how much energy is used when the incandescent bulb is left on for 12 hours

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

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 enclosed may help ....

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: 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.

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..

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.

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.

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]

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:

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''.

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