A spring is hung from the ceiling. A 0.573-kg block is then attached to the free end of the spring. When released from rest, the block drops 0.198 m before momentarily coming to rest, after which it moves back upward. (a) What is the spring constant of the spring

Answer:

Explanation:

Using one of the general gas equation to find the final volume of the R-134a.

According to pressure law; The volume of a given mas of gas is directly proportional to its temperature provided that the pressure remains constant.

VαT

V = kT

k = V/T

V1/T1 = V2/T2 = k

Given V1 = 0.14-m³ at T1 = –26.4°C = –26.4° + 273 = 246.6K

V2 = ? at T = 100°C = 100+273 = 373K

On substituting this values for T2;

0.14/246.6 = V2/373

373*0.14 = 246.6V2

V2 = 373*0.14 /246.6

V2 = 0.212m³

The final volume of R-134a is 0.212m³

Complete Question

A lawn mower has a flat, rod-shaped steel blade that rotates about its center. The mass of the blade is 0.65 kg and its length is 0.55 m. You may want to review (Pages 314 - 318) .

Part A What is the rotational energy of the blade at its operating angular speed of 3510 rpm

Part B

If all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?

Answer:

Part A  

    [tex]R = 1081 \ J[/tex]

Part B  

     [tex]h = 169.7 \ m[/tex]

Explanation:

From the question we are told that

  The mass of the blade is  [tex]m_b = 0.65 \ kg[/tex]

   The length is  [tex]l = 0.55 \ m[/tex]

   The angular speed is  [tex]w = 3510 rpm = 3510 * \frac{2 \pi }{60} = 367.6 \ rad/sec[/tex]

Generally the moment of inertia of the of this mower is mathematically evaluated as

         [tex]I = \frac{m_b * l^2 }{12}[/tex]

substituting values

         [tex]I = \frac{0.65 * 0.55^2 }{12}[/tex]

         [tex]I = 0.016 \ kg m^2[/tex]

Generally the rotational kinetic energy of the bland is  

        [tex]R = \frac{1}{2} * I * w^2[/tex]

substituting values

       [tex]R = \frac{1}{2} * 0.016 * 367.6^2[/tex]

     [tex]R = 1081 \ J[/tex]

At point where the gravitational potential energy is equal to the rotational kinetic energy  we have that

       [tex]P = R = m_b * h * g[/tex]

Where P is the  gravitational potential energy

substituting values

          [tex]1081 = 0.65 * 9.8 * h[/tex]

=>       [tex]h = 169.7 \ m[/tex]

Answer:

Explanation:

1.  [tex]V_{x}[/tex] = [tex]V_{0}[/tex] * cos[tex]\alpha[/tex] ⇒ 16*cos32 ≈ 13.6 m/s (13.56)

2. [tex]V_{y}[/tex] = [tex]V_{0}[/tex] * sin[tex]\alpha[/tex] ⇒ 16* sin32 ≈ 9.4 m/s

3. [tex]y_{max}[/tex] = [tex]\frac{v_{0}^2*sin^2\alpha}{2g}[/tex]= [tex]\frac{16^2*sin^232}{2*9.8}[/tex] (the g (gravity) depends on the country but i'll take the average g which is 9.2m/s^2)

[tex]y_{max}[/tex] ≈ 3.6677+1.5 ≈ 5.2m

4.  [tex]x_{max}[/tex] = [tex]\frac{v_{0}^2*sin(2\alpha)}{g}[/tex]=[tex]\frac{16^2*sin(2*32)}{9.8}[/tex] ≈ 23.5m (23.47)

5. -

answer 4 could be wrong, not certain about that one and i don't know 5

Answer:

a)   E = 1.58 10²¹ J , b) Oil = 4,236 107 liter ,  e)   T = 54.3 C

Explanation:

a) To calculate the energy that reaches Earth, let us combine that the power emitted by the Sun is distributed uniformly on a spherical surface

     I = P / A

     A = 4π r²

in this case the radius of the sphere is the distance from the Sun to Earth r = 1.5 10¹¹ m

     I = P / A

     I = P / 4π r²

let's calculate

     I = 3,828 10²⁵/4 pi (1.5 10¹¹)²

     I = 1.3539 10²W / m² = 135.4 W / m2

the energy that reaches the disk of the Earth is

    E = I A

the area of ​​a disc

    A = π r²

    E = I π r²

where r is the radius of the Earth 6.37 10⁶ m

     E = 135.4 π(6.37 10⁶)

     E = 1,726 10¹⁶ W

This is the energy per unit of time that reaches Earth

    t = 1 dai (24h / 1day) (3600s / 1h) = 86400 s

    E = 1,826 10¹⁶ 86400

     E = 1.58 10²¹ J

b) for this part we can use a direct proportions rule

      Oil = 1.58 10²¹ (1 / 37.3 10⁶)

      Oil = 4,236 10⁷ liter

c) to silence the surface temperature of the Earth we use the Stefan-Bolztman Law

       P = σ A e T⁴

       T = [tex]\sqrt[4]{P/Ae}[/tex]

nos indicate the refect, therefore the amount of absorbencies

       P_absorbed = 0.7 P

let's calculate

       T = REA (0.7 1.58 1021 / [pi (6.37 106) 2 1)

       T = RER (8,676 106)

       T = 54.3 C

b) Among the other factors that must be taken into account is the greenhouse effect, due to the absorption of gases from the atmosphere

Answer:

Increased blood circulation to the body.

Explanation:

plato/edmentum

Answer:

A , D , E

Explanation:

Solution:-

- Consider the two identical objects with mass ( m ).

- The stiffness of the springs are ( k1 and k2 ).

- Both the spring store 55.0 J of potential energy.

- We will apply the principle of energy conservation on both the systems. In both cases the spring stores 55.0 Joules of energy. Once released, the objects gain kinetic energy with a consequent loss of potential energy in either spring.

- The maximum speed ( v ) is attained when all the potential energy is converted to kinetic energy.

- Apply Energy conservation for spring with stiffness ( k1 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

- Apply Energy conservation for spring with stiffness ( k2 ).

                         ΔU = ΔEk

                         55.0 = 0.5*m*v^2

                         v = √ ( 110 / m )

Answer: Both objects will have the same maximum speed ( A )

- We are told that one spring is more stiff as compared to the other one. The measure of stiffness is proportionally quantified by the spring constant. To mathematically express we can write it as:

                         k1 > k2

Where,

                 k1: The stiff spring

                 k2: The flexible spring

Answer: The stiff spring has a larger spring constant than the flexible spring. ( D )

- We will assume that the spring with constant ( k1 ) undergoes a displacement ( x1 ) and the spring with constant ( k2 ) undergoes a displacement ( x2 ). The potential energy stored in both spring is the same. Hence,

                      U1 = U2

                      0.5*( k1 ) * ( x1 )^2 = 0.5*( k2 ) * ( x2 )^2

                      [ k1 / k2 ] = [ x2 / x1 ]^2

Since,

                     k1 > k2 , then [ k1 / k2 ] > 1    

Then,

                     [ x2 / x1 ]^2 > 1

                     [ x2 / x1 ] > 1

                     x2 > x1                  

Answer: The flexible spring ( x2 ) was compressed more than the stiff spring ( x1 ). ( E )

Answer:

Explanation:

The power rating of the battery isn't provided. But let us assume that it is one of the common batteries with ratings of 12 V and 50 A.h

Potential energy possessed by water at that height = mgh

m = mass of the water = ρV

ρ = density of water = 1000 kg/m³

V = volume of water = ?

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

Potential energy = ρVgh = 1000 × V × 9.8 × 0.5 = (4900V) J

Energy of the battery = qV

q = 50 A.h = 50 × 3600 = 180,000 C

V = 12 V

qV = 180,000 × 12 = 2,160,000 J

Energy = 2,160,000 J

At a 100% conversion rate, the energy of the water totally powers the battery

(4900V) = (2,160,000)

4900V = 2,160,000

V = (2,160,000/4900)

V = 440.82 m³

Hence, with our assumed power ratings for the battery (12 V and 50 A.h), 440.82 m³ of water at the given height of 50 cm would power the battery.

Incase the power ratings of the battery in the complete question is different, this solution provides you with how to obtain the correct answer, given any battery power rating.

Hope this Helps!!!

Answer:

a) [tex]E = 6.024\,USD[/tex], For m kilograms, it is 4184m J., 3600000 joules, b) [tex]i = 88.200\,A[/tex]

Explanation:

a) The amount of heat needed to warm water is given by the following expression:

[tex]Q_{needed} = m_{w}\cdot c_{w}\cdot (T_{f}-T_{i})[/tex]

Where:

[tex]m_{w}[/tex] - Mass of water, measured in kilograms.

[tex]c_{w}[/tex] - Specific heat of water, measured in [tex]\frac{J}{kg\cdot ^{\circ}C}[/tex].

[tex]T_{f}[/tex], [tex]T_{i}[/tex] - Initial and final temperatures, measured in [tex]^{\circ}C[/tex].

Then,

[tex]Q_{needed} = (1440\,kg)\cdot \left(4184\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (40^{\circ}C - 10^{\circ}C)[/tex]

[tex]Q_{needed} = 180748800\,J[/tex]

The energy needed in kilowatt-hours is:

[tex]Q_{needed} = 180748800\,J\times \left(\frac{1}{3600000}\,\frac{kWh}{J} \right)[/tex]

[tex]Q_{needed} = 50.208\,kWh[/tex]

The electric energy required to heat up the water is:

[tex]E = \frac{50.208\,kWh}{0.75}[/tex]

[tex]E = 66.944\,kWh[/tex]

Lastly, the cost of heating a hot tub is: (USD - US dollars)

[tex]E = (66.944\,kWh)\cdot \left(0.09\,\frac{USD}{kWh} \right)[/tex]

[tex]E = 6.024\,USD[/tex]

The heat needed to raise the temperature a degree of a kilogram of water is 4184 J. For m kilograms, it is 4184m J. Besides, a kilowatt-hour is equal to 3600000 joules.

b) The current required for the electric heater is:

[tex]i = \frac{Q_{needed}}{\eta \cdot \Delta V \cdot \Delta t}[/tex]

[tex]i = \frac{180748800\,J}{0.75\cdot (220\,V)\cdot (3.45\,h)\cdot \left(3600\,\frac{s}{h} \right)}[/tex]

[tex]i = 88.200\,A[/tex]

Answer:

A) 4D

Explanation:

The distance traveled by the cars before coming to rest can be determined by 3rd equation of motion:

2as = Vf² - Vi²

s = (Vf² - Vi²)/2a

where,

s = distance traveled

Vf = Final Speed = 0 m/s

Vi = Initial Speed

a = deceleration rate

First, we consider Car B and we assign a subscript 2 for it:

Vf₂ = 0 m/s  (As, car finally stops)

s₂ = D

a₂ = - a  (due to deceleration)

D = (0² - Vi₂²) /(-2a)

D = Vi₂²/2a    -------- equation (1)

Now, we consider Car A and we assign a subscript 1 for it:

Vf₁ = 0 m/s  (As, car finally stops)

s₁ = ?

a₁ = - a  (due to deceleration)

Vi₁ = 2 Vi₂  (Since, car A was initially traveling at twice speed of car B)

s₁ = (0² - Vi₁²) /(-2a)

s₁ = (2Vi₂)²/2a

s₁ = 4 (Vi₂²/2a)

using equation (1), we get:

s₁ = 4D

Therefore, the correct option is:

A) 4D

Answer:

a. 63.89°  in the north-southward manner

b.  2.2 sec

Explanation:

The goose is flying at 100 km/h

Air from east to west is 44 km/h

angle relative to the north-south direction for the bird to travel south will be

cos∅ = 44/100 = 0.44

∅ = [tex]cos^{-1}[/tex]0.44 = 63.89°  in the north-southward manner

Speed south relative to the ground will be v

Tan 63.89 = v/100

2.04 = v/100

v = 2.04 x 100 = 204 km/hr

to cover a distance of 450 m from north to south at this speed time will be

t = d/v = 450/204 = 2.2 sec

The ducks' flight path as observed by someone standing on the ground is the sum of the wind velocity and the ducks' velocity relative to the wind:

ducks (relative to wind) + wind (relative to Earth) = ducks (relative to Earth)

or equivalently,

[tex]\vec v_{D/W}+\vec v_{W/E}=\vec v_{D/E}[/tex]

(see the attached graphic)

We have

[tex]\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)[/tex]

[tex]\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)[/tex]

[tex]\vec v_{D/E}=v(\cos270^\circ\,\vec\imath+\sin270^\circ\,\vec\jmath)[/tex]

Then by setting components equal, we have

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0[/tex]

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]

We only care about the direction for this question, which we get from the first equation:

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}[/tex]

[tex]\cos\theta=-\dfrac57[/tex]

[tex]\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)[/tex]

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies both equations. We want

[tex]\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v[/tex]

which means θ must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

With smaller gravitational forces and therefor less vertical acceleration, the projectile launched on the moon ... with the same initial speed and direction ...

-- climbs faster,

-- spends more time climbing,

-- reaches a higher peak,

-- falls slower,

-- spends more time falling, and

-- covers more horizontal distance

than the projectile launched on the Earth.

This is not because of air resistance.  It would be true even if there were no air resistance on the Earth.  It's entirely a gravity thing.  

Answer:

[tex]T_{1}[/tex] = 14.88 N

Explanation:

Let's begin by listing out the given variables:

M = 2.7 kg, L = 3 m, m = 1.35 kg, d = 0.6 m,

g = 9.8 m/s²

At equilibrium, the sum of all external torque acting on an object equals zero

τ(net) = 0

Taking moment about [tex]T_{1}[/tex] we have:

(M + m) g * 0.5L - [tex]T_{2}[/tex](L - d) = 0

⇒ [tex]T_{2}[/tex] = [(M + m) g * 0.5L] ÷ (L - d)

[tex]T_{2}[/tex] = [(2.7 + 1.35) * 9.8 * 0.5(3)] ÷ (3 - 0.6)

[tex]T_{2}[/tex]= 59.535 ÷ 2.4

[tex]T_{2}[/tex] = 24.80625 N ≈ 24.81 N

Weight of bar(W) = M * g = 2.7 * 9.8 = 26.46 N

Weight of monkey(w) = m * g = 1.35 * 9.8 = 13.23 N

Using sum of equilibrium in the vertical direction, we have:

[tex]T_{1}[/tex] + [tex]T_{2}[/tex] = W + w   ------- Eqn 1

Substituting T2, W & w into the Eqn 1

[tex]T_{1}[/tex] + 24.81 = 26.46 + 13.23

[tex]T_{1}[/tex] = 14.88 N

Answer:

a)   v = 75 ft / s , b)  v = 55 ft / s , c)   Δx = 1000 ft

Explanation:

We can solve this exercise with the expressions of kinematics

a) average speed is defined as the distance traveled in a given time interval

        v = (x₂-x₁) / (t₂-t₁)

         v = (550 - 400) / (10 -8)

         v = 75 ft / s

b) we repeat the calculations for this interval

   v = (550 - 0) / (10 -0)

   v = 55 ft / s

c)  we clear the distance from the average velocity equation

     Δx = v (t₂ -t₁)

     Δx = 100 (20-10)

     Δx = 1000 ft

The average acceleration [tex]\vec a_{\rm ave}[/tex] over some time interval [tex][t_1,t_2][/tex] is equal to the ratio of the change in velocity [tex]\vec v_2-\vec v_1[/tex] over the duration of the interval [tex]t_2-t_1[/tex], or

[tex]\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}[/tex]

which can be split into the [tex]x[/tex] and [tex]y[/tex] components as

[tex]a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}[/tex]

[tex]a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}[/tex]

The magnitude of this average acceleration is

[tex]\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}[/tex]

and its direction is [tex]\theta[/tex] such that

[tex]\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ[/tex]

which corresponds to a direction of about 15.1º South of West.

Answer:The answer is 3000 N.

Force (F) is the multiplication of mass (m) and acceleration (a).

F = m · a

It is given:

mc = 1000 kg

mt = 2000 kg

total force: F = 4500 N 

total mass: m = mc + mt

Let's calculate acceleration which is common:

a = F/m = F/(mc + mt) = 4500/(1000 + 2000) = 4500/3000 = 1.5 m/s²

Now, when we know acceleration, let's calculate force on the truck:

Ft = mt · a = 2000 · 1.5 = 3000 N

Explanation:

Answer:(a)

Explanation:

Student must have known that insulators can only be charged when they are rubbed against each other. In this process, one becomes electrically negative while other becomes electrically positive such that both have the same magnitude. The one which gains electrons becomes electrically negative due to the transfer of electrons while others lose the electron becomes positive due to the transfer of an electron to another body.

Answer:

The moment of inertia is [tex]I = 1.8 \ kg m^2[/tex]

Explanation:

From the question we are told that

   The force applied is  [tex]F = 10 \ N[/tex]

    The distance of the knob to the hinge is  [tex]d = 0.9 \ m[/tex]

     The angular acceleration is  [tex]a = 5 \ rad/s[/tex]

The moment is mathematically represented as

        [tex]I = \frac{d Fsin(\theta)}{a}[/tex]

Here [tex]\theta = 90^o[/tex] This is because the force direction is perpendicular to the plane of the door

substituting values

          [tex]I = \frac{0.9 * 10 * sin (90)}{5}[/tex]

          [tex]I = 1.8 \ kg m^2[/tex]

Answer: Tech A is correct

Explanation:

Every vehicle has ignition system and without this system,it will not work. The battery of everything vehicle contain energy that start the vehicle and ignore it to start working. Electrical current move from the vehicle's battery and get to the induction coil, the induction coil increases the voltage in it so that the plug will be ignited. The spark plugs produce fire. The spark plug is connected to the ignition system. Once voltage is produced from the induction coil, electrical impulses move from induction coil to insulated plug wires. The spark plug need a very high voltage from the small voltage battery. Once the high voltage exceed the dielectric strength of the gases, spark jump the gap between the plug's fire end.

Answer:

t = 5.03 s

Explanation:

To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.

You use the following formula:

[tex]\omega = \frac{v}{r}[/tex]

w: angular speed

v: tangential speed

r: radius of the trajectory = 8 m

For  you have:

[tex]\omega=\frac{4m/s}{8m}=0.5\frac{rad}{s}[/tex]

For her brother:

[tex]\omega'=\frac{6m/s}{8m}=0.75\frac{rad}{s}[/tex]

Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:

[tex]\theta=\omega t\\\\\theta'=\omega ' t[/tex]

They encounter for θ = 2π-θ':

[tex]\omega t=2\pi-\omega' t[/tex]

You replace the values of the parameters in the previous equation and solve for t:

[tex]0.5t=2\pi-0.75t\\\\1.25t=2\pi\\\\t=5.026\approx5.03[/tex]

Hence, Sheila encounter her brother in 5.03 s

Complete question is;

A dart is inserted into a spring - loaded dart gun by pushing the spring in by a distance x. For the next loading, the spring is compressed a distance 2x. How much faster does the second dart leave the gun compared to the first?

Answer:

The second dart leaves the gun two times faster than the first one.

Explanation:

If we assume there was no energy loss during the spring - dart energy transfer, we can easily apply the principle of conservation of energy. So;

Potential energy = kinetic energy

Thus;

½kx² = ½mv²

Making velocity "v" the subject, we have;

v = √(kx²/m)

Since the initial distance is "x", thus initial launching velocity is;

v1 = √(kx²/m)

Since next distance is 2x, thus, second launch velocity is;

v2 = √(k(2x)²/m)

Expanding, we have;

v2 = √(4kx²/m)

v2 = 2√(kx²/m)

Comparing this to the one gotten for v1 earlier, we can see that it is double v1.

So, v2 = 2v1

Hence, The second dart leaves the gun two times faster than the first one.

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done  to compress the spring initially=[tex]\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J[/tex]

b.

By conservation law of energy

Initial energy of spring=Kinetic energy  of object

[tex]37.8=\frac{1}{2}(3)v^2[/tex]

[tex]v^2=\frac{37.8\times 2}{3}[/tex]

[tex]v=\sqrt{\frac{37.8\times 2}{3}}[/tex]

v=5.02 m/s

c.Work done by friction on the incline,[tex]w_{friction}=P.E-spring \;energy[/tex]

[tex]W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J[/tex]

Answer:

f = 1.96 revolutions per minute

Explanation:

The formula for the the frequency of revolution of a satellite, to develop an artificial gravity, with the help of centripetal acceleration is given as follows:

f = (1/2π)√(ac/r)

where,

f = frequency of rotation = ?

ac = centripetal acceleration= apparent gravity or artificial gravity = 2.2 m/s²

r = radius of station or satellite = diameter/2 = 104 m/2 = 52 m

Therefore,

f = (1/2π)√[(2.2 m/s²)/(52 m)]

f = (0.032 rev/s)(60 s/min)

f = 1.96 revolutions per minute

Answer:

i. 43.5 mH ii.  16 Ω. In phasor form Z = (8.33 + j13.66) Ω iii 58.64°

Explanation:

i. The resistance , R of the non-inductive load R = 125 V/15 A = 8.33 Ω

The reactance X of the inductor is X = 2πfL where f = frequency = 50 Hz.

So, x = 2π(50)L = 100πL Ω = 314.16L Ω

Since the current is the same when the 240 V supply is applied, then

the impedance Z = √(R² + X²) = 240 V/15 A

√(R² + X²) = 16 Ω

8.33² + X² = 16²

69.3889 + X² = 256

X² = 256 - 69.3889

X² = 186.6111

X = √186.6111

X = 13.66 Ω

Since X = 314.16L = 13.66 Ω

L = 13.66/314.16

= 0.0435 H

= 43.5 mH

ii. Since the same current is supplied in both circuits, the impedance Z of the circuit is Z = 240 V/15 A = 16 Ω.

So in phasor form Z = (8.33 + j13.66) Ω

iii. The phase difference θ between the current and voltage is  

θ = tan⁻¹X/R

= tan⁻¹(314.16L/R)

= tan⁻¹(314.16 × 0.0435 H/8.33 Ω)

= tan⁻¹(13.66/8.33)

= tan⁻¹(1.6406)

= 58.64°

Answer:

A13 mm is about 1/4 A5 mm

Explanation:

Find the attachment

Answer:

a

The free body diagram is shown on the first uploaded image

b

The tension on the rope is  [tex]T=39.16 \ N[/tex]  

Explanation:

From the  question we are told that

    The mass of the box is  m

    The tension on the box is  T

     The angle at which it is pulled is  [tex]\theta = 40^o[/tex]

     The force produced by the strong head wind is [tex]Fw = 30 \ N[/tex]

At equilibrium the net force acting on the block along the horizontal axis is zero i.e

     [tex]Tcos \theta -F_w = 0[/tex]

substituting values

     [tex]Tcos (40) -30 = 0[/tex]

     [tex]Tcos (40) = 30[/tex]

     [tex]T(0.76604)) = 30[/tex]

     [tex]T=39.16 \ N[/tex]      

Answer:

Explanation:

According to snell's law which states that the ratio of the sin of incidence (i) to the angle of refraction(n) is a constant for a given pair of media.

sini/sinr = n

n is the constant = refractive index

Since the diver shines light up to the surface of a flat glass-bottomed boat, the refractive index n = nw/ng

nw is the refractive index of water and ng is that of glass

sini/sinr = nw/ng

given i = 30°, nw = 1.33, ng = 1.5, r = angle the light leave the glass

On substitution;

sin 30/sinr = 1.33/1.5

1.5sin30 = 1.33sinr

sinr = 1.5sin30/1.33

sinr = 0.75/1.33

sinr = 0.5639

r = arcsin0.5639

r ≈35°

angle the light leave the glass is 35°

Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  [tex]Q =2.094 C[/tex]

Explanation:

From the question we are told that

    The diameter of the wire is  [tex]d = 0.205cm = 0.00205 \ m[/tex]

     The radius of  the wire is  [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]

     The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]

       The electric field change is mathematically defied as

         [tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]

Generally the charge is  mathematically represented as

       [tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]

Where A is the area which is mathematically represented as

       [tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]

 So

       [tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]

Therefore

      [tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]

substituting values

      [tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]

     [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]

From the question we are told that t =  5 sec

           [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]

          [tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]

         [tex]Q =2.094 C[/tex]

The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.

Given the following data:

Conversion:

Diameter of wire = 0.205 cm to m = 0.00205 meter.

Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]

To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:

First of all, we would find the area of the wire.

[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]

Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:

[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]

[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]

When t = 5 seconds:

[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]

Q = 2.094 Coulomb.

Find more information: https://brainly.com/question/18214726

Answer:

x = 3.76 cm

y = 3.76 cm

Explanation:

This composite shape can be modeled as a square (7.2 cm × 7.2 cm) minus a quarter circle in the lower left corner (3.6 cm radius) and a right triangle in the upper right corner (3.6 cm × 3.6 cm).

The centroid of a square (or any rectangle) is at x = b/2 and y = h/2.

The centroid of a quarter circle is at x = y = 4r/(3π).

The centroid of a right triangle is at x = b/3 and y = h/3.

Build a table listing each shape, the coordinates of its centroid (x and y), and its area (A).  Use negative areas for the shapes that are being subtracted.

Next, multiply each coordinate by the area (Ax and Ay), sum the results (∑Ax and ∑Ay), then divide by the total area (∑Ax / ∑A and ∑Ay / ∑A).  The result will be the x and y coordinates of the center of mass.

See attached image.