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Pavlova-9 [17]
2 years ago
5

Daniel takes his two dogs, Pauli the Pointer and Newton the Newfoundland, out to a field and lets them loose to exercise. Both d

ogs sprint away in different directions while Daniel stands still. From Daniel's point of view, Newton runs due north at 3.90 m/s, but from Pauli's point of view, Newton appears to be moving at 1.10 m/s due east. What is Pauli's velocity relative to Daniel, ????⃗ PD?
Physics
1 answer:
Cloud [144]2 years ago
4 0

Answer:

4.05 m/s

Explanation:

We shall represent the different velocity in vector form

Newton runs due north at 3.90 m/s, with respect to standing Daniel .

V_n = 3.9 j

Let Pauli runs with respect to standing Daniel with velocity X .

Then relative velocity of Newton with respect to running Pauli will be

3.9 j - X

Give that

relative velocity of Newton with respect to running Pauli = 1.1 i ( 1.1 due east )

So

3.9 j - X = 1.1 i

X =  -1.1 i + 3.9 j .

Magnitude of X

X² = 1.1 ² + 3.9²

X = 4.05 m/s

So Pauli runs with respect to standing Daniel with velocity 4.05 m /s .

Direction will be ,  west of north at angle θ ,

Tan θ = 1.1 / 3.9

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kirill [66]

Answer:

             U = 1 / r²

Explanation:

In this exercise they do not ask for potential energy giving the expression of force, since these two quantities are related

             

         F = - dU / dr

this derivative is a gradient, that is, a directional derivative, so we must have

          dU = - F. dr

the esxresion for strength is

         F = B / r³

let's replace

          ∫ dU = - ∫ B / r³  dr

in this case the force and the displacement are parallel, therefore the scalar product is reduced to the algebraic product

let's evaluate the integrals

            U - Uo = -B (- / 2r² + 1 / 2r₀²)

To complete the calculation we must fix the energy at a point, in general the most common choice is to make the potential energy zero (Uo = 0) for when the distance is infinite (r = ∞)

             U = B / 2r²

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5 0
2 years ago
To crossing over the flooded canal.
zubka84 [21]

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F. jumping

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4 0
2 years ago
A 900-kg car traveling east at 15.0 m/s collides with a 750-kg car traveling north at 20.0 m/s. The cars stick together. Assume
dalvyx [7]

Explanation:

It is given that,

Mass of the car 1, m_1=900\ kg

Initial speed of car 1, u_1=15i\ m/s (east)

Mass of the car 2, m_2=750\ kg

Initial speed of car 2, u_1=20j\ m/s (north)

(b) As the cars stick together. It is a case of inelastic collision. Let V is the common speed after the collision. Using the conservation of momentum as :

m_1u_1+m_2u_2=(m_1+m_2)V

900\times 15i +750\times 20j=(900+750)V

13500i+15000j=1650V

V=(8.18i+9.09j)\ m/s

The magnitude of speed,

|V|=\sqrt{8.18^2+9.09^2}

V = 12.22 m/s

(b) Let \theta is the direction the wreckage move just after the collision. It is given by :

tan\theta=\dfrac{v_y}{v_x}

tan\theta=\dfrac{9.09}{8.18}

\theta=48.01^{\circ}

Hence, this is the required solution.

4 0
2 years ago
A force of 6.0 N pulls a box 0.40 m along a frictionless plane that is inclined at 36°. What work is being done by the pulling f
lys-0071 [83]

Answer:

Expression of work done is

W = Fd cos\theta

Work done to move the sled is given as 1.94 J

Explanation:

As we know that the formula of work done is given as

W = Fd cos\theta

here we know that

F = 6 N

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\theta = 36 degree

so we will have

W = 6 \times 0.4 cos36

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7 0
2 years ago
What is the explanation for how a modern transmission electron microscope (TEM) can achieve a resolution of about 0.2 nanometers
IgorC [24]

Answer:

Explanation:

A simple light microscope uses light for imaging of objects where as a transmission electron microscope uses a monochromatic beam of electrons.

This beam is passed by a magnetic field which is very strong and thus act as a lens.

Its resolution of very high which is about 0.2 nanometers because of the separation between two atoms.

Because of this reason its resolution is about 1000 times greater than light microscope.

3 0
2 years ago
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