Consider the following distinct forces:______________________________.
1.A downward force of gravity.
2.A force exerted by the channel pointing from Q to O.
3.A force in the direction of motion.
4.A force pointing from O to Q.
Which of the above forces is (are) acting on the ball when it is within the frictionless channel at position 'Q'.
n Q?

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: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:

Explanation:

The force of a spring is described by Hooke's law:

F = kx

where k is the spring stiffness in N/m, and x is the displacement in m.

A spring force vs displacement graph is a line passing through the origin with a slope of k.

Final answer:

A spring force vs. displacement graph represents Hooke's law, showing the restorative force exerted by a spring as it is deformed. The slope of the straight line on the graph is the spring's force constant 'k'. The area under the line represents the work done on the spring.

Explanation:

Physical Meaning of Spring Force vs. Displacement

The physical meaning of a spring force vs. displacement graph is that it shows how the force exerted by a spring changes as the spring is stretched or compressed. This relationship is governed by Hooke's law, which states that the force (F) exerted by a spring is directly proportional to the displacement (x) from its equilibrium position, given by the equation F = -kx. Here, 'k' is the spring's force constant, representing its stiffness, with units of newtons per meter (N/m). The negative sign indicates that the force is restorative, meaning it acts in the opposite direction of displacement.

The graph typically portrays a straight line passing through the origin, with the slope equivalent to the spring's force constant 'k'. The steeper the slope, the greater the force constant, and vice versa. Additionally, the work done on the spring, which is equal to the energy stored in it, can be calculated from the area under the force versus displacement graph, given by W = (1/2)kx², where W is the work done and x is the displacement.

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: [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

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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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:

b c and e

Explanation:

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

Answer:

The unit of measurement that defines the space available in a rack are  the inches for the width and for the height the units of Rack (U), this unit is equivalent to 44.45 mm.  Standard racks are 19 inches wide and 42U high. A rack is a type of shelf where servers can be stacked on top of each other.

The unit of measurement that defines rack space is 'rack unit' or 'U', with each unit equal to 1.75 inches or 44.45 millimeters. Standard racks can have different heights, with 42U being the most common.

The unit of measurement that defines the space available in a rack is called 'rack unit' or 'U'. Each rack unit is equal to 1.75 inches or 44.45 millimeters. So, when we talk about the height of a standard rack, we refer to the number of rack units it can accommodate.

The height of standard racks can vary, but the most common height for a rack is 42U, which means it can accommodate equipment up to 73.5 inches or 1866.9 millimeters tall. Other common rack heights include 48U and 45U.

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Galileo's work differed from his predecessors by relying on experimental evidence and mathematical analysis.

The work of Galileo differed from his predecessors in a fundamental way in terms of his approach to studying the sky. Unlike his predecessors who relied on pure reasoning and philosophical arguments, Galileo used experimental evidence and mathematical analysis to understand the natural phenomena.

For example, Galileo used a telescope to observe the moons of Jupiter, which challenged the prevailing geocentric model of the universe.

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Galileo used telescopes for detailed astronomical observations, challenging prevailing geocentric views, while predecessors relied on eye observations and theories.

Galileo's work differed fundamentally from his predecessors in astronomy by introducing a new method of observation. While his predecessors primarily relied on eye observations and abstract reasoning to understand celestial phenomena, Galileo pioneered the use of telescopes for systematic astronomical observations. By observing celestial objects through telescopes, he made groundbreaking discoveries such as the moons of Jupiter, the phases of Venus, and the mountains and craters on the Moon.

These observations provided concrete evidence that supported the heliocentric model proposed by Copernicus, challenging the geocentric worldview dominant at the time. Galileo's empirical approach and emphasis on experimental evidence marked a departure from traditional reliance on philosophical arguments. His method paved the way for modern observational astronomy, emphasizing the importance of empirical data and observation through telescopes, revolutionizing our understanding of the cosmos.

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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:

a

Explanation:

Answer:

A

Explanation:

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:

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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Answer:

It is called the elementary charge

Explanation:

Charge is quantized; it comes in integer multiples of individual small units called the elementary charge, e, about 1.602×10−19 coulombs, which is the smallest charge which can exist freely

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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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:

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:

-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.

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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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:

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

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.

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:

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.

Answer: noise that you hear in passing

Explanation:

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:

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:

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:

The potential energy can be given as

E = mgh. m is mass, g = acceleration due to gravity = 9.8m/s, h is the heigh, given as 100.0m

E = m x 9.8 x 100 = (980m)J

E = (980m)/10^9GJ = (0.000000980m)GJ to 3 significant figures

Explanation:

Hydroelectric dams exploit storage of gravitational potential energy. A mass, m, raised a height, h against gravity, g = 9.8 m/s², is given a potential energy E = mgh. The result will be in Joules if the input is expressed in meters, kilograms, and seconds (MKS, or SI units).