Examine the symbols below:

Four schematics are shown. Symbol A shows two dots and a line draw from one not connected to the other. Symbol B shows two dots and a line draw from one connected to the other. Symbol C shows vertical lines in the pattern long, short, long, and short with a plus and minus symbol on it. Symbol D shows a dot in the center of two brackets with short horizontal dashes connected to them.

Which image represents an open switch in a circuit?

Symbol B
Symbol A
Symbol C
Symbol D

Answer:

a

Explanation:

Answer:

A

Explanation:

The final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:

Explanation:

So in this problem, the initial speed of the train is at 25.8 m/s before it comes to incline with constant slope. So the acceleration or the rate of change in velocity while moving on the incline is given as 1.66 m/s². So the final velocity need to be found after a time period of 8.3 s. According to the first equation of motion, v = u +at.

So we know the values for parameters u,a and t. Since, the train slows down on the slope, so the acceleration value will have negative sign with the magnitude of acceleration. Then

v = 25.8 + (-1.66×8.3)

v =12.022 m/s.

So the final velocity of the train after 8.3 s on the incline will be 12.022 m/s.

Answer:k = 10.83 N/m²

Explanation: The angular frequency (ω), spring constant (k) and mass is related by the formulae below

ω = √k/m

But ω = 2πf, where f = frequency.

f = number of oscillations /time taken

Number of oscillations = 14, time taken = 11s

f = 14/11 = 1.27Hz.

ω = 2×22/7×1.27

ω = 7.98 rad/s.

By substituting this parameters into ω = √k/m

Where ω = 7.98rad/s, m = 170g = 170/1000 = 0.17kg.

7.98 = √k/0.17

By squaring both sides

(7.98)² = k/ 0.17

k = (7.98)² × 0.17

k = 10.83 N/m²

Final answer:

To find the spring constant (k) of the spring system, we use the period of oscillation formula. After calculating the period of one oscillation (T = 0.7857 s), we solve for k = m/(T/2π)^2 and find that k is approximately 44.1 N/m.

Explanation:

The student asked how to find the spring constant of a spring if a 170 g air-track glider is attached to it, pushed in 11.2 cm, released, and makes 14 oscillations in 11.0 seconds. The spring constant (k) can be found using the formula for the period of a mass-spring system (T = 2π√(m/k)) where m is mass and k is the spring constant.

First, calculate the period of one oscillation by dividing the total time by the number of oscillations: T = 11.0 s / 14 = 0.7857 s.  Then rearrange the period formula to solve for k: k = m/(T/2π)2. Convert the mass to kilograms (m = 0.170 kg) and substitute the values to calculate k.

Doing the calculation: k = 0.170 kg / (0.7857 s / 2π)2 gives us the spring constant k. Upon solving, we find that the spring constant is approximately 44.1 N/m.

Answer:

Explanation:

Given

magnitude of centripetal acceleration is twice the magnitude of tangential acceleration

Suppose [tex]\theta [/tex] is theta angle rotated by electric drill

it is given that it starts from rest i.e. [tex]\omega _0=0[/tex]

suppose [tex]\omega [/tex] and [tex]\alpha [/tex] is the final angular velocity and angular acceleration

using rotational motion equation

[tex]\omega ^2-\omega _0^2=2\times \alpha \times \theta [/tex]

where [tex]\theta [/tex]=angle turned by drill

[tex]\omega _0[/tex]=initial angular velocity

[tex]\omega [/tex]=final angular velocity

[tex]\alpha [/tex]=angular acceleration

[tex]\omega ^2-0=2\times \alpha \times \theta [/tex]

[tex]\omega ^2=2\alpha \theta ---1[/tex]

It is also given that centripetal acceleration is twice the magnitude of tangential i.e.

[tex]\omega ^2r=\alpha \times r[/tex]

where r=radial distance of any point from axis of drill

i.e. [tex]\omega ^2=\alpha [/tex]

substitute this value to equation 1

we get

[tex]\theta =\frac{\omega ^2}{2\alpha }[/tex]

[tex]\theta =1\ rad[/tex]

Answer:

Four charges of equal magnitude sitting at the vertices of a square

Explanation:

We can arrive at such a situation by thinking of a simple example first, a configuration of two charges. The force acting on the middle point of a straight line joining the two points(charges) will be zero. That is, the net Electric field will be zero as they cancel out being equal in magnitude and opposite in direction.

Now, we can extend this idea to a square having charge q at each vertex. If we put 'p' at the geometric center, we can see that the Electric fields along the diagonals cancel out due to the charges at the diagonally opposite vertices(refer to the figure attached). Actually, the only requirement is that the diagonally opposite charges are equal.

We can further take this to 3 dimensions. Consider a cube having charges of equal magnitude at each vertex. In this case, the point 'p' will yet again be the geometric center as the Electric field due to the diagonally opposite charges will cancel out.

Answer: pebble has the greatest value of acceleration due to its low mass.

Explanation:

According to newton's second law of motion,

Force = mass × acceleration

For boulder,

Acceleration = Force/mass

Force acting on boulder is 200N

Its mass is 100kg

Acceleration = 200/100

Acceleration = 2m/s²

Similarly for pebble,

Force on pebble = 200N (the same as boulder)

Its mass is 130g = 0.13kg(has to be converted to the standard unit which is kg)

Its acceleration = 200/0.13

Acceleration of pebble = 1538.5m/s²

Since the question doesn't specify what to compare, we will compare their accelerations.

Therefore, pebble has the greatest value of acceleration due to its low mass.

Answer:

Explanation:

compressibility = 1 / bulk modulus of elasticity ( B )

B = 1 / 4.8 x 10⁻⁵ = Δp / Δv /v ( Δp is change in pressure , Δv is change in volume )

1 / 4.8 x 10⁻⁵ = 429  / Δv /v

Δv /v = 429 x 4.8 x 10⁻⁵

= 2059.2 x 10⁻⁵

= .021

v = m / d ( d is density and m is mass of the water taken )

taking log and then differentiating

Δv /v  = - Δd / d

- .021 = - Δd / d

Δd =.021  x d

= .021 x 998

= 20.9

new density

= 998 + 20.9

1018.9 kg/m3

The increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally is 20.6 kg/m³.

In this problem, we are asked to find the increase in the density of water when it is compressed from 1 atm to 430 atm pressure isothermally. The isothermal compressibility of water is given as 4.80 × 10−5 atm−1. The density of water at 1 atm pressure and 20°C is given as rho1 = 998 kg/m^3.

The formula that connects these quantities is Δρ = - ρ * β * ΔP, where Δρ is the change in density, β is the isothermal compressibility (in atm^−1), ΔP is the change in pressure (in atm), and ρ is the initial density (in kg/m^3).

Substituting the given values into the formula we get Δρ = - (998 kg/m³) * (4.80 × 10−5 atm−1) * (430 atm - 1 atm) = - ((998 kg/m³) * (4.80 × 10−5 atm−1) * (429 atm)) = -20.6 kg/m^3. The negative sign indicates an increase in density due to compression. So the increase in density of the water is 20.6 kg/m³.

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Answer:

a) 24.0 N.m b) 3.6*10⁻² rad/s² c) 1.07 m/s²

Explanation:

a) If the force that produces the torque is perpendicular to the tethering wire, we can determine its magnitude just as follows:

τ = F*r = 0.800 N * 30.0 m = 24.0 N*m (1)

b)  We can express the torque we found above, using the rotational form of Newton´s 2nd Law, as follows:

τ = I* α (2)

where I is the rotational inertia regarding an axis passing through the center of the circle and α is the angular acceleration of the airplane.

If we consider the airplane as a point mass, the rotational inertia I can be calculated as follows:

I = m*r² = 0.750 Kg * (30.0)² m² = 675 Kg*m²

From (1) and (2), we can solve for α, as follows:

[tex]\alpha = \frac{T}{I} = \frac{24.0 N*m}{675.0 kg*m2} = (3.6e-2) rad/s2[/tex]

⇒α = 3.6*10⁻² rad/s²

c) Applying the definition of the angular velocity, and the definition of an angle, we can find the following realtionship between the linear and angular velocity:

v = ω*r

Dividing both sides by Δt, we can extend this relationship to the linear and angular acceleration, as follows:

a = α*r  

a = 3.6*10⁻² rad/s²* 30.0 m = 1.07 m/s²

a. The torque that the net thrust produces about the center of the circle is 24 Newton.

b. The angular acceleration of the airplane is equal to [tex]0.036 \;rad/s^2[/tex]

c. The linear acceleration of the airplane tangent to its flight path is[tex]1.08 \;m/s^2[/tex]

Given the following data:

a. To find the torque that the net thrust produces about the center of the circle:

Mathematically, the torque produced by a perpendicular force is given by:

[tex]T = Fr[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]T = 0.8 \times 30[/tex]

Torque, T = 24 Newton

b. To find the angular acceleration of the airplane, we would use Newton's Second Law of rotational motion:

[tex]T = I\alpha[/tex]

But, [tex]I = mr^2[/tex]

[tex]I = 0.750 \times 30^2\\\\I = 0.750 \times 900[/tex]

Moment of inertia, I =  [tex]675\;Kgm^2[/tex]

[tex]\alpha = \frac{T}{I} \\\\\alpha = \frac{24}{675}\\\\\alpha = 0.036 \;rad/s^2[/tex]

c. To find the linear acceleration of the airplane tangent to its flight path:

[tex]a = r\alpha \\\\a = 30 \times 0.036[/tex]

Linear acceleration, a = [tex]1.08 \;m/s^2[/tex]

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To find the magnitude and direction of the resultant vector R, we can use graphical methods. First, find the components of vectors A and B along the x and y axes. Then, use the Pythagorean theorem to find the magnitude of R and the inverse tangent function to find the direction of R.

To find the magnitude and direction of the resultant Vector B, with a magnitude of 4 units, points along the negative y-axis, so its x-component is 0 and its y-component is -4.

To find the components of R, we can simply add the corresponding components of A and B: Rx = Ax + Bx

= 3 + 0 = 3, Ry = Ay + By = 0 + (-4) = -4.

Using the Pythagorean theorem, we can find the magnitude of R: R = sqrt(Rx^2 + Ry^2)

= sqrt(3^2 + (-4)^2) = sqrt(9 + 16) = sqrt(25) = 5 units.

To find the direction of R, we can use the inverse tangent function: Rtheta = atan(Ry/Rx)

= atan((-4)/3)

= atan(-4/3) = -53.13 degrees.

However, since vector B points along the negative y-axis, the direction of R is 90 degrees minus the calculated angle: Rtheta = 90 - 53.13 = 36.87 degrees.

Therefore, the magnitude of R is 5 units and it points at an angle of 36.87 degrees north of the x-axis.

Answer:

Radio Waves:

Radio waves being the lowest-energy form of light and are known to be produced by electrons spiraling around magnetic fields. These Magnetic fields are generated by stars, including our sun, and many weird celestial objects like black holes and neutron stars.

Explanation:

All electromagnetic radiation is light, but we can only see a small portion of this radiation that is the portion we call visible light. Cone-shaped cells in our eyes act as receivers tuned to the wavelengths in this narrow band of the spectrum

Visible light and radio waves are both examples of electromagnetic waves, but visible light has shorter wavelengths that our eyes can detect, while radio waves have longer wavelengths that are outside the range of what our eyes can see.

Visible light and radio waves are both examples of electromagnetic waves, but they have different wavelengths. Visible light has a shorter wavelength than radio waves, which is why we can see it. Our eyes are sensitive to the wavelengths of visible light, but not to the longer wavelengths of radio waves.

The human eye can detect wavelengths within a certain range, known as the visible spectrum, which includes the colors of the rainbow. Radio waves have much longer wavelengths, ranging from meters to kilometers. These longer wavelengths are outside the range of what our eyes can detect, so we cannot see radio waves.

However, even though we can't see radio waves, we can still use them for various purposes, such as wireless communication, radar, and satellite transmission.

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Explanation:

We have equation of motion v = u + at

     Initial velocity, u = 33 m/s

     Final velocity, v = 29 m/s    

     Time, t = 2.50 s

     Substituting

                      v = u + at  

                      29 = 33 + a x 2.50

                      a = -1.6 m/s²

We have equation of motion v² = u² + 2as

Initial velocity, u = 33 m/s  

Acceleration, a = -1.6 m/s²  

Final velocity, v = 29 m/s  

Substituting  

v² = u² + 2as

29² = 33² + 2 x -1.6 x s

s = 77.5 m  

Distance traveled during that time is 77.5 m  

Final answer:

The car will travel 77.5 meters while decelerating from 33.0 m/s to 29.0 m/s over a period of 2.50 seconds by using the average speed during the deceleration period.

Explanation:

The question deals with calculating the distance covered by a car while decelerating from 33.0 m/s to 29.0 m/s over 2.50 seconds. We can solve this by finding the car's average speed during this period and multiplying it by the time duration. Here's how we can solve it:

Step-by-Step Calculation

Find the average speed: (Initial speed + Final speed) / 2 = (33.0 m/s + 29.0 m/s) / 2 = 31.0 m/s.

Multiply the average speed by the time taken to decelerate to find the distance covered: Distance = Average speed * Time = 31.0 m/s * 2.50 s = 77.5 meters.

Therefore, the car will travel 77.5 meters during that time.

Answer:

Explanation:

consider the principle of moment

when a system is in equilibrium, the clockwise moment (torque) about the pivot is equal to the counterclockwise moment ( torque). Since the plank is uniform the weight of the plank act at the middle which = 6.1 m / 2 = 3.05 m

the distance that can support the weight of the man = d

mass of the man = 70

70 × d = 33 × ( 3.05 - 1.6)

d = 47.85 / 60 = 0.798 m, if the man work beyond this point he will fall.

Answer:

a) Emission

b) Reflection

c) Absorption

d) Transmission

Explanation:

a) The screen of computers (most of them LED or LCD) are a bunch of tiny lights that take power of our pc to produce light so they EMMIT light.

b) Sun's light contains all the visible colors of electromagnetic spectrum, when sun light hits an object, due their chemical characteristics the object absorbs a bunch of wavelengths and reflects others, the reflected light is the color we perceive with our eyes. So, for a banana, it absorbs almost all the other colors and reflects only yellow light.

c) As explained on b) a banana absorbs all the other colors except yellow so blue is absorbed on it.

d) When a light hits transparent medium, some rays are reflected and other passes through the medium so that rays are called transmitted and  that is the case of Sunlight passing through a glass door.

Final answer:

Correct words are emission for light from a computer screen, reflection for yellow light on a banana, absorption for blue light on a banana, and transmission for sunlight through a glass door. These terms describe how light interacts with different materials through emission, reflection, absorption, and transmission.

Explanation:

From the list of terms provided, the correct words to complete the statements about the behavior of light are:

a. Light coming from your computer screen is an example of emission.

b. Yellow light hitting a yellow banana is an example of reflection.

c. Blue light hitting a yellow banana is an example of absorption.

d. Sunlight passing through a glass door is an example of transmission.

When we talk about light waves interacting with materials, we discuss concepts such as reflection, absorption, and transmission. Reflection occurs when light bounces off an object, such as a yellow light reflecting off a yellow banana. Absorption occurs when an object takes in the light energy, which might later be transformed into other forms of energy like heat, as is the case when a yellow banana absorbs blue light. Lastly, transmission refers to the passage of light through a transparent or semi-transparent material, such as sunlight passing through a glass door.

Answer:c

Explanation:

Although the Carnot engine is the most efficient it is not feasible in real life. Carnot engine contains processes that are reversible which makes it difficult to build in real life.

For a system to be in equilibrium a system must be in equilibrium with its surroundings in each step and every step which allows it to be infinitely slow.

Due to this slow rate Power of the engine is zero as the work is being done at an infinitely slow rate.

Answer:

good

Explanation:

Answer:

In The Funeral of St. Bonaventure, Francisco de Zurbarán used the principle of _contrast_ to create emphasis and focal point

Explanation:

Contrast relates to the combination of opposing components and effects as an art theory. Lighter and darker colours, polished and rough materials, large and small shapes. Contrast can be used for the production of diversity visual interest and excitement.

A) Angular speed: 0.42 rev/s

B) Frequency: 0.42 Hz

C) Tangential speed: 26.4 cm/s

D) Distance travelled: 528 cm

Explanation:

A)

In this problem, the ladybug is rotating together with the record.

The angular velocity of the ladybug, which is defined as the rate of change of the angular position of the ladybug, in this problem is

[tex]\omega = 25 rpm[/tex]

where here it is measured in revolutions per minute.

Keeping in mind that

1 minute = 60 seconds

We can rewrite the angular speed in revolutions per second:

[tex]\omega = 25 \frac{rev}{min} \cdot \frac{1}{60 s/min}=0.42 rev/s[/tex]

B)

The relationship between angular speed and frequency of revolution for a rotational motion is given by the equation

[tex]\omega = 2 \pi f[/tex] (1)

where

[tex]\omega[/tex] is the angular speed

f is the frequency of revolution

For the ladybug in this problem,

[tex]\omega=0.42 rev/s[/tex]

Keeping in mind that [tex]1 rev = 2\pi rad[/tex], the angular speed can be rewritten as

[tex]\omega = 0.42 \frac{rev}{s} \cdot 2\pi = 2\pi \cdot 0.42[/tex]

And re-arranginf eq.(1), we can find the frequency:

[tex]f=\frac{\omega}{2\pi}=\frac{(2\pi)0.42}{2\pi}=0.42 Hz[/tex]

And the frequency is the number of complete revolutions made per second.

C)

For an object in circular motion, the tangential speed is related to the angular speed by the equation

[tex]v=\omega r[/tex]

where

[tex]\omega[/tex] is the angular speed

v is the tangential speed

r is the distance of the object from the axis of rotation

For the ladybug here,

[tex]\omega = 2\pi \cdot 0.42 rad/s[/tex] is the angular speed

r = 10 cm = 0.10 m is the distance from the center of the record

So, its tangential speed is

[tex]v=(2\pi \cdot 0.42)(0.10)=0.264 m/s = 26.4 cm/s[/tex]

D)

The tangential speed of the ladybug in this motion is constant (because the angular speed is also constant), so we can find the distance travelled using the equation for uniform motion:

[tex]d=vt[/tex]

where

v is the tangential speed

t is the time elapsed

Here we have:

v = 26.4 cm/s (tangential speed)

t = 20 s

Therefoe, the distance covered by the ladybug is

[tex]d=(26.4)(20)=528 cm[/tex]

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Answer:

Speed s= 19

Explanation:

Take note of the following parameters:

Speed= s,

Number of cars= N,

Number N of cars that can pass a given spot per minute=

N(s)=88s/16+16(s19)2

The principle of differentiation is  here:

We let N(s) = N

N = 86s / (17 + 17((s/19)^2))

17N = 86s /(1 + s²/19²)

17N = 361* 86s /(361 + s²)

17N = 31046s /(361 + s²)

Next step;

Differentiate with respect to s

17N = 31046s /(361 + s²)

Remember the quotient rule [u/v]’ = (vu’ - uv’) / v²

Therefore,

u = 31046s =====> du/ds = u’ = 31046

v = (361 + s²) ====>dv/ds = v’ = 2s

17N = 31046s /(361 + s²)

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

The maximum when dN/ds = 0

17 dN/ds = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

17 * 0 = ( 31046(361 + s²) - 31046s(2s) ) / (361 + s²)²

( 31046(361 + s²) - 31046s(2s) ) = 0

31046 [ (361 + s²) - 2s² ] = 0

31046 (361 - s²) = 0

(361 - s²) = 0

(19 - s)(19 + s) = 0

either s = -19 or s = 19, but s > 0

s = 19

The speed at which the greatest number of cars travel safely on that road is; s = 19

We are given the function to represent Number N of cars that can pass a given spot per minute as;

N(s) = 88s/(16 + 16(s/19)²)

where;

s is Speed

N is number of cars

Differentiating the function gives;

N' = -3971(s² - 361)/(2(s² + 361)²

Now, the speed at the greatest number of cars would be gotten when N' = 0. Thus;

-3971(s² - 361)/(2(s² + 361)² = 0

Cross multiply to get;

-3971(s² - 361) = 0

divide both sides by -3971 to get;

s² - 361 = 0

s = √361

s = 19

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Answer:

b c and e

Explanation:

Answer: C, D, and E are all molecules!

Answer:

specific gravity = 0.8

specific gravity of  solution  = 2

Explanation:

given data

rectangular block above water  = 0.400 in

rectangular block below water = 1.60 in

material floats below water = 0.800 in

solution

first we get here specific gravity of block  that is

specific gravity = block vol below ÷ total block vol × specific gravity  water   ..............1

put here value we get

specific gravity =  [tex]\frac{1.60}{1.60+0.400}[/tex]  × 1

specific gravity = 0.8

and now we get here specific gravity of  solution  that is express as

specific gravity of  solution  = total block vol ÷ block vol below × specific gravity  block   ........................2

put here value we get

specific gravity of  solution  = [tex]\frac{1.60+0.400}{0.800}[/tex] × 0.8

specific gravity of  solution  = 2

C. added an electron to the outer electron shell.

Explanation:

Atoms consist of three particles:

- Protons: they are located in the nucleus, they have positive charge of [tex]+e[/tex], and mass of [tex]1.67\cdot 10^{-27}kg[/tex]

- Neutrons: they are also located in the nucleus, they have no electric charge, and mass similar to that of the proton

- Electrons: they orbit around the nucleus, they have negative charge of [tex]-e[/tex], and mass around 1800 smaller than the proton

Normally, atoms are neutral (no electric charge), because they have an equal number of protons and electrons.

However, sometimes atoms can give off or take electrons from other atoms. We have two cases:

Therefore, in order to form a negative ion, the atom must have

C. added an electron to the outer electron shell

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Answer:

The time taken for a signal to reach the Earth after it is transmitted from the spacecraft is (500x) seconds

Explanation:

Distance of spacecraft from Earth = x AU = x × 149.9×10^6 Km = (149.9×10^6x) Km

Speed of light = 299,800Km/s

Time taken for a signal to reach the Earth after it is transmitted from the spacecraft = distance of spacecraft from Earth ÷ speed of light = (149.9×10^6x)Km ÷ 299,800Km/s = (500x) seconds

Answer:

Explanation:

Given

Train travels towards south with a velocity if [tex]v_t=25\ m/s[/tex]

Rain makes an angle of [tex]\theta =66^{o}[/tex]  with vertical

If an observer sees the drop fall perfectly vertical i.e. horizontal component of rain velocity is equal to train velocity

suppose [tex]v_r[/tex] is the velocity of rain with respect to ground then

[tex]v_r\sin\theta =v_t[/tex]

[tex]v_r\times \sin (66)=25[/tex]

[tex]v_r=27.36\ m/s[/tex]

Therefore velocity of rain drops is 27.36 m/s              

Answer:

Option (B)

Explanation:

A heterogeneous mixture is usually defined as a combination of two or more chemical substances. It can be also elements as well as compounds. These contrasting components can be easily separated from one another by means of physical process. These are comprised of substances that are not even everywhere. There is variation in it.

Thus, the correct answer is option (B).

Answer:

B: a heterogeneous mixture

Explanation:

If the components of a substance can be separated by the physical means then the substance is a mixture and its component have not undergone any kind of chemical change with their original molecular structure.

The mixtures are of two types homogeneous and heterogeneous.

Answer:

Maximum velocity of rod will be equal to 26.315 m/sec

Explanation:

We have given voltage of the battery in the circuit V = 3.3 volt

Magnetic field B = 0.66 Tesla

Length of the rod l = 0.19 m

We know that emf is given by e = BVl

We have to find the maximum velocity of the rod

Here velocity is maximum

So [tex]e=v_{max}Bl[/tex]

So [tex]3.3=v_{max}\times 0.66\times 0.19[/tex]

[tex]v_{max}=26.315m/sec[/tex]

So maximum velocity of rod will be equal to 26.315 m/sec

Answer:

C. R1 only

Explanation:

As the wire is cut at x, there will be no current through the resistors R2 and R3. Then the current will only go from a to b through the R1 resistor.

Another way to think about this is that once the wire is cut at x, there is now infinite resistance at the point of cutting; therefore, the current can no longer flow through R2 and R3 resistors, but now it only flows through the R1 resistor.

Therefore, only choice C is correct.

Answer:

-5.9 rad/s^{2}

Explanation:

radius (r) = 30 cm = 0.3 m

mass (m) = 1.05 kg

initial speed (u) = 77 rpm

final speed (v) = 0 rpm

time (t) = 1.37 s

angular acceleration =[tex]\frac{(final speed-initial speed)rad/s}{time}[/tex]

therefore

initial speed (U) = 77 rpm = 77 x (2π/60) = 8.06 rad/s

final speed (v) = 0 rpm = 0 rad/s

angular acceleration = [tex]\frac{0-8.06}{1.37}[/tex] = -5.9 rad/s^{2}

Answer:

-5.886 rad/s^2.

Explanation:

radius, r = 30 cm

= 0.3 m

mass, m

= 1.05 kg

initial speed, wo = 77 rpm

Converting from rpm to rad/s,

= 77 rpm * 2pi rad * 1 min/60 s

= 8.063 rad/s

final speed, wi = 0 rad/s

time, t = 1.37 s

angular acceleration = Δw/Δt

= (wi - wo)/t

= 8.063/1.37

= -5.886 rad/s^2.

The angles to the incident direction are the boats inside the harbor most protected against wave action will be  23.57 °

Diffraction is the phenomenon that occur when a wave of light encounter an obstacle or a slit generally.

considering wide opening of harbor as thickness d

the ocean wave as light source (coherent )

boats inside the harbor as screen where diffraction pattern is going to happen

so , destructive interference  should happen (to minimize the amplitude of wave ) in order to save the boats from its effect

n * lambda = d sin (theta )

n=1  ( first  order minima )

sin(theta ) = lambda / d

sin( theta ) = 20 /50

sin(theta) = 2/5

theta = sin inverse (2/5)

theta = 23.57 °

The angles to the incident direction are the boats inside the harbor most protected against wave action will be  23.57 °

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Answer:

range becomes 4 times

Explanation:

We know that the range of a projectile is given as:

[tex]R=\frac{u^2.\sin(2\theta)}{g}[/tex]

where:

[tex]R=[/tex] range of the projectile

[tex]u=[/tex] initial velocity of projectile

[tex]\theta=[/tex] initial angle of projection form the horizontal

g = acceleration due to gravity

When the initial velocity of launch is doubled:

[tex]R'=\frac{(2u)^2.\sin(2\theta)}{g}[/tex]

[tex]R'=\frac{4u^2.\sin(2\theta)}{g}[/tex]

[tex]R'=4R[/tex]

range becomes 4 times

Doubling the initial launch speed of a projectile, while keeping the angle of launch fixed, results in the range being quadrupled.

The range R of a projectile motion is given by the formula R = ((v^2)*sin(2*theta))/g, where v is the initial launch speed, theta is the launch angle, and g is the acceleration due to gravity. If the initial launch speed v is doubled, the new range R' would be R' = ((2v)^2)*sin(2*theta))/g, which simplifies to R' = 4R.

Thus, if you double the initial launch speed, the range of the projectile is quadrupled, assuming the launch angle is fixed.

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Answer:

0.62 s

Explanation:

given,

maximum height of the flea = 0.49 m

velocity at maximum height = 0 m/s

now, calculating initial velocity

using equation of motion

[tex]v^2 = u^2 + 2 g h[/tex]

[tex]0^2 = u^2 - 2\times 9.8 \times 0.49[/tex]

[tex]u^2 = 9.604[/tex]

[tex]u = 3\ m/s[/tex]

now, calculating time he take to reach at the highest point

v = u + g t

0 = 3 - 9.8 x t

9.8 t = 3

t = 0.31 s

Time it will be in air will be twice the time it took to reach to the maximum height.

Time for which it was in air = 2 x 0.31 s = 0.62 s

Answer: [tex]f_{r}[/tex] = 16.49N

Explanation: The object is placed on an inclined plane at an angle of 37° thus making it weight have two component,

[tex]W_{x}[/tex] = horizontal component of the weight = mgsinФ

[tex]W_{y}[/tex] = vertical component of weight = mgcosФ

Due to the way the object is positioned, the horizontal component of force will accelerate the object thus acting as an applied force.

by using newton's law of motion, we have that

mgsinФ - [tex]f_{r}[/tex] = ma

where m = mass of object=5kg

a = acceleration= unknown

Ф = angle of inclination = 37°

g = acceleration due to gravity = 9.8[tex]m/s^{2}[/tex]

[tex]f_{r}[/tex] = frictional force = unknown

we need to first get the acceleration before the frictional force which is gotten by using the equation below

[tex]v^{2} = u^{2} + 2aS[/tex]

where v = final velocity = 2m/s

u = initial velocity = 0m/s (because the object started from rest)

a= unknown

S= distance covered = length of plane = 5m

[tex]2^{2} = 0^{2} + 2*a*5\\\\4= 10 *a\\\\a = \frac{4}{10} \\a = 0.4m/s^{2}[/tex]

we slot in a into the equation below to get frictional force

mgsinФ - [tex]f_{r}[/tex] = ma

3 * 9.8 * sin 37 - [tex]f_{r}[/tex] = 3* 0.4

17.9633 - [tex]f_{r}[/tex] =  1.2

[tex]f_{r}[/tex] = 17.9633 - 1.2

[tex]f_{r}[/tex] = 16.49N