What is the speed of a 0.145kg baseball if its kinetic energy is 109J?

Concerning the work done by a conservative force, which of the following statements, if any, are true? It can always be expresse

Vera_Pavlovna [14]

Answer:

It is independent of the path of the body and depends only on the starting and ending points.

Explanation:

In Physics we define a conservative force as a force that is independent of the path of the body and depends only on the starting and ending points.

For conservative forces we can write;

KEi + PEi = KEf +PEf

where;

KEi= initial kinetic energy

PEi= initial potential energy

KEf= final kinetic energy

PEf= final potential energy

This equation is known as the principle conservation of mechanical energy . It applies only to conservative forces where friction is negligible. The term KE + PE is also known as the total mechanical energy of the system.

3 0

4 years ago

What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You c

Maksim231197 [3]

Answer:

The age of the universe would be 9.9 billion years

Explanation:

We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.

Thus, the time it has taken for the galaxies to reach their current separations is:

$\displaystyle{t=D/v}$

and from Hubble's Law:

$v =H_0D$

Therefore:

$\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}$

With the given value for the Hubble's constant we have:

$H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s$

and thus,

$t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years$

6 0

3 years ago

The law of reflection is quite useful for mirrors and other flat, shiny surfaces. (This sort of reflection is called specular re

Sever21 [200]

Answer:

Explanation:

Suppose initially the plane was horizontal and light was reflected back at some angle θ from the normal .

Now the reflecting surface is twisted so that is becomes inclined at angle alpha .

The reflected light will be deviated from its original direction by angle

2 x alpha .

Similarly when the reflecting surface is further twisted so that it becomes inclined at angle beta then again the reflected beam will deviated by angle

2 x beta

Hence angle between these two reflected beam

= 2 beta - 2 alpha

= 2 ( β - α )

So, angular separation between the rays reflected from the two surfaces

= 2 ( β - α ) .

5 0

3 years ago

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin

tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

cos 40 = v₀ₓ / V₀

v₀ₓ = v₀ cos 40

v₀ₓ = 50.0 cos 40

v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

Y Axis y

$p_{oy}$ = 0

After the break

X axis

$p_{fx}$ = m₂ v₂

Axis y

$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

8 0

3 years ago

(PLEASE HELP FAST 20 POINTS)

Lera25 [3.4K]

Everything we see or do in everyday life that involves electricity in any way is the result of electrons moving from one place to another, or from one object to another. <em> (last choice)</em>

4 0

3 years ago