How high will a 1.30-kg rock go from the point of release if thrown straight up by someone who does 80.0 J of work on it? Neglect air resistance.

Answer:

48 m

Explanation:

Initial speed of laser,u=12 m/s

Acceleration due to gravity=-[tex]1.5m/s^2[/tex]

Because laser goes against to gravity when it is thrown in upward direction.

Final velocity of laser=v=0

We know that

[tex]v^2-u^2=2as[/tex]

Substitute the values then we get

[tex]0-(12)^2=2(-1.5)s[/tex]

[tex]-144=-3s[/tex]

[tex]s=\frac{144}{3}[/tex]

[tex]s=48 m[/tex]

Hence, the maximum height reached by the laser= 48 m

The maximum height reachedby the laser is 48 m.

To calculate the maximu height reached by the laser, we use the formula below.

Formula:

Where:

Make H the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, The maximum height reachedby the laser is 48 m.

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Answer:

In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force

Explanation:

Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter than if you land on water. Using the impulse-momentum theorem as a guide, the statement in bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force is correct.

The impulse-theorem defined as the change in momentum of an object will be equal to the impulse exerted on it.

Mathematically, the theorem can be described as:

The impulse is defined as the product between the force exerted (F) and the duration of the collision Δt,

I = FΔt

The change in momentum is equal to the product between the mass of the object (m) and the change in velocity (Δυ):

Δp = mΔv

So, the theorem can be written as,

FΔt = mΔv

This theorem can also be proved using Newton's second law, we know that,

F = ma

where, F - Force,

m - mass

a - acceleration

Here, we have to alter it with the acceleration change.

a = Δv /Δt,

when we substitute this with the force,

F = mΔv /Δt,

FΔt =mΔv.

The above formula is the exact formula of impulse- momentum theorem.

By using this concept, according to the force and velocity, If you land on sand your stopping time is much shorter than if you land on water, in bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force.

Hence, Option B is the correct answer.

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Answer:

The question is incomplete, below is the complete question

"Two vectors A and B are added together to form a vector C. The relationship between the magnitudes of the vectors is given by A + B = C. Which one of the following statements concerning these vectors is true?

a. A and B must be displacements.

b. A and B must have equal lengths.

c.A and B must point in opposite direction.

d. A and B must point in the same direction.

e.A and B must be at right angles to each other."

answer:

d. A and B must point in the same direction.

Explanation:

a.false:From vector analysis, all forms of vector with the same unit can be added, and we add vector component by component. hence this defile option (a).

b. false: in addition of vectors, length is not a criteria to consider before carrying out the operation, hence vectors of different lengths can be  added

c. false: this is against the claim as vectors in opposite direction can give rise to a new vector of negative value.

d. true: this is true as vector in the same direction add up to give rise to another vector.

e.false: this is not a true assumption

The relationship A + B = C only holds for vectors A and B that are collinear. Generally, the magnitude of the resultant vector C is derived using the triangle rule of vector addition, and is not a simple sum of the magnitudes of vectors A and B.

The relationship between the magnitudes of vectors A, B, and C is not always expressed as A + B = C. This can only be true in the special case where the vectors A and B are collinear (aligned along the same line and direction).

In general, according to the triangle rule of vectors addition, if we add two vectors A and B (not necessary to be in the same direction), the resultant vector C can be represented by the diagonal of a parallelogram which is constructed using A and B as two adjacent sides. Meaning, |C|= √ (A² + B² + 2ABcosθ), where θ is the angle between A and B.

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The weight of an oil tank with the mass of 50 slugs on the Earth's surface would be 1,610 pounds or 7164.2732 Newtons. On the moon, the weight would be 268.5 pounds, but its mass would remain the same (50 slugs) because mass is independent of location.

To begin, the weight can be determined using the formula Weight = mass x gravitational force. On the Earth's surface, the gravitational force is approximately 32.2 ft/s2.

A) The weight of the oil tank on Earth can be calculated by multiplying its mass (50 slugs) by the gravitational force. Therefore, it would be 50 slugs * 32.2 ft/s2 = 1,610 pounds. To convert to Newtons, since 1 pound = 4.44822 Newtons, the weight is 1,610 pounds * 4.44822 = 7164.2732 Newtons.

B) On the moon, the gravitational force is a sixth of that on Earth, i.e., 32.2 ft/s2 / 6 = 5.37 ft/s2. The mass of the oil tank would remain the same, 50 slugs, because mass doesn't change with location. However, its weight on the moon would be 50 slugs * 5.37 ft/s2 = 268.5 pounds.

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To find the weight of 50 slugs on Earth, multiply by 32.17 ft/s^2 to get 1608.5 pounds and convert to 7157.48 Newtons. On the Moon, the weight is 1/6th, resulting in 268.083 pounds, with the mass still being 50 slugs.

To determine the weight of a tank of oil with a mass of 50 slugs on Earth's surface, we use the U.S. Customary system. The acceleration due to gravity (g) in this system is 32.17 feet per second per second. The formula to convert mass in slugs to weight in pounds is: Weight (lb) = mass (slugs) \\u00d7 acceleration due to gravity (ft/s^2).

A) On Earth, the weight in pounds is:
Weight (lb) = 50 slugs \\u00d7 32.17 ft/s^2 = 1608.5 pounds.
To convert this to Newtons, we use the conversion factor 1 pound = 4.44822 Newtons, resulting in:
Weight (N) = 1608.5 lb \\u00d7 4.44822 N/lb = 7157.48 Newtons.

B) On the Moon, where gravity is 1/6th that of Earth, the mass remains the same. Therefore, the weight in pounds is:
Weight (lb) = 50 slugs \\u00d7 (32.17 ft/s^2 / 6) = 268.083 pounds.
The mass in slugs remains the same at 50 slugs as mass is independent of gravity.

Answer:

Royal proclamation was issued in 1763 by the king George III which forbade the american colonist from settling west of Appalachia. This proclamation was issued to avoid the conflicts between the native Americans and the new settlers.

Explanation:

Answer:

The mass density of the string is (0.3/L)kg/m

Explanation:

Mass density of the string = Mass/Length

Mass = 0.3kg

The length of the string is unknown so it is assumed to be L meter(s)

Therefore, mass deny of the string = 0.3kg/Lm = (0.3/L)kg/m

Answer:

w = 5.3311 rad/sec

Explanation:

n = 100 has a k = 540 N/m

k depends on the number of coils by:

[tex]k = \frac{G*d^4}{8*n*D^3}[/tex]

By the design equation we see that the spring stiffness k has an inverse relationship with number of coils n.

Hence, when n = 25 coils ; k = 4* 540 = 2160 N/m

The relationship between angular frequency and k is:

[tex]w = \sqrt{\frac{k}{m} }\\\\Hence,\\\\w = \sqrt{\frac{2160}{76} }\\\\w = 5.3311 rad/sec[/tex]

The angular frequency is approximately 5.34 rad/s.

To determine the angular frequency of the motion, we need to follow these steps:

Calculate the spring constant of one of the 25-coil springs.

Since the original spring has 100 coils and a spring constant of 540 N/m, the spring constant of one 25-coil spring, k(short), is:

k(short) = 4 * k(original) = 4 * 540 N/m

k(short) = 2160 N/m

Calculate the angular frequency, ω, using the formula:

Where:

Thus,

The angular frequency of the motion is approximately 5.34 rad/s.

Answer: 0.05470atm

Explanation:

The vapor pressure is determined by Clausius Clapeyron equation:

[tex]ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}(\frac{1}{T_1}-\frac{1}{T_2})[/tex]

where,

[tex]P_1[/tex] = initial pressure at  = 1 atm (standard atmospheric pressure

[tex]P_2[/tex] = final pressure at [tex]15^oC[/tex] = ?

[tex]\Delta H_{vap}[/tex] = enthalpy of vaporisation = 38.56 kJ/mol = 38560 J/mol

R = gas constant = 8.314 J/mole.K

[tex]T_1[/tex]= initial temperature = [tex]78.4^oC=273+78.4=351.4K[/tex]

[tex]T_2[/tex] = final temperature =[tex]15^oC=273+15=288K[/tex]

Now put all the given values in this formula, we get

[tex]\log (\frac{P_2}{1atm})=\frac{38560}{2.303\times 8.314J/mole.K}[\frac{1}{351.4K}-\frac{1}{288K}][/tex]

[tex]\log (\frac{P_2}{1atm})=-1.262[/tex]

[tex]P_2=0.05470atm[/tex]

Thus the vapor pressure of ethanol at [tex]15^0C[/tex] is 0.05470atm

Answer:

It corresponds to a distance of 100 parsecs away from Earth.

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax.

It is defined in a analytic way as it follows:

[tex]\tan{p} = \frac{1AU}{d}[/tex]

Where d is the distance to the star.

[tex]p('') = \frac{1}{d}[/tex] (1)  

Equation (1) can be rewritten in terms of d:

[tex]d(pc) = \frac{1}{p('')}[/tex] (2)

Equation (2) represents the distance in a unit known as parsec (pc).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (1AU).

For the case of   ([tex]p('') = 0.01[/tex]):

[tex]d(pc) = \frac{1}{0.01}[/tex]

[tex]d(pc) = 100[/tex]

Hence, it corresponds to a distance of 100 parsecs away from Earth.

Summary:

Notice how a small parallax angle means that the object is farther away.

Key terms:

Parsec: Parallax of arc second

A stellar parallax of 0.01 arcsecond measured using telescopes on Earth corresponds to a distance of 100 parsecs, or about 326 light-years.

When measuring stellar parallax from Earth with a precision of about 0.01 arcsecond, we can calculate the corresponding distance in space using the concept that a parallax of 1 arcsecond is equivalent to a distance of 1 parsec. To find the distance for a parallax of 0.01 arcsecond, we use the formula: distance (parsecs) = 1 / parallax angle (arcseconds).

Thus, a parallax of 0.01 arcsecond corresponds to:
distance = 1 / 0.01 arcseconds = 100 parsecs. This means the star is 100 parsecs away, which is approximately 326 light-years, since 1 parsec equals 3.26 light-years.

Additional Context

Parallax measurements are vital for determining distances to the nearest stars in astronomy, and with ground-based telescopes, we can measure out to about 100 parsecs. However, due to atmospheric turbulence, accuracy diminishes beyond 40 parsecs, and ground-based parallax measurement becomes unfeasible beyond 100 parsecs. Space missions like Hipparcos have improved our ability to measure stellar parallax, pushing accurate distance measurements out beyond 500 parsecs.

The displacement of the object between t = 1.0 s and t = 3.0 s, using the equation x = –4t + t², is found to be 0 m.

The displacement of the object between t = 1.0 s and t = 3.0 s can be found by substituting these time values into the given equation x = –4t + t2 and then subtracting the earlier position from the later one.

For t = 1.0 s; x1 = -4(1) + (1)² = -3 m
For t = 3.0 s; x2 = -4(3) + (3)² = -3 m

The displacement can be calculated as Δx = x2 - x1 = -3 - (-3) = 0 m. So, the displacement of the object between t = 1.0 s and t = 3.0 s is 0 m.

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Answer:

spring constant will be 547.619 N /m

Explanation:

We have given that force exerted if uniformly from 0 to 230 N

So exerted force F = 230 N

String is stretched by 0.420 m due to applying force

So x = 0.420 m

We have to find the spring constant in N/m

We know that stretched force is given by

[tex]F=Kx[/tex] , here K is spring constant and x is stretched length

So [tex]230=K\times 0.420[/tex]

[tex]K=\frac{230}{0.420}=547.619N/m[/tex]

So spring constant will be 547.619 N /m

The equivalent spring constant of the bow is 547.62 N/m.

To determine the equivalent spring constant of the bow, we can use Hooke's law which states that the force exerted by a spring is directly proportional to the displacement of the spring from its equilibrium position.

Given that the bow string is pulled back by 0.420 m and the force increases uniformly from 0 to 230 N, we can use the formula for spring force: F = kx, where F is the force, k is the spring constant, and x is the displacement.

Plugging in the given values, we have:

230 N = k * 0.420 m

Solving for k, we divide both sides of the equation by 0.420 m:

k = 230 N / 0.420 m = 547.62 N/m

Therefore, the equivalent spring constant of the bow is 547.62 N/m.

Answer:

2.50 m/s

Explanation:

This question can be solved using momentum conservation equation

combined mass of crow and feeder = 450+670=1120 gm

let the recoil speed of feeder be v m/s

Then applying momentum conservation we get;

1120×1.5 = 670×v

v= 2.50 m/s

the speed at which the feeder initially recoils backwards = 2.50 m/s

Final answer:

By applying the conservation of momentum principle, the feeder initially recoils at a speed of approximately 1.007 meters per second when a 450-gram crow pushes off of it with a takeoff speed of 1.5 m/s.

Explanation:

The question you've asked is rooted in the principle of conservation of momentum in physics. When the crow pushes off from the feeder, both the crow and the feeder will have momenta that are equal in magnitude and opposite in direction, assuming no external forces act on the system. The situation described can be solved using the formula for conservation of linear momentum, which states that the total momentum of a closed system remains constant if no external forces are acting on it.

To find the speed at which the feeder recoils, we use the fact that the initial momentum of the system (before the crow pushed off) was zero since the feeder was at rest. Consequently, the momentum of the crow after pushing off and the momentum of the feeder must cancel each other out:

Mass of crow imes Velocity of crow = Mass of feeder imes Velocity of feeder

(450 g) imes (1.5 m/s) = (670 g) imes Velocity of feeder

We now need to solve the Velocity of the feeder:

Velocity of feeder = (450 g imes 1.5 m/s) / (670 g)

Velocity of feeder = (675 g imes m/s) / (670 g)

Velocity of feeder \ 1.007 m/s

Note that we converted the masses to kilograms in the calculation (which isn't shown above) because the unit of mass in the formula for momentum (p = mv) should be in kilograms to match with the unit for velocity in meters per second (m/s). The feeder recoils at a speed of approximately 1.007 meters per second.

Answer: Each half life is 8 minutes long.

Explanation:

Expression for rate law for first order kinetics is given by:

[tex]t=\frac{2.303}{k}\log\frac{a}{a-x}[/tex]

where,

k = rate constant  

t = age of sample  = 24 minutes

a = initial amount of the reactant  = 10 grams

a - x = amount left after decay process  = 1.25 grams

[tex]24min=\frac{2.303}{k}\log\frac{10}{1.25}[/tex]

[tex]k=\frac{2.303}{24}\log\frac{10}{1.25}[/tex]

[tex]k=0.0866min^{-1}[/tex]

for completion of half life:  

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

[tex]t_{\frac{1}{2}}=\frac{0.693}{k}[/tex]

[tex]t_{\frac{1}{2}}=\frac{0.693}{0.0866min^{-1}}=8.00min[/tex]

Thus half life is 8 minutes long

Answer:

The answer to your question is letter A.     r = 1.07 x 10⁻¹⁴ m

Explanation:

Data

F = 2 N

d = ?

q = 1.6 x 10 ⁻¹⁹ C

k = 8.987 Nm²/C²

Formula

                 [tex]F = K\frac{q1q2}{r^{2}}[/tex]

Solve for r

                [tex]r = \sqrt{\frac{kq1q2}{F}}[/tex]

Substitution

                [tex]r = \sqrt{\frac{8.987 x 10^{9}x1.6 x 10^{-19} x 1.6 x 10x^{-19}}{2}}[/tex]

Simplification

                r = [tex]\sqrt{\frac{2.3 x 10^{-28}}{2}}[/tex]

                r = [tex]\sqrt{1.15 x 10^{-24}}[/tex]

Result

                r = 1.07 x 10⁻¹⁴ m

Answer:

A

Explanation:

Using F = kq1q2/r^2

         r^2 = kq1q2/F

          r = square root of (kq1q2/F)

          r =  square root of(8.99 x 10^9 x 1.6 x 10^-19 x 1.6 x 10^-19/2)

          r = square root of(1.15072 x 10^-28)

           r = 1. 07 x 10^-14m

Answer

given,

initial speed of merry-go-round = 0 rad/s

final speed of merry-go-round = 1.5 rad/s

time = 7 s

Radius of the disk = 6 m

Mass of the merry-go-round = 25000 Kg

Moment of inertia of the disk

[tex]I = \dfrac{1}{2}MR^2[/tex]

[tex]I = \dfrac{1}{2}\times 25000\times 6^2[/tex]

   I = 450000 kg.m²

angular acceleration

[tex]\alpha = \dfrac{\omega_f-\omega_0}{t}[/tex]

[tex]\alpha = \dfrac{1.5-0}{7}[/tex]

[tex]\alpha =0.214\ rad/s^2[/tex]

we know,

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

[tex]\tau= 450000\times 0.214[/tex]

[tex]\tau=96300\ N.m[/tex]

The required value of frictional torque produced by the bearings is 96300 N-m.

Given data:

The final angular speed of merry-go round is, [tex]\omega_{2}=1.5 \;\rm rad/s[/tex].

The time interval is, t = 7.0 s.

The radius of merry-go round is, r = 6.0 m.

The mass of merry-go round is, m = 25,000 kg.

The frictional torque of an object undergoing rotational motion is the product of moment of inertia and angular acceleration. So,

[tex]T_{f} = I \times \alpha[/tex] ....................................................(1)

Here, I is the moment of inertia of merry-go round and its value is,

[tex]I =\dfrac{mr^{2}}{2}\\\\I =\dfrac{25000 \times (6.0)^{2}}{2}\\\\I =450000 \;\rm kg.m^{2}[/tex]

And angular acceleration is,

[tex]\alpha = \dfrac{\omega_{2}-\omega_{1}}{t}\\\\\alpha = \dfrac{1.5-0}{7}\\\\\alpha = 0.214 \;\rm rad/s^{2}[/tex]

Then the frictional torque is calculated from equation (1) as,

[tex]T_{f} = 450000 \times 0.214\\\\T_{f}=96300 \;\rm N.m[/tex]

Thus, we can conclude that the required value of frictional torque produced by the bearings is 96300 N-m.

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Answer:

You are going to see the AC functions shifted down

Explanation:

Because you had a positive DC signal first, you had to readjust the position to the middle, otherwise, the sum of AC and DC would have been shifted up; because of this, your cursor has been moved down and ones the DC disappears from channel B the signal will be shifted down.

Note: Changing from DC to AC in channel B will kill all the DC signal

Answer:

In no case does the amplitude appear, so we would not have to change the period of the system when we change to the amplitude

Explanation:

In the harmonic movement when the amplitude of oscillation increases, body speed also increases, so the frequency remains constant.

When we solve the different case of harmonic movement

Simple pendulum 2pi f = Ra l / g

Spring mass 2pi f = RA k / m

2pi torsion pendulum f = RA I / k

In no case does the amplitude appear, so we would not have to change the period of the system when we change to the amplitude

Answer:

The frequency does not depend on the amplitude for any (ideal) mechanical or electromagnetic waves.

In electromagnetism we have that the relation is:

Velocity = wavelenght*frequency.

So the amplitude of the wave does not have any effect here.

For a mechanical system like an harmonic oscillator (that can be used to describe almost any oscillating system), we have that the frequency is:  

f = (1/2*pi)*√(k/m)

Where m is the mass and k is the constant of the spring, again, you can see that the frequency only depends on the physical properties of the system, and no in how much you displace it from the equilibrium position.

This happens because as more you displace the mass from the equilibrium position, more will be the force acting on the mass, so while the "path" that the mass has to travel is bigger, the mas moves faster, so the frequency remains unaffected.

The distance covered by the proton is 48.4 m

Explanation:

The electric field produced by an electrically charged infinite plane is given by

[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]

where in this case,

[tex]\sigma = -1.10\cdot 10^{-6} C/m^2[/tex] is the surface charge density

[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity

Substituting,

[tex]E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C[/tex]

And the direction is towards the plane (because the charge is negative).

The electric force on the proton due to this field is

[tex]F=qE[/tex]

where

[tex]q=1.6\cdot 10^{-19}C[/tex] is the proton charge

Substituting,

[tex]F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N[/tex]

where the direction is toward the plane.

Now we can calculate the proton's acceleration using Newton's second law:

[tex]a=\frac{F}{m}[/tex]

where

[tex]m=1.67\cdot 10^{-27}kg[/tex] is the proton mass

Substituting,

[tex]a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2[/tex]

Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:

[tex]v^2-u^2=2as[/tex]

where

v = 0 is the final velocity

[tex]u=2.40\cdot 10^7 m/s[/tex] is the initial velocity

a is the acceleration

s is the distance covered

And solving for s,

[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m[/tex]

Therefore, the closest answer is 48.4 m.

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Answer:

The affirmation is only true if the truck is traveling at 50 mi/h.

Explanation:

Hi there!

Let´s use the equation of average velocity (AV) to solve this problem:

AV = Δx / t

Where:

Δx = traveled distance.

t = time.

For the car, its average velocity (vc) will be:

vc = 120 miles / t

For the truck:

vtr = 100 miles / t

If we solve both equations for t and then equalize them (since the time is the same for both vehicles):

vc = 120 miles / t

t = 120 mi / vc

vtr = 100 miles / t

t = 100 mi / vtr

120 mi / vc = 100 mi / vtr

multiply both sides by vc and divide by 100:

120 mi / 100 mi = vc / vtr

1.2 = vc / vtr

1.2 · vtr = vc

The car travels 1.2 times faster than the truck.

Let´s see at which velocity of the truck, the car is traveling 10 mi/h faster. In this case, vc = 10 mi/h + vtr:

1.2 · vtr = 10 mi/h + vtr

1.2 vtr - vtr = 10 mi/h

0.2 vtr = 10 mi/h

vtr = 10 mi/h / 0.2

vtr = 50 mi/h

The affirmation is only true if the truck is traveling at 50 mi/h.

Answer:

c. It is rubbed with another object, and electrons move onto the rod.

Explanation:

A rubber rod is negatively charged through friction when gains electrons, this occurs when electrons are transferred from another material by simple contact between them. When these two materials are rubbed, the difference in electronic affinity causes one of the materials to be positively charged and the other negatively charged.

The percent decrease in the number of errors committed by the baseball player is 33.33%, calculated by finding the difference in errors between the two years and dividing by the number of errors in the first year.

To calculate the percent decrease in the number of errors committed by the baseball player, we need to first find the difference in the number of errors between the two years and then divide that by the number of errors made in the first year. Finally, we'll multiply the result by 100 to get the percentage.

Here's the step-by-step calculation:

Therefore, the baseball player had a 33.33% decrease in the number of errors committed compared to the previous year.

Answer:

[tex]\omega_f = 3.584\ rad/s[/tex]

Explanation:

given,

turntable rotate to, θ = 5 rad

time, t = 2.8 s

initial angular speed  = 0 rad/s

final angular speed = ?

now, using equation of rotational motion

[tex]\theta = \omega_i t + \dfrac{1}{2}\alpha t^2[/tex]

[tex]5 = 0+ \dfrac{1}{2}\alpha\times 2.8^2[/tex]

[tex]\alpha= \dfrac{10}{2.8^2}[/tex]

       α = 1.28 rad/s²

now, calculation of angular velocity

[tex]\omega_f = \omega_i + \alpha t[/tex]

[tex]\omega_f =0 +1.28\times 2.8[/tex]

[tex]\omega_f = 3.584\ rad/s[/tex]

hence, the angular velocity at the end is equal to 3.584 rad/s

The angular velocity at the end of that time 10rad/s

In order to get the angular velocity [tex]\omega[/tex], we will use the equation of motion expressed as [tex]\omega = \omega_0 + \alpha t[/tex]

[tex]\alpha[/tex] is the angular acceleration

t is the time taken

[tex]\omega_0[/tex] is the initial angular velocity

Get the angular acceleration [tex]\alpha[/tex]

[tex]\theta = \omega_0 t+ \frac{1}{2} gt^2[/tex]

[tex]5=0+\frac{1}{2} \alpha (2.8)\\2\times5 = 2.8 \alpha \\10 = 2.8 \alpha\\\alpha = \frac{10}{2.8}\\\alpha = 3.57rad/s^2[/tex]

Get the angular velocity [tex]\omega[/tex]

[tex]\omega = \omega_0 + 3.57(2.8)\\\omega = 0+10\\\omega =10rad/s\\[/tex]

Hence the angular velocity at the end of that time 10rad/s

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Answer:

Swallowed fish speed=4 m/s

Explanation:

Given

m₁(fish mass)=20 kg

v₁(fish speed)=1 m/s

m₂(Swallowed fish mass)=5 kg

To find

v₂(Swallowed fish speed)

Solution

From law of conservation of momentum

[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v\\ (20kg)(1 m/s)+(5 kg)v_{2}=(20kg+5kg)0m/s\\20kg.m/s+5kgv_{2}=0\\v_{2}=(20kg.m/s)/5kg\\v_{2}=4m/s[/tex]

As I neglected the negative sign because it only shows that swallowed fish is moving in opposite direction

The smaller fish was moving at a velocity of 4 m/s towards the larger fish before being eaten

The question involves the principle of conservation of momentum to solve how fast a 5-kg fish was moving prior to being eaten by a 20-kg fish to bring both to a halt. In the scenario, the 20-kg fish is moving at 1 m/s towards the smaller fish. To find the speed of the smaller fish before being swallowed, we assume no external forces are acting on the system.

Momentum is calculated using the formula p = mv, where p is momentum, m is mass, and v is velocity. Initially, the momentum of the 20-kg fish is (20 kg)(1 m/s) = 20 kg·m/s, and let's say the smaller fish has a velocity v. Since they come to a halt, the final momentum is 0. By setting the total initial momentum equal to the final momentum, we have:

(20 kg)(1 m/s) + (5 kg)(-v) = 0
⇒ 20 - 5v = 0
⇒ v = 4 m/s

Answer:

The average force between the ball and bat during contact is 3006.72  N.

Explanation:

Given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the ball, u = 27 m/s

Time of contact between bat and ball is 2.5 ms, [tex]t=2.5\times 10^{-3}\ s[/tex]

After striking the bat, it pops straight up to a height of 31.5 m. The final velocity of the ball is given by using third equation of motion as :

[tex]v^2-u^2=2as[/tex]

a = -g

And initially, u = 0

[tex]v=\sqrt{2gs}[/tex]

[tex]v=\sqrt{2\times 9.8\times 31.5}[/tex]

v = -24.84 m/s (as it pops straight up)

Let F is the average force between the ball and bat during contact. It is given by :

[tex]F=\dfrac{m(v-u)}{t}[/tex]

[tex]F=\dfrac{0.145\times (24.84 -(-27))}{2.5\times 10^{-3}}[/tex]

F = 3006.72  N

So, the average force between the ball and bat during contact is 3006.72  N. Hence, this is the required solution.                                  

Final answer:

Two vectors with different magnitudes can add up to zero only if they are antiparallel. For three or more vectors of different magnitudes, they can sum to zero if they form a closed shape. The maximum distance achieved by adding two vectors is the sum of their lengths if they are in the same direction.

Explanation:

When dealing with vectors, it is possible for two vectors with different magnitudes to add up to zero if they are equal in magnitude but opposite in direction, making them antiparallel. This is due to the definition of vector addition, which considers both magnitude and direction.

For example, if you take a step of 5 units to the north (vector A) and then a step of 5 units to the south (vector B), you will end up back at your starting point as these two vectors cancel each other out. This can only happen if the two vectors are directly opposite to each other (antiparallel).

However, in the more general case with three or more vectors, they can add to zero even if they have different magnitudes, as long as they form a closed shape like a triangle or a polygon when placed tip-to-tail. The result of adding multiple vectors together in such a manner would be zero, bringing you back to your starting point.

Moreover, the maximum distance you can end up from the starting point when adding two vectors A and B is indeed the sum of their lengths, but only when the vectors are aligned and pointing in the same direction. When they point in opposite directions, the maximum distance will be the difference in their lengths.

Answer:

The time = 74.61 s.

Explanation:

From Newton's Third law of motion,

The force needed to stop is equal and opposite to the propelling force from the engine of the freight car.

F = ma ........................ Equation 1

Where F = force needed to stop the the freight car, m = mass of the freight car, a = acceleration of the freight car.

a = F/m....................... Equation 2.

Given: F = - 1150 N(Opposite in direction to the motion of the car)  m = 26000 kg.

Substituting into equation 2

a = -1150/26000

a = -0.04423 m/s²

Also

v = u + at

t = (v-u)/a.................... Equation 3

Where t = time needed for the force to stop the car, v = final velocity, u = initial velocity, a = acceleration.

Given: u = 3.3 m/s v = 0 m/s (brought to rest), a = -0.04423 m/s²

Substituting into equation 3

t = (0-3.3)/-0.04423

t = -3.3/0.04423

t = 74.61 sec.

Thus the time = 74.61 s

Answer:

Option b

b. Vega

Explanation:

The angle due to the change in position of a nearby object against the background stars it is known as parallax (see the image below).

The parallax angle can be used to find out the distance by means of triangulation. Making a triangle between the nearby star, the Sun and the Earth (as is shown in the image below), knowing that the distance between the Earth and the Sun (150000000 Km), is defined as 1 astronomical unit (AU).

[tex]\tan{p} = \frac{1AU}{d}[/tex]

Where d is the distance to the star.

[tex]p('') = \frac{1}{d}[/tex]  (1)  

Equation (1) can be rewritten in terms of d:

[tex]d(pc) = \frac{1}{p('')}[/tex]  (2)

Equation (2) represents the distance in a unit known as parsec (pc).

For the case of Sirius  ([tex]p('') = 0.38[/tex]):

[tex]d(pc) = \frac{1}{0.38}[/tex]

[tex]d(pc) = 2.63[/tex]

Hence, the object is 2.63 parsecs away from Earth.  

For the case of Procyon  ([tex]p('') = 0.29[/tex]):

[tex]d(pc) = \frac{1}{0.29}[/tex]

[tex]d(pc) = 3.44[/tex]

Hence, the object is 3.44 parsecs away from Earth.  

For the case of Vega  ([tex]p('') = 0.13[/tex]):

[tex]d(pc) = \frac{1}{0.13}[/tex]

[tex]d(pc) = 7.69[/tex]

Hence, the object is 7.69 parsecs away from Earth.  

Therefore, Vega is the star farther away.

Summary:

Notice how a small parallax angle means that the object is farther away.

Answer:

The synthesis of a large collection of information that contains well-tested and verified hypotheses about certain aspects of the world is known as a scientific THEORY

A scientific theory is the synthesis of well-tested and verified hypotheses about certain aspects of the world.

Scientific theory represents the pinnacle of scientific knowledge, signifying the meticulous synthesis of an expansive body of information that encompasses thoroughly tested and validated hypotheses pertaining to specific facets of our world. This robust and comprehensive explanation is not conjecture; instead, it is firmly rooted in a wealth of empirical evidence that has undergone rigorous scrutiny through a multitude of scientific experiments and meticulous observations.

A scientific theory, therefore, stands as an enduring framework that not only elucidates natural phenomena but also endows us with a profound comprehension of the intricate workings of the natural world. It serves as the bedrock upon which scientific understanding and further inquiry are built, fostering a deeper appreciation of the complexities that underlie our universe.

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Answer:

The breast bone is anterior to the spine. The collarbone is superior to the shoulder blade. The elbow is proximal to the wrist. The navel is distal to the chin.

Explanation:

The spine, also known as the backbone, is posterior to the breast bone.

The collarbone (also known as the clavicle), which is part of the shoulder girdle, lies superior in position to the shoulder blade (also known as the scapula).

The elbow is the joint between the arm and the wrist. it is proximal to the wrist.

The navel, located on the anterior abdominal wall, is distal to the chin that is part of the face.

Answer:

714s

Explanation:

t=H/v=500000m/700m/s=714s

The space shuttle edeavor is launched to altitude of 500,000 m above the surface of earth. The shuttle travels at an average rate of 700 m/s. Time taken by space shuttle is 714.28 m/s to reach its orbit.

Given the data to find the time taken by space shuttle,

Distance covered by the space shuttle = 500,000 meter

Speed of space shuttle = 700 m/s

Mathematically, it can be calculated as follows :

distance = speed × time

As we have to find the time taken, the formula will be altered.

Computation:

Time taken = Distance / Speed

Time taken by space shuttle = Distance cover by space shuttle / Speed of space shuttle

Time taken by space shuttle = 500,000 / 700

Time taken by space shuttle = 714.28 m/s

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