1) Focal length
We can find the focal length of the mirror by using the mirror equation:
(1)
where
f is the focal length
is the distance of the object from the mirror
is the distance of the image from the mirror
In this case,
, while
(the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
from which we find
2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.
3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.
4) The radius of curvature of a mirror is twice its focal length, so for the mirror in our problem the radius of curvature is:
First, let's determine the gravitational force of the Earth exerted on you. Suppose your weight is about 60 kg.
F = Gm₁m₂/d²
where
m₁ = 5.972×10²⁴ kg (mass of earth)
m₂ = 60 kg
d = 6,371,000 m (radius of Earth)
G = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(5.972×10²⁴ kg)/(6,371,000 m )²
F = 589.18 N
Next, we find the gravitational force exerted by the Sun by replacing,
m₁ = 1.989 × 10³⁰<span> kg
Distance between centers of sun and earth = 149.6</span>×10⁹ m
Thus,
d = 149.6×10⁹ m - 6,371,000 m = 1.496×10¹¹ m
Thus,
F = ( 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²)(60 kg)(1.989 × 10³⁰ kg)/(1.496×10¹¹ m)²
F = 0.356 N
Ratio = 0.356 N/589.18 N
<em>Ratio = 6.04</em>
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
Answer:
the direction that should be walked by Ricardo to go directly to Jane is 23.52 m, 24° east of south
Explanation:
given information:
Ricardo walks 27.0 m in a direction 60.0 ∘ west of north, thus
A= 27
Ax = 27 sin 60 = - 23.4
Ay = 27 cos 60 = 13.5
Jane walks 16.0 m in a direction 30.0 ∘ south of west, so
B = 16
Bx = 16 cos 30 = -13.9
By = 16 sin 30 = -8
the direction that should be walked by Ricardo to go directly to Jane
R = √A²+B² - (2ABcos60)
= √27²+16² - (2(27)(16)(cos 60))
= 23.52 m
now we can use the sines law to find the angle
tan θ = 
= By - Ay/Bx -Ax
= (-8 - 13.5)/(-13.9 - (-23.4))
θ = 90 - (-8 - 13.5)/(-13.9 - (-23.4))
= 24° east of south
The change in horizontal velocity is (4.7 - 8.1) = -3.4 m/s
The change in vertical velocity is (3.2 + 3.3) = 6.5 m/s
These are the components of velocity DELIVERED to the ball by the player's pretty head during the collision.
The magnitude of the change in velocity is √(-3.4² + 6.5²) = 7.336 m/s .
The magnitude of the ball's change in momentum is (m · v) = (0.44 · 7.336) = 3.228 kg-m/s .
==> The change in the ball's momentum is exactly the <em>impulse</em> during the collision. . . . . . <em>3.228 kg-m/s</em> .
==> The direction of the impulse is the direction of the change in momentum: (-3.4)i + (6.5)j
The direction is arctan (6.5 / -3.4) = -62.39°
That's clockwise from the +x axis, which is roughly "southeast". The question wants it counterclockwise from the +x axis. That's (360-62.39) =
<em>Direction of the impulse = 297.61°</em>
<em></em>
We know that impulse is equivalent to the <u>change in momentum</u>, and that's how I approached the solution. Impulse is also (<u>force x time</u>) during the collision. We're given the time in contact, but I didn't need to use it. I guess I would have needed to use it if we were interested in the FORCE she exerted on the ball with her head, but we didn't need to find that.
