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You say goodbye to your friend at the intersection of two perpendicular roads. At time t=0 you drive off North at a (constant) speed v and your friend drives West at a (constant) speed ????. You badly want to know: how fast is the distance between you and your friend increasing at time t?

Answers

Answer 1

Answer:

Answer:

$d'=√(v^2+w^2)$

Explanation:

Rate of Change

When an object moves at constant speed v, the distance traveled at time t is

$x=v.t$

We know at time t=0 two friends are at the intersection of two perpendicular roads. One of them goes north at speed v and the other goes west at constant speed w (assumed). Since both directions are perpendicular, the distances make a right triangle. The vertical distance is

$y=v.t$

and the horizontal distance is

$x=w.t$

The distance between both friends is computed as the hypotenuse of the triangle

$d^2=x^2+y^2$

We need to find d', the rate of change of the distance between both friends.

Plugging in the above relations

$d^2=(v.t)^2+(w.t)^2$

$d^2=v^2.t^2+w^2.t^2=(v^2+w^2)t^2$

Solving for d

$d=√((v^2+w^2)t^2)$

$d=√((v^2+w^2)).t$

Differentiating with respect to t

$\boxed{d'=√(v^2+w^2)}$

Answer 2

Answer:

Final answer:

The problem is solved using Pythagoras' Theorem, representing the two travel paths forming a right triangle. The rate at which the distance increases between two points moving perpendicularly can be found by differentiating the resulting equation, which yields the expression sqrt[(v^2)+(u^2)].

Explanation:

The question is about the rate at which the distance between you and your friend is increasing at time t. It's a typical problem in kinematics. Because the roads are perpendicular to each other, we can solve the problem using Pythagoras' Theorem which states that in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

Let's denote the distance you've traveled as D1 = v*t (because distance = speed * time) and the distance your friend has travelled D2 = u*t. The distance between you can be computed using Pythagoras' Theorem as D = sqrt(D1^2 + D2^2). Hence, D = sqrt[(v*t)^2 + (u*t)^2]. Differentiating D with respect to t using the chain rule will give us the rate at which the distance between you is increasing, which is sqrt[(v^2)+(u^2)].

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Assume we are given an electric field set up by an unknown charge distribution. U0 is the amount of work needed to bring a point charge of charge q0 in from infinity to a point P. If the charge q0 is returned to infinity, how much work would it take to bring a new charge of 4 q0 from infinity to point P?

Answers

Answer:

$4U_0$

Explanation:

We are given that

Amount of work needed to bring a point charge q0 from infinity to a point P= $U_0$

We know that potential at point P= $V=(U)/(q)$

$U=Vq$

Where U=Amount of work needed to bring a point charge q from infinity to a point P

Initially , $V=(U_0)/(q_0)$

New charge, q= $4q_0$

Then, work done,U= $(U_0)/(q_0)* (4q_0)=4U_0$

Hence, the amount of work needed to bring a new charge 4q0 from infinity to point P= $4U_0$

Answer:

$4U_(0)$

Explanation:

V = u / q,

Work = P = V

1U / 1/4 = 4U

1. Two charges Q1( + 2.00 μC) and Q2( + 2.00 μC) are placed along the x-axis at x = 3.00 cm and x=-3 cm. Consider a charge Q3 of charge +4.00 μC and mass 10.0 mg moving along the y-axis. If Q3 starts from rest at y = 2.00 cm, what is its speed when it reaches y = 4.00 cm?

Answers

Answer:

speed when it reaches y = 4.00cm is

v = 14.9 g.m/s

Explanation:

given

q₁=q₂ =2.00 ×10⁻⁶

distance along x = 3.00cm= 3×10⁻²

q₃= 4×10⁻⁶C

mass= 10×10 ⁻³g

distance along y = 4×10⁻²m

r₁ = $\sqrt{3^(2) +2^(2) }$ = $√(13)$ = 3.61cm = 0.036m

r₂ = $\sqrt{4^(2) + 3^(2) }$ = $√(25)$ = 5cm = 0.05m

electric potential V = $(kq)/(r)$

change in potential ΔV = $V_(1) - V_(2)$

ΔV = $(2kq_(1) )/(r_(1)) - (2kq_(2) )/(r_(2) )$ , where $q_(1) = q_(2)=$ 2.00μC

ΔV = $2kq((1)/(r_(1)) - (1)/(r_(2) ))$

ΔV = 2 × 9×10⁹ × 2×10⁻⁶ × $((1)/(0.036) - (1)/(0.05) )$

ΔV= 2.789×10⁵

$(1)/(2)mv^(2)$ = ΔV × q₃

$(1)/(2)$ ˣ 10×10⁻³ ×v² = 2.789×10⁵× 4 ×10⁻⁶

v² = 223.12 g.m/s

v = 14.9 g.m/s

The speed of the charge q₃ when it starts from rest at y = 2 cm and reaches y = 4 cm is; v = 14.89 m/s

We are given;

Charge 1; q₁ = 2.00 μC = 2 × 10⁻⁶ C

Charge 2; q₂ = 2.00 μC = 2 × 10⁻⁶ C

Distance of charge 1 along x = 3 cm = 3 × 10⁻² m

Distance of charge 2 along x = -3 cm = -3 × 10⁻² m

Charge 3; q₃ = +4.00 μC = 4 × 10⁻⁶ C

mass; m = 0.01 g

distance of charge 3 along y = 4 cm = 4 × 10⁻² m

q₃ starts from rest at y = 2 × 10⁻² m and reaches y = 4 × 10⁻² m.

Thus;

Distance of charge 1 from the initial position of q₃;

r₁ = √((3 × 10⁻²)² + ((2 × 10⁻²)²)

r₁ = 0.0361 m

Distance of charge 2 from the final position of q₃;

r₂ = √((3 × 10⁻²)² + ((4 × 10⁻²)²)

r₂ = 0.05 m

Now, formula for electric potential is;

V = kq/r

Where k = 9 × 10⁹ N.m²/s²

Thus,change in potential is;

ΔV = V₁ - V₂

Now, Net V₁ = 2kq₁/r₁

Net V₂ = 2kq₂/r₂

Thus;

ΔV = 2kq₁/r₁ - 2kq₂/r₂

ΔV = (2 × 9 × 10⁹)[(2 × 10⁻⁶/0.0361) - (2 × 10⁻⁶/0.05)]

ΔV = 277229.92 V

Now, from conservation of energy;

½mv² = q₃ΔV

Thus;

½ × 0.01 × v² = 4 × 10⁻⁶ × 277229.92

v² = 2 × 4 × 10⁻⁶ × 277229.92/0.01

v = √(221.783936)

v = 14.89 m/s

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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.

Which of the following is a description of the Remote Associates Test (RAT)?

Answers

Answer:

The description is outlined throughout the clarification section following, and according to the given word.

Explanation:

Throughout the 1960s, Sarnoff Mednick created the RAT as a tool used for testing imaginative convergent thought. Through each RAT test query lists a set of terms, which demands that we have a single additional term that will tie any of the others around.
Those other words may also be related in something like a variety of ways, such as through creating a compound word or perhaps a semantic connexon.

Calculate the ionization potential for C+5 ( 5 electrons removed for the C atom) and in addition compute the wavelength of the transition from n=3 to n= 2.