65. If Jill was traveling at 300 miles in 4 hours due South what was his velocity in miles per second,

What inertia is present in streched rybber

MrRissso [65]

<h2>You input potential (stored) energy into the rubber band system when you stretched the rubber band back. Because it is an elastic system, this kind of potential energy is specifically called elastic potential energy.</h2><h2>Hope it helps..</h2>

4 0

2 years ago

a shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. th

Savatey [412]

Answer:

15.7 m

Explanation:

The range (horizontal distance) of the projectile is determined only by its horizontal motion.

The horizontal motion is a motion with constant speed, which is equal to the initial horizontal velocity of the object:

$v_x = v cos \theta$

where

v = 12.0 m/s is the initial velocity

$\theta=51.0^{\circ}$ is the angle between the direction of v and the horizontal

Substituting,

$v_x = (12.0 m/s)(cos 51.0^{\circ} )=7.55 m/s$

We know that the projectile hits the ground in a time of

t = 2.08 s

so the horizontal distance covered is

$d = v_x t = (7.55 m/s)(2.08 s)=15.7 m$

8 0

2 years ago

Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron a

victus00 [196]

Answer:

diameter = 21.81 ft

Explanation:

The gravitational force equation is:

$F=\frac{GMm}{R^{2} }$

Where:

F => Gravitational force or force of attraction between two masses
M => Mass of asteroid 1
m => Mass of asteroid 2
R => Distance between asteroids 1 and 2 (from center of gravity)

We also know that the asteroids are identical so their masses are identical:

M=m

Since R is the distance between centers of the two asteroids and their diameters are identical (see attachment), we can conclude that:

R=d=2r

We don´t know the mass of the asteroids but we know they are composed of pure iron, so we can relate their masses to their density:

m=ρV

This is going to be helpful because the volume of a sphere is:

$\frac{4}{3}\pi r^{3}$

And know we can write our original force of gravity equation in terms of the radius of the asteroids:

$F=\frac{GMm}{R^{2} } =\frac{Gmm}{(2r)^{2} } =\frac{Gm^{2} }{4r^{2} }$
$F=\frac{G ( \frac{4}{3}\pi r^{3}ρ)^{2} }{4r^{2} }$
$F= \frac{G(16)\pi ^{2} r^{6} ρ^{2}}{(9)(4)r^{2} } =\frac{G(16)\pi ^{2} r^{4}ρ^{2} }{36}$