x 10⁻³ rad/s²

Therefore,

θ₁ = (0 rad/s)(2.8 x 10³ s) + (1/2)(2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)²

θ₁ = 8.78 x 10³ radians

Now we find final angular velocity (ωf) by using 1st equation of motion:

ωf = ωi + αt

ωf = 0 rad/s + (2.24 x 10⁻³ rad/s²)(2.8 x 10³ s)

ωf = 6.272 rad/s

FOR UNIFORM ANGULAR SPEED:

For uniform angular speed we use following equation:

θ₂ = ωt

where,

θ₂ = Angular Displacement during uniform motion = ?

ω = Uniform Angular Speed = ωf = 6.272 rad/s

t = Time Taken = 1.57 x 10³ s

Therefore,

θ₂ = (6.272 rad/s)(1.57 x 10³ s)

θ₂ = 9.85 x 10³ radians

FOR DECELERATED MOTION:

Now, we use 3rd equation of motion for decelerated motion:

2αθ₃ = ωf² - ωi²

where,

α = Angular deceleration = - 2.01 x 10⁻³ rad/s²

θ₃ = Angular Displacement during decelerated motion = ?

ωf = Final Angular Speed = 2.99 rad/s

ωi = Initial Angular Speed = 6.272 rad/s

Therefore,

2(-2.01 x 10⁻³ rad/s²)θ₃ = (2.99 rad/s)² - (6.272 rad/s)²

θ₃ = (- 30.4 rad²/s²)/(-4.02 x 10⁻³ rad/s²)

θ₃ = 7.56 x 10³ radians

FOR TOTAL ANGULAR DISPLACEMENT:

Total Angular Displacement = θ = θ₁ + θ₂ + θ₃

θ = 8.78 x 10³ radians + 9.85 x 10³ radians + 7.56 x 10³ radians

θ = 26.19 x 10³ radians

Answer:

The horizontal acceleration of the block is 4.05 m/s².

Explanation:

The horizontal acceleration can be found as follows:

[tex] F = m \cdot a [/tex]

[tex] Fcos(\theta) - \mu_{k}N = m\cdot a [/tex]

[tex] Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)] = m\cdot a [/tex]  

[tex] a = \frac{Fcos(\theta) - \mu_{k}[mg - Fsen(\theta)]}{m} [/tex]

Where:

a: is the acceleration

F: is the force exerted by the rope = 28.2 N

θ: is the angle = 30°

[tex]\mu_{k}[/tex]: is the kinetic coefficient = 0.12

m: is the mass = 5 kg

g: is the gravity = 9.81 m/s²

[tex] a = \frac{28.2 N*cos(30) - 0.12[5 kg*9.81 m/s^{2} - 28.2 N*sen(30)]}{5 kg} = 4.05 m/s^{2} [/tex]

Therefore, the horizontal acceleration of the block is 4.05 m/s².

I hope it helps you!

Answer:

0.0665 days

Explanation:

We are given;

The mean distance from the Earth's center to the moon;a1 = 385000 km

The mean distance from the Earth's center to the space craft;a2 = 6965 km

Formula for kepplers third law is;

T² = 4π²a³/GM

However, the proportion of both distances would be;

(T1)²/(T2)² = (a1)³/(a2)³

Where;

T1 is the period of orbit of the moon around the earth. T1 has a standard value of 27.322 days

T2 is the period of the space craft orbit.

Making T2 the subject, we have;

T2 = √((T1)²×(a2)³)/(a1)³)

Thus, plugging in the relevant values;

T2 = √(27.322² × 6965³)/(385000)³

T2 = 0.0665 days

Answer:

-72.0°C

Explanation:

PV = nRT

Since n, number of moles, is constant:

PV / T = PV / T

(4.65×10⁶ Pa) V / (21 + 273.15) K = (1.06×10⁶ Pa) (3V) / T

T = 201.16 K

T = -72.0°C