Which of the following has the greatest inertia ?

Suppose your friend claims to have discovered a mysterious force in nature that acts on all particles in some region of space. H

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

we substitute the value of B = 2

U = 1 / r²

5 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

An object is located 50.0 cm from a concave mirror. The magnitude of the mirror focal length is 25.0 cm. What is the image dista

ioda

Answer:

Correct answer: C. 50 cm

Explanation:

Given data:

The distance of the object from the top of the concave mirror o = 50.0 cm

The magnitude of the concave mirror focal length 25.0 cm.

Required : Image distance d = ?

If we know the focal length we can calculate the center of the curve of the mirror

r = 2 · f = 2 · 25 = 50 cm

If we know the theory of spherical mirrors and the construction of figures then we know that when an object is placed in the center of the curve, there is also a image in the center of the curve that is inverted, real and the same size as the object.

We conclude that the image distance is 50 cm.

We will now prove this using the formula:

1/f = 1/o + 1/d => 1/d = 1/f - 1/o = 1/25 - 1/50 = 2/50 - 1/50 = 1/50

1/d = 1/50 => d = 50 cm

God is with you!!!

6 0

3 years ago

Read 2 more answers

PLEASE ANSWER, HURRY! THANKS :) A green object will absorb ____________________ light and reflect ____________________ light. Ex

olga55 [171]

Answer:

It will absorb yellow and reflect green

8 0

3 years ago

Which statement is not true about radiation

Mekhanik [1.2K]

I don’t know i’m sorry

4 0

2 years ago