A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.

Answers

Answer 1

Answer:

Resistance of the armature coil = 6.61 ohms

Back emf developed at normal speed = 98.62 V (Approx.)

Current drawn by the motor at one-third normal speed = 12.73 A

Explanation:

Given:

Potential difference V = 117 V

Current = 17.7 A

Motor drawn current = 2.78 A

Find:

Resistance of the armature coil

Back emf developed at normal speed

Current drawn by the motor at one-third normal speed

Computation:

A] Resistance of the armature coil R = V/ I

Resistance of the armature coil = 117 / 17.7

Resistance of the armature coil = 6.61 ohms

B] Back emf developed at normal speed  = V- IR

Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)

Back emf developed at normal speed = 117 V - 18.37

Back emf developed at normal speed = 98.62 V (Approx.)

C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)

Current drawn by the motor at one-third normal speed = 17.7 - 4.97

Current drawn by the motor at one-third normal speed = 12.73 A


Related Questions

A 771.0-kg copper bar is melted in a smelter. The initial temperature of the copper is 300.0 K. How much heat must the smelter produce to completely melt the copper bar? For solid copper, the specific heat is 386 J/kg • K, the heat of fusion is 205 kJ/kg, and the melting point is 1357 K.

Answers

Answer:

4.73 × 10^5

Explanation:

Where is the center of mass of homogeneous body which has a regular ​

Answers

Following the definition of the center of mass, "In physics, the center of mass of a distribution of mass in space is the unique point where the weighted relative position of the distributed mass sums to zero."

(see explanation below)

Liquid plastic is frozen in a physical change that increases its volume. What can be known about the plastic after the change?
(A) Its mass will increase.
(B) Its density will increase.
(C) Its mass will remain the same.
(D) Its density will remain the same.

Answers

Answer:

c

Explanation:

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same, therefore the correct answer is option C.

What is the matter?

Anything which has mass and occupies space is known as matter ,mainly there are four states of matter solid liquid gases, and plasma.

These different states of matter have different characteristics according to which they vary their volume and shape.

It is known about plastic that its mass will remain the same when liquid plastic is frozen, by increasing its volume.

Liquid plastic is frozen in a physical change that increases its volume,it can be known about the plastic that Its mass will remain the same,  therefore the correct answer is C.

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A scientist who studies the tiny microorganisms of the environment .

geologist
meteorologist
microbiologist
entomologist

Answers

Question:- A scientist who studies the tiny microorganisms of the environment

Answer:- Microbiologist

Explanation:-

Microbiologist means a person who studies micro sized living organisms

Microbiologist word is combination of two words Micro and biologist

Micro stands for objects which cannot be seen with the naked eyes and are very small in sizeBiologist a person who studies living forms.

Answer:

microbiologist i think

hope it helps

please mark Brainliest if you think the answer is correct

27. The part of the Earth where life exists .

Mesosphere
Stratosphere
Troposphere
Biosphere

Answers

Answer:

Biosphere is the part of the earth where life exists.

Megan accelerates her skateboard from 0 m/s to 8 m/s in 2 seconds. What is the magnitude of the acceleration of the skateboard?
O 8 m/s^2
O 16 m/s^2
O 2 m/s^2
O 4 m/s^2​

Answers

Answer:

chk picture for eqn

Explanation:

A rectangular loop of wire with sides 0.129 and 0.402 m lies in a plane perpendicular to a constant magnetic field (see part a of the drawing). The magnetic field has a magnitude of 0.888 T and is directed parallel to the normal of the loop's surface. In a time of 0.172 s, one-half of the loop is then folded back onto the other half, as indicated in part b of the drawing. Determine the magnitude of the average emf induced in the loop.

Answers

Answer:

[tex]0.2677\ \text{V/m}[/tex]

Explanation:

A = Area of loop = [tex]0.129\times0.402[/tex]

B = Magnetic field = [tex]0.888\ \text{T}[/tex]

t = Time taken = [tex]0.172\ \text{s}[/tex]

Electric field is given by

[tex]E=B\dfrac{dA}{dt}\\\Rightarrow E=0.888\times\dfrac{0.129\times 0.402}{0.172}\\\Rightarrow E=0.2677\ \text{V/m}[/tex]

The emf induced is [tex]0.2677\ \text{V/m}[/tex].

what is the value of x if x-36=5?​

Answers

Answer:

Therefore, the value of x is 41

Explanation:

x=5+36

x=41

A 50-turn coil has a diameter of . The coil is placed in a spatially uniform magnetic field of magnitude so that the face of the coil and the magnetic field are perpendicular. Find the magnitude of the emf, , induced in the coil if the magnetic field is reduced to zero unfiformly

Answers

Answer:

EMF = 51.01 Volt

Explanation:

A 50-turn coil has a diameter of 15 cm. The coil is placed in a spatially uniform magnetic field of magnitude 0.500~\text{T}0.500 T so that the plane of the coil makes an angle of 30^\circ30 ​∘ ​​ with the magnetic field. Find the magnitude of the emf induced in the coil if the magnetic field is reduced to zero uniformly in 0.100~\text{s}0.100 s

We have,

Number of turn in the coil, N = 50

The diameter of the coil, d = 15 cm

Radius, r = 7.5 cm = 0.075 m

Initial magnetic field, [tex]B_i=0.5\ T[/tex]

The plane of the coil makes an angle of 30° with the magnetic field.

The magnetic field reduced to zero in 0.1 seconds

We need to find the emf induced in the coil. We know that, emf is equal to the rate of change of magnetic flux. So,

[tex]\epsilon=\dfrac{BNA\cos\theta}{t}\\\\\epsilon=\dfrac{0.5\times 50\times \pi \times 0.075\cos(30)}{0.1}\\\\\epsilon=51.01\ V[/tex]

So, the induced emf in the coil is 51.01 V.


In the diagram, the amplitude of the wave is shown by:

A
B
C
D

Answers

Answer:

A.

Explanation:

Amplitude measures how much a wave rises or falls. This is illustrated by A.

In the diagram, the amplitude of the wave is shown by A.

What is Amplitude?The amplitude of a periodic variable is a measure of its change in a single period. The amplitude of a non-periodic signal is its magnitude compared with a reference value.

There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amplitude of a variable is simply a measure of change relative to its central position. In contrast, magnitude is a measure of the distance or quantity of a variable irrespective of its direction.

Amplitude is a property that is unique to waves and oscillations.

Therefore, in the diagram, the amplitude of a wave is shown by A.

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A chair of weight 85.0 N lies atop a horizontal floor; the floor is not frictionless. You push on the chair with a force of F = 40.0 N directed at an angle of 35.0deg below the horizontal and the chair slides along the floor.
Using Newton's laws, calculate n, the magnitude of the normal force that the floor exerts on the chair.

Answers

Answer:

 N = 107.94 N

Explanation:

For this exercise we must use Newton's second law.

Let's set a reference system with the x-axis parallel to the ground and the y-axis vertical

X axis

        Fₓ = ma

ej and

       N -F_y - W = 0

let's use trigonometry to decompose the applied force

     cos -35 = Fₓ / F

     sin -35 = F_y / F

     Fₓ = F cos -35

     F_y = F sin -35

     Fₓ = 40.0 cos -35 = 32.766 N

     F_y = 40.0 sin -35 = -22.94 N

we substitute

     N = Fy + W

     N = 22.94 + 85

     N = 107.94 N

1 hallar el trabajo mecanico de un cuerpo que tiene una fuerza de 250 newton y recorre 750 metros

2 hallar la potencia necesaria para levantar un transformador de masa 2500kg,una altura de 4 metros en un tiempo de 30 segundos
porfa es para hoy

Answers

Answer: TRACK

Explanation:

Differences between LED and CFL bulb..​

Answers

Explanation:

CFL bulbs were made to take the place of incandescent bulbs, which generate light as a result of heat. ... LED (light-emitting diode) is a type of bulb that produces light using a narrow band of wavelengths. LED lighting is more energy efficient than CFL bulb.

A 2000 kg truck has put its front bumper against the rear bumper of a 2500 kg SUV to give it a push. With the engine at full power and good tires on good pavement, the maximum forward force on the truck is 18,000 N.
What is the maximum possible acceleration the truck can give the SUV?
At this acceleration, what is the force of the SUV's bumper on the truck's bumper?

Answers

Answer:

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

Explanation:

Concepts and reason

The concept required to solve this problem is Newton’s second law of motion.

Initially, write an expression for the force according to the Newton’s second law of motion. Later, rearrange the expression for the acceleration. Finally, substitute the value of the acceleration obtained to find the new force.

Fundamentals

According to the Newton’s second law of motion, the net force is equal to the product of the mass and the acceleration of an object. The expression for the Newton’s second law of motion is as follows:

F = maF=ma

Here, m is mass and a is the acceleration.

(a)

Rearrange the equation F = maF=ma for a.

a = \frac{F}{m}a=  

m

F

 

Substitute 18,000 N for F and \left( {2300{\rm{ kg + 2400 kg}}} \right)(2300kg+2400kg) for m in the equation a = \frac{F}{m}a=  

m

F

 .

\begin{array}{c}\\a = \frac{{18,000{\rm{ N}}}}{{\left( {2300{\rm{ kg + 2400 kg}}} \right)}}\\\\ = \frac{{18,000{\rm{ N}}}}{{\left( {4700{\rm{ kg}}} \right)}}\\\\ = 3.83{\rm{ m/}}{{\rm{s}}^2}\\\end{array}  

a=  

(2300kg+2400kg)

18,000N

 

=  

(4700kg)

18,000N

 

=3.83m/s  

2

 

 

(b)

Substitute 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 for a and 2400 kg for m in the equation F = maF=ma .

\begin{array}{c}\\F = \left( {2400{\rm{ kg}}} \right)\left( {3.83{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 9120{\rm{ N}}\\\end{array}  

F=(2400kg)(3.83m/s  

2

)

=9120N

 

Ans: Part a

The maximum possible acceleration the truck can give the SUV is 3.83{\rm{ m/}}{{\rm{s}}^2}3.83m/s  

2

 .

Part b

The net magnitude of the force of the SUV's bumper on the truck's bumper is 9120 N.

The maximum possible acceleration the truck can give the SUV is equal to 4 m/s².

The force of the SUV's bumper on the truck's bumper is 10000N

What is acceleration?

Acceleration of an object can be described as as the change in the velocity of an object w.r.t. time. The acceleration is a vector quantity, contains both magnitude and direction. Acceleration is the second derivative of position w.r.t. time and the first derivative of velocity w.r.t. time.

According to Newton's second law of motion, the force is equal to the product of acceleration and mass of an object.

F = ma

And, a =  F/m

Given, the mass of the ruck , m = 2000 Kg

The mass of the SUV, M = 2500 Kg

The total mass of the both = 2000 + 2500 = 4500 Kg

The maximum force on the trick , F = 18000 N

The maximum acceleration of the truck can give the SUV:

[tex]a_{max} = \frac{F_{max}}{m+M}[/tex]

a = 18000/4500

a = 4 m/s²

The force of the SUV's bumper on the truck's bumper will be:

[tex]F_{max} -f= ma_{max}[/tex]

[tex]f= 18000-2000\times 4[/tex]

[tex]f =10000N[/tex]

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You attach a 2.30 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched by 0.500 m and release it from rest. Assume the weight slides on a horizontal surface with negligible friction. The weight reaches a speed of zero again 0.400 s after release (for the first time after release). What is the maximum speed of the weight (in m/s)

Answers

Answer: [tex]3.92\ m/s[/tex]

Explanation:

Given

Mass of the attached object is [tex]m=2.3\ kg[/tex]

Spring is stretched by [tex]A=0.5\ m[/tex]

Speed reaches zero after [tex]t=0.4\ s[/tex]

Speed is zero at the extremities of the S.H.M motion that is

[tex]\Rightarrow \dfrac{T}{2}=0.4\\\\\Rightarrow T=0.8\ s[/tex]

Time period of motion is [tex]0.8\ s[/tex] which can also be given by

[tex]\Rightarrow \omega T=2\pi\\\\\Rightarrow \omega=\dfrac{2\pi }{T}\\\\\Rightarrow \omega =\dfrac{2\pi }{0.8}\\\\\Rightarrow \omega=\dfrac{5\pi }{2}[/tex]

Maximum speed for S.H.M. is [tex]v_{max}=A\omega[/tex]

[tex]\Rightarrow v_{max}=0.5\times 2.5\pi\\\Rightarrow v_{max}=3.92\ m/s[/tex]

A 10 n force is applied horizontally on a box to move it 10 m across a frictionless surface. How much work was done to move the box?

Answers

Given from question
Force = 10 N
Displacement = 10 m
Work done = ?
We know that
Work done = force X displacement
So 10 X 10
100
Work done = 100J answer

Answer:

[tex]\boxed {\boxed {\sf 100 \ J}}[/tex]

Explanation:

We are asked to calculate the work done to move a box.

Work is the product of force and distance or displacement.

[tex]W= F*d[/tex]

A 10 Newton force is applied horizontally on the box. Since the surface is frictionless, there is no force of friction, and the net force is 10 Newtons. The force moves the box 10 meters.

F= 10  N d= 10 m

Substitute the values into the formula.

[tex]W= 10 \ N * 10 \ m[/tex]

Multiply.

[tex]W= 100 \ N*m[/tex]

Let's convert the units. 1 Newton meter is equal to 1 Joule, therefore our answer of 100 Newton meters is equal to 100 Joules.

[tex]W= 100 \ J[/tex]

100 Joules of work was done to move the box.

what do we mean by thrust?

Answers

Answer:

the answer is push example: she thrust her hand into her pocket

the lamp cord is 85cm long and comprises cupper wire. Calculate the wire‘s resistance?
radius of a wire is 1.8mmm,Use value of resistivity for Cu as 1.75 × 10-8Ωm.

Answers

Answer:

R = 0.0015Ω

Explanation:

The formula for calculating the resistivity of a material is expressed as;

ρ = RA/l

R is the resistance

ρ is the resistivity

A is the area of the wire

l is the length of the wire

Given

l = 85cm = 0.85m

A = πr²

A = 3.14*0.0018²

A = 0.0000101736m²

ρ = 1.75 × 10-8Ωm.

Substitute into the formula

1.75 × 10-8 = 0.0000101736R/0.85

1.4875× 10-8 = 0.0000101736R

R = 1.4875× 10-8/0.0000101736

R = 0.0015Ω


Physics question plz help ASAP

Answers

The Correct answer is D Hope this helps :)

what additional load will be required to cause the extension of 2.0cm when an elastic wire extend by 1.0cm when a load of 20g range from it

Answers

Answer:

The additional load is 20g

00
Use base units to check whether the following equations are balance
(a) pressure = depth x density gravitational field strength,
(b) energy = mass x (speed of light).
dod to molt solid

Answers

Answer:

In a column of fluid, pressure increases with depth as a result of the weight of the overlying fluid. Thus a column of fluid, or an object submerged in the fluid, experiences greater pressure at the bottom of the column than at the top. This difference in pressure results in a net force that tends to accelerate an object upwards.

The pressure at a depth in a fluid of constant density is equal to the pressure of the atmosphere plus the pressure due to the weight of the fluid, or p = p 0 + ρ h g , p = p 0 + ρ h g , 14.4

Granite: 2.70 × 10 32.70 × 10 3

Lead: 1.13 × 10 41.13 × 10 4

Iron: 7.86 × 10 37.86 × 10 3

Oak: 7.10 × 10 27.10 × 10 2

Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m. What is the work done on the ball by the 34.4 N tension force in the string during one half-revolution of the ball

Answers

Answer:

the work done on the ball is 0

Explanation:

Given the data in the question;

Katie swings a ball around her head at constant speed in a horizontal circle with circumference 2.1 m.

circle circumference = 2πr = 2.1 m

radius r will be; r = 2.1 m / 2π = 0.33 m

Tension force = 34.4 N

one half revolution means, displacement of the ball is;

d = 2r = 2 × 0.33 = 0.66 m

Now, Work done = force × displacement × cosθ

we know that, the angle between the tension force on string and displacement of object is always 90.

so we substitute

Work done = 34.4 N × 0.66 m × cos(90)

Work done = 34.4 N × 0.66 m × 0

Work done = 0 J

Therefore,  the work done on the ball is 0

A cycle track is 500 metres long. A cyclist completes 10 laps (that is, he rides completely round the track 10 times).
a) How many kilometres has the cyclist travelled?
b) On average it took the cyclist 50 second to complete one lap (that it, to ride round just one).
(i) What was the average speed of the cyclist?
(ii) How long in minutes and seconds dit it take the cyclist to complete the 10 laps?
c) Near the end of the run the cyclist put on a spurt. During this spurt it took the cyclist 2 seconds to increase speed from 8 m/s to 12 m/s. What was the cyclist's acceleration during this spurt?

Answers

Explanation:

a) D_t = 500m*10laps

D_t = 5000m or 5km

b)

I) v = d/t

v = 500m/50s

v = 10m/s

ii) 10m/s = 5000m/t

t = 5000m/10m/s

t = (500s)*(1min/60s)

t = 8'20" or 8 mins and 20 sec

c) v_f = v_0 + a*t

v_f-v_0 = a*t

a = (v_f-v_0)/t

a = (12m/s-8m/s)/2s

a = (4m/s)/2s

a = 2m/s²

Part(a),

The distance travelled by the cyclist is 5 km.

Part(b),

(i) The average speed is 10 m/s

(ii) The time taken to cover 10 laps is 8 minutes and 20 seconds.

Part(c),

The acceleration is 2 m/s²

What is speed?

Speed is defined as the ratio of the time distance travelled by the body to the time taken by the body to cover the distance. Speed is the ratio of the distance travelled by time. The unit of speed in miles per hour.

a) The distance will be calculated as

D = 500m*10laps

D = 5000m or 5km

b) The average speed is calculated as,

I) v = d/t

v = 500m/the 50s

v = 10m/s

The time will be calculated as,

ii) 10m/s = 5000m/t

t = 5000m/10m/s

t = (500s)*(1min/60s)

t = 8'20" or 8 mins and 20 sec

The acceleration is calculated as,

c) vf = v0 + at

vf-v0 = at

a = (vf-v0)/t

a = (12m/s-8m/s)/2s

a = (4m/s)/2s

a = 2m/s²

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The mass flow rate through a centrifugal compressor is 1 kg/s. If air enters at 1 bar and 288k and leaves at 200 kN/m² and 370k, determine the power of the compressor. Take Cp = 1.103 kJ (kg.K), R = 287 kJ (kg.k)​

Answers

We have that the power of the compressor is

[tex]H_p=24.242hp[/tex]

From the question we are told that:

Flow rate [tex]W=1kg/s[/tex]

inlet Pressure [tex]P_1=1 bar[/tex]

inlet Temperature [tex]T_1= 288k[/tex]

Outlet Temperature [tex]T_2= 370k[/tex]

Outlet Pressure [tex]P_2=200 kN/m^2=2bars[/tex]

[tex]Cp = 1.103 kJ[/tex]

[tex]R = 287 kJ (kg.k)​[/tex]

Generally, the equation for Adiabatic head  is mathematically given by

[tex]H=\frac{ZRT_1}{Cp-1/K}[\frac{P_2}{P_1}^{(Cp-1)/Cp}-1][/tex]

Where

[tex]Z=Compressibility\ factor[/tex]

[tex]Z=0.99[/tex]

Therefore

[tex]H=\frac{(0.99)(287)(288)}{(1.103)-1/(1.103)}[\frac{(200)}{(1)}^{((1.103)-1)/(1.103)}-1][/tex]

[tex]H=560925.5958 J/kg[/tex]

[tex]H=5.6*10^5J/kg[/tex]

Generally, the equation for centrifugal compressor power  is mathematically given by

[tex]H_p=\frac{WH}{E*33000}[/tex]

Where

E is efficiency (adiabatic)

[tex]E=70\%=0.7[/tex]

Therefore

[tex]H_p=\frac{(1)(5.6*10^5)}{0.7*33000}[/tex]

[tex]H_p=24.242hp[/tex]

In conclusion

The power of the compressor is

[tex]H_p=24.242hp[/tex]

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A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be:_______.
A. in the positive y direction
B. in the negative y direction
C. in the positive x direction
D. in the negative x direction

Answers

C maybe don’t count on it 100%

A single force acts on a particle situated on the positive x axis. The torque about the origin is in the negative z direction. The force might be in the negative y direction. Thus, option B is correct.

To determine the force that could generate a torque in the negative z direction, we need to consider the right-hand rule for cross products. The torque vector, denoted by τ, is given by the cross product of the position vector, r, and the force vector, F:

[tex]τ = r × F[/tex]

In this case, the position vector, r, points along the positive x-axis. The negative z-direction torque indicates that the force vector must be perpendicular to both the position vector and the negative z-axis.

Using the right-hand rule, we can determine that the force vector must be in the negative y-direction, which is option B.

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The temperature of a body falls from 30°C to 20°C in 5 minutes. The air
temperature is 13°C. Find the temperature after a further 5 minutes.

Answers

Answer:

15.88°C I am not 100% sure this is right but I am 98% sure this IS right

Consider two closely spaced and oppositely charged parallel metal plates. The plates are square with sides of length L and carry charges Q and -Q on their facing surfaces. What is the magnitude of the electric field in the region between the plates

Answers

Answer:

  E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]

Explanation:

In this exercise you are asked to calculate the electric field between two plates, the electric field is a vector

         E_ {total} = E₁ + E₂

         E_ {total} = 2 E

where E₁ and E₂ are the fields of each plate, we have used that for the positively charged plate the field is outgoing and for the negatively charged plate the field is incoming, therefore in the space between the plates for a test charge the two fields point in the same direction

to calculate the field created by a plate let's use Gauss's law

          Ф = ∫ E . dA = q_{int} /ε₀

As a Gaussian surface we use a cylinder with the base parallel to the plate, therefore the direction of the electric field and the normal to the surface are parallel, therefore the scalar product is reduced to the algebraic product.

           E 2A = q_{int} / ε₀

where the 2 is due to the surface has two faces

indicate that the surface has a uniform charge for which we can define a surface density

           σ = q_{int} / A

           q_{int} = σ A

we substitute

           E 2A = σ A /ε₀

           E = σ / 2ε₀  

therefore the total field is

           E_ {total} = σ /ε₀

let's substitute the density for the charge of the whole plate

           σ= Q / L²

           

            E_ {total} = [tex]\frac{Q }{L^2 \epsilon_o}[/tex]

Although human beings have been able to fly hundreds of thousands of miles into outer space, getting inside the earth has proven much more difficult. The deepest mines ever drilled are only about 10 miles deep. To illustrate the difficulties associated with such drilling, consider the following: The density of steel is about 7900 kilograms per cubic meter, and its breaking stress, defined as the maximum stress the material can bear without deteriorating, is about 2.0×1092.0×109 pascals. What is the maximum length of a steel cable that can be lowered into a mine? Assume that the magnitude of the acceleration due to gravity remains constant at 9.8 meters per second per second.
Use two significant figures in your answer, expressed in kilometers.

Answers

Answer:

26 km

Explanation:

Let's say our "cable" has a cross section of 1 m²

Then each meter of cable would weight 7900(9.8) = 77420 N

A Pascal is a Newton per square meter

2 x 10⁹ / 77420 = 25840 m or about 26 km or about 16 miles

A light bulb, attached (by itself) to an ideal 10 V battery, becomes hotter as time goes on. The bulb's filament is made of tungsten, a metal, which becomes more resistive as its temperature increases. Which statement below is true?

a. As time goes on and the bulb grows hotter, the voltage across the bulb increases so the bulb would grow brighter.
b. As time goes on and the bulb grows hotter, the current through the bulb increases so the bulb would grow brighter.
c. As time goes on and the bulb grows hotter, the voltage across the bulb decreases so the bulb would grow dimmer.
d. As time goes on and the bulb grows hotter, the current through the bulb decreases so the bulb would grow dimmer.

Answers

Answer:

the correct one is d

Explanation:

The filament of the bulb fulfills the law of ohm

          V = i R

indicate that the filament resistance increases with temperature, as the voltage remains constant if the resistance of the filament increases the current should decrease,

When examining the different statements, the correct one is d

how much is need to lift a load of 100n placed at a distance of 29 cm from fulcrum if effort is applied at 60cm from the fulcrum on opposite side of the load? calculate mechanical advantage and velocity ratio of the lever​

Answers

Answer:

206.8965517 n

Explanation:

First, we need to see that 60:29 is 2.078965517:1. Then we need to multiply the energy put 29 cm from the fulcrum by 2.078965517, giving us the end result of our answer.

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