An object moves at constant speed in a circle. Which of the following is true:

Why do water and table salt have very different properties?

Alexandra [31]

Answer:

D)

Explanation:

6 0

2 years ago

A car accelerates from rest at 3 m/s for 6 seconds. The velocity of the car after 6 seconds is what ?

Artist 52 [7]

$vf = vi + at$
then solve for v

so
$v = a(t)$
so v = (3m/s^2)(6s)

so v = 18m/s

5 0

2 years ago

Which statement explains how it is possible to carry books to school without changing the kinetic or potential energy of the boo

Mamont248 [21]

Answer:

a. by moving the book without acceleration and keeping the height of the book constant

Explanation:

FOR CONSTANT KINETIC ENERGY:

The kinetic energy of a body depends upon its speed according to its formula:

ΔK.E = (1/2)mΔv²

So, for Δv = 0 m/s

ΔK.E = 0 J

So, for keeping kinetic energy constant, the books must be moved at constant speed without acceleration.

FOR CONSTANT POTENTIAL ENERGY:

The potential energy of a body depends upon its height according to its formula:

ΔP.E = mgΔh

So, for Δh = 0 m/s

ΔP.E = 0 J

So, for keeping potential energy constant, the books must be moved at constant height.

So, the correct option is:

<u>a. by moving the book without acceleration and keeping the height of the book constant</u>

8 0

2 years ago

Match the following items.

Ray Of Light [21]

Answer:

Je ne Sachez que Qu’est-ce que le

8 0

2 years ago

A two-stage rocket moves in space at a constant velocity of +4300 m/s. The two stages are then separated by a small explosive ch

ololo11 [35]

Answer:

$v_{1f}$ = +3,394 103 m / s

Explanation:

We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.

The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s

The moment before the explosion

p₀ = (m₁ + m₂) v₀

After the explosion

pf = m₁ $v_{1f}$ + m₂ $v_{2f}$

p₀ = [texpv_{f}[/tex]

(m₁ + m₂) v₀ = m₁ $v_{1f}$ + m₂ $v_{2f}$

Let's calculate the final speed (v1f) of the first stage

$v_{1f}$ = ((m₁ + m₂) v₀ - m₂ $v_{2f}$ ) / m₁

$v_{1f}$ = ((2100 +1160) 4300 - 1160 5940) / 2100

$v_{1f}$ = (14,018 10 6 - 6,890 106) / 2100

$v_{1f}$ = 7,128 106/2100

$v_{1f}$ = +3,394 103 m / s

come the same direction of the final stage, but more slowly

4 0

3 years ago