If 10 J of work (electric potential energy) is used in pushing 2 C of charge into an electric field, its electric potential relative to its starting position is a. less than 5 V c. more than 20 V. d. 20 V

Answer:

0.56

Explanation:

Let the coefficient of friction is μ.

m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s

By the free body diagram,

Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N

Friction force, f = μ N = 36.49 μ

Net force acting on the block,

Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ

Fnet = 21.07 - 36.49μ

Net acceleartion, a = Fnet / m

a = (21.07 - 36.49μ) / 4.3

use second equation of motion

s = ut + 1/2 a t^2

2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3

By solving we get

μ = 0.56

Answer:

13.18 m/s

Explanation:

Let the velocity of sports utility car is

-u as it is moving in opposite direction.

mc = 1200 kg, uc = 31.1 m/s

ms = 2830 kg, us = - u = ?

Using conservation of momentum

mc × uc + ms × us = 0

1200 × 31.1 - 2830 × u = 0

u = 13.18 m/s

a. The particle has position vector

[tex]\vec r(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)t\right)\,\vec\imath+\left(\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\right)\,\vec\jmath[/tex]

[tex]\vec r(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\vec\jmath[/tex]

b. Its velocity vector is equal to the derivative of its position vector:

[tex]\vec v(t)=\vec r'(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\vec\jmath[/tex]

c. At [tex]t=7.00\,\mathrm s[/tex], the particle has position

[tex]\vec r(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(7.00\,\mathrm s)\,\vec\imath+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)^2\,\vec\jmath[/tex]

[tex]\vec r(7.00\,\mathrm s)=\left(56.0\,\vec\imath+49.0\,\vec\jmath\right)\,\mathrm m[/tex]

That is, it's 56.0 m to the right and 49.0 m up relative to the origin, a total distance of [tex]\|\vec r(7.00\,\mathrm s)\|=\sqrt{(56.0\,\mathrm m)^2+(49.0\,\mathrm m)^2}=74.4\,\mathrm m[/tex] away from the origin in a direction of [tex]\theta=\tan^{-1}\dfrac{49.0\,\mathrm m}{56.0\,\mathrm m}=41.2^\circ[/tex] relative to the positive [tex]x[/tex] axis.

d. The speed of the particle at [tex]t=7.00\,\mathrm s[/tex] is the magnitude of the velocity at this time:

[tex]\vec v(7.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\vec\imath+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(7.00\,\mathrm s)\,\vec\jmath[/tex]

[tex]\vec v(7.00\,\mathrm s)=\left(8.00\,\vec\imath+14.0\,\vec\jmath\right)\dfrac{\rm m}{\rm s}[/tex]

Then its speed at this time is

[tex]\|\vec v(7.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+(14.0\dfrac{\rm m}{\rm s}\right)^2}=16.1\dfrac{\rm m}{\rm s}[/tex]

(a) The vector position of the particle at any time t; is s(t) = (8i)t + (j)t²

(b) the velocity of the particle at any time is v = 8i + (2j)t

(c) the position of the particle at t = 7.00 s is 56i + 49 j at a distance of 74.4 m

(d) the speed of the particle at t = 7.00 is 16.12 m/s

The given parameters;

acceleration of the particle, a = 2 j m/s²

initial velocity of the particle, v = 8 i m/s

(a) The vector position of the particle at any time t;

[tex]s = v_0t + \frac{1}{2} at^2\\\\s(t) = (8 \ i) t \ + (0.5 \times 2 \ j)t^2\\\\s(t) = (8 \ i)t \ + (j)t^2[/tex]

(b) the velocity of the particle at any time

[tex]velocity = \frac{\Delta \ displacement }{\Delta \ time} \\\\s' = v = 8i \ + (2j)t[/tex]

(c) the position of the particle at t = 7.00 s

[tex]s(t) = (8 \ i)t \ + (j)t^2\\\\s(7) = (8 \ i)\times 7 \ \ + \ \ (j)\times 7^2\\\\s(7) = 56i \ \ + \ \ 49j\\\\|s| = \sqrt{(56)^2 + (49)^2} = 74.4 \ m\\\\|s| = 74.4 \ m[/tex]

(d) the speed of the particle at t = 7.00

[tex]v(t) = 8i + (2j)t\\\\v(7) = 8i \ + (2j \times 7)\\\\v(7) = 8i + 14j\\\\|v| = \sqrt{8^2 + 14^2} \\\\|v| = \sqrt{260} \\\\|v| = 16.12 \ m/s[/tex]

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Answer:

  162 km

Explanation:

A diagram can be helpful.

Using the law of cosines, we can find the magnitude of the distance (c) to satisfy ...

  c^2 = a^2 +b^2 -2ab·cos(C)

where C is the internal angle of the triangle of vectors and resultant. Its value is ...

  180° -39.8° -59.9° = 80.3°

Filling in a=76 and b=156, we get ...

  c^2 = 76^2 +156^2 -2·76·156·cos(80.3°) ≈ 26116.78

  c ≈ √26116.78 ≈ 161.607

The magnitude of the total displacement is about 162 km.

_____

Please note that in the attached diagram North is to the right and East is up. That alteration of directions does not change the angles or the magnitude of the result.

Answer:

Magnitude of total displacement = 162.87 km

Explanation:

Let east be x axis and north be y axis.

The first is 76 km at 39.8◦ west of north.

Displacement 1 = 76 km at 39.8◦ west of north =  76 km at 129.8◦ north of east.

Displacement 1 = 76 cos129.8 i + 76 sin 129.8 j = -48.65 i + 58.39 j

The second is 156 km at 59.9◦ east of north.

Displacement 2 = 156 km at 59.9◦ east of north =  156 km at 31.1◦ north of east.

Displacement 2 = 156 cos31.1 i + 156 sin 31.1 j = 133.58 i + 80.58 j

Total displacement = Displacement 1 + Displacement 2

Total displacement = -48.65 i + 58.39 j + 133.58 i + 80.58 j = 84.93 i + 138.97 j

[tex]\texttt{Magnitude of total displacement =}\sqrt{84.93^2+138.97^2}=162.87km[/tex]

Magnitude of total displacement = 162.87 km

Answer:

The cost of energy is 0.29$.

Explanation:

Given that,

Voltage V= 12 V

Charge Q= 160 A-h

We need to calculate the energy

Using formula of energy

[tex]E = Pt[/tex]

[tex]E =VIt[/tex]

Where, E = energy

I= current

V = Potential

t = time

Put the value into the formula

[tex]E = 12\times160\times3600[/tex]

[tex]E =6.9\times10^{6} J[/tex]

The cost of this energy at 0.15/kwh

[tex]Total\ cost = Energy\times 0.15 kwh[/tex]

Here, Energy = power x time

In units,

[tex]J=watt\times s[/tex]

But ,  Energy in 1 hour is:

[tex]E = 1920\ J[/tex]

So, The cost of this energy is

[tex] Total\ cost =1920\dfrac{\times0.15}{1000}[/tex]

[tex]total\ cost = 0.29\$[/tex]

Hence, The cost of energy is 0.29$.

For a battery rated at 12V and 160 A-h, the total energy stored is about 6912000 Joules or 1.92 kWh. The cost of this energy, at a rate of $0.15 per kWh, would therefore be $0.288.

To find out the amount of energy stored in the battery, we first understand that the battery's rated capacity is given as Ampere-hours (A-h). It is a measure of the electric charge that a battery can deliver over time. We can find the energy stored in this battery by multiplying the charge by the battery's voltage. With a rating of 12 V and 160 A-h, the total energy would be calculated as Energy (E) = Voltage (V) x Charge (Q), where Q (the charge) is 160 A-h, which must be converted to seconds by multiplying by 3600 (the number of seconds in one hour). So, Q = 160 A-hr x 3600 = 576000 Coulombs. Once we substitute these values into the formula, the total energy stored would be E = 12 V * 576000 C = 6912000 J.

Now, to find the cost of this energy, we need to convert this energy (which is in Joules) to kWh by dividing by 3.6 million (as 1 kWh = 3.6 million Joules), leading to E = 6912000 J / 3600000 = 1.92 kWh. The cost of this energy would then be calculated by multiplying the energy by the cost per unit of energy, here given as $0.15 per kWh. Therefore, Cost = 1.92 kWh * $0.15/kWh = $0.288.

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Answer:

it would be like a perfect triangle but the sides instead of being straight lines they would all be curved

Explanation:

A line with a constant bendiness ... if continued far enough ... forms a circle.

Answer:

Option (D)

Explanation:

n = 0.35,

Temperature of hot reservoir,

T1 = 427 C = 700 K

Temperature of cold reservoir, T2 = ?

Use the formula for efficiency of Carnot cycle

n = 1 - T2 / T1

0.35 = 1 - T2 / 700

0.65 = T2 / 700

T2 = 700 × 0.65

T2 = 455 K

Answer:

28.0 cm

Explanation:

Displacement is final position minus initial position.

Δx = xf − xi

Δx = -64.9 cm − -92.9 cm

Δx = 28.0 cm

The magnitude of displacement is 28 cm.

We have a snail moving on a railway track.

We have to determine snail's displacement.

The length of the straight line joining the initial and final position of an object is called its displacement.

According to the question -

Initial Position x[i] = -92.9 cm

Final Position x[f] = -64.9 cm

Since, the motion is one dimensional, the net displacement can be calculated by the formula -

|Δx| = x[f] - x[i] = - 64.9 - (- 92.9) = - 64.9 + 92.9 = |- 28| = 28 cm

Hence, the magnitude of displacement is 28 cm.

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(a) 24.6 Nm

The torque produced by the net thrust about the center of the circle is given by:

[tex]\tau = F r[/tex]

where

F is the magnitude of the thrust

r is the radius of the wire

Here we have

F = 0.795 N

r = 30.9 m

Therefore, the torque produced is

[tex]\tau = (0.795 N)(30.9 m)=24.6 N m[/tex]

(b) [tex]0.035 rad/s^2[/tex]

The equivalent of Newton's second law for a rotational motion is

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

where

[tex]\tau[/tex] is the torque

I is the moment of inertia

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

If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

[tex]I=mr^2 = (0.741 kg)(30.9 m)^2=707.5 kg m^2[/tex]

And so we can solve the previous equation to find the angular acceleration:

[tex]\alpha = \frac{\tau}{I}=\frac{24.6 Nm}{707.5 kg m^2}=0.035 rad/s^2[/tex]

(c) [tex]1.08 m/s^2[/tex]

The linear acceleration (tangential acceleration) in a rotational motion is given by

[tex]a=\alpha r[/tex]

where in this problem we have

[tex]\alpha = 0.035 rad/s^2[/tex] is the angular acceleration

r = 30.9 m is the radius

Substituting the values, we find

[tex]a=(0.035 rad/s^2)(30.9 m)=1.08 m/s^2[/tex]

Fermi level determines charge carrier population according to the integral over density of states times Fermi Dirac distribution. Therefore, saying that the position of Fermi level is higher (lower), relative to middle of band gap, is just the mathematical description of saying you have more electrons (holes), relative to the intrinsic state, in the system. Dopants in the system increase the concentration of electrons or holes.

Your weight on a neutron star can be calculated based on your given weight on Earth using the Universal Law of Gravitation equation. This allows us to determine the gravitational force or weight for any celestial body given we know its mass and radius.

The topic at hand is fundamentally related to Physics since it pertains to the gravitational effects of a massive body. From the question, we are given your weight on Earth which is 665 N. The weight on Earth can be related to the weight of the neutron star by the equation of universal gravitation:

W = mg = GMm/r²

Where:

Given that the diameter of the neutron star is 18km, its radius becomes 9km = 9000m. Substituting the given values into the formula yields a theoretical weight on a neutron star.

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We find this acceleration to be approximately 4.09×[tex]10^{12} m/s^2[/tex]. Thus, your weight would be around 2.77×[tex]10^{14}[/tex] N on the neutron star.

To calculate your weight on the surface of a neutron star, you need to find the acceleration due to gravity on that neutron star using Newton's Universal Law of Gravitation.

The formula for gravitational force is:

F = G x (m1 x m2) /[tex]R^2[/tex]

Where G is the gravitational constant, m1 and m2 are the masses, and R is the radius of the object.

Given:

Mass of the neutron star, M = 1.99×[tex]10^{30}[/tex] kg

Radius, R = 18.0 km = 18,000 meters

Gravitational constant, G = 6.67×[tex]10^{-11} N⋅m^2/kg^2[/tex]The acceleration due to gravity on the neutron star, g_s, can be calculated as:

g_s = G x (M / [tex]R^2[/tex])

Substituting in the given values:

g_s = (6.67×[tex]10^{-11}[/tex] N⋅[tex]m^2/kg^2[/tex] x 1.99×[tex]10^{30}[/tex] kg) / [tex](18,000 m)^2[/tex]

g_s ≈ 4.09×[tex]10^{12}[/tex] m/[tex]s^2[/tex]

Your weight on Earth is given as 665 N. Since weight (W) is mass (m) times the acceleration due to gravity (g):

m = W / g = 665 N / 9.810 m/[tex]s^2[/tex] ≈ 67.8 kg

So, your weight on the neutron star would be:

Weight on neutron star = mass x g_s = 67.8 kg x 4.09×[tex]10^{12} m/s^2[/tex] ≈ 2.77×[tex]10^{14}[/tex] N

The change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC is approximately 0.21 meters.

To calculate the change in length of a column of water for a temperature increase of 1.00ºC, we need to use the coefficient of thermal expansion for water. The coefficient of thermal expansion for water is approximately 0.00021 per degree Celsius.

To find the change in length, we multiply the height of the column (1.00 km) by the coefficient of thermal expansion (0.00021) and the temperature increase (1.00ºC).

Change in length = 1.00 km × 0.00021 × 1.00ºC = 0.00021 km = 0.21 m

Therefore, the change in length of a column of water 1.00 km high for a temperature increase of 1.00ºC is approximately 0.21 meters.

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Final answer:

The speed of the meteorite just before striking the sand is 0 m/s.

Explanation:

To find the speed of the meteorite just before striking the sand, we can use the principle of conservation of energy. Let's consider the initial energy when the meteorite is 850 km above the Earth's surface and the final energy when it comes to rest in the sand. Initially, the meteorite has gravitational potential energy and no kinetic energy. Finally, when it comes to rest in the sand, it has no potential energy and only kinetic energy.

Using the equation for gravitational potential energy, PE = mgh, where m is the mass of the meteorite, g is the acceleration due to gravity (9.8 m/s²), and h is the height above the Earth's surface, we can calculate the initial potential energy. PE = (2.0 × 10¹³ kg)(9.8 m/s²)(850,000 m) = 1.61 × 10¹⁹ J.

The final kinetic energy of the meteorite just before striking the sand can be calculated using the equation KE = (1/2)mv², where m is the mass of the meteorite and v is its velocity. We know that the meteorite comes to rest in a distance of 3.25 m, so its final velocity is 0. Using this information, we can solve for the final kinetic energy. 0 = (1/2)(2.0 × 10¹³ kg)v². By rearranging the equation, we find v = 0 m/s.

Therefore, the speed of the meteorite just before striking the sand is 0 m/s.

Answer:

A) 3.11 Hrs

Explanation:

Wave speed is given by the formula

[tex]Speed = \frac{wavelength}{Time\: period}[/tex]

now we will have

[tex]speed = \frac{40}{0.80}[/tex]

[tex]v = 50 m/s[/tex]

now the time taken by the wave to move the distance 560 km is given as

[tex]t = \frac{distance}{speed}[/tex]

[tex]t = \frac{560\times 10^3}{50}[/tex]

[tex]t = 11200 s[/tex]

t = 3.11 hours

Answer: [tex]1.079(10)^{-6}Earth-years[/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 asteroid in this case) orbiting a greater body in space (the Sun 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]G[/tex] is the Gravitational Constant and its value is [tex]6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex]

[tex]M=1.989(10)^{30}kg[/tex] is the mass of the Sun

[tex]a=2.47(Earth-radius)=2.47(6371000m)=15736370m[/tex]  is orbital radius of the orbit the asteroid describes around the Sun.

Now, if we want to find the period, we have to express equation (1) as written below and substitute all the values:

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

[tex]T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(1.989(10)^{30}kg)}(15736370m)^{3}}[/tex]   (3)

[tex]T=34.0428s[/tex]   (4) This is the period in seconds, but we were asked to find it in Earth years. So, we have to make the conversion:

[tex]T=34.0428s.\frac{1Earth-hour}{3600s}.\frac{1Earth-day}{24Earth-hour}.\frac{1Earth-year}{365Earth-day}[/tex]  

Finally:

[tex]T=1.079(10)^{-6}Earth-years[/tex]  This is the period of the asteroid around the Sun in Earth years.

The asteroid's orbital period is about 3.88 years

[tex]\texttt{ }[/tex]

Centripetal Acceleration can be formulated as follows:

[tex]\large {\boxed {a = \frac{ v^2 } { R } }[/tex]

a = Centripetal Acceleration ( m/s² )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

[tex]\texttt{ }[/tex]

Centripetal Force can be formulated as follows:

[tex]\large {\boxed {F = m \frac{ v^2 } { R } }[/tex]

F = Centripetal Force ( m/s² )

m = mass of Particle ( kg )

v = Tangential Speed of Particle ( m/s )

R = Radius of Circular Motion ( m )

Let us now tackle the problem !

[tex]\texttt{ }[/tex]

Given:

Orbital Radius of Earth = R

Orbital Radius of Asteroid = R' = 2.47 R

Orbital Period of Earth = T = 1 year

Unknown:

Orbital Period of Asteroid = T' = ?

Solution:

Firstly , we will use this following formula to find the orbital period:

[tex]F = ma[/tex]

[tex]G \frac{ Mm}{R^2}=m \omega^2 R[/tex]

[tex]G M = \omega^2 R^3[/tex]

[tex]\frac{GM}{R^3} = \omega^2[/tex]

[tex]\omega = \sqrt{ \frac{GM}{R^3}}[/tex]

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

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

[tex]\texttt{ }[/tex]

Next , we could find the asteroid's orbital period by using above formula:

[tex]T' : T = 2\pi\sqrt{ \frac{(R')^3}{GM}} : 2\pi\sqrt{ \frac{R^3}{GM}}[/tex]

[tex]T' : T = \sqrt{(R')^3} : \sqrt{R^3}[/tex]

[tex]T' : 1 = \sqrt{(2.47R)^3} : \sqrt{R^3}[/tex]

[tex]T' : 1 = \sqrt{2.47^3}\sqrt{R^3} : \sqrt{R^3}[/tex]

[tex]T' : 1 = \sqrt{2.47^3} : 1[/tex]

[tex]T' = \sqrt{2.47^3}[/tex]

[tex]T' \approx 3.88 \texttt{ years}[/tex]

[tex]\texttt{ }[/tex]

[tex]\texttt{ }[/tex]

Grade: High School

Subject: Physics

Chapter: Circular Motion

[tex]\texttt{ }[/tex]

Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

According to Coulomb's Law, the magnitude of the electric force between two charged particles is inversely proportional to the square of the distance between them. Therefore, if the distance between the two particles is halved, the electric force becomes four times as strong, i.e., it quadruples.

The phenomenon you're asking about is governed by Coulomb's Law. This law states that the magnitude of the electric (or Coulomb) force between two charged particles is directly proportional to the product of the magnitudes of their charges, and inversely proportional to the square of the distance between them.

So, if the distance between these two charges is halved, the electric force between them would be affected by the square of that change in distance. Therefore, instead of being half as strong, the electric force becomes four times as strong, i.e., it quadruples. Hence, the correct answer to your question is: it quadruples (option 5). The inverse square law, which this is an example of, is a fundamental principle in physics that applies to other forces too, like gravity.

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Answer:

32 cm

Explanation:

f = focal length of the converging lens = 16 cm

Since the lens produce the image with same size as object, magnification is given as

m = magnification = - 1

p = distance of the object from the lens

q = distance of the image from the lens

magnification is given as

m = - q/p

- 1 = - q/p

q = p                                    eq-1

Using the lens equation, we get

1/p + 1/q = 1/f

using eq-1

1/p + 1/p = 1/16

p = 32 cm

To create a real image with the same size as the object using a converging lens, the object should be placed at a distance equal to twice the focal length of the lens.

Given the focal length (f) of the lens is 16 cm, the object should be placed at a distance of 2 * 16 cm = 32 cm from the lens.

1. The lens formula is given by:

   1/f = 1/v - 1/u

  Where:

  - f is the focal length of the lens.

  - v is the image distance.

  - u is the object distance.

2. In this case, we want a real image of the same size as the object, which means that the magnification (M) is equal to 1. Magnification is given by:

  M = -v/u

  Since M = 1, we have v = -u.

3. For a real image formed by a converging lens, v is negative. So, we have v = -u = -2f.

4. Substituting the focal length (f = 16 cm) into the equation, we find that u = -2 * 16 cm = -32 cm. The negative sign indicates that the object is placed on the same side as the object. Therefore, the object should be placed 32 cm from the lens to produce the desired image.

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Answer:

0.953c

Explanation:

T₀ = lifetime of the particle at rest = 2.52 μs

T = Observed lifetime of the particle while in motion = 8.32 μs

v = speed of particle

c = speed of light

Using the formula for time dilation

[tex]T = \frac{T_{o}}{\sqrt{1-(\frac{v}{c})^{2}}}[/tex]

inserting the values

[tex]8.32 = \frac{2.52}{\sqrt{1-(\frac{v}{c})^{2}}}[/tex]

v = 0.953c

Final answer:

To calculate the speed at which a particle's lifetime appears dilated to 8.32 μs, one uses the time dilation formula from special relativity. The calculated speed is approximately 0.995c, which does not match the provided choices. A careful review of the calculation and rounding effects might lead to one of the options being the nearest correct answer.

Explanation:

To find the required speed for the particle's lifetime to be observed as 8.32 μs, we need to apply the concept of time dilation from Einstein's theory of special relativity. Time dilation is given by the equation τ' = τ / √(1 - v²/c²), where τ' is the dilated lifetime, τ is the proper lifetime (lifetime at rest), v is the velocity of the particle, and c is the speed of light.

The problem gives us τ' = 8.32 μs and τ = 2.52 μs. We substitute these values into the equation and solve for v:

8.32 μs = 2.52 μs / √(1 - v²/(3.00 × 10⁸ m/s)²)

We first square both sides, rearrange the equation, and then take the square root to find v:

(8.32 μs / 2.52 μs)² = 1 / (1 - v²/c²)

(3.30)² = 1 / (1 - v²/c²)

10.89 = 1 + v²/c²

v²/c² = 10.89 - 1

v²/c² = 9.89

v = √(9.89) ∑ c

v ≈ 0.995c

Thus the required speed of the particle for its lifetime to be observed as 8.32 μs is approximately 0.995c. However, this speed is not listed among the multiple-choice options, suggesting either a calculation error or that none of the provided answers are correct. It's important to review the calculation carefully and consider rounding effects that might lead to one of the provided answers being the nearest correct choice.

Answer:

The speed of the man is 4.54 m/s.

Explanation:

Given that,

Mass of man=8100 g

Mass of stone = 79 g

Speed  = 4.5 m/s

We need to calculate the speed of the man

Using momentum of conservation

[tex](m_{1}+m_{2})V=m_{1}v_{1}+m_{2}v_{2}[/tex]

Where,

[tex]m_{1}[/tex]=mass of man

[tex]m_{2}[/tex]=mass of stone

[tex]v_{1}[/tex]=velocity of man

[tex]v_{2}[/tex]=velocity of stone

Put the value in the equation

As the stone is away from the man

So, the speed of stone is zero

[tex]8179\times4.5=8100\times v+0[/tex]

[tex]v=\dfrac{8179\times4.5}{8100}[/tex]

[tex]v = 4.54\ m/s[/tex]

Hence, The speed of the man is 4.54 m/s.

Answer:

C) the reactants have more potential energy than the products.

Explanation:

Since there is a difference between the potential energy of the reactants and the products, that energy has to leave the formula and into its surroundings, meaning that there will be a liberation of energy, increasing the temperature of the surroundings and or in the closed system in which the reaction is taking place, that is an exhotermic reaction.

The reactants have more potential energy than the products due to which it releases heat energy.

There is a difference between the potential energy of the reactants and the products. That reactions which release thermal energy is also called exothermic reaction which only occurs when there is more potential energy in reactants than products.

So we can conclude that reactants have more potential energy than the products due to which it releases heat energy.

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Answer:

109.13 rad/s^2

Explanation:

m = 140 g = 1.4 kg, r = 0.537 m, f = 2.27 rps

The centripetal acceleration is given by

a = r ω^2

a = r x (2 π f)^2

a = 0.537 x ( 2 x 3.14 x 2.27)^2

a = 109.13 rad/s^2

Answer:

(A)chimney

Explanation:

bc all the smoke is going into the chimney

To find the magnitude and direction of the resultant displacement of a grasshopper's jumps, calculate the components using trigonometry and then determine the magnitude and angle of the resultant vector.

The magnitude of the resultant displacement: First, calculate the x and y components of each jump using trigonometry. Then sum up all the x components and y components separately to find the resultant components. Finally, use these components to calculate the magnitude.

The direction of the resultant displacement: Use inverse trigonometric functions to find the angle the resultant displacement makes with respect to due west.

Resultant displacement: (a) Magnitude ≈ 17.09 cm, (b) Direction ≈ 116.88° south of west.

To find the resultant displacement, we can first break down each displacement vector into its horizontal (x) and vertical (y) components. Then we'll sum up all the x-components and y-components separately to get the total horizontal and vertical displacement. Finally, we'll use these values to find the magnitude and direction of the resultant displacement.

Let's start by breaking down each displacement vector:

(1) 31.0 cm due west:

  - Horizontal component (x): -31.0 cm

  - Vertical component (y): 0 cm

(2) 32.0 cm, 34.0° south of west:

  - Horizontal component (x): 32.0 cm * cos(34.0°)

  - Vertical component (y): -32.0 cm * sin(34.0°)

(3) 16.0 cm, 57.0° south of east:

  - Horizontal component (x): 16.0 cm * cos(57.0°)

  - Vertical component (y): -16.0 cm * sin(57.0°)

(4) 16.0 cm, 75.0° north of east:

  - Horizontal component (x): 16.0 cm * cos(75.0°)

  - Vertical component (y): 16.0 cm * sin(75.0°)

Now let's calculate the components:

For (2):

- Horizontal component (x): 32.0 cm * cos(34.0°) ≈ 26.61 cm

- Vertical component (y): -32.0 cm * sin(34.0°) ≈ -17.09 cm

For (3):

- Horizontal component (x): 16.0 cm * cos(57.0°) ≈ 7.96 cm

- Vertical component (y): -16.0 cm * sin(57.0°) ≈ -13.57 cm

For (4):

- Horizontal component (x): 16.0 cm * cos(75.0°) ≈ 4.14 cm

- Vertical component (y): 16.0 cm * sin(75.0°) ≈ 15.43 cm

Now, let's sum up all the x-components and y-components:

Total horizontal displacement (x):

= -31.0 cm + 26.61 cm + 7.96 cm + 4.14 cm

= 7.71 cm

Total vertical displacement (y):

= 0 cm - 17.09 cm - 13.57 cm + 15.43 cm

= -15.23 cm

Now, to find the magnitude (R) of the resultant displacement:

R = sqrt((Total horizontal displacement)^2 + (Total vertical displacement)^2)

 = sqrt((7.71 cm)^2 + (-15.23 cm)^2)

 ≈ sqrt(59.44 cm^2 + 232.52 cm^2)

 ≈ sqrt(291.96 cm^2)

 ≈ 17.09 cm

Now, to find the direction of the resultant displacement:

θ = atan(Total vertical displacement / Total horizontal displacement)

   = atan(-15.23 cm / 7.71 cm)

   ≈ atan(-1.975)

This gives us an angle of approximately -63.12° with respect to the horizontal. Since the angle is negative, we'll add 180° to it to find the angle with respect to due west:

θ = -63.12° + 180°

  ≈ 116.88°

So, the magnitude of the resultant displacement is approximately 17.09 cm, and its direction is approximately 116.88° south of west.

Answer:

The total momentum before and after collision is 72000 kg-m/s.

Explanation:

Given that,

Mass of car = 1200 kg

Velocity of car = 10 m/s

Mass of truck = 2000 kg

Velocity of truck = 30 m/s

Using conservation of momentum

The total momentum before the collision is equal to the total momentum after collision.

[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})V[/tex]

Where, [tex]m_{1}[/tex]=mass of car

[tex]v_{1}[/tex] =velocity of car

[tex]m_{1}[/tex]=mass of truck

[tex]v_{1}[/tex] =velocity of truck

Put the value into the formula

[tex]1200\times10+2000\times30=(1200+2000)V[/tex]

[tex]V=\dfrac{1200\times10+2000\times30}{(1200+2000)}[/tex]

[tex]V = 22.5\ m/s[/tex]

Now, The total momentum before collision is

[tex]P=m_{1}v_{1}+m_{2}v_{2}[/tex]

[tex]P=1200\times10+2000\times30[/tex]

[tex]P=72000\ kg-m/s[/tex]

The total momentum after collision is

[tex]P=(m_{1}+m_{2})v_{2}[/tex]

[tex]P=(1200+2000)\times22.5[/tex]

[tex]P= 72000 kg-m/s[/tex]

Hence, The total momentum before and after collision is 72000 kg-m/s.

Answer:c

Explanation:

(a) 0.5 kg m/s

Before the collision, only the first body is moving, so only the first body contributes to the total momentum.

The first body has

m = 400 g = 0.4 kg (mass)

v = 1.25 m/s (velocity, towards east direction)

So its momentum is

[tex]p_x=mv = (0.4 kg)(1.25 m/s)=0.5 kg m/s[/tex]

And since the body is moving along the east direction, this is also the easterly component of the total momentum before the collision.

(b) Zero

Before the collision, we have:

- The first body moving along the east direction --> so its northerly component is zero

- The second body at rest --> this means that it does not contribute to the momentum, since it is zero

This means that the northerly component of the total momentum before the collision is zero.

(c) 0.5 m/s at 53.1 degrees south of east

The law of conservation of momentum states that each component of the total momentum must be conserved.

- Along the easterly direction:

[tex]p_x = p_{1x} + p_{2x}[/tex]

where

[tex]p_x = 0.5 kg m/s[/tex] is the easterly component of the total momentum

[tex]p_{1x} = m v cos \theta = (0.4 kg)(1.00 m/s) cos 36.9^{\circ} =0.32 kg m/s[/tex] is the easterly component of the momentum of the first body after the collision

[tex]p_{2x}[/tex] is the easterly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision

Solving the equation we find

[tex]p_{2x} = p_x - p_{1x} = 0.5 kg m/s - 0.32 kg m/s = 0.18 kg m/s[/tex]

- Along the northerly direction:

[tex]p_y = p_{1y} + p_{2y}[/tex]

where

[tex]p_y = 0 kg m/s[/tex] is the northerly component of the total momentum

[tex]p_{1y} = m v sin \theta = (0.4 kg)(1.00 m/s) sin 36.9^{\circ} =0.24 kg m/s[/tex] is the northerly component of the momentum of the first body after the collision

[tex]p_{2y}[/tex] is the northerly component of the momentum of the second body (mass m = 600 g = 0.6 kg) after the collision

Solving the equation we find

[tex]p_{2y} = p_y - p_{1y} = 0 - 0.24 kg m/s = -0.24 kg m/s[/tex]

So now we find the momentum of the 600 g body after the collision:

[tex]p_2=\sqrt{p_{2x}^2 + p_{2y}^2}=\sqrt{(0.18)^2+(-0.24)^2}=0.3 kg m/s[/tex]

and so its final speed is

[tex]v=\frac{p_2}{m}=\frac{0.3 kg m/s}{0.6 kg}=0.5 m/s[/tex]

and the direction is

[tex]\theta=tan^{-1} (\frac{p_{2y}}{p_{2x}})=tan^{-1} (\frac{-0.24}{0.18})=-53.1^{\circ}[/tex]

so 53.1 degrees in the south-east direction.

Answer:

The answer is 30 N a foot.

Explanation:

The power in kW transmitted at 620 rpm is 40.88 kW.

A gear is component used to transmit motion. They are used in cars, trucks, machines, bicycles, etc.

A helical gear has 30 teeth and a pitch diameter of 280 mm. Gear module is 8 and pressure angle normal to tooth is 20°. The force normal to the tooth surface is 4500 N. Speed is 620 RPM.

The torque obtained on the gear will be

T = Force x Radius

T = 4500 x (0.280/2)

T = 630 N.m

Power Transmitted P = T ω = 2πNT/60

Plug the values, we get

P = 2π x 620 x 630 /60

P = 40882.8 W

P = 40.88 kW

Thus, the power in kW transmitted at 620 rpm is 40.88 kW.

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Answer:

Increase the radius of curvature of surface.

Explanation:

Radius of curvature is the measurement that is two times the focal length, for a given lens. It lies on either side of the lens.

Focal length is the distance which is half of the radius of curvature. Radius of curvature is a measure of the radius of the circle . Focal length is the distance between the center of curvature of the lens and the point where all the rays are brought to a focus for a distant object.

A perfect electric dipole doesn't feel a net force in a uniform electric field.

To determine whether this statement is true or false, we need to know about the behavior of perfect dipole in an electric field.

Thus, we can conclude that a perfect electric dipole doesn't feel a net force in a uniform electric field.

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Answer:

Distance, d = 61.13 ft

Explanation:

It is given that,

Initial speed of the car, u = 50 mi/h = 73.34 ft/s

Finally, it stops i.e. v = 0

Deceleration of the car, [tex]a=-44\ ft/s^2[/tex]

We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :

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

[tex]s=\dfrac{v^2-u^2}{2a}[/tex]

[tex]s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}[/tex]

s = 61.13 ft

So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.

Answer:

The distance covered by car before stopping is 61.13 ft.

Explanation:

Given data:

Initial Speed of car is, [tex]u=50 \;\rm mi/h = 50 \times 1.467 =73.35 \;\rm ft/s[/tex].

Deceleration of car is, [tex]a=-44\;\rm ft/s^{2}[/tex]. (Negative sign shows negative acceleration)

Applying the second kinematic equation of motion as,

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

Here, s is the distance covered and v is the final speed. Since, car stops finally, v = 0.

Solving as,

[tex]0^{2}=73.35^{2}+2(-44) \times s\\88s =73.35^{2}\\s \approx 61.1 \;\rm ft[/tex]

Thus, distance covered by car before stopping is 61.1 ft.

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