A certain particle at rest has a lifetime of 2.52 mu s. what must be the speed of the particle for its lifetime observed to be 8.32 mu s? 0.985c 0.953c 0.913c 0.302c None of the above

Explanation:

You need to find the least common multiple (LCM) of 30 and 50.  First, write the prime factorization of each:

30 = 2×3×5

50 = 2×5²

The LCM must contain all the factors of both, so:

LCM = 2×3×5²

LCM = 150

It will take 150 minutes (or 2 hours and 30 minutes) before two trains depart at the same time again.

The next time that two trains, one on each line, will depart from Center Station at the same time is at 10:30 A.M. This is calculated by finding the least common multiple of the two train schedules.

This question requires finding the least common multiple (LCM) of the two train schedules, which represents the time duration when both trains will depart again at the same time. The LCM of 30 and 50 is 150 minutes. This means, from 8:00 A.M., it will be the next 150 minutes or 2 hours and 30 minutes when both trains will depart from Center Station at the same time. Therefore, the answer is 10:30 A.M.

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Answer:

141

Explanation:

The atomic number (Z) corresponds to the number of protons:

Z = p

while the mass number (A) corresponds to the number of protons+neutrons:

A = p + n

So the number of neutrons in a nucleus is equal to the difference between mass number and atomic number:

n = A - Z

For the initial nucleus of Uranium, Z = 92 and A = 235, so the initial number of neutrons is

n = 235 - 92 = 143

An alpha particle carries 2 protons and 2 neutrons: so, when the isotope of Uranium emits an alpha particle, it loses 2 neutrons. Therefore, the number of neutrons after the decay will be

n = 143 - 2 = 141

Answer:

Density of meteorite = 2.44 g/cm³

Explanation:

Apparent mass = Mass of solid - Mass of water displaced

Mass of water displaced = Mass of solid - Apparent mass

                                         = 15.6 - 9.2 = 6.4 g

Density of water = 1 g/cm³

Volume of water displaced [tex]=1\times 6.4=6.4cm^3[/tex]

Volume of meteorite = Volume of water displaced = 6.4 cm³

[tex]\texttt{Density of meteorite}=\frac{\texttt{Mass of meteorite}}{\texttt{Volume of meteorite}}=\frac{15.6}{6.4}=2.44g/cm^3[/tex]

Density of meteorite = 2.44 g/cm³

Answer:

Instantaneous speed of contact point will be ZERO

Explanation:

As we know that disc is rolling without slipping on horizontal surface

So here the speed of center of the disc is given as

v = 5 m/s

now at the contact point the tangential speed will be in reverse direction

[tex]v_t = R\omega[/tex]

now we know that net contact speed with respect to its lower surface must be zero

[tex]v_{net} = v - v_t = 0[/tex]

so net velocity of contact point with respect to its lower surface must be ZERO here

The instantaneous speed of the point of a disc that is rolling without slipping and makes contact with the surface is zero. This is because that point is momentarily at rest relative to the surface at that instant.

The instantaneous speed of the point of the disc that makes contact with the surface is zero. This is because for a disc rolling without slipping on a surface, the point of contact at any instant is momentarily at rest relative to the surface. This can be understood by imagining the point of contact as the 'pivot' or point about which the disc rotates while rolling. As the disc rolls, the pivot point changes, but whichever point is in contact with the surface at a given instant is not moving relative to the surface. Hence, the instantaneous speed of that point is zero.

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Answer:

Wavelength, [tex]\lambda=\dfrac{c}{f}[/tex]

Explanation:

In an electromagnetic wave both electric and magnetic field propagate simultaneously. Radio waves, microwaves, gamma rays etc are some of the examples of electromagnetic waves.

If f is the frequency of electromagnetic wave, c is the speed of light, then the relationship between the frequency f, wavelength and the speed is given by :

[tex]\lambda=\dfrac{c}{f}[/tex]

Hence, the correct option that shows the wavelength of electromagnetic wave in free space is(a) " c/f ".

Answer:

9.47 rad/s^2

Explanation:

Diameter = 15 cm, radius, r = diameter / 2 = 7.5 cm = 0.075 m, u = 0, v = 7.1 m/s,

s = 35.4 m

let a be the linear acceleration.

Use III equation of motion.

v^2 = u^2 + 2 a s

7.1 x 7.1 = 0 + 2 x a x 35.4

a = 0.71 m/s^2

Now the relation between linear acceleration and angular acceleration is

a = r x α

where,  α is angular acceleration

α = 0.71 / 0.075 = 9.47 rad/s^2

Answer: [tex]4.487(10)^{11}m[/tex]

Explanation:

This problem can be solved using the Third Kepler’s Law of Planetary motion:

“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.  

This law states a relation between the orbital period [tex]T[/tex] of a body (the exoplanet in this case) orbiting a greater body in space (the star in this case) with the size [tex]a[/tex] of its orbit:

[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex] (1)  

Where:

[tex]T=3.87Earth-years=122044320s[/tex] is the period of the orbit of the exoplanet (considering [tex]1Earth-year=365days[/tex])

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]  

[tex]M=3.59(10)^{30}kg[/tex] is the mass of the star

[tex]a[/tex] is orbital radius of the orbit the exoplanet describes around its star.

Now, if we want to find the radius, we have to rewrite (1) as:

[tex]a=\sqrt[3]{\frac{T^{2}GM}{4\pi^{2}}}[/tex] (2)  

[tex]a=\sqrt[3]{\frac{(122044320s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(3.59(10)^{30}kg)}{4\pi^{2}}}[/tex] (3)  

Finally:

[tex]a=4.487(10)^{11}m[/tex] This is the radius of the exoplanet's orbit

Given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.

Given the data in the question;

To determine the radius of the exoplanet's orbit, we use the equation from Kepler's Third Law:

[tex]T^2 = \frac{4\pi^2 }{GM}r^3\\[/tex]

Where, T is the period of the orbit of the exoplanet, G is the Gravitational Constant, M is the mass of the star and r is orbital radius.

We make "r", the subject of the formula

[tex]r = \sqrt[3]{\frac{T^2GM}{4\pi ^2} }[/tex]

We substitute our given values into the equation

[tex]r = \sqrt[3]{\frac{(122044320s)^2*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} } \\\\r = \sqrt[3]{\frac{(1.48948*10^{16}s^2)*(6.67430 * 10^{-11} m^3/kg s^2)*(3.59*10^{30}kg)}{4*\pi ^2} }\\\\r = \sqrt[3]{\frac{3.5689*10^{36}m^3}{4*\pi ^2} }\\\\r = \sqrt[3]{9.04*10^{34}m^3}\\\\r = 4.488*10^{11}m[/tex]

Therefore, given the orbital period and mass of the exoplanet, its radius of orbit is 4.488 × 10¹¹m.

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Answer:

The angular speed of the disc be at this time is 46.65 rad/s.

Explanation:

Given that,

Angular acceleration [tex]\alpha= 3.11\ rad/s^2[/tex]

Time t =15.0 s

We will calculate the angular speed of the disc

A disc initially at rest.

So, [tex]\omega=0[/tex]

Using rotational kinematics equation

[tex]\omega'=\omega+\alpha\ t[/tex]

Where, [tex]\omega[/tex] = initial angular speed

[tex]\omega'[/tex] =final angular speed

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

Put the value in the equation

[tex]\omega'=0+3.11\times15[/tex]

[tex]\omega'=46.65\ rad/s[/tex]

Hence, The angular speed of the disc be at this time is 46.65 rad/s.

Explanation:

The escape velocity [tex]V_{esc}[/tex] is given by the following equation:

[tex]V_{esc}=\sqrt{\frac{2GM}{R}}[/tex]   (1)

Where:

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

[tex]M[/tex]  is the mass of the asteroid

[tex]R[/tex]  is the radius of the asteroid

On the other hand, we know the density of the asteroid is [tex]\rho=3.84(10)^{8}g/m^{3}[/tex] and its volume is [tex]V=2.17(10)^{12}m^{3}[/tex].

The density of a body is given by:

[tex]\rho=\frac{M}{V}[/tex]  (2)

Finding [tex]M[/tex]:

[tex]M=\rhoV=(3.84(10)^{8} g/m^{3})(2.17(10)^{12}m^{3})[/tex]  (3)

[tex]M=8.33(10)^{20}g=8.33(10)^{17}kg[/tex]  (4)  This is the mass of the spherical asteroid

In addition, we know the volume of a sphere is given by the following formula:

[tex]V=\frac{4}{3}\piR^{3}[/tex]   (5)

Finding [tex]R[/tex]:

[tex]R=\sqrt[3]{\frac{3V}{4\pi}}[/tex]   (6)

[tex]R=\sqrt[3]{\frac{3(2.17(10)^{12}m^{3})}{4\pi}}[/tex]   (7)

[tex]R=8031.38m[/tex]   (8)  This is the radius of the asteroid

Now we have all the necessary elements to calculate the escape velocity from (1):

[tex]V_{esc}=\sqrt{\frac{2(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(8.33(10)^{17}kg)}{8031.38m}}[/tex]   (9)

Finally:

[tex]V_{esc}=117.626m/s[/tex] This is the minimum initial speed the rocks need to be thrown in order for them never return back to the asteroid.

To calculate the minimum speed to escape the gravitational pull of the asteroid, you'll first determine the asteroid's mass using its density and volume. Find its radius assuming it's a sphere. Plug these into the escape speed equation sqrt(2*G*M/R) to determine the velocity.

The escape velocity of an object from another object’s gravitational pull can be determined by the formula: V_esc = sqrt(2*G*M/R). Here, V_esc is the escape speed, G is the universal gravitational constant, M is the mass of the object (in this case, the asteroid), and R is the radius of the object. To find M in this case, we can multiply the given density (p) of the asteroid by its volume (V), and convert this to kg. Once we know the mass, we can find the radius of the asteroid (assuming it is spherical), using the formula for the volume of a sphere (V = 4/3 * pi * R^3).

Once we have determined all the values above, we can substitute them into the escape speed formula to find the minimum initial speed a rock would need to be thrown in order never to fall back to the asteroid.

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Answer: [tex]1.893(10)^{27}kg [/tex]

Explanation:

This problem can be solved by the Third Kepler’s Law of Planetary motion, which states:

“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.

In other words, this law stablishes a relation between the orbital period [tex]T[/tex] of a body (moon, planet, satellite) orbiting a greater body in space with the size [tex]a[/tex] of its orbit.

This Law is originally expressed as follows:

[tex]T^{2}=\frac{4\pi^{2}}{GM}a^{3}[/tex]    (1)

Where;

[tex]T=7.16days=618624s[/tex]  is the period of the orbit Ganymede describes around Jupiter

[tex]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

[tex]M[/tex] is the mass of Jupiter  (the value we need to find)

[tex]a=10.7(10)^{8}m[/tex]  is the semimajor axis of the orbit Ganymede describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find [tex]M[/tex], we have to express equation (1) as written below and substitute all the values:

[tex]M=\frac{4\pi^{2}}{GT^{2}}a^{3}[/tex]    (2)

[tex]M=\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(618624s)^{2}}(10.7(10)^{8}m)^{3}[/tex]    (3)

Finally:

[tex]M=1.8934(10)^{27}kg[/tex]   This is the mass of Jupiter

Answer:

B. The algebraic sum of the two electric potentials is determined at a distance r/2 from each of the charges, making sure to include the signs of the charges.

Explanation:

Total electric potential is the sum of all the electric potential. And because electric potential is a scalar quantity you have to account for the signs.

Answer:

79.5 m

Explanation:

Let t be the time taken to hit the surface of water and x be the horizontal distance traveled.

use II equation of motion in Y axis direction

h = uy t + 1/2 g t^2

- 60 = - 100 Sin 35 x t - 1/2 x 9.8 x t^2

-60 = - 57.35 t - 4.9 t^2

4.9 t^2 + 57.35 t - 60 = 0

[tex]t = \frac{-57.35\pm \sqrt{57.35^{2} + 4 \times 4.9 \times 60}}{2\times 4.9}[/tex]

By solving we get

t = 0.97 second

The horizontal distance traveled is

x = ux t

x = 100 Cos 35 x 0.97

x = 79.5 m

Final answer:

To find how far from the cliff a projectile hits the water, one must use projectile motion principles to calculate the time of flight based on vertical movement and then determine the horizontal distance traveled during this time.

Explanation:

The question involves calculating how far from the foot of a vertical cliff a projectile hits the water when it is shot from an elevation with a given initial speed and angle. To solve this, we need to use the concepts of projectile motion, specifically focusing on the horizontal distance traveled by a projectile. The key equations involve splitting the initial velocity into its horizontal and vertical components, calculating the time of flight based on the vertical motion, and then using this time to find the horizontal distance traveled.

Given the projectile has a speed of 100 m/s and is fired at an angle of 35.0° below the horizontal from a height of 60.0 m, the calculation involves several steps:

Determine the initial horizontal and vertical velocity components.

Calculate the time of flight using the vertical motion equations.

Finally, compute the horizontal distance traveled using the time of flight.

Due to the complexity and the need for specific formulae and calculations, a detailed step-by-step solution would be necessary to find the exact distance.

Answer:

18.4 m

Explanation:

(a)

The known variables in this problem are:

u = 1.40 m/s is the initial vertical velocity (we take downward direction as positive direction)

t = 1.8 s is the duration of the fall

a = g = 9.8 m/s^2 is the acceleration due to gravity

(b)

The vertical distance covered by the life preserver is given by

[tex]d=ut + \frac{1}{2}at^2[/tex]

If we substitute all the values listed in part (a), we find

[tex]d=(1.40 m/s)(1.8 s)+\frac{1}{2}(9.8 m/s^2)(1.8 s)^2=18.4m[/tex]

Using the equation of motion, the calculation shows the life preserver was released from approximately 18.4 meters above the water.

y = vot + 1/2(at²)

Substituting the known values, we get:

y = (1.40 m/s)(1.8 s) + 1/2(9.8 m/s²)(1.8 s)²

y = 2.52 m + 15.876 m

y = 18.396 m

The life preserver was released approximately 18.4 meters above the water.

Answer:

1.42 s

Explanation:

The equation for free fall of an object starting from rest is generally written as

[tex]s=\frac{1}{2}at^2[/tex]

where

s is the vertical distance covered

a is the acceleration due to gravity

t is the time

On this celestial body, the equation is

[tex]s=10.04 t^2[/tex]

this means that

[tex]\frac{1}{2}g = 10.04[/tex]

so the acceleration of gravity on the body is

[tex]g=2\cdot 10.04 = 20.08 m/s^2[/tex]

The velocity of an object in free fall starting from rest is given by

[tex]v=gt[/tex]

In this case,

g = 20.08 m/s^2

So the time taken to reach a velocity of

v = 28.6 m/s

is

[tex]t=\frac{v}{g}=\frac{28.6 m/s}{20.08 m/s^2}=1.42 s[/tex]

It takes approximately [tex]\( 1.424 \)[/tex] seconds for the rock to reach a velocity of 28.6 m/s on this celestial body.

We need to use the given equation and the relationship between position, velocity, and acceleration.

The equation for the position [tex]\( s \)[/tex] as a function of time [tex]\( t \)[/tex] is given by:

[tex]\[s = 10.04t^2\][/tex]

Step 1: Find the Acceleration

This equation is similar to the general form of the kinematic equation for free fall under constant acceleration:

[tex]\[s = \frac{1}{2} a t^2\][/tex]

Comparing the two equations:

[tex]\[10.04t^2 = \frac{1}{2} a t^2\][/tex]

We can solve for the acceleration [tex]\( a \)[/tex]:

[tex]\[10.04 = \frac{1}{2} a\][/tex]

[tex]\[a = 2 \times 10.04 = 20.08 \, \text{m/s}^2\][/tex]

Step 2: Use the Acceleration to Find the Time

The velocity [tex]\( v \)[/tex] of an object in free fall under constant acceleration is given by:

[tex]\[v = at\][/tex]

We need to find the time [tex]\( t \)[/tex] when the velocity [tex]\( v \)[/tex] is 28.6 m/s:

[tex]\[28.6 = 20.08 t\][/tex]

Solving for [tex]\( t \)[/tex]:

[tex]\[t = \frac{28.6}{20.08}\][/tex]

[tex]\[t \approx 1.424 \, \text{seconds}\][/tex]

Answer:

Yes

Explanation:

The velocity measured by Jennifer and Johnny is

[tex]v_m = 34.87 m/s[/tex]

The actual velocity is

[tex]v=34.15 m/s[/tex]

We can calculate the % error of the students measurement as follows:

[tex]Err = \frac{v_m - v}{v}\cdot 100 = \frac{34.87 m/s-34.15 m/s}{34.15 m/s}\cdot 100 =0.021 \cdot 100 = 2.1 \%[/tex]

Which is lower than the 2.5% maximum error required, so the two students will pass the test.

Final answer:

Jennifer and Johnny's measurement has a percentage error of approximately 2.11%, which is less than the maximum allowed error of 2.5%. Therefore, they will pass their final project.

Explanation:

To determine if Jennifer and Johnny passed their final project, we need to calculate the percentage error of their measured velocity. The percentage error is calculated using the formula:

Percentage Error = |(Actual Value - Experimental Value) / Actual Value| × 100%

First, let's find the absolute difference between the actual velocity (34.15 m/s) and the measured velocity (34.87 m/s):

|34.15 m/s - 34.87 m/s| = |(-0.72 m/s)| = 0.72 m/s

Now, we calculate the percentage error:

Percentage Error = (0.72 m/s / 34.15 m/s) × 100% ≈ 2.11%

Since the percentage error they obtained (2.11%) is less than the maximum allowed error of 2.5%, Jennifer and Johnny will pass their final project.

Answer:

Gravitational potential energy = 524.85 kJ

Explanation:

Refer the figure.

Gravitational potential energy = mgh

Mass, m = 75 kg

Acceleration due to gravity, g = 9.81 m/s²

We have

        [tex]sin14.6=\frac{h}{2830}\\\\h=713.36m[/tex]

Gravitational potential energy = 75 x 9.81 x 713.36 = 524854.62 J = 524.85 kJ

Answer:

The distance between the charges is 9.5 cm

(c) is correct option

Explanation:

Given that,

Charge [tex]q= 10^{-6}\ C[/tex]

Force F = 1 N

We need to calculate the distance between the charges

Using Coulomb's formula

[tex]F = \dfrac{kq_{1}q_{2}}{r^2}[/tex]

Where, q = charge

r = distance

F = force

Put the value into the formula

[tex]1=\dfrac{9.0\times10^{9}\times(-10^{-6})^2}{r^2}[/tex]

[tex]r=\sqrt{9\times10^{9}\times(-10^{-6})^2}[/tex]

[tex]r=0.095\ m[/tex]

[tex]r= 9.5\ cm[/tex]

Hence, The distance between the charges is 9.5 cm

Answer:

[tex]r = 3.36 \times 10^{-7} m[/tex]

Explanation:

As per Ampere's law of magnetic field we know that

line integral of magnetic field along closed ampere's loop is equal to the product of current enclosed and magnetic permeability of medium

So it is given as

[tex]\int B. dl = \mu_0 i_{en}[/tex]

here we can say that enclosed current is given as

[tex]i_{en} = \frac{i}{\pi R^2} (\pi r^2)[/tex]

now from ampere'e loop law for any point inside the wire we will have

[tex]B.(2\pi r) = \mu_o (\frac{ir^2}{R^2}[/tex]

[tex]B = \frac{\mu_0 i r}{2\pi R^2}[/tex]

now we know that magnetic field inside the wire is 84% of the field at its surface

so we will have

[tex]0.84 \frac{\mu_o i}{2\pi R} = \frac{\mu_o i r}{2\pi R^2}[/tex]

so we have

[tex]r = 0.84 R[/tex]

now we know

[tex]\frac{\mu_o i}{2\pi R} = 1[/tex]

here i = 2 A

[tex]R = 2\times 10^{-7} m[/tex]

so now we have

[tex]r = 3.36 \times 10^{-7} m[/tex]

The point  ( r ) within the wire where the field strength equals 84% of the field strength at the wire surface is : 0.84 R

Given data :

Radius of wire = R

current in the wire = 2A

magnetic field strength = 1 T

we will apply Ampere's law

i) Ampere's law applied inside the wire

B₁ (2πr ) = μ₀I ( r² / R² )

ii) Ampere's law applied at the surface

B₂ ( 2πr ) = μ₀ I

Resolving equations above

Therefore : B₁ / B₂ = 0.84  also r / R = 0.84

Hence ( r ) = 0.84 R

Therefore we can conclude that The point  ( r ) within the wire where the field strength equals 84% of the field strength at the wire surface is : 0.84 R

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Answer:

see attachment

Explanation:

The electric potential varies inversely with the distance. So, the third charge should be placed at a distance of 3 cm from the origin on the x-axis.

The work done on an electric charge to shift it from infinity to a point is known as electric potential at that point. And its expression is,

[tex]V = \dfrac{kq}{r}[/tex]

here, k is the coulomb's constant.

Given data:

The magnitude of two point charges are, [tex]+2.0 \;\rm \mu C[/tex]  and  [tex]-6.0 \;\rm \mu C[/tex].

The location of each charge on the x-axis is -1.0 cm and +2.0 cm.

Let the third charge ( [tex]+3.0 \;\rm \mu C[/tex] ) be placed at a distance of x. Then the electric potential at origin is,

[tex]V = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}[/tex]

Since, potential at origin is zero (V = 0). Then,

[tex]0 = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}\\\\\dfrac{k \times 6.0}{0.02} = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times 3.0}{x}\\\\\dfrac{6.0}{0.02} = \dfrac{2.0}{0.01} +\dfrac{3.0}{x}\\\\x = 0.03 \;\rm m =3 \;\rm cm[/tex]

Thus, we can conclude that the third charge should be placed at a distance of 3 cm from the origin on the x-axis.

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The angular acceleration of the barrel ride is 0.0714 rad/s^2.

To find the angular acceleration, we first need to convert the rotational speed from rev/s to radians per second (rad/s) because the standard unit for angular speed and acceleration is in rad/s and rad/s² respectively. We know that 1 revolution is equal to 2π radians, therefore 0.520 rev/s equals 0.520 x 2π rad/s.

Angular acceleration (α) is calculated using the formula α = Δω / Δt, where Δω is the change in angular velocity and Δt is the time it takes for the change. As the ride started from rest, the change in angular velocity was simply its final angular velocity. Thus, by putting the values in the formula, we will get the angular acceleration during the 7.26 s.

The angular acceleration of the barrel ride at the amusement park can be calculated using the formula:

α = (Δω) / t

Plugging in the values from the question, we have:

Δω = 0.520 rev/sec

t = 7.26 s

Therefore, the angular acceleration during that time is α = (0.520 rev/sec) / (7.26 s) = 0.0714 rad/s^2.

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The angular acceleration of the barrel ride at the amusement park, which sped up from rest to 0.520 rev/sec in 7.26 s, is 0.45 rad/s².

To answer this question the first step we should do is to convert the given rate of 0.520 revolutions per second to radians per second, as our target unit is rad/s². We know that 1 revolution is equal to 2π radians, so multiplying the revolution rate by 2π, we get: 0.520 rev/sec * 2π rad/rev = 3.27 rad/sec.

Next, we need to calculate the angular acceleration using the formula α = (ωf- ωi) / t. Where ωi is the initial angular velocity, ωf is the final angular velocity, and t is the time. The ride starts from rest, so the initial angular velocity is 0, the final angular velocity is 3.27 rad/sec and the time is 7.26 seconds.

Substituting these values into the equation, we get: α = (3.27 rad/sec - 0 rad/sec) / 7.26 s = 0.45 rad/s².

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1) 5 s

The vertical position of the object is given by

[tex]y(t) = h - \frac{1}{2}gt^2 =  400 - 16 t^2[/tex]

where

h=400 ft represents the initial height

g = 32 ft/s^2 is the acceleration of gravity

t is the time

We want to find the time t at which the object reaches the ground, so the time t at which

y(t) = 0

By substituting this into the equation, we find

[tex]0 = 400 - 16t^2\\t=\sqrt{\frac{400}{16}}=5 s[/tex]

2) 160 ft/s

The object is released from rest, so the initial velocity is zero

u = 0

The final vertical velocity can be found by using

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

where

v is the final velocity

a = 32 ft/s^2 is the acceleration of gravity

h = 400 ft is the vertical distance covered

Solving for v, we find

[tex]v=\sqrt{u^2 +2ay}=\sqrt{2(32 ft/s)(400 ft)}=160 ft/s[/tex]

Final answer:

The object takes approximately 5 seconds to hit the ground and has an impact velocity of 160 feet per second.

Explanation:

Time to Hit the Ground and Velocity at Impact

To find out how long it takes the object to hit the ground, we need to solve the equation s(x) = 400 - 16x2 for the moment when the height s(x) is zero (s(x) = 0). Setting the equation to zero, we get 0 = 400 - 16x2, which simplifies to x2 = 25 after dividing both sides by 16. Taking the square root of both sides gives us x ≈ 5 seconds, which is the time the object takes to hit the ground.

For the object's velocity at the moment of impact, we use the formula v = gt, where g is the acceleration due to gravity (32 feet per second squared), and t is the time in seconds. Thus, the velocity at impact is v = 32 * 5 = 160 feet per second.

Explanation:

It is given that,

Initial velocity of the bird, u = 13 m/s

Final speed of the bird, v = 10.5 m/s

Time taken, t = 4.20 s

(a) Acceleration of the bird is given by :

[tex]a=\dfrac{v-u}{t}[/tex]

[tex]a=\dfrac{10.5\ m/s-13\ m/s}{4.20\ s}[/tex]

[tex]a=-0.59\ m/s^2[/tex]

So, The direction of acceleration is opposite to the direction of motion.

(b) We need to find the bird’s velocity after an additional 1.50 s has elapsed i.e. t = 4.2 + 1.5 = 5.7 s. Let v' is the new final velocity.

It can be calculated using first equation of motion as :

[tex]a=\dfrac{v'-u}{t}[/tex]

v' = u + at

[tex]v'=(-0.59)\times 5.7+13[/tex]

v' = 9.64 m/s

Hence, this is the required solution.

To calculate the deflection at point C of the beam, we can use the deflection equation for beams under a uniformly distributed load and a point load.

To calculate the deflection at point C of the beam, we will use the formula for deflection under a uniformly distributed load and a point load. The deflection equation for beams is given by

δ = (5wL⁴ - PL³) / (384EI)

where δ is the deflection, w is the uniformly distributed load, L is the span length, P is the point load, E is the modulus of elasticity, and I is the moment of inertia of the beam. Substituting the given values into the equation, we can calculate the deflection at point C.

Answer:

a scientist examines the results and answers the lab question- last choice

Answer:

D. a scientist examines the results and answers the lab question

Answer:

(a) 5.725 Ω

(b) 1.3 Ω

Explanation:

(a)

E = emf of the battery = 15.0 Volts

V = terminal voltage of the battery = 12.2 Volts

P = Power delivered to external load resistor "R"  = 26.0 W

R = resistance of external load resistor

Power delivered to external load resistor is given as

[tex]P = \frac{V^{2}}{R}[/tex]

26.0 = 12.2²/R

R = 5.725 Ω

(b)

r = internal resistance of the battery

i = current coming from the battery

Power delivered to external load resistor is given as

P = i V

26.0 = i (12.2)

i = 2.13 A

Terminal voltage is given as

V = E - ir

12.2 = 15 - (2.13) r

r = 1.3 Ω

Final answer:

The external load resistor R is calculated to be 5.72 Ohms using the power and terminal voltage, and the internal resistance of the battery is 1.31 Ohms as determined from the emf, terminal voltage, and the current flowing through R.

Explanation:

To solve for external load resistor R and the internal resistance of the battery, we can use the formulas related to electric power and the relation between emf, terminal voltage, and internal resistance.

(a) Value of R:

The power delivered to the resistor (P) is given by:

P = V2 / R

Where V is the terminal voltage and R is the resistance.

Substituting the given values:

26.0 W = (12.2 V)2 / R

Hence, R = (12.2 V)2 / 26.0 W = 5.72 Ω (Ohms)

(b) Internal resistance of the battery:

We know that terminal voltage V is emf - (current * internal resistance).

The current I flowing through R can be calculated using the power:

I = P / V = 26.0 W / 12.2 V = 2.13 A

Now, using the emf (E) of the battery and terminal voltage (V):

E = V + Ir

15.0 V = 12.2 V + (2.13 A × r)

We solve for r, the internal resistance:

r = (15.0 V - 12.2 V) / 2.13 A = 1.31 Ω (Ohms)

Answer:

(a) 2.512 m

(b) 144 m

(c) 904.32 m

Explanation:

radius, r = 4.8 m

(a) for 30 degree

As we know that in 360 degree it rotates a complete round that means circumference.

In 360 degree, it rotates = 2 x π x r

in 30 degree, it rotates = 2 x π x r x 30 / 360

                                      = 2 x 3.14 x 4.8 x 30 / 360

                                      = 2.512 m

(b) for 30 rad

As we know that in one complete rotation, it rotates by 2π radian.

so,

for 2π radian it rotates = 2 x π x r

for 30 radian, it rotates = 2 x π x r x 30 / 2 π = 144 m

(c) For 30 rev

In one complete revolution, it travels = 2 x π x r

in 30 rev, it travels = 2 x π x r x 30 = 2 x 3.14 x 4.8 x 30 = 904.32 m

[tex]x(t)=7.00\,\mathrm m-\left(9.00\dfrac{\rm m}{\rm s}\right)t+\left(3\dfrac{\rm m}{\mathrm s^2}\right)t^2[/tex]

a. The particle has velocity at time [tex]t[/tex],

[tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=-9.00\dfrac{\rm m}{\rm s}+\left(6\dfrac{\rm m}{\mathrm s^2}\right)t[/tex]

so that after [tex]t=1\,\mathrm s[/tex] it will have velocity [tex]\boxed{-3.00\dfrac{\rm m}{\rm s}}[/tex].

b. The sign of the velocity is negative, so it's moving in the negative [tex]x[/tex] direction.

c. Its speed is 3.00 m/s.

d. The particle's velocity changes according to

[tex]\dfrac{\mathrm d^2x(t)}{\mathrm dt^2}=6\dfrac{\rm m}{\mathrm s^2}[/tex]

which is positive and indicates the velocity/speed of the particle is increasing.

e. Yes. The velocity is increasing at a constant rate. Solving for [tex]\dfrac{\mathrm dx(t)}{\mathrm dt}=0[/tex] is trivial; this happens when [tex]\boxed{t=1.50\,\mathrm s}[/tex].

f. No, the velocity is positive for all [tex]t[/tex] beyond 1.50 s.

Answer:

1.34 seconds

Explanation:

h(t) = 57t - 16t²

Velocity is the derivative of position with respect to time:

v(t) = dh/dt

v(t) = 57 - 32t

When v = 14.25:

14.25 = 57 - 32t

32t = 42.75

t = 1.34

Answer:

[tex]9.96\cdot 10^{-10}J[/tex]

Explanation:

The capacitance of the parallel-plate capacitor is given by

[tex]C=\epsilon_0 k \frac{A}{d}[/tex]

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

[tex]A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2[/tex] is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

[tex]C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F[/tex]

Now we can calculate the energy of the capacitor, given by:

[tex]U=\frac{1}{2}CV^2[/tex]

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

[tex]U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J[/tex]

Answer:

Magnetic field, [tex]B=2.55\times 10^{-14}\ T[/tex]

Explanation:

It is given that,

Speed of proton, [tex]v=4\times 10^6\ m/s[/tex]

Mass of the proton, [tex]m=1.67\times 10^{-27}\ kg[/tex]

Charge on proton, [tex]q=1.6\times 10^{-19}\ C[/tex]

We need to find the magnetic field strength required to just balance the weight of the proton and keep it moving horizontally.

The Lorentz force is given by :

[tex]F=q(v\times B)=qvB\ sin90[/tex].............(1)

The weight of proton,

[tex]W=mg[/tex]..............(2)

From equation (1) and (2), we get :

[tex]mg=qvB[/tex]

[tex]B=\dfrac{mg}{qv}[/tex]

[tex]B=\dfrac{1.67\times 10^{-27}\ kg\times 9.8\ m/s^2}{1.6\times 10^{-19}\ C\times 4\times 10^6\ m/s}[/tex]

[tex]B=2.55\times 10^{-14}\ T[/tex]

Hence, this is the required solution.

Final answer:

The magnetic field strength required to balance the weight of a proton and keep it moving horizontally is 3.07 x 10^-4 Tesla, calculated by setting the magnetic force (qvB) equal to the gravitational force (mg) and solving for B with given values for q, v, m, and g.

Explanation:

To calculate the magnetic field strength required to balance the weight of a proton and keep it moving horizontally, we can use the relationship between the magnetic force and the gravitational force acting on the proton. The magnetic force that will balance the weight of the proton is given by FB = qvB sin(θ), where q is the charge of the proton, v is its velocity, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. Because the proton is moving at a right angle to the magnetic field, sin(θ) will be 1.

The weight of the proton (gravitational force) is given by W = mg, where m is the mass of the proton and g is the acceleration due to gravity. Setting the magnetic force equal to the weight gives us:

FB = W
qvB = mg
B = mg/qv

Substituting the given values for mass m = 1.67 x 10-27 kg, charge q = 1.60 x 10-19 C, speed v = 4.00 x 106 m/s, and the acceleration due to gravity g = 9.81 m/s2, we can find the magnetic field strength:

B = (1.67 x 10-27 kg * 9.81 m/s2) / (1.60 x 10-19 C * 4.00 x 106 m/s)
B = 3.07 x 10-4 T

The magnetic field strength required is 3.07 x 10-4 Tesla.