1. A 5 Ohm resistor is connected to a 9 Volt battery. How many Joules of thermal energy are produced in 7 minutes?

2. The current in a flashlight powered by 4.5 Volts is 0.5 A. What is the power delivered to the flashlight?

3. If the flashlight in the previous problem is left on for 3 minutes, how much electric energy is delivered to the bulb?

4. A 50 Ohm resistance causes a current of 5 milliamps to flow through a circuit connected to a battery. What is the power in the circuit?

5. How many Joules of electric energy are delivered to a 60 Watt lightbulb if the bulb is left on for 2.5 hours?

Answer:

20cal/s

Explanation:

Question:

There are two questions. The first one has been answered:

From the formular, Power = Q/t = (kA∆T)/l

the amount heat depends on the duration of time interval, length of the iron rod, the thermal conductivity of iron and the temperature difference between the ends of the rod.

The amount of heat that flows through the rod by conduction during a given time interval does not depend upon the mass of the iron rod (D).

Second question:

The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Solution:

Power = 10cal/s

Power = energy per unit time = Q/t

Where Q = energy

Power = (kA∆T)/l

k = thermal conductivity of iron

A = area

Area = πr^2

r = radius

Diameter = d = 2r

r = d/2

Area = (πd^2)/4

Length = l

∆T = change in temperature

10 = (kA∆T)/l

For a steel rod with length doubled and diameter doubled:

Let Length (L) = 2l

Diameter (D)= 2d

Area = π [(2d)^2]/4 = (π4d^2)/4

Area = 4(πd^2)/4

Using the formula Power = (kA∆T)/l, insert the new values for A and l

Power = [k × 4(πd^2)/4 × ∆T]/2l

Power = [4k((πd^2)/4) ∆T]/2l

Power = [(4/2)×k((πd^2)/4) ∆T]/l

Power = [2k(A) ×∆T]/l = 2(kA∆T)/l

Power of a steel that has its length doubled and diameter doubled = 2(kA∆T)/l

Recall initial Power = (kA∆T)/l = 10cal/s

And ∆T is the same

2[(kA∆T)/l] = 2 × 10

Power of a steel that has its length doubled and diameter doubled = 20cal/s

Answer: A.

tracking training through a leaming records store LRS.

Explanation:

An LRS uses xAPI to collect learner data, or experiences, from both online and offline sources. These experiences are reported back to the LRS in the form of xAPI statements, where they are stored. These statements can then be retrieved for reporting and interpretation of the learner data.

Answer:

Explanation:

Let v₁ and v₂ be velocities at lowest and topmost position . Let r be the radius of the circle .

Let N₁ and N₂ be the normal reaction force .

At the top position

centripetal force = N₂ + mg ;  so

N₂ + mg  = m v₂² / r

At the bottom  position

centripetal force = N₁ - mg ;  so

N₁ - mg  = m v₁² / r

subtracting these two equations

N₁ - mg - N₂ - mg = m v₁² / r  - m v₂² / r

N₁ - N₂ - 2mg = 1/r (m v₁²   - m v₂²  )

N₁ - N₂ - 2mg = 1/r x mg x 2r  ( loss of potential energy = gain of kinetic energy )

N₁ - N₂ =  2mg +  2mg

= 4 mg .

Answer is Weaker. If it is talking about the objects' gravitational forces.

Answer:147 joules

Explanation:

Mass=m=5kg

Acceleration due to gravity=g=9.8m/s^2

Height=h=3 meter

Potential energy=m x g x h

Potential energy=5 x 9.8 x 3

Potential energy=147 joules

Answer:

Question 3: 2.25 watts

Question 4: 405 joules

Explanation:

question 3:

Current =0.5 amps

Voltage =4.5 volts

Power= current x voltage

Power=0.5 x 4.5

power=2.25 watts

Question 4

Current =0.5 amps

Voltage =4.5v

Time=3 minutes

Time =3x60

Time =180 seconds

Energy=current x voltage x time

Energy =0.5 x 4.5 x 180

Energy =405 joules

Answer:

t = 2.94 x 10⁶ years

Explanation:

The equation used in the case of constant speed is:

s = vt

t = s/v

where,

s = distance = 12 light years

s = (12 light years)(9.461 x 10¹² km/light year) =  113.532 x 10¹² km

v = speed = 440 km/hr

t = time passed = ?

Therefore,

t = (113.532 x 10¹² km)/(440 km/hr)

t = 2.58 x 10¹¹ hr

Now, converting it to years:

t = (2.58 x 10¹¹ hr)(1 year/8766 hr)

t = 2.94 x 10⁶ years

Answer:

The third image

Explanation:

The one with the thumb pointing to the right

Answer:

3, correct on Edge 2020

Answer:

Explanation:

We shall use Ampere's circuital law to find magnetic field at required point.

The point is outside the circumference of two given wires so whole current will be accounted for .

Ampere's circuital law

B = ∫ Bdl = μ₀ I

line integral will be over circular path of radius r = 41 cm .

Total current  I  = 5A -3A = 2A .

∫ Bdl = μ₀ I

2π r B = μ₀ I

2π x .41  B = 4π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ x 2 / .41

= 9.75 x 10⁻⁷ T . It will be along - ve Y - direction.

Answer:

The temperature is [tex]T = -106 ^oC[/tex]

Explanation:

From the question we are told that

   The temperature is [tex]T_1 = T_t= T_a=15^oC[/tex]

   The  diameter is  [tex]d_1 = 0.5001 cm[/tex]

    The diameter of the hole [tex]d_2 = 0.500 \ cm[/tex]

    The coefficient of linear expansion for aluminum is [tex]\alpha _1 = 25 *10^{-6} \ ^oC^{-1}[/tex]

    The coefficient of linear expansion for  titanium is [tex]\alpha _2 = 8.5 *10^{-6} \ ^o C^{-1}[/tex]

According to the law of linear expansion

     [tex]d = d_o (1 + \alpha \Delta T )[/tex]

Where [tex]d_o[/tex] represents the original diameter

  So for aluminum

          [tex]d_a = d_1 (1 + \alpha_1 (T- T_a) )[/tex]

Where [tex]d_a[/tex] is the new diameter of aluminum

          [tex]T_a[/tex] is the new temperature of the aluminum

So for titanium

      [tex]d_t = d_2 (1 + \alpha_1 (T- T_t) )[/tex]

Where [tex]d_t[/tex] is the new diameter of  titanium

          [tex]T_t[/tex] is the new temperature of the aluminum

So for the aluminum rivets to fit into the holes

     [tex]d_a = d_t[/tex]

=>  [tex]d_1 (1 + \alpha_1 (T- T_a) ) = d_2 (1 + \alpha_2 (T- T_t) )[/tex]

       Making T the subject of the formula

     [tex]T = \frac{(d_1 - d_2 ) + (d_2 *\alpha_2 T_t) - d_1 \alpha_1 * T_a }{d_2 \alpha_2 - d_1 \alpha_1 }[/tex]

    Substituting values

     [tex]T = \frac{(0.501 - 0.500 ) + (0.500 *(8.5*10^{-6}) * 15) - 0.500* (25*10^{-6}) * 15 }{0.500 * (8.5 *10^{-6}) - 0.501 * (25 *10^{-6}) }[/tex]

    [tex]T = -106 ^oC[/tex]

Answer:

The center of mass move with the velocity of -3.487 m/s.

Explanation:

Given values of block A.

Mass of block A, (M1) = 4 kg

Speed of block A, (V1) = 2 m/s

Given values of block B.

 Mass of block B, (M2) = 8.4 kg

Speed of block B, (V2) = -6.1 m/s

Below is the formula to find the velocity of center of mass.

[tex]Velocity = \frac{M1V1 + M2V2}{M1 + M2} \\[/tex]

[tex]= \frac{4 \times 2 + 8.4 \times (-6.1) }{4 + 8.4} \\[/tex]

[tex]= \frac{- 43.24}{12.4}\\[/tex]

[tex]= - 3.487 m/s[/tex]

Answer:

The thickness of the door is 0.4230 m

Explanation:

Given;

mass of bullet, m = 0.009 kg

initial velocity of the bullet, u = 803 m/s

final velocity of the bullet, v = 617 m/s

average resistive force of the door on the bullet, F = 5620 N

Apply Newton's second law of motion;

Force exerted by the door on the bullet = Force of the moving bullet

F = ma

where;

F is applied force

m is mass

a is acceleration

Also, Force exerted by the door on the bullet = Force of the moving bullet

[tex]F =ma, \ But \ a =\frac{dv}{dt} = \frac{u-v}{t} \\\\F = \frac{m(u-v)}{t}[/tex]

where;

v is the final velocity of the bullet

u is initial velocity of the bullet

t is time

We need to calculate the time spent by the bullet before it passes through the door.

[tex]t = \frac{m(u-v)}{F} \\\\t = \frac{0.009(803-617)}{5620} = 0.0002979 \ s[/tex]

Distance traveled by the bullet within this time period = thickness of the door

This distance is equivalent to the product of average velocity and time

[tex]S = (\frac{u+v}{2}) t[/tex]

where;

s is the distance traveled

[tex]S = (\frac{u+v}{2}) t\\\\S = (\frac{803+617}{2}) 0.0002979\\\\S = 0.4230 \ m[/tex]

Therefore, the thickness of the door is 0.4230 m

Answer:171 N

Explanation:

Answer:

171 N.

Explanation:

Answer:

a) 1504.8 J

b) 991.76 J

c) 0J

d) 0J

Explanation:

(a) The work done by the force P on the box is given by the following formula:

[tex]W_P=Px[/tex]

P: applied force = 171N

x: distance in which the for P is applied = 8.80m

you replace the values of P and x and obtain:

[tex]W_P=(171N)(8.80m)=1504.8J[/tex]

(b) The work don by the friction force is:

[tex]W_f=F_fx=\mu N x=\mu Mg x[/tex]

μ = coefficient of kinetic friction = 0.250

M: mass of the box = 46.0kg

g: gravitational constant = 9.8 m/s^2

[tex]W_f=(0.250)(46.0kg)(9.8m/s^2)(8.80m)=991.76J[/tex]

(c) The Normal force is

[tex]N=Mg=(46.0kg)(9,8m/s^2)=450.8N[/tex]

but this force does not do work on the box because the direction is perpendicular to the direction of the force P.

[tex]W_N=0J[/tex]

(d) the same as before:

[tex]W_g=0J[/tex]

Answer:

Option A, Sea Breeze

Explanation:

Ssea breeze is a wind that blows from the ocean or any water body to the nearby land mass. This breeze is cold as compared to the air on land. The water in water bodies has high specific heat capacity and hence takes longer time to cool as compared to the surrounding objects. The warmer air over the land rises upward thereby reducing the pressure on land and hence the sea breeze starts flowing from region of high pressure (i.e above the water body) towards the low pressure region that is the land.

Hence, option A is correct

Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

[tex]v(t)=ate^{-6t}[/tex]   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

[tex]\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))][/tex]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

[tex]a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}[/tex]

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

[tex]v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a[/tex]

hence, the maximum speed is v_max = ((1/6)e^-1)a

Given Information:

Power = P = 100 Watts

Voltage = V = 220 Volts

Required Information:

a) Current = I = ?

b) Resistance = R = ?

Answer:

a) Current = I = 0.4545 A

b) Resistance = R = 484 Ω

Explanation:

According to the Ohm’s law, the power dissipated in the light bulb is given by

[tex]P = VI[/tex]

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and P is the power dissipated in the light bulb.

Re-arranging the above equation for current I yields,

[tex]I = \frac{P}{V} \\\\I = \frac{100}{220} \\\\I = 0.4545 \: A \\\\[/tex]

Therefore, 0.4545 A current is flowing through the light bulb.

According to the Ohm’s law, the voltage across the light bulb is given by

[tex]V = IR[/tex]

Where V is the voltage across the light bulb, I is the current flowing through the light bulb and R is the resistance of the light bulb.

Re-arranging the above equation for resistance R yields,

[tex]R = \frac{V}{I} \\\\R = \frac{220}{0.4545} \\\\R = 484 \: \Omega[/tex]

Therefore, the resistance of the bulb is 484 Ω

Answer:

bulb will burn out!

Explanation:

Answer:

The angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Explanation:

Initial angular speed of a CD player is 0 and final angular speed is 41.8 rad/s. Time to change the angular speed is 3 s.

It is required to find the angular acceleration. The change in angular speed of the CD player divided by time taken is called its angular acceleration. It can be given by :

[tex]a=\dfrac{\omega_f-\omega_i}{t}\\\\a=\dfrac{41.8-0}{3}\\\\a=13.94\ rad/s^2[/tex]

So, the angular acceleration of CD player is [tex]13.93\ rad/s^2[/tex].

Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

     The mass of the bat is [tex]m_b = 0.800 \ kg[/tex]

      The bat length is  [tex]L_b = 0.900 \ m[/tex]

      The distance of the bat's center of mass to the handle end is  [tex]z_c = 0.600 \ m[/tex]

      The moment of inertia of the bat is    [tex]I = 0.0530 \ kg \cdot m^2[/tex]

The objective of the solution is to find  x   which is the distance from the handle of the bat to the point where the baseball hit the bat

Generally the velocity change at the end of the bat is mathematically represented as

         [tex]\Delta v_e = \Delta v_c - \Delta w* z_c[/tex]

         Where  [tex]\Delta v_c[/tex] is the velocity change at the center of the bat  which is mathematically represented as

                [tex]\Delta v_c = \frac{Impulse}{m_b }[/tex]

We are told that the impulse is  J so

              [tex]\Delta v_c = \frac{J}{m_b }[/tex]

And   [tex]\Delta w[/tex] is the change in angular velocity which is mathematically represented as

         [tex]\Delta w = \frac{J (z -z_c)}{I}[/tex]

Now we have that

           [tex]\Delta v_e = \frac{J}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]

Before a swing when the bat is at rest the velocity change a the end of the bat handle is zero  and the impulse will be  1

   So  

            [tex]0 = \frac{1}{m_b } - [\frac{J (x- z_c)}{I} ] * z_c[/tex]

=>           [tex]x = \frac{I}{m_b z_c} + m_b[/tex]

substituting values

            [tex]x = \frac{0.530}{0.800 * 0.600} + 0.600[/tex]

           [tex]x = 0.710 \ m[/tex]

Answer:

26.92 N

Explanation:

The normal reaction of the ball is due to two force component acting on it.

However ; the acceleration is in polar coordinate which is given by the relation:

[tex]a^ { ^ \to} = (r- r \omega^2) \hat {e_r} + ( r \theta + 2 r \omega ) \hat {e_ \theta}[/tex]

[tex]a_{\theta} = r \theta + 2 r \omega[/tex]

Given that :

ω = 5 rad/s

mass m = 6 kg

θ = 30 ◦

r = 0.9 m

speed v =  0.4 m/s

[tex]a_{\theta} = 0 + 2(-0.4)*5[/tex]

[tex]a_{\theta}= -4 \ m/s[/tex]

The normal force reaction (N) that the arm applies to the ball at this instant is :

N = mg cos θ  + [tex]ma_{\theta}[/tex]

N = (6 × 9.8× cos 30) + (6 ×(-4))

N = 26.92 N

Answer:

A.

Explanation:

Refraction light rays are bent as they pass into a different medium where its speed is different. As refracted light rays pass from a fast medium to a slow medium, the light ray bends toward the normal to the boundary between the two medium. Light refracts as it travels at an angle into a medium with a different refractive index.

Answer:

Time of flight=3.5 seconds

Speed at maximum height is 0

Explanation:

Φ=60°

initial velocity=u=20m/s

Acceleration due to gravity=g=9.8 m/s^2

Total time of flight=T

Final speed=v

question 1:

T=(2 x u x sinΦ)/g

T=(2 x 20 x sin60)/9.8

T=(2 x 20 x 0.8660)/9.8

T=34.64/9.8

T=3.5 seconds

Question 2

Speed at maximum height is 0

Answer:

Total internal reflection (TIR) is the phenomenon that involves the reflection of all the incident light off the boundary. TIR only takes place when both of the following two conditions are met: the light is in the more dense medium and approaching the less dense medium.

Explanation: Hope i helped!!!

Answer:

coefficient of friction (μ) and normal force (N)

Answer: How hard the surfaces push together and the types of surfaces involved

Explanation:

Answer:

the answer is a

Explanation:

Answer:

A.  falling water

Explanation:

I got it right on Edgenuity. Good luck on your quiz.

In hydroelectric plants, water falls on turbine and makes it rotate. In generator, this mechanical energy transforms into electrical energy.

Hydroelectric power is generated by turbines that turn the potential energy of falling or swiftly flowing water into mechanical energy, which is then used to power generators. The most popular renewable energy source in the early 21st century was hydroelectricity, which in 2019 accounted for more than 18% of the world's total power producing capacity.

Water is gathered or stored at a higher elevation during the production of hydroelectric power and then transported through substantial pipes or tunnels (penstocks) to a lower elevation; the difference between these two elevations is referred to as the head. The falling water turns turbines as it nears the bottom of the pipelines. In turn, the turbines power generators, which transfer the mechanical energy of the turbines into electricity.

Learn more about  hydroelectric power here:

https://brainly.com/question/15228003

#SPJ2

Answer:

A) equal to the battery's terminal voltage.

Explanation:

When the capacitor is fully charged after long hours of charging , its  potential becomes equal to the emf of the battery and its polarity is opposite to that of battery . Hence net emf becomes equal . The capacitor itself becomes a battery which is connected in the circuit with opposite polarity . This results in the net emf and  current becoming zero . There is no charging current when the capacitor is fully charged .

Answer:

-7.23 rad/s

Explanation:

Given that

Mass of the cockroach, m = 0.157 kg

Radius of the disk, r = 14.9 cm = 0.149 m

Rotational Inertia, I = 5.92*10^-3 kgm²

Speed of the cockroach, v = 2.92 m/s

Angular velocity of the rim, w = 3.89 rad/s

The initial angular momentum of rim is

Iw = 5.92*10^-3 * 3.89

Iw = 2.3*10^-2 kgm²/s

The initial angular momentum of cockroach about the axle of the disk is

L = -mvr

L = -0.157 * 2.92 * 0.149

L = -0.068 kgm²/s

This means that we can get the initial angular momentum of the system by summing both together

2.3*10^-2 + -0.068

L' = -0.045 kgm²/s

After the cockroach stops, the total inertia of the spinning disk is

I(f) = I + mr²

I(f) = 5.92*10^-3 + 0.157 * 0.149²

I(f) = 5.92*10^-3 + 3.49*10^-3

I(f) = 9.41*10^-3 kgm²

Final angular momentum of the disk is

L'' = I(f).w(f)

L''= 9.41*10^-3w(f)

Using the conservation of total angular momentum, we have

-0.068 = 9.41*10^-3w(f) + 0

w(f) = -0.068 / 9.41*10^-3

w(f) = -7.23 rad/s

Therefore, the speed of the lazy Susan after the cockroach stops is -7.23 and is directed in the opposite direction of the initial lazy Susan angular speed

b)

The mechanical energy of the cockroach is not converted as it stops

Answer:

340 W

Explanation:

Power = change in energy / change in time

P = ΔKE / Δt

P = ½ mv² / Δt

P = ½ (90 kg) (15 m/s)² / (30 s)

P = 337.5 W

Rounded to 2 significant figures, the power is 340 W.

Answer:

b) 20 kJ

Explanation:

Efficiency of carnot engine = (T₁ - T₂ ) / T₁  Where T₁ is temperature of hot source  and T₂ is temperature of sink .

T₁ = 270 + 273 = 543K

T₂ = 50 + 273 = 323 K

Putting the given values of temperatures

efficiency = (543 - 323) / 543

= .405

heat input = 50 KJ

efficiency = output work / input heat energy

.405 = output work / 50

output work = 20.25 KJ.

= 20 KJ .