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Tambaya 1 Rahoto
What will be the weight of a man of mass 60kg standing in a lift if the lift is descending vertically at 3ms2 ?
Bayanin Amsa
To find the apparent weight of a man of mass 60 kg standing in a descending lift, we first need to understand the concept of apparent weight. Apparent weight is the force that the man feels as his weight due to the reaction of the lift floor on him. When the lift accelerates, the apparent weight changes from his actual weight.
In this case, the lift is descending with a constant velocity of 3 m/s2. Since the acceleration is downward, it means the lift is accelerating negatively compared to an upward acceleration.
The formula to find the apparent weight (Wapparent) when in a lift is:
Wapparent = m(g - a)
Where:
Substituting these values into the formula, we get:
Wapparent = 60 (9.8 - 3)
Calculating further:
Wapparent = 60 × 6.8
Wapparent = 408 N
The closest option to 408 N in the answers provided is 420 N. Therefore, the correct answer is 420 N.
Tambaya 2 Rahoto
Bifocal lens is used to correct the eye defect of
Bayanin Amsa
Bifocal lenses are primarily used to correct the eye defect known as presbyopia. As people age, the lens of the eye naturally loses its flexibility, making it difficult to focus on objects that are close up. This condition is known as presbyopia. A bifocal lens is designed with two different optical powers to accommodate this need. The upper part of the lens is usually crafted for distance vision, while the lower segment is designed for near vision tasks, such as reading.
Astigmatism is a different eye condition caused by irregular curvature of the cornea or lens, resulting in blurred or distorted vision at all distances. This condition is typically corrected with cylindrical lenses rather than bifocals.
Hypermetropia, commonly known as farsightedness, is a condition where distant objects can be seen more clearly than near ones. Simple convex lenses are usually used for this correction.
Myopia, or nearsightedness, is a condition where nearby objects are seen clearly, while distant objects appear blurry. Concave lenses are generally used to correct this condition.
In summary, bifocal lenses are specifically designed to address the challenges of focusing at different distances simultaneously, making them ideal for managing presbyopia.
Tambaya 3 Rahoto
The gravitational force between two objects masses 1024 kg and 1027 kg is 6.67N. Calculate the distance between them [ G = 6.6 x 10−11 Nm2 kg−2 ]
Bayanin Amsa
To calculate the distance between two objects based on the gravitational force acting between them, we need to use the formula for gravitational force:
F = (G * m1 * m2) / r²
Where:
We need to compute r by rearranging the formula:
r² = (G * m1 * m2) / F
Therefore, the distance r is:
r = √((G * m1 * m2) / F)
Substitute the given values into the equation:
r = √((6.6 x 10-11 Nm²/kg² * 1024 kg * 1027 kg) / 6.67 N)
Calculating inside the square root:
G * m1 * m2 = 6.6 x 10-11 * 1024 * 1027 = 6.6 x 1040 Nm²
Then divide by the force:
6.6 x 1040 Nm² / 6.67 N = 0.99 x 1040 m²
Finally, calculate the square root:
r = √(0.99 x 1040)
r ≈ 1.0 x 1020 m
Therefore, the distance between the two objects is approximately 1.0 x 1020 m.
Tambaya 4 Rahoto
A thick glass tumbler cracks when boiling water is poured into it because
Bayanin Amsa
When boiling water is poured into a thick glass tumbler, the inner surface of the glass is suddenly exposed to a much higher temperature compared to the outer surface. Glass is a poor conductor of heat, which means it does not transfer heat quickly. As a result, the inside of the tumbler becomes hot and attempts to **expand quickly**, while the outside remains cooler and does not expand at the same rate.
**This uneven expansion** creates tension between the inner and outer layers of the glass. The inner surface tries to expand but is constrained by the cooler, rigid outer surface, which isn't expanding as much or as quickly. This stress and tension can lead to cracking.
Therefore, the correct reason a thick glass tumbler cracks when boiling water is poured into it is because **the inside expands more rapidly than the outside.**
Tambaya 5 Rahoto
At a pressure of 105 Nm−2 , a gas has a volume of 20m3 . Calculate the volume at 4 x 105 Nm−2 at constant temperature.
Bayanin Amsa
In order to solve this problem, we can apply **Boyle's Law**, which states that the **pressure** and **volume** of a gas are inversely proportional at a constant temperature. Mathematically, this is expressed as:
P1V1 = P2V2
Where:
Rearranging the formula to solve for V2:
V2 = (P1V1) / P2
Substituting the given values:
V2 = (105 Nm-2 x 20 m3) / (4 x 105 Nm-2)
By calculating:
V2 = (2100 m3) / 4 x 105
V2 = 5 m3
Therefore, at a pressure of 4 x 105 Nm-2, the volume of the gas is 5 m3.
Tambaya 6 Rahoto
The charge of magnitude 1.6 x 10 −19 C is placed in a uniform electric field of intensity 1200Vm−1 . Calculate its acceleration, if the mass of the charge is 9.1 x 10−31 kg
Bayanin Amsa
To calculate the acceleration of a charge in an electric field, we start by determining the force acting on the charge. The force \( F \) experienced by a charge \( q \) in a uniform electric field \( E \) is given by the equation:
F = q * E
We are given:
Substituting these values into the equation for force:
F = 1.6 x 10-19 C * 1200 V/m
This results in:
F = 1.92 x 10-16 N
Next, we use Newton’s second law of motion to find the acceleration \( a \) of the charge. This law is given as:
F = m * a
Rearranging for \( a \), we have:
a = F / m
We know:
Substituting these values in the equation for acceleration:
a = \(\frac{1.92 x 10^{-16} N}{9.1 x 10^{-31} kg}\)
Calculating the above expression gives:
a ≈ 2.11 x 1014 ms-2
Therefore, the acceleration of the charge is approximately 2.11 x 1014 ms-2.
Tambaya 7 Rahoto
An example of a non-rechargeable cell is
Bayanin Amsa
A non-rechargeable cell, commonly known as a primary cell, is a type of chemical battery that is designed to be used once until the chemical reactions that produce electricity are exhausted. After this point, the cell cannot be reversed or recharged.
In the given examples, the dry leclanche cell is a well-known example of a non-rechargeable cell. It is commonly used in everyday devices like remote controls, wall clocks, and torches. This cell type utilizes zinc and manganese dioxide as electrodes and relies on a moist paste of ammonium chloride for the electrolyte.
The other examples, such as nickel iron, mercury cadmium, and lead-acid, involve rechargeable cells (secondary cells) that are specifically designed to endure multiple charges and discharges throughout their useful life. Thus, unlike the dry leclanche cell, these can be recharged after use.
Therefore, the dry leclanche cell is an ideal example of a non-rechargeable cell because it can only be used once. After depletion, it cannot be recharged or reused.
Tambaya 8 Rahoto
Find the amount of current required to deposit 0.02kg of metal in a given electrolysis for 120 seconds. [electro chemical equivalent of the metal = 1.3 x 10−7 kgC−1 ]
Bayanin Amsa
To determine the amount of current required, we need to use Faraday's laws of electrolysis. The first law states that the mass of the substance deposited at an electrode is directly proportional to the quantity of electricity (or charge) that passes through the electrolyte.
Here, we have:
According to Faraday's first law of electrolysis, the mass (\( m \)) can be calculated by the formula:
m = z \times I \times t
Where:
Rearranging the formula to solve for current \( I \):
I = \(\frac{m}{z \times t}\)
Substituting the given values into the formula:
I = \(\frac{0.02 \, \text{kg}}{1.3 \times 10^{-7} \, \text{kg/C} \times 120 \, \text{s}}\)
Calculating the denominator:
I = \(\frac{0.02}{1.56 \times 10^{-5}}\)
Solving for \( I \):
I = 1282.05 \, \text{A}
Thus, the appropriate amount of current required to deposit 0.02 kg of metal in 120 seconds is approximately 1.3 x 103 A.
Tambaya 9 Rahoto
The device for measuring the angle of dip is
Bayanin Amsa
The device used for measuring the angle of dip is the dip circle.
Let me explain this in simple terms:
The angle of dip, also known as the magnetic inclination, is the angle made by the Earth's magnetic field lines with the horizontal plane. It varies depending on where you are on the Earth's surface. In some places, magnetic field lines are nearly vertical, while in others they are more horizontal.
A dip circle is a specialized scientific instrument used to measure this angle. It usually consists of a magnetic needle that is free to rotate in the vertical plane.
When using a dip circle, you align it so that its plane is parallel to the direction of the Earth's magnetic field. Then, you read the angle at which the magnetic needle stabilizes. This is the angle of dip. The instrument's mechanism allows for accurate measurement of this angle by compensating for any external influences or inclinations.
Tambaya 10 Rahoto
Bayanin Amsa
To solve this problem, we need to understand the relationship between pressure, volume, and temperature of a gas. The relevant law here is the **Combined Gas Law**, which is expressed as:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
In the given problem:
Applying the Combined Gas Law:
(P1 * V1) / 300 = (2 * P1 * V2) / 400
Simplifying this equation:
V1/300 = 2V2/400
Multiply both sides by 400 to clear the fraction:
400 * V1 / 300 = 2 * V2
Which further simplifies to:
(4/3) * V1 = 2 * V2
Dividing both sides by 2:
(2/3) * V1 = V2
This shows that the final volume, V2, is **2/3 of the initial volume, V1**. Therefore, the volume of the gas will **decrease by 1/3**.
Tambaya 11 Rahoto
The dimension of power is
Bayanin Amsa
The dimension of power in physics is expressed in terms of the base units of mass (M), length (L), and time (T). Power is the rate at which work is done or energy is transferred over time, and it has the unit of watt (W) which is equivalent to one joule per second.
To derive the dimension of power:
1. Work has the dimension of energy, which is force applied over a distance. The dimension of work (or energy) is M L2 T-2 because force has the dimension M L T-2 and distance adds another L.
2. Since power is work done per unit time, you would divide the dimension of work by time (T).
Thus, the dimensional formula for power is:
M L2 T-3
Tambaya 12 Rahoto
Which of these gas laws is equivalent to workdone
Bayanin Amsa
To understand which of these gas laws is equivalent to work done, we must first understand the basic concept of work in the context of gases. For gases, work is done when there is a change in volume under pressure, typically expressed as W = P ΔV, where W is work, P is pressure, and ΔV is the change in volume.
Let's consider the given gas laws:
Among these, Boyle's law relates directly to work done because it involves a change in volume at constant temperature, implying that work occurs as a gas expands or compresses. The equation P₁V₁ = P₂V₂ is foundational for calculating work done in reversible processes, which aligns with the expression for work done on a gas, W = P ΔV. Thus, **Boyle's law** is most directly connected to the concept of work done on a gas.
Tambaya 13 Rahoto
The capacitance of a capacitor, C, is inversely proportional to
Bayanin Amsa
The capacitance of a capacitor is primarily determined by three key factors: the area of the plates, the distance between the plates, and the dielectric material used between the plates.
Capacitance (C) is calculated using the formula:
\(C = \frac{\varepsilon A}{d}\)
Where:
Let's analyze the relationship:
In summary, the capacitance of a capacitor is inversely proportional to the distance between the plates. Hence, you increase capacitance by decreasing the distance between the plates.
Tambaya 14 Rahoto
An accumulator is 90% efficient. If it gives out 2700J of energy while discharging, how much energy does it take in?
Bayanin Amsa
In order to find out how much energy the accumulator takes in, given that it is 90% efficient and gives out 2700J of energy, we can use the formula for efficiency:
Efficiency = (Useful Energy Output / Total Energy Input) × 100%
Given:
Efficiency = 90%
Useful Energy Output = 2700J
We need to calculate the Total Energy Input (how much energy the accumulator takes in). Rearranging the formula to solve for Total Energy Input, we get:
Total Energy Input = Useful Energy Output / Efficiency
Substitute the known values:
Total Energy Input = 2700J / 0.9
Calculate the input:
Total Energy Input = 3000J
Therefore, the accumulator takes in 3000J of energy.
Tambaya 15 Rahoto
Convert 60ºC to degree Fahrenheit
Bayanin Amsa
To convert temperatures from Celsius to Fahrenheit, we use the formula:
F = (C × 9/5) + 32
Here, F represents the temperature in Fahrenheit, and C represents the temperature in Celsius.
Let's use this formula to convert 60ºC to Fahrenheit:
F = (60 × 9/5) + 32
First, multiply 60 by 9/5:
60 × 9/5 = 108
Next, add 32 to 108:
108 + 32 = 140
Therefore, 60ºC is equal to 140ºF.
Tambaya 16 Rahoto
The energy of light of frequency 2.0 x 1015 Hz is (h = 6.63 x 10−34 Js)
Bayanin Amsa
To determine the energy of light given its frequency, we can utilize the formula:
E = h × f
Where:
E is the energy of the photon in joules (J)
h is Planck's constant, approximately 6.63 × 10-34 J·s
f is the frequency of light in hertz (Hz)
Given the frequency f = 2.0 × 1015 Hz, we can substitute the known values into our equation:
E = 6.63 × 10-34 J·s × 2.0 × 1015 Hz
To simplify the calculation, multiply the numerical parts and then add the indices of 10:
E = (6.63 × 2.0) × (10-34 × 1015)
E = 13.26 × 10-19 J
This can be approximated to 1.33 × 10-18 J. Thus, the energy of light with the given frequency is 1.33 × 10-18 J.
Tambaya 17 Rahoto
The acceleration of a free fall due to gravity is not a constant everywhere on the Earth's surface because
Bayanin Amsa
The elliptical shape of the Earth: The Earth is not a perfect sphere; it is slightly flattened at the poles and bulging at the equator. This shape causes variations in gravitational acceleration.
Tambaya 18 Rahoto
A light ray passing from air into water at an angle of 30º from the normal in air would
Bayanin Amsa
When light passes from one medium to another, such as from air to water, it bends or refracts. This phenomenon is described by Snell's Law, which states: n₁ * sin(θ₁) = n₂ * sin(θ₂), where:
The refractive index of air is approximately 1, and the refractive index of water is approximately 1.33. Given the angle of incidence in air is 30º:
Using Snell's Law:
1 * sin(30º) = 1.33 * sin(θ₂)
You will find:
sin(θ₂) = sin(30º) / 1.33
sin(θ₂) ≈ 0.5 / 1.33
sin(θ₂) ≈ 0.375
Now, solve for θ₂ by taking the inverse sine (arcsin):
θ₂ ≈ arcsin(0.375)
θ₂ ≈ 22.09º
Thus, when a light ray passes from air into water at an angle of 30º from the normal in air, it will make an angle less than 30º from the normal in water, approximately 22.09º. This is because the light ray bends toward the normal as it enters a denser medium (water).
Tambaya 19 Rahoto
I
6 X + 6 H2 O → C6 H12 O6 + 6O2
III chlorophyll II IV
Use the diagram above to answer question that follows
The part labelled I is
Bayanin Amsa
The part labelled I in the diagram refers to **sunlight**.
Here's a simple explanation:
The given chemical equation is a representation of **photosynthesis**, a process by which green plants, algae, and some bacteria convert light energy, typically from the sun, into chemical energy stored in glucose (C6H12O6) and release oxygen (O2) as a by-product.
In the context of the equation:
- **6CO2 (Carbon Dioxide) + 6H2O (Water) → C6H12O6 (Glucose) + 6O2 (Oxygen)**
The arrow indicates the transformation that occurs during the process. The **chlorophyll** (labelled in the diagram) indicates the presence of chlorophyll pigments in the chloroplasts of plant cells which are essential for **absorbing sunlight**.
Since **sunlight** is the source of energy that powers this transformation, it is the correct component for the part labelled I in the diagram.
Tambaya 20 Rahoto
In a Hare's apparatus, the height of water and a liquid X are 0.3m and 0.5m respectively. The relative density of x is?
Bayanin Amsa
For Hare's apparatus
Relative density = hwhl
Given: height of liquid = 0.5cm, height of water = 0.3cm
Relative density = 0.30.5 = 0.6
Tambaya 21 Rahoto
The total number of ATP produced during glycolysis is
Bayanin Amsa
During the process of glycolysis, a single glucose molecule is broken down into two molecules of pyruvate. During this metabolic pathway, there is a net gain of adenosine triphosphate (ATP) molecules. To understand how many ATP molecules are produced, let's break it down step by step.
1. **Initial ATP Investment:** Glycolysis initially requires an investment of 2 ATP molecules to phosphorylate glucose and convert it into a more reactive form during the early stages of the glycolytic pathway.
2. **ATP Production:** As glycolysis progresses, a total of 4 ATP molecules are produced. This occurs in the later steps of the pathway where adenosine diphosphate (ADP) is phosphorylated to form ATP. This is known as substrate-level phosphorylation.
3. **Net ATP Gain:** To find out the net gain of ATP through glycolysis, simply subtract the initial ATP investment from the total ATP produced:
Net ATP = Total ATP produced - Initial ATP investment
Net ATP = 4 ATP - 2 ATP
Net ATP = 2 ATP
Thus, the net total number of ATP produced during glycolysis is 2 molecules.
Tambaya 22 Rahoto
Inbreeding is highly discouraged in humans because it may
Bayanin Amsa
Inbreeding is the process where closely related individuals, like cousins or siblings, mate and produce offspring. **This practice is highly discouraged in humans for several reasons, but a significant concern is the potential for an outbreak of hereditary diseases.**
Here’s why inbreeding is problematic:
Therefore, **to promote genetic diversity and reduce the risk of hereditary diseases in offspring, inbreeding is discouraged in human populations**. This way, offspring are less likely to inherit harmful genetic combinations that can lead to health problems.
Tambaya 23 Rahoto
When thermal energy in a solid is increased, the change in state is called
Bayanin Amsa
When the thermal energy in a solid is increased, the solid particles gain energy and begin to vibrate more vigorously. As the temperature rises, these particles eventually have enough energy to overcome the forces holding them in their fixed positions. This leads to a change of state from a solid to a liquid. This process is known as melting.
To further understand this, imagine an ice cube. As it absorbs heat, it gains energy, and the ice (which is a solid) starts to turn into water (which is a liquid). This transition is what we refer to as melting.
Thus, the term that describes this change of state, when a solid is heated and turns into a liquid, is melting.
Tambaya 24 Rahoto
Which of the following is the best as shaving mirror?
Bayanin Amsa
When selecting the best type of mirror for shaving, the key consideration is how the mirror reflects light and creates an image. For the purpose of shaving, it is important to have a mirror that magnifies the face and provides a clear view.
The best option for a shaving mirror is a concave mirror. Here is why:
Other types of mirrors, like convex and plane mirrors, and parabolic mirrors, do not provide the same level of magnification or focused reflecting properties, making them less suitable for shaving purposes.
Tambaya 25 Rahoto
What is the inductance reactance of a coil of 7H when connected to a 50Hz a.c circuit?
Bayanin Amsa
To determine the inductive reactance of a coil, we use the formula:
Inductive Reactance (XL) = 2πfL
Where:
Given:
Substituting the given values into the formula:
XL = 2 × π × 50 × 7
Calculating this:
XL = 2 × 3.14159 × 50 × 7
XL ≈ 2 × 3.14159 × 350
XL ≈ 2 × 1099.557
XL ≈ 2199.114
Therefore, the inductive reactance of the coil is approximately 2200Ω.
Tambaya 26 Rahoto
Bayanin Amsa
In a series resonant circuit, the current flowing in the circuit is at its maximum. Let me explain why:
In a series resonant circuit, we have a resistor (R), inductor (L), and capacitor (C) connected in series with an AC source. At a particular frequency called the resonant frequency, these circuits exhibit some unique characteristics. This resonant frequency is determined by the values of the inductor and capacitor and is given by the formula:
f₀ = 1 / (2π√(LC))
At the resonant frequency:
Thus, in a series resonant circuit, when it is operating at its resonant frequency, the current flowing is at its maximum.
Tambaya 27 Rahoto
The value of R in the above circuit to make the galvanometer measure 2A is
Bayanin Amsa
Given: Ig = 50mA = 0.05A, I to be measured = 2A, r = 2Ω , Is = I - Ig = 2 - 0.05 = 1.95A
Shunt(R) = IgIs x r
R = 0.051.95 x 10 = 0.2564Ω
Tambaya 28 Rahoto
Rainbow is formed when sunlight undergoes
Bayanin Amsa
A rainbow is formed through a combination of three processes: reflection, refraction, and dispersion. Let's break down each process to understand how a rainbow forms:
1. Refraction: When sunlight enters a raindrop, it bends or changes direction. This bending of light is known as **refraction**. Different colors of sunlight bend by different amounts because they have different wavelengths.
2. Reflection: Once inside the raindrop, the light gets reflected off the inside surface of the drop. This reflection sends the light back out of the raindrop at different angles.
3. Dispersion: As the light exits the raindrop, it bends again (refraction). Because each color bends by a different amount, the sunlight is spread out into its component colors, creating a spectrum. This spreading into a spectrum is called **dispersion**.
All three processes contribute to the formation of a rainbow. The combination of **refraction, reflection, and dispersion** results in the beautiful arc of colors that we see in the sky.
Tambaya 29 Rahoto
Which of the following materials has a very large energy gap band?
Bayanin Amsa
An insulator is a material that has a very large energy gap between its valence band and conduction band. To understand this, let's first consider the concept of energy bands: In materials, electrons exist in different energy levels. These levels form bands called the valence band and the conduction band. A material is classified based on the size of the energy gap between these bands.
Thus, insulators have a very large energy gap band, making them poor conductors of electricity.
Tambaya 30 Rahoto
Mouth part adapted for piercing and sucking is found in
Bayanin Amsa
The mouthpart adapted for piercing and sucking is found in the mosquito. Mosquitoes have a specialized mouth structure called a proboscis. This proboscis is long and slender, allowing mosquitoes to puncture the skin of their hosts and suck blood. The proboscis is a complex structure that contains several needle-like parts that make the piercing and sucking process efficient and effective.
Tambaya 31 Rahoto
Bile is a greenish alkaline liquid which is stored in the
Bayanin Amsa
Bile is a greenish alkaline liquid that plays a crucial role in the digestive process, particularly in the digestion and absorption of fats. It is produced in the liver, but it is not stored there. Instead, the bile is transported to a small organ where it is concentrated and stored until the body needs it for digestion. This organ is the gall bladder.
The gall bladder stores the bile and releases it into the small intestine when food, especially fatty food, enters the digestive tract. This helps in breaking down the fats into smaller droplets, making it easier for enzymes to digest them.
To sum up, the gall bladder is the organ responsible for storing bile.
Tambaya 32 Rahoto
Calculate the quantity of heat for copper rod whose thermal capacity is 400Jk−1 for a temperature change of 60ºC to 80ºC
Bayanin Amsa
To calculate the quantity of heat absorbed or released by a substance, we can use the formula:
Q = C × ΔT
where:
Given:
First, calculate the change in temperature:
ΔT = Final temperature - Initial temperature = 80°C - 60°C = 20°C
Now, substitute the values into the formula to find the quantity of heat:
Q = 400 J/°C × 20°C
Calculate the answer:
Q = 8000 J
Since the options provided are in kilojoules (KJ), we need to convert joules (J) to kilojoules (1 KJ = 1000 J):
Q = 8000 J ÷ 1000 = 8 KJ
Therefore, the quantity of heat for the copper rod, given the specified conditions, is 8 KJ.
Tambaya 33 Rahoto
Bayanin Amsa
When you insert a sheet of an insulating material between the plates of an air capacitor, the capacitance will increase.
Here's why:
C = ε₀ * (εr) * (A/d)
Therefore, inserting an insulating material as a dielectric enhances the capacitor's ability to store charge, ultimately resulting in an increase in capacitance.
Tambaya 34 Rahoto
The unit of impedance is
Bayanin Amsa
The unit of impedance is Ohm, which is symbolized by the Greek letter Ω (Omega). In electrical circuits, impedance (Z) is a measure of opposition that a circuit offers to the passage of electric current when a voltage is applied. It is similar to resistance but extends to alternating currents (AC) and contains the effects of resistance as well as reactance (which accounts for capacitors and inductors).
Just like resistance, the unit of impedance is the ohm because they measure similar concepts; however, impedance also accounts for phase shifts between voltage and current, which are not considered in simple resistance. Ohm's Law is used in AC circuits as Z = V/I, where Z is impedance, V is voltage, and I is current. This relationship shows why the unit of impedance is the same as that of resistance.
Tambaya 35 Rahoto
An object is placed 25cm in front of a convex mirror has its image formed 5cm behind the mirror. what is the focal length of the convex mirror
Bayanin Amsa
Object distance (u) = -25 cm (negative because the object is in front of the mirror)
Image distance (v) = +5 cm (positive because the image is behind the convex mirror)
Using 1f = 1u + 1v
1f = 1−25 + 15
f = 254 = 6.250cm.
Tambaya 36 Rahoto
The moon's acceleration due to gravity is 16 of the earth's value. The weight of a bowling ball on the moon would be
Bayanin Amsa
To determine the weight of a bowling ball on the moon, we need to understand the relationship between weight, gravity, and mass.
Weight is the force exerted by gravity on an object. On Earth, this force depends on the object's mass and the acceleration due to gravity, which is approximately 9.8 m/s². Weight can be calculated using the formula:
Weight = Mass x Gravity
On the moon, the acceleration due to gravity is only 1/6 of Earth’s gravity. This means the gravitational pull on the moon is much weaker compared to the Earth. If we take the Earth's gravity to be 9.8 m/s², the moon's gravity would be:
Moon's Gravity = (9.8 m/s²) x (1/6) ≈ 1.63 m/s²
Given that the weight of an object is directly proportional to the gravitational force, the weight of an object on the moon would be substantially less than its weight on Earth. Thus, the weight of the bowling ball on the moon would be:
Weight on Moon = (Mass) x (1.63 m/s²) = 1/6 of its weight on Earth
Therefore, the weight of a bowling ball on the moon is 1/6 of its weight on Earth.
Tambaya 37 Rahoto
Which of the following structures enables the exchange of gases in insects?
Bayanin Amsa
In insects, the structure responsible for the exchange of gases is the tracheae. Insects have a unique respiratory system where air is taken in through tiny openings called spiracles located on the surface of their body.
The air then travels directly into a network of tubes known as the tracheae. The tracheae branch out extensively throughout the insect's body, allowing oxygen to diffuse directly to the insect's tissues and cells. The carbon dioxide produced in the cells travels back through the tracheae and exits the body through the spiracles.
Other structures like the skin, Malpighian tubules, and flame cells have different functions:
Thus, the correct answer is the tracheae as they specifically enable the exchange of gases in insects.
Tambaya 38 Rahoto
The energy stored in the above capacitor is
Bayanin Amsa
The energy stored in the capacitor = 12 q2C
Where C = 2F, q = 3C
= 12 322 = 94 = 2.25J
Tambaya 39 Rahoto
When a cell of e.m.f 3.06V is connected, the balance of a potentiometer is 75cm, Calculate the new balance of a cell of e.m.f 2.295V
Bayanin Amsa
To solve this problem, we first need to understand the principle behind a potentiometer. A potentiometer is a device used to measure the electromotive force (e.m.f) of a cell by comparing it with a known voltage. The balance length on a potentiometer corresponds to a proportional measurement of the e.m.f.
Let's denote:
- \( V_1 \): the e.m.f of the first cell = 3.06V
- \( l_1 \): the balance length for the first cell = 75 cm
- \( V_2 \): the e.m.f of the second cell = 2.295V
- \( l_2 \): the balance length for the second cell (which we need to find)
The basic relationship for a potentiometer is given by:
\( V_1 / V_2 = l_1 / l_2 \)
Substituting the given values:
\( 3.06 / 2.295 = 75 / l_2 \)
We need to solve for \( l_2 \):
\( l_2 = (2.295 \times 75) / 3.06 \)
Now, calculating the above expression:
\( l_2 = 171.975 / 3.06 \approx 56.26 \) cm
Therefore, the new balance length for the cell with an e.m.f of 2.295V is approximately 56.26 cm.
Tambaya 40 Rahoto
Two capacitors of 0.0003μF and 0.0006μF are connected in series, find their combined capacitance.
Bayanin Amsa
When capacitors are connected in series, the formula to find their combined capacitance \(C_{\text{total}}\) is given by:
\[ \frac{1}{C_{\text{total}}} = \frac{1}{C_1} + \frac{1}{C_2} \]
where \(C_1\) and \(C_2\) are the capacitances of the individual capacitors. In this case, \(C_1 = 0.0003 \, \mu\text{F}\) and \(C_2 = 0.0006 \, \mu\text{F}\).
First, calculate the reciprocal of each capacitance:
\[ \frac{1}{C_1} = \frac{1}{0.0003} \]
\[ \frac{1}{C_2} = \frac{1}{0.0006} \]
Calculating each value:
\[ \frac{1}{0.0003} = \frac{10^6}{3} \] and \[ \frac{1}{0.0006} = \frac{10^6}{6} \]
Now, add these values together:
\[ \frac{1}{C_{\text{total}}} = \frac{10^6}{3} + \frac{10^6}{6} = \frac{10^6 \times 2}{6} + \frac{10^6 \times 1}{6} = \frac{10^6 \times 3}{6} = \frac{10^6}{2} \]
Finally, take the reciprocal of the resulting value to find \(C_{\text{total}}\):
\[ C_{\text{total}} = \frac{2}{10^6} = 0.0002 \, \mu\text{F} \]
So, the combined capacitance of the two capacitors in series is 0.0002 μF.
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