Lamp is placed in the lamp holder. The switch is closed. The lamp glows brightly for a short time and then the lamp does not work. Explain these observations

Answers

Answer 1

Solution :

It is given that the lamp glows brightly for a shorter period of time when the switch is closed on it the switch is put on. But after the some time the lamp goes off and it stops working.

This is because as soon as we on the switch, the current start flowing to the lamp which makes the filament of the lamp to glow, but due to some issue, the current stop flowing even when the switch is on and this stops the lamp from glowing and hence the lamp does not work.


Related Questions

What is an example of a series circuit

Answers

Answer:

Explanation

The most famous and common example is Christmas tree lights. You can't tell easily by looking at them whether they are in series or parallel. But you sure know the difference when one of them burns out. When that happens, the whole string goes dead. No matter what you do (other than find out which bulb burned out) will not fix the problem.

Another example is anything that is temperature controlled. For example a furnace is controlled by a thermostat. When the room temperature reaches a certain point, the thermostat is constructed in a certain way so that it forms an open circuit and no current can flow through it. The furnace motor turns off and the furnace stops pumping hot air into a room.

Parallel incident rays appear to bounce like they have all originated from the same point. What is this point called?

A. cross point

B. midpoint

C. bounce point

D. focal point

Answers

B midpoint because the rays appear to bounce like they have all originated from the same point.

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The current in a resistor is 5 A and the voltage between its terminals is 40 V. Calculate the resistance.





An iron wire has a resistance of 24 Ω. If the voltage across its ends is 12 V, calculate the current in the wire.​

Answers

Answer:

1=8 ohms 2=0.5 Amps

Explanation:

What happens when a negatively charged object A is brought near a neutral object B?
A.
Object B gets a negative charge.
Ο
O o
B.Object B gets a positive charge.
O C.
Object B stays neutral but becomes polarized.
D.
Object A gets a positive charge.
O
E.
Object A loses all its charge.
Reset
Next

Answers

Answer:

A.

Explanation:Object b will get a negative charge .

A car is travelling at 60m/s. and brakes to a speed of 14m/s, in 2.7 seconds. What is the deceleration?​

Answers

Answer:

by using v = u + at equation we can find "a"

14 = 60 - 2.7a

2.7a = 60 - 14

2.7a = 46

decceleration = 17.03

Rewrite the false statements correctly
1.If an object sinks in one liquid and floats on another liquid,it implies that the density of second liquid is less than the first liquid.
2.The immersed volume of body in a liquid depends on density of the liquid.
3.Relative density of a body is usually expressed in kgm^-3​

Answers

Explanation:

1. if an object sjnks in one liquid and floats on another liquid it implies that the density of second liquid is greater than the density of first liquid

what is the speed of a wave with a wavelength of 3.0 m and a period of 0.40 s?

Answers

The formula you need for this is
v = f λ
(velocity = frequency • wavelength)
A built-in reminder for this relationship is the units:
meters / second = meters • 1/seconds (aka hz, frequency)
therefore v = 3m / 0.4 seconds = 7.5 m/s

How do you use the periodic table to recall the ionic charge of an alkali metal, an alkaline earth metal, or aluminum?


The positive charge is the group number.

The negative charge is the group number.

The positive charge is the period number.

The negative charge is the period number.

Answers

Answer:

the positive charge is the period number

Explanation:

I might be wrong

Answer:

The positive charge is the group number.

Explanation:

Which energy store is increased when an object is heated?

Answers

Answer:

Kinetic Energy

Explanation:

which forces are capable of affecting particles or objects from large distance

Answers

Answer:

only long-range force that affects all particles is the gravitational force.

Explanation:

In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.

The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.

Consequently the only long-range force that affects all particles is the gravitational force.

Use the universal law of gravitation to solve the following problem.


Hint: mass of the Earth is = 5.97 x 1024 kg


A scientific satellite of mass 1300 kg orbits Earth 200 km above its surface. If Earth has a radius of 6378 km, what is the force of gravity acting on the scientific satellite?


a. Write out the formula for this problem.


b. Plug in the values from this problem into the formula.


c. Solve the problem, writing out each step.


d. Correct answer

Answers

Answer:

a.

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

b.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N

Explanation:

a. The formula for finding the force of gravity, F, acting object on an object is given as follows;

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

Where;

F = The force acting between the Earth and the object

G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M = The mass of the Earth = 5.97 × 10²⁴ kg

m = The mass of the object

r = The distance between the center of the Earth and the object

b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;

The given mass of the satellite, m = 1,300 kg

The distance between the center of the Earth and the center of the satellite,  r = The length of the radius of the Earth + The height of orbit of the satellite

The given height of orbit of the satellite, h = 200 km

∴ r = R + h = 6,378 km + 200 km = 6,578 m

Therefore, by plugging in the values, we get;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c. Solving the above equation gives;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton

1. A vehicle of mass 1500 kg braked to a standstill from a
velocity of 24 m/s in 12 s.
i. Show that the deceleration of the vehicle was 2.0 m/s2.
ii. Calculate the resultant force on the vehicle.

Answers

Explanation:

i. Vi=24

Vf=0

t= -2

a=vf-vi/t =0-24/12 = -2m/S2

ii. F=ma = 1500×-2= -3000 N

A ship anchored at sea is rocked by waves that have crests Lim apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answers

Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

used to measure temperature
used to measure force
prefix that means 1/100
prefix that means 1,000
prefix that means 1/1,000

Answers

Answer:

prefix that means 1/100 = Centi

prefix that means 1,000 = Kilo

prefix that means 1/1000 = Milli

Explanation:

A 15 cm length of wire is moving perpendicularly
through a magnetic field of strength 1.4 T at the rate
of 0.12 m/s. What is the EMF induced in the wire?
A. OV
C. 0.025 v
B. 0.018 V
D. 2.5 V

Answers

Answer: C or B

Explanation:

The EMF induced in the wire moving perpendicularly through a magnetic field is 0.025V. The correct option is C.

What is EMF?

The EMF is the electro motive force which causes the current to induce in the object moving in the magnetic field.

Given is the length of wire L =15cm =0.15m, magnetic field strength B = 1.4T and velocity of wire V =0.12 m/s

EMF is related to the length of wire, magnetic field strength  and velocity of wire proportionally.

ε = B x L x V

Plug the values, we get

ε = 1.4 x 0.15 x 0.12

ε = 0.025 Volts

Thus, the correct option is C.

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In this graph, calculate the speed of
segment A in m/s?

Answers

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

What type of electromagnetic waves do heat lamps give off?

A. infrared

B. ultraviolet

C. microwaves

D. radio waves

Answers

Microwaves c…………….nnsjksisisysgxgd

a sphere of mass 5kg and volume 2×10-5completely immersed in water find the buoyant force exerted water​

Answers

Answer:

Buoyant force exerted water​ =  0.196 Newton

Explanation:

Given:

Mass of sphere ball = 5 kg

Volume = 2 x 10⁻⁵

Find:

Buoyant force exerted water​

Computation:

Buoyant force exerted water​ = Gravity due to acceleration x volume of object x density of given liquid

Buoyant force exerted water​ = 9.8 x 2 x 10⁻⁵ x 1000

Buoyant force exerted water​ =  0.196 Newton

At the local grocery store, you push a 14.5-kg shopping cart. You stop for a moment to add a bag of dog food to your cart. With a force of 12.0 N you now accelerate the cart from rest through a distance of 2.29 m in 3.00 s. What was the mass of the dog food?

Answers

Answer:

The mass of the dog food added is 9.03 kg

Explanation:

Given;

mass of the shopping cart, m₁ = 14.5 kg

let the mass of the bag added = m₂

the force applied, F = 12 N

initial velocity of the cart-bag system, u = 0

distance traveled by the system, d = 2.29 m

time of motion of the system, t = 3.0 s

The acceleration of the system is calculated as;

[tex]d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2[/tex]

The total mass of the system (M) is calculated as follows;

F = Ma

M = F/a

M = (12)/(0.51)

M = 23.53 kg

The mass of the dog food added is calculated as;

m₂ = M - m₁

m₂ = 23.53 kg - 14.5 kg

m₂ = 9.03 kg

Explain: What happens to the velocity of a stream as the size of the sediment increases?

Answers

Answer:

Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.

in a series circuit, how does the voltage supplied by the battery compare to the voltages on each load?

Answers

Answer:

In a series circuit, how does the voltage supplied by the battery compare to the voltage on each load? The voltage of the battery is equal to the voltage of each load added together. ... The voltage across the two resistors must both have the same voltage of the battery.

Explanation:

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Answer:

The voltage of the battery is equal to the voltage of each load added together. The voltage across the two resistors must both have the same voltage of the battery.

>3

Which type of wave causes particles of matter to vibrate in a direction
perpendicular to the direction of its motion?
O A. Sound
B. Transverse
C. Longitudinal
D. Compression

Answers

It’s transverse so the answer is B

Answer:

C.) Longitudinal

please help me..im begging you​

Answers

Answer: The equations in column A is matched with gas laws in column B as follows:

21. PV = nRT : (g) Ideal gas law

22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law

23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law

24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law

25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law

26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law

27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion

Explanation:

(A) Ideal gas law: It states that the product of pressure and volume is directly proportional to the product of number of moles and temperature.

So, PV = nRT

where,

P = pressure

V = volume

n = no. of moles

R = gas constant

T = temperature

Boyle's law: At constant temperature, the pressure of a gas is inversely proportional to volume.

So, [tex]P_{1}V_{1} = P_{2}V_{2}[/tex]

Charles' law: At constant pressure, the volume of a gas is directly proportional to temperature. So,

[tex]V \propto T\\\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}\\V_{1}T_{2} = V_{2}T_{1}[/tex]

Gay-Lussac's law: At constant volume, the pressure of a gas is directly proportional to temperature.

So,  [tex]P_{1}T_{2} = P_{2}T_{1}[/tex]

Avogadro's law: At same temperature and pressure, the volume of gas is directly proportional to moles of gas.

So, [tex]V_{1}n_{2} = V_{2}n_{1}[/tex]

Combined gas law: When Boyle's law, Charles' law, and Gay-lussac's law are combined together then it is called combined gas law. So,

[tex]\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\or, P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex]

Graham's law of effusion: It states that the rate of effusion of a gas is inversely proportional to the square root of mass of its particles.

[tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex]

Thus, we can conclude that equation in column A is matched with gas laws in column B as follows:

21. PV = nRT : (g) Ideal gas law

22. [tex]V_{1}n_{2} = V_{2}n_{1}[/tex] : (f) Avogadro's law

23. [tex]P_{1}V_{1}T_{2} = P_{2}V_{2}T_{1}[/tex] : (e) Combined Gas Law

24. [tex]P_{1}T_{2} = P_{2}T_{1}[/tex] : (d) Gay-Lusaac's law

25. [tex]V_{1}T_{2} = V_{2}T_{1}[/tex] : (c) Charles' law

26. [tex]P_{1}V_{1} = P_{2}V_{2}[/tex] : (b) Boyle's law

27. [tex]\frac{v_{1}}{v_{2}} = \frac{\sqrt{MM_{1}}}{MM_{2}} = \frac{\sqrt{p_{1}}}{p_{2}}[/tex] : (a) Graham's Law of effusion

The force of friction acting on a sliding crate is 223 N.
How much force must be applied to main- tain a constant velocity?

Answers

Answer:

Friction Opposes Motion of an Object.

Now

To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force

Net force = Pushing Force - Frictional Force

Recall

Net Force; F=Ma

Ma = P - Fr

Now the question asked for How Much force Must be applied to Maintain a Constant velocity.

In a Constant Velocity Motion... Acceleration do not change... Its Zero

So Putting this into the formula above

M(0) = P - Fr

0=P - Fr

Fr = P.

This means

That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force

Since Frictional Force; Fr =223N

The Applied Force(Pushing Force) Must be equal to 223N too.

A educação física, enquanto componente curricular da educação básica. Qual a tarefa que educação física?? Alguem me ajuda por fvr ??

Answers

Answer:

como assim qual a tarefa que educação física? se você me explicar melhor eu consigo te responder !!

Explanation:

A girl travels 50m in 12s and then another 30m in 5s .Calculate her average speed?​

Answers

Answer:

4.71m/s

Explanation:

Average speed = Total distance travelled ÷ Total time taken.

80/17=4.71

4.71m/s

Answer:

Average speed = Total distance travelled ÷ Total time taken.

In this question,

Total distance travelled = 50m + 30m

                                                = 80m.

Total time taken = 12s + 5s

                                     = 17s

So, Average speed would be 80 ÷ 17

= [tex]\frac{80}{17}[/tex]

 = 4.71 m/s. or 4.71 meter per second.

In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

The force of the wall on the car and the car on the wall are equal

The force of the wall on the car is greatest

The force of the car on the wall is greatest

There is not enough information to tell

Answers

Answer:

A...................................

The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.

The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.

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The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

a. The force of the wall on the car and the car on the wall are equal

b. The force of the wall on the car is greatest

c. The force of the car on the wall is greatest

d. There is not enough information to tell"

A person skateboards at 3.25 m/s for 55.0 s. How far did he travel?

Answers

178.75 m/s or 399.852 mph

HELP, SCIENCE QUESTION I AM STUCK
6. Which of the following is NOT part of a circuit?
A. rim B. load C. power source D. conductor

Answers

The answer is A because it is not part of the circuit

help me please
only if you really know​

Answers

Assume R is measured in meters (m) and M in kilograms (kg). Then

R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²

so t ² is indeed proportional to R ²/(GM).

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