What is the magnitude of the acceleration vector which causes a particle to move from velocity −5i−2j m/s to −6i+ 7j m/s in 8 seconds. Answer in m/s.

Final answer:

Gauss' Law in differential form can be used to show that the electric field outside a long non-conducting hollow cylinder decreases as 1/r.

Explanation:

Gauss' Law in differential form can be used to show that the electric field outside a long non-conducting hollow cylinder decreases as 1/r. To do this, we need to consider a Gaussian surface in the form of a cylindrical shell with radius r and length L. The charge enclosed by this surface is the charge on a length L of the cylinder, which is given by Qenc = λL, where λ is the linear charge density. According to Gauss' Law in differential form, the electric field through the Gaussian surface is given by E⋅dA = λenc/ε₀, where E is the electric field, dA is the area vector of the Gaussian surface, λenc is the charge enclosed by the surface, and ε₀ is the permittivity of free space. Since the Gaussian surface is a cylindrical shell, the area vector is perpendicular to the electric field and its magnitude is equal to the area of the cylindrical shell, which is 2πrL. Substituting these values in the equation, we get E(2πrL) = λL/ε₀. Solving for E, we find that the electric field outside the cylinder is given by E = λ/(2πε₀r). Since λ is constant for a long hollow cylinder, we can replace it with the charge density ρ multiplied by the length of the cylinder, L. Therefore, the electric field outside the cylinder decreases as 1/r, where r is the distance from the center of the cylinder.

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

Answer:

Magnitude of force, F = 11 Newtons

Explanation:

Charge 1, [tex]q_1=5.3\ \mu C=5.3\times 10^{-6}\ C[/tex]

Charge 2, [tex]q_2=11.2\ \mu C=11.2\times 10^{-6}\ C[/tex]

The separation between the charges is, r = 0.22 m

We have to find the magnitude of the force between two point charges. It can be calculated using the formula as :

[tex]F=k\dfrac{q_1q_2}{r^2}[/tex]

[tex]F=9\times 10^9\times \dfrac{5.3\times 10^{-6}\ C\times 11.2\times 10^{-6}\ C}{(0.22\ m)^2}[/tex]

F = 11.03 N

or

F = 11 Newtons

Hence, this is the required solution.

Answer:

Minimum elastic modulus of fiber = 455.64 GPa

Explanation:

Contents of composite material = Epoxy and Unidirectional fibers

Elastic modulus of epoxy = 3.5 GPa

Elastic modulus of composite material = 320 GPa

Volume fraction of fiber = 70 %

Volume fraction of epoxy = 100 - 70 = 30%

Elastic modulus of composite material = 3.5 x 0.3 + Elastic modulus of fiber x 0.7 = 320

0.7 x Elastic modulus of fiber = 320 - 1.05 = 318.95

Elastic modulus of fiber = 455.64 GPa

Minimum elastic modulus of fiber = 455.64 GPa

Answer:

Speed of car A is half the speed of car B.

Explanation:

Kinetic energy, [tex]KE=\frac{1}{2}mv^2[/tex]

Car A has twice the mass of car B.

         [tex]m_A=2m_B[/tex]

Car A has the half the kinetic energy of car B

         [tex]KE_A=\frac{1}{2}KE_B[/tex]

We have

          [tex]KE_A=\frac{1}{2}KE_B\\\\\frac{1}{2}m_Av_A^2=\frac{1}{2}\times \frac{1}{2}m_Bv_B^2\\\\\frac{1}{2}\times 2m_Bv_A^2=\frac{1}{2}\times \frac{1}{2}m_Bv_B^2\\\\v_A^2=\frac{v_B^2}{4}\\\\v_A=\frac{v_B}{2}[/tex]

Speed of car A is half the speed of car B.

Explanation:

It is given that, a mass attached to a spring oscillates and completes 53 full cycles in 28 s. Frequency of the system is given by the number of oscillations or vibrations per second. Here, the frequency of this system is given by :

[tex]f=\dfrac{no\ of\ oscillations}{time}[/tex]

[tex]f=\dfrac{53}{28}[/tex]

f = 1.89 Hertz

The relationship between the frequency and the time period of the spring is inverse i.e.

[tex]T=\dfrac{1}{f}[/tex]

[tex]T=\dfrac{1}{1.89\ Hz}[/tex]

T = 0.52 seconds

Hence, this is the required solution.

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:

[tex]I = 94.33 kg m^2[/tex]

Explanation:

Let say the rod is slightly pulled away from its equilibrium position

So here net torque on the rod due to its weight is given as

[tex]\tau = mg dsin\theta[/tex]

since rod is pivoted at distance of 1.4 m from centre of gravity

so its moment of inertia about pivot point is given as

[tex]Inertia = I[/tex]

now we have

[tex]I \alpha = mg d sin\theta[/tex]

now for small angular displacement we will have

[tex]\alpha = \frac{mgd}{I}\theta[/tex]

so angular frequency of SHM is given as

[tex]\omega = \sqrt{\frac{mgd}{I}}[/tex]

now we will have

[tex]\frac{2\pi}{9} = \sqrt{\frac{4.8(9.8)1.4}{I}[/tex]

[tex]I = 94.33 kg m^2[/tex]

Final answer:

The moment of inertia from the pivot is approximately [tex]72.4 kg\(\cdot\)m2.[/tex]

Explanation:

To generate an accurate answer for the moment of inertia of the stick about the pivot when it is used as a physical pendulum, we can use the formula for the period of a physical pendulum, which is:

[tex]T = 2\(\pi\)\(\sqrt{I/(mgd)}\),[/tex]

where T is the period, I is the moment of inertia, m is the mass of the rod, g is the acceleration due to gravity (approximately 9.8 m/s2), and d is the distance from the pivot to the center of mass. Given that T = 9 seconds, m = 4.8 kg, and d = 1.4 m, we can rearrange the formula to solve for I:

[tex]I = T2mgd/(4\(\pi\)2).[/tex]

Plugging the given values into this equation:

[tex]I = 92 * 4.8 kg * 9.8 m/s^2 * 1.4 m / (4 * \(\pi\)2).[/tex]

Calculating this yields the moment of inertia I:

[tex]I \(\approx\) 72.4 kg\(\cdot\)m2.[/tex]

Answer:

a) Current = 11 mA

b) Energy = 66 mJ

c) Power = 101.54 W

Explanation:

a) Voltage, V = IR

   Voltage, V = 6 V, Resistance, R = 550 Ω

   Current, I [tex]=\frac{6}{550}=0.011A=11mA[/tex]

b) Energy = Current x Voltage = 6 x 0.011 = 0.066 J = 66 mJ

c) [tex]\texttt{Power=}\frac{Energy}{Time}=\frac{0.066}{0.65\times 10^{-3}}=101.54W[/tex]    

Answer:

Final velocity will be 1.7 m/s in the initial direction of running back

Explanation:

Since there is no external force on the system of two line man so here we can say that momentum of the two person will remain constant

So we can say

initial momentum of the two persons = final momentum of them

[tex]m_1v_{1i} + m_2v_{2i} = (m_1 + m_2) v[/tex]

now from the above equation we will have

[tex]v = \frac{m_1v_{1i} + m_2v_{2i}}{m_1 + m_2}[/tex]

now plug in all the values in the above equation

[tex]v = \frac{118(1) + 74(-6)}{118 + 74}[/tex]

[tex]v = - 1.7 m/s[/tex]

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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Explanation:

The boat's velocity relative to the water is 6.10 m/s, and the water's velocity relative to the shore is 1.95 m/s.  We want the boat's velocity relative to the shore to be perpendicular to the shore.

If we say that θ is the angle the boat makes with the shore, then:

6.10 cos θ - 1.95 = 0

6.10 cos θ = 1.95

cos θ = 1.95 / 6.10

θ = 71.4°

The boat should be steered 71.4° relative to the shore, or 18.6° relative to the vertical.

The angle at which the boat must be steered can be determined by using the trigonometric function sine (sin), taking the river current speed as the opposite side and the boat's speed in still water as the hypotenuse in a right triangle. Calculate θ = arcsin(1.95 m/s / 6.10 m/s) to find the angle.

To find the angle the boat must be steered, we need to consider the velocity of both the boat and the river current. Firstly, consider that the boat's velocity (6.10 m/s) and the river current velocity (1.95 m/s) form a right triangle. The speed in still water is the hypotenuse while the river current speed forms the opposite side of the triangle.

To find the angle θ which is the angle between the boat's direction and the river current, we use the trigonometric function sine (sin), defined as the length of the opposite side divided by the length of the hypotenuse. Therefore, sin(θ) = (river current speed) / (boat speed in still water), sin(θ) = 1.95 m/s / 6.10 m/s. To find the angle θ, find the inverse of sin(θ).

Therefore, θ = arcsin(1.95 m/s / 6.10 m/s), which can be calculated using a scientific calculator to get an accurate degree measurement for the angle.

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Explanation:

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / F

F = 50 / cos 28°

F = 56.6 lbs

56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs.

Newton's Second Law states that The resultant force acting on an object is proportional to the rate of change in momentum.

As given in the problem if a ramp leading to the entrance of a building is inclined upward at an angle of 7 degrees. A suitcase is to be pulled up the ramp by a handle that makes an angle of 35 degrees horizontal.

The angle of the handle relative to the horizontal is 35°.  The angle of the ramp to the horizontal is 7°.  So the angle of the handle relative to the ramp is 28°.

cos 28° = 50 / Force

Force  = 50 / cos 28°

Force  = 56.6 lbs

Thus, the 56.6 lbs force must be applied in the direction of the handle so that the component of the force parallel to the ramp is 50 lbs

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(a) [tex]4.3\cdot 10^9 J[/tex]

The kinetic energy of an object is given by:

[tex]K=\frac{1}{2}mv^2[/tex]

where

m is the mass of the object

v is its speed

For the ship in the problem, we have

[tex]m=6.40\cdot 10^7 kg[/tex] is the mass

[tex]v=11.6 m/s[/tex] is the speed

So its kinetic energy is

[tex]K=\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=4.3\cdot 10^9 J[/tex]

(b) [tex]-4.3\cdot 10^9 J[/tex]

According to the work-kinetic energy theorem, the work done on an object is equal to the change in kinetic energy of the object:

[tex]W= \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2[/tex]

where

W is the work done

m is the mass of the object

v is the final speed of the object

u is the initial speed of the object

Here we want to find the work done to stop the ship, so the final speed of the ship is v=0, while the initial speed is u=11.6 m/s. So the work done will be

[tex]W= 0 -\frac{1}{2}(6.40\cdot 10^7 kg)(11.6 m/s)^2=-4.3\cdot 10^9 J[/tex]

(c) [tex]1.3\cdot 10^6 N[/tex]

The work done on an object can be also written as follows

[tex]W=Fd[/tex]

where

F is the magnitude of the force

d is the displacement of the object

Here we know:

[tex]W=-4.3\cdot 10^9 J[/tex] is the work done

d = 3.20 km = 3200 m is the displacement of the ship

So we can solve the formula to find F, the force exerted on the ship to stop it:

[tex]F=\frac{W}{d}=\frac{-4.3\cdot 10^9 J}{3200 m}=-1.3\cdot 10^6 N[/tex]

and the negative sign simply means that the force is opposite to the displacement of the ship (in fact, the force acts against the motion of the ship).

The ship's kinetic energy is 5.77 × 10^9 J. The work required to stop it is -5.77 × 10^9 J. The magnitude of the force required to stop it is 1.80 × 10^3 N.

(a) The kinetic energy of the ship can be calculated using the formula:

Ek = (1/2)mv2

where m is the mass of the ship and v is its velocity. Plugging in the given values, we get a kinetic energy of 5.77 × 109 J.

(b) To stop the ship, work needs to be done to bring it to a complete stop. The work done is equal to the change in kinetic energy of the ship. Initially, the ship had a kinetic energy of 5.77 × 109 J. When it comes to a stop, its kinetic energy becomes zero. Therefore, the work required to stop the ship is -5.77 × 109 J (negative since work is done on the ship).

(c) The magnitude of the constant force required to stop the ship can be calculated using the work-energy theorem:

Work = Force × Distance

Given that the displacement of the ship is 3.20 km, or 3.20 × 106 m, and the work done on the ship is -5.77 × 109 J, we can solve for the force:

Force = Work / Distance = -5.77 × 109 J / (3.20 × 106 m) = -1.80 × 103 N

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Answer:

(A)chimney

Explanation:

bc all the smoke is going into the chimney

Answer:

The number of protons 6.19 more than electron.

Explanation:

Given that,

Charge [tex]Q=+9.9\times10^{-6}\ C[/tex]

We know that,

formula of charge

[tex]Q = nc[/tex]

Where,

Q = total charge

n = number of protons

e = charge of electron

Put the value into the formula

[tex]n = \dfrac{Q}{e}[/tex]

[tex]n=\dfrac{9.9\times10^{-6}}{1.6\times10^{-19}}[/tex]

[tex]n=6.1875\times10^{13}\ C[/tex]

According to statement of question

Divide the answer by [tex]10^{13}[/tex]

[tex]n=\dfrac{6.1875\times10^{13}}{10^{13}}[/tex]

[tex]n=6.19\ C[/tex]

Hence, The number of protons 6.19 more than electron.

Answer:

initial: 1654.6 J, final: 0 J, change: -1654.6 J

Explanation:

The length of the slide is

L = 8.80 ft = 2.68 m

So the height of the child when he is at the top of the slide is (with respect to the ground)

[tex]h = L sin \theta = (2.68 m)sin 25.0^{\circ}=1.13 m[/tex]

The potential energy of the child at the top is given by:

[tex]U = mgh[/tex]

where

m = 63.0 kg is the mass of the child

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

h = 1.13 m

Substituting,

[tex]U=(63.0 kg)(9.8 m/s^2)(2.68 m)=1654.6 J[/tex]

At the bottom instead, the height is zero:

h = 0

So the potential energy is also zero: U = 0 J.

This means that the change in potential energy as the child slides down is

[tex]\Delta U = 0 J - (1654.6 J) = -1654.6 J[/tex]

The potential energy at the top is 700.9 J. At the bottom is 0 J. Change is -700.9 J (loss).

The length of the slide [tex]\( L = 8.80 \, \text{ft} \)[/tex] and the angle with the horizontal [tex]\( \theta = 25.0^\circ \)[/tex], we can use the sine function to find the vertical height h from the top to the bottom of the slide:

[tex]\[ h = L \sin \theta \][/tex]

[tex]\[ h = 8.80 \, \text{ft} \times \sin 25.0^\circ \][/tex]

First, calculate [tex]\( \sin 25.0^\circ \)[/tex]:

[tex]\[ \sin 25.0^\circ \approx 0.4226 \][/tex]

Now, calculate h:

[tex]\[ h = 8.80 \times 0.4226 \approx 3.719 \, \text{ft} \][/tex]

Potential energy PE is given by:

[tex]\[ PE = mgh \][/tex]

where [tex]\( m = 63.0 \, \text{kg} \), \( g = 9.81 \, \text{m/s}^2 \)[/tex] , and h is the height

Convert h from feet to meters (1 ft = 0.3048 m):

[tex]\[ h = 3.719 \, \text{ft} \times 0.3048 \, \text{m/ft} = 1.133 \, \text{m} \][/tex]

The potential energy at the top:

[tex]\[ PE_{\text{top}} = 63.0 \, \text{kg} \times 9.81 \, \text{m/s}^2 \times 1.133 \, \text{m} \][/tex]

[tex]\[ PE_{\text{top}} \approx 63.0 \times 9.81 \times 1.133 \approx 700.9 \, \text{J} \][/tex]

Since the child is at the bottom of the slide, where h = 0, the potential energy at the bottom [tex](\( PE_{\text{bottom}} \))[/tex] is 0 Joules.

The change in potential energy

The change in potential energy [tex](\( \Delta PE \))[/tex] is:

[tex]\[ \Delta PE = PE_{\text{bottom}} - PE_{\text{top}} \][/tex]

[tex]\[ \Delta PE = 0 - 700.9 \, \text{J} \][/tex]

[tex]\[ \Delta PE = -700.9 \, \text{J} \][/tex]

Answer:

0.042 m^3

Explanation:

side, a = 0.5 m, m = 100 kg

density of fluid, d = 1.2 g/cm^3 = 1200 kg/m^3

Volume of cube, V = side^3 = (0.5)^3 = 0.125 m^3

density of solid = mass of solid / volume of solid cube

density of solid, D = 100 /  0.125 = 800 kg/m^3

Let v be the volume of cube immersed in fluid.

According to the principle of flotation

Weight of cube = Buoyant force acting on the cube

V x D x g = v x d x g

0.125 x 800 = v x 1200

v = 0.083 m^3

Volume above the level of fluid = V - v = 0.125 - 0.083 = 0.042 m^3

Answer:

The period of oscillation is 3.14 sec.

Explanation:

Given that,

Spring constant = 20 N/m

Mass = 5.0 kg

We need to calculate the time period

Using formula of time period

[tex]T =2\pi\sqrt{\dfrac{m}{k}}[/tex]

m= mass

k = spring constant

[tex]T =2\times3.14\sqrt{\dfrac{5.0}{20}}[/tex]

[tex]T =3.14\ sec[/tex]

Hence, The period of oscillation is 3.14 sec.

[tex]a=5000\dfrac{mm}{s}=5\dfrac{m}{s}[/tex]

[tex]f=10{Hz}\Longrightarrow t=\dfrac{1}{10}s[/tex]

[tex]a_{max}=\dfrac{50\frac{m}{s}}{\frac{s}{10}}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{50\dfrac{m}{s^2}}{\sqrt{2}}\approx\boxed{35.4\dfrac{m}{s^2}}

[/tex]

Hope this helps.

r3t40

The maximum displacement or amplitude of the oscillating machine, given the acceleration of 5,000 mm/s² and a frequency of 10 Hz, is calculated using principles of simple harmonic motion as approximately 13 mm.

The question you asked involves the principles of Physics, in particular, the topics of simple harmonic motion and amplitude are applied. The machine's oscillation modelled as a simple spring-mass system operates under the principles of simple harmonic motion. The maximum displacement the machine undergoes, also known as the amplitude, can be calculated using the given data of acceleration and frequency.

Acceleration in a simple harmonic system is given by the formula a = -ω²x, where ω represents angular frequency, x represents displacement, and a represents acceleration. The maximum displacement, or amplitude, can be described as A = a_max/ω².

To find ω, we use the given frequency and recall that ω=2πf, so substituting 10 Hz for f, we get ω=20π rad/sec. To find the acceleration, we need to convert 5000 mm/s² to m/s² which is 5 m/s².

Substituting these values back into the amplitude equation A = a_max/ω², we get A = 5/400π² which calculates approximately as 0.013 m or 13 mm.

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Answer:

Explanation:

When two equal and opposite force acting at a line of action, they form a couple. It always gives a turning effect to the body.

The turning force is said to be torque .

Torque is the product of either force and the perpendicular distance between the forces.

Torque = force x perpendicular distance

Its SI unit is Nm. It is a vector quantity.

Torque is a physical quantity that measures the effectiveness of a force in changing or accelerating a rotation. The equation for the magnitude of torque is t = rF sin θ.

Torque is a physical quantity that measures the effectiveness of a force in changing or accelerating a rotation. It incorporates the magnitude, direction, and point of application of the force. The equation for the magnitude of torque is given by t = rF sin θ, where t is torque, r is the distance from the pivot point to the point of force application, F is the magnitude of the force, and θ is the angle between the force and the lever arm.

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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.

The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C with 6 negatively charged electrons on it.

If the charge on any negatively charged oil droplet is always a whole-number multiple of the fundamental charge of a single electron [tex]10^{-19[/tex] C.  and Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons.

Given that :

Number of electrons = 6

Given charge = −1.60× [tex]10^{-19}[/tex]C

The total charge can be calculated as:

Total charge = Number of electrons × Fundamental charge

Total charge  = 6 × −1.60× [tex]10^{-19}[/tex]C

                       = 9.6 ×[tex]10^{-19}[/tex]

The droplet has the charge of 9.6 ×[tex]10^{-19}[/tex] C.

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Using the findings of Millikan's oil drop experiment, the charge on an oil droplet with 6 negatively charged electrons would be -9.60×10^−19 Coulombs.

According to the data from Millikan’s oil drop experiment, the charge of a single electron is identified to be −1.60×10−19 C. It was found that the charge on any negatively charged oil droplet is always a whole-number multiple of this fundamental charge. Therefore, if Millikan was measuring the charge on an oil droplet with 6 negatively charged electrons on it, he would calculate the charge as

-1.60×10−19 C (charge of a single electron) multiplied by 6 (number of electrons) resulting in a total charge of -9.60×10^−19 C.

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Answer:

Maximum height reached by the bullet = 4.01 m

Explanation:

Horizontal displacement = 92 m

Time taken = 6.4 s

Horizontal velocity

           [tex]=\frac{92}{6.4}=14.375m/s[/tex]

We have angle of projection = 32°

Horizontal velocity = u cos 32 = 14.375

                      u = 16.95 m/s

Vertical velocity = u sin θ = 16.95 x sin 32 = 8.98 m/s

Time of flight till it reaches maximum height = 0.5 x 6.4 = 3.2s

Now we have vertical motion of bullet

        S = ut + 0.5 at²

       Vertical velocity = u = 16.95 m/s

       a = -9.81 m/s²

        t = 3.2s

Substituting

        S = 16.95 x 3.2 - 0.5 x 9.81 x 3.2² = 4.01 m

Maximum height reached by the bullet = 4.01 m

Answer:

Range, R = 5090.1 meters

Explanation:

It is given that,

Velocity of cannon, v = 240 m/s

A cannon is shot at a 30 degree angle above the horizontal. The range of this shot is given by the formula as follows :

[tex]R=\dfrac{v^2\ sin2\theta}{g}[/tex]

g = acceleration due to gravity

[tex]R=\dfrac{(240\ m/s)^2\ sin2(30)}{9.8\ m/s^2}[/tex]

R = 5090.1 meters

So, the range of this shot is 5090.1 meters. Hence, this is the required solution.

Answer:

The value of A will increase as you increase the charges.

Explanation:

A is actually kq1*q2. So if you increase the charges, A will also increase. The big increase comes if you make r smaller.

Increased charges is a direct variation. As the charge increases the value of A increases. As the charges decrease, the value of A decreases.

r is an inverse variation. As r increases the force decreases. As r decreases the force increases. R does not affect a.

Answer:

Power required, P = 175 watts

Explanation:

It is given that,

Weight of a student, F = mg = 200 N

The student climbs a flight of stairs of height, h = 7 m

Time taken, t = 8 s

We have to find the power required to perform this task. Work done per unit time is called the power required. Mathematically, it is given by :

[tex]P=\dfrac{W}{t}[/tex]

W = work done

t = time taken

[tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{200\ N\times 7\ m}{8\ s}[/tex]

P = 175 watts

Hence, the power required to complete this task is 175 watts.

Explanation:

The speed [tex]v[/tex] of a wave is given by the following equation:

[tex]v=\frac{\lambda}{T}[/tex] (1)

Where [tex]\lambda[/tex] is the wavelength and [tex]T[/tex] the period of the wave.

In addition we know there is an inverse relation between the period of a wave and its frequency [tex]f[/tex] :

[tex]f=\frac{1}{T}[/tex]  (2)

Substituting (2) in (1):

[tex]v=\lambda.f[/tex]  (3)

Now we can find the velocity of the wave with the known given values applying equation (3):

[tex]v=(10m)(2Hz)[/tex]  (4)

Finally:

[tex]v=20 \frac{m}{s}[/tex]  

(a) The pitcher must throw the ball at 27.7 m/s

The momentum of an object is given by:

[tex]p=mv[/tex]

where

m is the mass of the object

v is the object's velocity

Let's calculate the momentum of the bullet, which has a mass of

m = 2.70 g = 0.0027 kg

and a velocity of

[tex]v=1.50\cdot 10^3 m/s[/tex]

Its momentum is:

[tex]p=mv=(0.0027 kg)(1.50\cdot 10^{3} m/s)=4.05 kg m/s[/tex]

The pitcher must throw the baseball with this same momentum. The mass of the ball is

m = 0.146 kg

So the velocity of the ball must be

[tex]v=\frac{p}{m}=\frac{4.05 kg m/s}{0.146 kg}=27.7 m/s[/tex]

So, the pitcher must throw the ball at 27.7 m/s.

(b) a. The bullet has greater kinetic energy

The kinetic energy of an object is given by

[tex]K=\frac{1}{2}mv^2[/tex]

where m is the mass of the object and v is its speed.

For the bullet, we have:

[tex]K=\frac{1}{2}(0.0027 kg)(1.50\cdot 10^3 m/s)^2=3037.5 J[/tex]

For the ball:

[tex]K=\frac{1}{2}(0.146 kg)(27.7 m/s)^2=56.0 J[/tex]

So, the bullet has greater kinetic energy.

Final answer:

The baseball's speed must be 27.74 m/s to match the momentum of the bullet, and the bullet has greater kinetic energy than the baseball.

Explanation:

Calculating Baseball's Speed and Comparing Kinetic Energies

To validate the pitcher's claim that a 0.146-kg baseball can have the same momentum as a 2.70-g bullet traveling at 1.50 x 10³ m/s, we must use the formula for momentum (p = mv), which gives us:

Momentum of bullet = (2.70 g) x (1.50 x 10³ m/s) = (0.0027 kg) x (1500 m/s) = 4.05 kg m/s.

Therefore, the baseball's speed v needed to have the same momentum is calculated by rearranging the formula to v = p/m, which gives us:

Baseball's speed v = 4.05 kg m/s / 0.146 kg = 27.74 m/s.

To determine which has the greater kinetic energy, we use the kinetic energy formula KE = (1/2)mv². Calculating the kinetic energy of both the bullet and the baseball:

KE of bullet = (1/2)(0.0027 kg)(1500 m/s)² = 3037.5 J.

KE of baseball = (1/2)(0.146 kg)(27.74 m/s)² = 55.90 J.

Comparing the results shows that the bullet has greater kinetic energy than the baseball.