If you travel 2 km north, then travel 5 km south, what is your displacement?

Answers

Answer 1

Answer:

Answer:

➢This is a vector addition problem which requires magnitude and direction as the answer. First is to resolve the southbound vector and the northbound vector. Since they are opposite in directions their vector sum is their algebraic sum. 3 km north + 5 km south = 2 km south.

We then add 2 km west and 2 km south using Pythagorean theorem since west and south form a right angle. (2 km)^2 west + (2 km)^2 south gives (4 + 4) km^2 southwest = 8 (km)^2 45 degrees south of west

Extracting the square root of 8 gives us about 2.83 km 45 degrees south of west.

Explanation:

I hope it will help you...

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An early model of the atom, proposed by Rutherford after his discovery of the atomic nucleus, had a positive point charge +Ze (the nucleus) at the center of a sphere of radius R with uniformly distributed negative charge −Ze. Z is the atomic number, the number of protons in the nucleus and the number of electrons in the negative sphere. Show that the electric field inside this atom is : Ein=Ze4πϵ0(1r^2−rR^3). b. What is the electric field at the surface of the atom? Is this the expected value? Explain.c. A uranium atom has Z = 92 and R = 0.10 nm. What is the electric field at r = R/2?

Answers

Answer:

Part a)

$E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))$

Part b)

$E = 0$

Yes it is the expected value of electric field at the surface of an atom

Part c)

$E = 4.64 * 10^(13) N/C$

Explanation:

Since negative charge of electrons in uniformly distributed in the atom while positive charge is concentrated at the nucleus

So the electric field due to positive charge of the nucleus is given as

$E = (kq)/(r^2)$

$E_1 = (Ze)/(4\pi \epsilon_0 r^2)$

now charge due to electrons inside a radius "r" is given as

$q = (-Ze r^3)/(R^3)$

now we will have electric field given as

$E_2 = ((-Zer^3)/(R^3))}{4\pi\epsilon_0 r^2}$

now net electric field is given as

$E = E_1 + E_2$

$E = (Ze)/(4\pi \epsilon_0 r^2) - (Zer)/(4\pi \epsilon_0 R^3)$

$E = (Ze)/(4\pi\epsilon_0)((1)/(r^2) - (r)/(R^3))$

Part b)

At the surface of an atom

$r = R$

$E = 0$

Yes it is the expected value of electric field at the surface of an atom

Part c)

If Z = 92

R = 0.10 nm

$r = (R)/(2)$

so we will have

$E = 92(1.6 * 10^(-19)) * (9 * 10^9)((4)/(R^2) - (1)/(2R^2))$

$E = (4.64 * 10^(-7))/((0.10 * 10^(-9))^2)$

$E = 4.64 * 10^(13) N/C$

A particle located at the position vector m has a force N acting on it. The torque about the origin is

Answers

Final answer:

The torque about a given origin when a force N is acting on a particle at the position vector m is given by the cross product of the position and force vectors. It's represented by the SI unit Newton-meters, and for multiple particles, the total angular momentum is the vector sum of their individual angular momenta.

Explanation:

The torque about a given origin, when a force N is acting on a particle located at the position vector m, is calculated using the cross product of the position vector and the force vector. This can be written as τ = m x N. The SI unit of torque is Newton-meters (N.m).

As an example, if you apply a force perpendicularly at a distance from a pivot point, you will create a torque relative to that point. Similarly, the torque on a particle is also equal to the moment of inertia about the rotation axis times the angular acceleration.

If we consider multiple particles, the total angular momentum of these particles about the origin is the vector sum of their individual angular momenta. This is calculated by the expression for the angular momentum Ỉ = ŕ x p for each particle, where ŕ is the vector from the origin to the particle and p is the particle's linear momentum.

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Final answer:

The torque on a particle at a position vector m with force N acting on it is calculated by taking the cross-product of the position vector and the force. This principle is the same even in systems with multiple particles. The SI unit of torque is Newton-meters (N·m), which should not be confused with Joules (J).

Explanation:

The torque on a particle located at a position vector m with a force N acting on it is calculated by taking the cross-product of the position vector and the force. In terms of physics, torque (τ) is a measure of the force that can cause an object to rotate about an axis, and it is calculated as the product of the force and the distance from the axis of rotation to the point where force is applied. Hence, the formula for torque is τ = r x F where r is the position vector (or distance from the origin to the point where the force is applied) and F is the force. Remember, this equation gives a vector result with a direction perpendicular to the plane formed by r and F and a magnitude equal to the product of the magnitudes of r and F and the sine of the angle between r and F.

The same principle applies to systems where multiple particles are present. The total angular momentum of the system of particles about a particular point is the vector sum of the individual angular momenta about that point. Torque is the time derivative of angular momentum.

The SI unit for torque is Newton-meters (N·m), which should not be confused with Joules (J), as both have the same base units but represent different physical concepts. In this context, a net force of 40N acting at a distance of 0.800m from the origin would generate a torque of 32 N·m at the origin.

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Go to his profile and roast the mess out of him plzz 403665fl 50 points

Answers

Answer:

Explanation:

A sinusoidal wave is travelling on a string under tension T = 8.0(N), having a mass per unit length of 1 = 0.0128(kg/m). It’s displacement function is D(x,t) = Acos(kx - t). It’s amplitude is 0.001m and its wavelength is 0.8m. It reaches the end of this string, and continues on to a string with 2 = 0.0512(kg/m) and the same tension as the first string. Give the values of A, k, and , for the original wave, as well as k and  the reflected wave and the transmitted wave.

Answers

Answer:

Explanation:

What is the density, in Mg/m3, of a substance with a density of 0.14 lb/in3? (3 pts) What is the velocity, in m/sec, of a vehicle traveling 70 mi/hr?

Answers

Answer:

$276.74* 10^8Mg/m^3$

31.29 m/sec

Explanation:

We have given density of substance $0.14lb/in^3$

We have convert this into $Mg/m^3$

We know that 1 lb = 0.4535 kg. so 0.14 lb = 0.14×0.4535 = 0.06349 kg

We know that 1 kg = 1000 g ( 1000 gram )

So 0.06349 kg = 63.49 gram

And we know that 1 gram = 1000 milligram

So 63.49 gram $=63.49* 10^3\ Mg$

We know that $1 in^3=1.6387* 10^(-5)m^3$

So $0.14in^3=0.14* 1.6387* 10^(-5)=0.2294* 10^(-5)m^3$

So $0.14lb/in^3$ =\frac{63.49\times 10^3}{0.2249\times 10^{-5}}=276.74\times 10^8lb/m^3[/tex]

In second part we have to convert 70 mi/hr to m/sec

We know that 1 mi = 1609.34 meter

So 70 mi = 70×1609.34 = 112653.8 meter

1 hour = 3600 sec

So 70 mi/hr $=(70* 1609.34meter)/(3600sec)=31.29m/sec$

Select all the statements regarding electric field line drawings that are correct. Group of answer choices:
1. Electric field lines are the same thing as electric field vectors.
2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space.
3. The number of electric field lines that start or end at a charged particle is proportional to the amount of charge on the particle.
4. The electric field is strongest where the electric field lines are close together.

Answers

Answer:

All statement are correct.

Explanation:

1. Electric field lines are the same thing as electric field vectors, electric field are mathematically vectors quantity. These vectors point in the direction in which a positive test charge would move.

2. Electric field line drawings allow you to determine the approximate direction of the electric field at a point in space. Yes it is correct tangent drawn at any point on these lines gives the direction of electric filed at that point.

3. The number of electric field lines that start or end at a charged particle is proportional to the magnitude of charge on the particle, is a correct statement.

4.The electric field is strongest where the electric field lines are close together, again a correct statement as relative closeness of field lines indicate a stronger strength of electric field.

Hence we can say that all the statement are correct.

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