(a) calculate the focal length of the mirror formed by the convex side of a shiny spoon that has a 2.41 cm radius of curvature.

The molecules of O2 that are  present in 3.90 L flask at  a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules  of O2

Step  1:  used the ideal gas equation to calculate the moles of O2

that is Pv=n RT  where;

P(pressure)= 1.00 atm

V(volume) =3.90 L

n(number of moles)=?

R(gas constant) = 0.0821 L.atm/mol.K

T(temperature) = 273 k

by  making n the subject of the formula by  dividing  both side by RT

n= Pv/RT

n=[( 1.00 atm x 3.90 L)  /(0.0821 L.atm/mol.k  x273)]=0.174  moles

Step 2: use the Avogadro's  law constant  to calculate  the number of molecules

that  is  according to Avogadro's law

                          1  mole =  6.02 x10^23  molecules

                            0.174 moles=? molecules

by  cross  multiplication

the number of  molecules

= (0.174  moles x  6.02 x10^23  molecules)/ 1 mole  =1.047 x 10^23 molecules of O2

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The motorcycle had covered a distance of 16 meters in half the total time.

a) To calculate the acceleration, we can use the formula:

a = (v - u) / t

where a is the acceleration, v is the final velocity, u is the initial velocity (which is 0 since the motorcycle starts from rest), and t is the time.

Given:

u = 0 m/s (initial velocity)

v = ? (final velocity)

t = 4 s (time)

s = 64 m (distance)

Using the equation of motion:

s = ut + 1/2at^2

We can rearrange the equation to solve for acceleration:

a = 2s / t^2

a = 2(64) / (4)^2

a = 128 / 16

a = 8 m/s^2

Therefore, the acceleration of the motorcycle is 8 m/s^2.

b) To find the final velocity, we can use the formula:

v = u + at

v = 0 + (8)(4)

v = 32 m/s

Therefore, the final velocity of the motorcycle is 32 m/s.

c) To determine the time at which the motorcycle had covered half the total distance, we divide the total distance by 2 and use the formula:

s = ut + 1/2at^2

32 = 0 + 1/2(8)t^2

16 = 4t^2

t^2 = 4

t = 2 s

Therefore, the motorcycle had covered half the total distance at 2 seconds.

d) To calculate the distance covered in half the total time, we use the formula:

s = ut + 1/2at^2

s = 0 + 1/2(8)(2)^2

s = 0 + 1/2(8)(4)

s = 0 + 16

s = 16 m

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\(\\ \sf\longmapsto T=2a\)

\(\\ \sf\longmapsto mg=2a\)

\(\\ \sf\longmapsto (2+m_2)(10)=2(3)=6\)

\(\\ \sf\longmapsto m_2+2=\dfrac{6}{10}=0.6\)

\(\\ \sf\longmapsto m_2=0.6+2\)

\(\\ \sf\longmapsto m_2=2.6kg\)

The mass of 0.88 Kg should be suspended on the other end of the string in order to get an acceleration of 3 m/s².

The tension is produced when the rope on the pulley experiences a pulling force on the ends of the rope.

The two bodies of different masses m₁ and m₂  are connected at the ends of a string and mass m₂ is placed on a horizontal, frictionless plane and frictionless pulley.

m₁a + m₂a = m₂g

(m₁ + m₂) a = m₂g

Given the mass placed at the horizontal plane, m₁ = 2 Kg

The acceleration obtained, a = 3 m/s²

The acceleration due to gravity acting, g = 9.8m/s²

(m₁ + m₂) a = m₂g

(2 + m₂) × 3 = m₂ ×9.8

6 + 3m₂ = 9.8 m₂

9.8 m₂ - 3m₂ = 6

6.8 m₂ = 6

m₂ = 0.882 Kg

Therefore, the mass of 0.882 Kg should be suspended to obtain an acceleration of 3 m/s².

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N₂O: Nitrous oxide

oxygen

\({\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\\)

\(\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}\)

\(\\\)

\({\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\\)

\(\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}\)

\(\\\)

\({\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\\)

\(\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\\)

\(\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ \)

\(\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}\)

\(\\\)

☯ As we know that,

\(\\\)

\(\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}\)

\(\\\)

\(\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }\)

\(\\\)

\(\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }\)

\(\\\)

\(\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }\)

\(\\\)

\(\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ \)

\(\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}\)

\(\\\)

\(\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}\)

\(\\\)

\(\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ s = 0.04 \: m }\)

\(\\\)

☯ For left penetration (s₂)

\(\\\)

\(\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }\)

\(\\\)

\(\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}\)

\(\\\)

\(\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }\)

\(\\\)

\(\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}\)

\(\\\)

\(\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ \)

The difficulties in measuring extragalactic distances as well as phenomenon confused scientists for several decades, causing them to overestimate distances and other factors are discussed below.

By simply looking at stars, we are not able  to know their distance from Earth. It is not possible to determine a star's distance only from its apparent brightness. The issue becomes even more difficult when attempting to determine how far away galaxies are.

Astronomers were known to have created a multitude of techniques for measuring extragalactic distances in order to overcome the challenge. To measure waves in space that existed before any star was born, these techniques include charting the locations of millions of galaxies, searching for the characteristic light from specific types of atoms, and identifying objects or events with known brightness.

Cepheids are helpful for determining distances to relatively nearby galaxies, but they lose accuracy at greater separations due to the fact that it gets harder to distinguish individual stars as we go away from the Milky Way.

Molecules are converted into masers, which are the microwave equivalent of lasers, by certain stars and black holes. Astronomers can use these very bright objects, which produce light in a predictable manner, to determine the separation between adjacent galaxies for star masers.

Therefore, determining the extent to which the universe's expansion has redshifted the light spectra of galaxies in order to determine their distance. Due to the acceleration of the universe's expansion, this measurement is said to be useful only for galaxies that are close to the Milky Way, but it loses accuracy for galaxies that are farther away.

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Explanation:

time = Charge / Current.

time = 80/5

time = 16 seconds.

Answer:

Examples of electromagnetic waves include radio waves, microwaves, infrared, visible light, ultraviolet, x-rays, and gamma rays. ... Microwaves are used to cook your food. Infrared waves are used in remote controls and are emitted from all warm objects, allowing them to be used to create heat-sensitive cameras.

Explanation:

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The potential energy of the roller coaster is 176,400 J (joules).

The potential energy of an object is given by the formula PE = mgh, where PE is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height or vertical position of the object.

In this case, the roller coaster is at a peak of 20m and has a mass of 900kg. The acceleration due to gravity, g, is approximately 9.8 \(m/s^2\).

Using the formula, we can calculate the potential energy:

PE = mgh

= (900 kg)(9.8 \(m/s^2\))(20 m)

= 176,400 J

Therefore, the potential energy of the roller coaster is 176,400 J (joules).

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Answer:

a. 20 seconds

b. 160 seconds

Explanation:

a. Frequency is given by \(f=\frac{1}{T}\), where \(T\) is the period in seconds. Rearranging, we get \(T=\frac{1}{f}\). Given that frequency is 0.05 Hz, we have:

\(T=\frac{1}{0.05}=\boxed{20\text{ seconds}}\)

b. Since we just found that it takes 20 seconds to complete one rotation (period), it will take \(20\cdot 8=\boxed{160\text{ seconds}}\) to complete 8 rotations.

Answer:

Ex; If the robot traveled 20 cm around the octagon then the circle circumference = 20 cm

Use the equation circumference = 2 π radius to calculate radius(C = 2πr)

          If circumference = 20cm;  radius = circumference / ( 2 π) = 20/6.28= 3.18 cm. Use your scale to express your answer in meters, km, etc

now set the protractor distance measure as 3.18 cm, place the protractor's metal spike at the octagon center and draw the circle.

Explanation:

The statement best describes the primary difference between scientific investigation and technological design is Scientific investigation involves experiments, and technological design involves products. Option D

It can be said that both scientific investigation and technolgical design require problem-solving, creativity, and rigorous methods.

However, both of them have different objectives. While Scientific investigation seeks to expand knowledge and understanding of the natural world. Research involvs hypothesis testing and different experiments;

on the other hand, Technological design, wants to create or improve products, processes, or systems to meet specific needs or solve particular problems.

It can then be defined as the application of scientific knowledge for practical purposes.

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Answer:

The speed after 2 seconds is 14.4 m/s

Explanation:

Vertical Launch Upwards

In a vertical launch upwards, an object is launched vertically up from a height H without taking into consideration any kind of friction with the air.

The speed changes from an initial value vo to a final value vf, given by:

\(v_f=v_o-g.t\)

Where g is the gravitational acceleration (\(9.8\ m/s^2\))

The tennis ball is shot up with an initial speed of vo=34 m/s. We are required to find the final speed at t=2 s:

\(v_f=34-9.8*2\)

\(v_f=14.4\ m/s\)

The speed after 2 seconds is 14.4 m/s

the velocity after 2 seconds of launch is 14.4 m/s

the initial velocity(a) is 34m/s

the time(t) is 2 seconds

acceleration(a) due to gravity is 9.8g

the formula is

v= u - at

v= 34 -(9.8)(2)

= 34 - 19.6

= 14.4 m/s

hence the final velocity is 14.4 m/s

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Answer:

200J

Explanation:

0.5 *4 * 10 * 10

=200J

Answer:

Explanation:

s = ½at²

a = 2s/t²

a = 2(11.6) / 9.8²

a = 0.2415660...

a = 0.24 m/s²

Acceleration of the wagon is 0.241 m/s²

Force exerted on the wagon, F = 81.17 N

Displacement of the wagon, S = 11.6 m

Time taken by the wagon to travel this distance, t = 9.8 s

Using the second equation of motion,

S = ut + 1/2 at²

where u is the initial velocity and a is the acceleration of the wagon.

Since the wagon was initially at rest, u = 0.

So,

S = 1/2 at²

Therefore acceleration of the wagon,

a = 2S/t²

a = 2 x 11.6/(9.8)²

Acceleration, a = 0.241 m/s²

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The remaining battery capacity would be 76%, which is still above the 25% limit on discharge depth.

If the electric bicycle can go 100 miles on a full charge, and the owner wants to limit the discharge depth to 25%, then they can only use up to 25 miles of the battery's capacity before needing to recharge.

Since the owner's commute is 12 miles round trip, they can make one round trip without depleting the battery more than 25% of its capacity.

Therefore, the owner can make 2 round trips (2 x 12 = 24 miles) before needing to recharge, but they will still have some battery capacity left.

The remaining battery capacity would be (100 - 24)/100 x 100% = 76%, which is still above the 25% limit on discharge depth. So the owner could make a few more round trips before needing to fully recharge the battery.

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Answer:

if an astronaut weighs 981 N on Earth and only 160 N on the Moon, then what is his mass on the Moon?

ANSWER 256n

The mass of the astronaut on the moon as compared to the earth will be  \(M_m=97.85\ lg\)

The mass of any substance or body is defined as how much quantity of matter is present.

Now it is given in the question:

Weight of the astronaut on earth  \(W_E=981\ N\)

Weight of the astronaut on earth  \(W_M=160\ N\)

The mass of the astronaut on the moon will be calculated as:

Weight on the moon will be  given as:

\(W_M=M_M\times g_m\)

Weight on the earth will be given as:

\(W_E=M_E\times g_e\)

The ratio of the gravity of the earth to the moon is given as

\(\dfrac{g_e}{g_m} =\dfrac{9.81}{1.62} =6\)

The mass of the earth will be calculated as

\(W_E=M_E\times ge\)

\(M_E=\dfrac{981}{9.81} =100\ kg\)

Now taking the ratio of the weight of the earth to the moon :

\(\dfrac{W_E}{W_M} =\dfrac{M_E\times g_e}{M_M\times g_m}\)

\(M_M= \dfrac{W_M\times M_E}{W_E} \times \dfrac{g_e}{g_m}\)

Now by putting the value in the formula we get;

\(M_M=\dfrac{160\times 100}{981} \times 6=97.85\ kg\)

Thus the mass of the astronaut on the moon as compared to the earth will be  \(M_m=97.85\ lg\)

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The magnitude of vector B is determined as 8.05 units.

The magnitude of a vector is the scalar value of a vector that can  be used to describe the unit value of the vector.

The magnitude of a vector can also be described as the resultant of the vector because it is usually the single value of the vector that can be used to represent the vector.

The magnitude of vector B is calculated as follows;

|B| = √(x² + y² + z²)

where;

|B| = √(4.3² + 4.5² + 5.1²)

|B| = 8.05 units

Thus, we can conclude that vector B with x, y, and z components of 4.3, 4.5, and 5.1 units, respectively, has a magnitude of 8.05 units.

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The efficiency of motor is 90.58%.To determine the efficiency of the motor, we need to calculate the input power and the output power, and then divide the output power by the input power

The input power can be calculated using the formula:

Input Power = Voltage × Current

Given that the voltage is 230 V and the current is 24 A, we have:

Input Power = 230 V × 24 A

Input Power = 5520 W (or 5.52 kW)

The output power of the motor is given as 5 kW (since it is a 5 kW motor).

Now, we can calculate the efficiency:

Efficiency = (Output Power / Input Power) × 100%

Efficiency = (5 kW / 5.52 kW) × 100%

Efficiency ≈ 90.58%

Therefore, the efficiency of this motor is approximately 90.58%.

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Answer: A solar eclipse results when the moon passes in between the earth and the sun hiding the sun fully or partly for some time. A lunar eclipse occurs when the earth passes in between the moon and the sun casting its shadow on the moon and thus hiding it fully or partly for some time.

while writing a scientific notation

one digit should be left of the decimal

Answer:

If the particles of the medium vibrate in a direction parallel to the direction of propagation of the wave, it is called a longitudinal wave. In longitudinal waves, the particle movement is parallel to the direction of wave propagation.

Explanation:

Answer

the medium does not change the speed of the on they should change.

Explanation:

The speed of the waves is constant for a given medium, depending on the physical properties of the medium,

When a wave is strapped on a wall of a medium it does not change the properties of the medium, the wave changes direction, but since the medium does not change the speed of the on they should change.

The speed of the roller coaster at the bottom of the hill if the track was frictionless is 34.04 m/s.

Given that the weight of the roller coaster is 2000 kg and the initial vertical drop of the ride is 59.3 m. We are to find the speed of the roller coaster at the bottom of the hill if the track was frictionless.We know that the roller coaster will lose potential energy due to the vertical drop. Assuming there is no friction, the potential energy will be converted into kinetic energy at the bottom of the hill.Considering the conservation of energy between the potential and kinetic energy, we can set the initial potential energy equal to the final kinetic energy. We can use the formula to calculate potential energy, which is PE = mgh where m = 2000 kg, g = 9.8 m/s², and h = 59.3 m. Therefore,PE = 2000 kg × 9.8 m/s² × 59.3 m = 1,157,924 JWe can use the formula to calculate kinetic energy, which is KE = 1/2mv² where m = 2000 kg and v is the final velocity. Therefore,KE = 1/2 × 2000 kg × v².The total energy remains constant as we know there is no friction. Therefore the final kinetic energy will be equal to the initial potential energy,1,157,924 J = 1/2 × 2000 kg × v²v² = (2 × 1,157,924 J) / 2000 kgv² = 1157.924v = √1157.924v = 34.04 m/s.

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