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how much power would a light bulb have if it used 30 joules of energy every 3 seconds

Given data

          mass (m) = 0.2 Kg ,

       velocity (v) =300 m/s ,

      calculate momentum (p) = m × v  

                                                = 0.2 × 300

                                              P  = 60 Kg. m/s

Momentum of the bullet is 60 Kg. m/s..

To read and understand electrical blueprints, you need to be familiar with the various electric symbols used in the drawings.

Hence, the correct option is D.

These symbols represent different electrical components and devices, such as switches, outlets, lights, transformers, and more. Understanding these symbols is crucial for interpreting the layout and design of electrical systems in buildings and structures.

While having a background in electrical engineering (A) or knowledge of building construction (B) can certainly be helpful in interpreting electrical blueprints, the most critical aspect is understanding the meaning and representation of electric symbols (D) as they directly convey information about the electrical components and connections in a given system. Knowing how to sketch a plan view (C) is useful but is not specifically essential for reading electrical blueprints.

Therefore, To read and understand electrical blueprints, you need to be familiar with the various electric symbols used in the drawings.

Hence, the correct option is D.

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a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²

Final answer:

The magnitude of the car's acceleration is calculated using the change in velocity over the time taken. With a change in velocity of 12.0 m/s over 8.0 seconds, the car's acceleration is 1.5 m/s².

Explanation:

To find the magnitude of the car's acceleration, we can use the formula for acceleration, which is the change in velocity (Δv) divided by the time (Δt) it takes for that change. In this case, the car's velocity increases from 18.0 m/s to 30.0 m/s over 8.0 seconds.

We calculate acceleration (a) as follows:

a = Δv / Δt

Δv = final velocity - initial velocity = 30.0 m/s - 18.0 m/s = 12.0 m/s

Δt = 8.0 s

Thus, a = 12.0 m/s / 8.0 s = 1.5 m/s².

The magnitude of the car's acceleration is 1.5 m/s².

Answer:

The type of current described by the given statements is D.C. (or direct Current).

Explanation:

There are two different types of currents. They are Alternating Current (AC) and Direct current (DC) . Alternating current varies alternately, as the name suggests. DC is used for devices that need low voltage, like the chargers of a mobile phone. Domestic power supply is  AC .

The type of current described by these statements is direct current (DC).

The type of current described by these statements is direct current (DC). Direct current flows in one direction from the negative terminal to the positive terminal. It is commonly created by batteries. However, it is generally not found in U.S. electrical outlets, which typically provide alternating current (AC). Alternating current changes direction periodically.

Direct current (DC) is an electric current that flows consistently in one direction. In a DC circuit, electrons move from the negative terminal to the positive terminal of a voltage source, such as a battery. DC is used in many electronic devices while alternating current (AC) periodically changes direction and is typically used for power distribution.

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T = half life of cobalt-60 = 5.3 years

λ = decay constant for cobalt-60 = ?

decay constant is given as

λ = 0.693/T

inserting the values

λ = 0.693/5.3 = 0.131

N₀ = initial amount of cobalt-60 = 10 g

t = time of decay = 21.2 years

N = final amount of cobalt-60 after time "t"

final amount of cobalt-60 after time "t" is given as

N = N₀ [tex]e^{-\lambda t}[/tex]

inserting the values

N = 10 [tex]e^{-(0.131) (21.2)}[/tex]

N = 10 [tex]e^{-(0.131) (21.2)}[/tex]

N = 10 x 0.0622 g

N = 0.63 g

hence

C. 0.63 g

Given data

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

                We know that, efficiency (η) = Net work done ÷ Heat supplied

                                                           η =   W ÷ Q₁  

                                                           W = η × Q₁

               First determine the efficiency ( η ) = ?

                                Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

                                                                        = 33.3% = 0.333

               Now, Work done is W = η × Q₁

                                                    = 0.33 × 3600

                                                   W = 1188 J

Work done by the engine is 1188 J

A. Formula: F=ma or F/m=a

10,000N/1,267kg≈7.9m/[tex]s^{2}[/tex]

B. Formula: a=[tex]\frac{V-V_{0} }{t}[/tex] and s=d/t

speed= 394.6/15

s=26.3m/s

a=[tex]\frac{26.3-0}{15}[/tex]

a=1.75m/[tex]s^{2}[/tex]

C. 7.9-1.75=difference of 6.15m/[tex]s^{2}[/tex]

D. The force that most likely caused this difference is friction forces

Answer:

Part a)

[tex]a = 7.89 m/s^2[/tex]

Part b)

[tex]a = 3.51 m/s^2[/tex]

Part c)

[tex]\Deltra a = 4.38 m/s^2[/tex]

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

[tex]F_f = 5552.8 N[/tex]

Explanation:

Part a)

As we know by newton's II law

[tex]F = ma[/tex]

here we know that

[tex]m = 1267 kg[/tex]

[tex]F = 10,000 N[/tex]

Now we have

[tex]a = \frac{F}{m}[/tex]

[tex]a = \frac{10,000}{1267}[/tex]

[tex]a = 7.89 m/s^2[/tex]

Part b)

distance covered by the car

[tex]d = 394.6 m[/tex]

t = 15 s

now by kinematics we have

[tex]d = \frac{1}{2}at^2[/tex]

[tex]394.6 = \frac{1}{2}a(15^2)[/tex]

[tex]a = 3.51 m/s^2[/tex]

Part c)

Difference of acceleration is given as

[tex]\Delta a = a_{expected} - a_{real}[/tex]

[tex]\Delta a = 7.89 - 3.51 [/tex]

[tex]\Deltra a = 4.38 m/s^2[/tex]

Part d)

This difference in acceleration  is due to some frictional force on the surface.

Part e)

Now for magnitude of force is given as

[tex]F - F_f = ma[/tex]

[tex]10,000 - F_f = ma[/tex]

[tex]10,000 - F_f = 1267\times 3.51[/tex]

[tex]F_f = 5552.8 N[/tex]

Graphite is great lubricant because each carbon atom forms weak covalent bonds with three other carbon atoms.

Diamond and graphite both are entirely made up of carbon but both the elements are completely different because of the bonding of carbon atoms. In graphite the carbon atoms form weak covalent with the other three carbon atom due to which there is less or no force between the atoms, and makes the graphite very soft and slippery, and graphite act as lubricant.

Answer:

Carbon atoms form strong bonds within each graphite layer but weak bonds between layers.

Explanation:

Graphite is known as solid lubricant which means it is a lubricant which is in the solid state form.

Here we know that lubricant has function to make the relative movement very smooth or easy.

So here in structure of graphite it is formed in such a way that all carbon atoms in graphite layer are bonded strongly with each other. While two layers of graphite are weakly bonded to each other

So here two layers of graphite can easily slide over each other which is useful for the function of lubricants

So correct answer will be

Carbon atoms form strong bonds within each graphite layer but weak bonds between layers.

Answer: 9 × 10¹⁶ m²/s² (square of speed of light in vacuum)

Explanation:

Einsteins' energy mass relation:

E = mc²

Where, E is energy, m is mass and c is the speed of light.

The ratio of rest energy to rest mass:

[tex]\frac{E_o}{m_o} = c^2[/tex]

speed of light in vacuum is 3 × 10⁸ m/s

Hence, the numerical quantity of the ratio of rest energy/ rest mass = (3 × 10⁸ m/s)² = 9 × 10¹⁶ m²/s².

You haven't told us the "wattage" rating of the bulb.  We'll just have to call it ' W ' .

The bulb uses energy at the rate of W watts, or 0.001W kilowatts.

In 12 hours, it uses 0.012W kilowatt-hours of energy.

= = = = =

W watts = W Joules/second

1 hour = 3600 seconds

12 hours = (12 x 3600) seconds

Energy = (W Joule/sec) x (12 x 3600 sec)

Energy =  43,200W Joules

Final answer:

The energy used by a 100W incandescent bulb is 4,320,000 joules

Explanation:

Calculating Energy Use of an Incandescent Bulb

To calculate the energy used by an incandescent bulb left on for 12 hours, we need to know the power rating of the bulb. However, if we take a typical 100W incandescent bulb as an example, we can use the formula for electrical energy (Energy = Power *Time) to find out the energy consumption in watt-hours (Wh) and then convert it to joules (J). The power rating is usually mentioned on the bulb itself.

For a 100W bulb left on for 12 hours:

Energy in watt-hours = Power*Time

Energy in watt-hours = 100 W * 12 h = 1200 wh

Since 1 watt-hour is equal to 3600 joules,

Energy in joules == 4320000 J

Therefore, a 100W incandescent bulb uses 4320000 joules of energy when left on for 12 hours.

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Final answer:

The tension force is the force transmitted through a flexible connector, like a rope or cable, when it is pulled on both ends. To solve for the tension force, consider the forces acting on the object connected by the rope and apply Newton's laws of motion.

Explanation:

In physics, tension force refers to the force that is transmitted through a flexible connector, such as a rope or a cable, when it is pulled on both ends.

To solve for the tension force, you need to consider the forces acting on the object connected by the rope or cable. This can be done by applying Newton's laws of motion.

For example, if you know the weight of the object and the angle at which the rope or cable is inclined, you can use trigonometry to calculate the tension force.

Answer:

Centripetal force

Explanation:

The object that keeps an object in a circle is called uniform circular motion. In this type of motion, the distance covered by an object is equal to the circumference of the circle. The mathematical form of centripetal force is given by,

From second law of motion,

F = m × a

a is the centripetal acceleration

[tex]F=\dfrac{mv^2}{r}[/tex]

So, the correct option is (a) '' centripetal force''.

POWER OF CIRCUIT

P=VI

While,

VOLTAGE OF CIRCUIT

V=IR

THEN,

P=I^R

P/R=I^

2.00/30.0=I^

I=0.258A.

Answer:0.258

Explanation:

The Arctic fox is well adapted to its environment with seasonal coat color changes for camouflage and longer, denser winter fur for insulation against the cold.

The Arctic fox is a highly adapted species that thrives in the cold climates of the Arctic region. These foxes have developed several adaptations that make them well-suited for survival in their environment. Changing coat color with the seasons is one such adaptation, allowing Arctic foxes to maintain camouflage against snowy backgrounds in winter and rocky terrains in other seasons.

Furthermore, the longer fur that these animals grow in winter helps them to trap heat more effectively, ensuring they stay warm despite the freezing temperatures. This fur is not just longer, but also denser, adding an extra layer of insulation. Such physical adaptations, alongside behavioral ones like burrowing and hunting tactics, have equipped Arctic foxes to be masterful inhabitants of their harsh and unrelenting environment.