Which of the following describes half-life? Choose which apply.

How high would you have to lift a 1000kg car to give it a potential energy of:

Elza [17]

Given parameters:

Mass of the car = 1000kg

Unknown:

Height = ?

To find the heights for the different amount potential energy given, we need to understand what potential energy is.

Potential energy is the energy at rest due to the position of a body.

It is mathematically expressed as:

P.E = mgh

m is the mass

g is the acceleration due to gravity = 9.8m/s²

h is the height of the car

Now the unknown is h, height and we make it the subject of the expression to make for easy calculation.

h = $\frac{P.E }{mg}$

For 2.0 x 10³ J;

h = $\frac{2000}{1000 x 9.8}$ = 0.204m

For 2.0 x 10⁵ J;

h = $\frac{200000}{9.8 x 1000}$ = 20.4m

For 1.0kJ = 1 x 10³J;

h = $\frac{1000}{9.8 x 1000}$ = 0.102m

5 0

3 years ago

Ax = 4.3 m

Dimas [21]

Answer:

the answer will be option no b plss mark me brainliest

7 0

3 years ago

Who said the greater the temperature the greater the volume

Serjik [45]

The greater the temperature, the greater the volume - this is Charles's law, said by Jacques Charles, a French inventor, scientist, and mathematician.

3 0

3 years ago

Read 2 more answers

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$