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AleksandrR [38]
1 year ago
15

After an incandescent lamp is turned on, the temperature of its filament rapidly increases from room temperature to its operatin

g temperature. As the temperature of the filament increases, what happens to the resistance of the filament and the current through the filament?
(1) The resistance increases and the current decreases.
(2) The resistance increases and the current increases.
(3) The resistance decreases and the current decreases.
(4) The resistance decreases and the current increases
Physics
2 answers:
Strike441 [17]1 year ago
8 0
I think the correct answer from the choices listed above is option 1. As the temperature of the filament increases, the resistance of the filament and the current through the filament would be that the <span> resistance increases and the current decreases. Hope this answers the question.</span>
IrinaVladis [17]1 year ago
6 0

Answer: (1) The resistance increases and the current decreases.

Explanation:  

When the temperature of the filament increases, the vibrational energy of the constituent atoms increases which leads to increase in inter-atomic collision. Thus, the resistance would increase. The increases in resistance would obstruct the flow of charges more leading to decrease in the value of the current.

Hence, when the temperature of the filament increase, the resistance increases and current decreases.

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Official (Closed) - Non Sensitive
Pavlova-9 [17]

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s    (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s    (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s     (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

<u>t = 319.47 s</u>

5 0
2 years ago
How do air mass conditions ahead of the squall line support the development of new cell?
IRISSAK [1]
<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>
7 0
2 years ago
What is Yusef most likely finding?
Ratling [72]

Explanation:

yusef adds all of the values in his data set and then divide by the number of values in the set. the actual density of iron is 7.874 g/ml .

4 0
1 year ago
A 54 kg man holding a 0.65 kg ball stands on a frozen pond next to a wall. He throws the ball at the wall with a speed of 12.1 m
Inessa [10]

Answer:

The velocity of the man is 0.144 m/s

Explanation:

This is a case of conservation of momentum.

The momentum of the moving ball before it was caught must equal the momentum of the man and the ball after he catches the ball.

Mass of ball = 0.65 kg

Mass of the man = 54 kg

Velocity of the ball = 12.1 m/s

Before collision, momentum of the ball = mass x velocity

= 0.65 x 12.1 = 7.865 kg-m/s

After collision the momentum of the man and ball system is

(0.65 + 54)Vf = 54.65Vf

Where Vf is their final common velocity.

Equating the initial and final momentum,

7.865 = 54.65Vf

Vf = 7.865/54.65 = 0.144 m/s

4 0
1 year ago
he drawing shows two perpendicular, long, straight wires, both of which lie in the plane of the paper. The current in each of th
AleksandrR [38]

Answer:

The magnitudes of the net magnetic fields at points A and B is 2.66 x 10^{-6} T

Explanation:

Given information :

The current of each wires, I = 4.7 A

dH = 0.19 m

dV = 0.41 m

The magnetic of straight-current wire :

B= μ_{0}I/2πr

where

B = magnetic field (T)

μ_{0} = 1.26 x 10^{-6} (N/A^{2})

I = Current (A)

r = radius (m)

the magnetic field at points A and B is the same because both of wires have the same distance. Based on the right-hand rule, the net magnetic field of A and B is canceled each other (or substracted). Thus,

BH = μ_{0}I/2πr

     = (1.26 x 10^{-6})(4.7)/(2π)(0.19)

     = 4.96 x 10^{-6} T

BV = μ_{0}I/2πr

     = (1.26 x  10^{-6})(4.7)/(2π)(0.41)

     = 2.3 x 10^{-6} T

hence,

the net magnetic field = BH - BV

                                     = 4.96 x 10^{-6} - 2.3 x 10^{-6}

                                     = 2.66 x 10^{-6} T

4 0
1 year ago
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