Is nucluer energy renewable or non-renewable and why?

A badger is running at a speed of 1 m/s. If the badger moves that was for 2600 seconds, how far will the badger travel?

katen-ka-za [31]

Answer:

It would be 2600

Explanation:

M/S stands for meters per second. If it moved 1 meter for 2600 seconds, than it would be 2600. You just multiply 2600 by 1! I hope this helps :D

8 0

2 years ago

The atoms of a molecule come from two or more?

AlladinOne [14]

Compounds. two or more compounds.

7 0

3 years ago

A brother and sister are standing next to each other at rest on a surface of frictionless ice. The brother’s mass is exactly twi

a_sh-v [17]

The sister suddenly pushes her brother. As a result, the sister moves with kinetic energy k. The resulting kinetic energy of the brother is k/2.

<h3>What is kinetic energy?</h3>

The ability of an object to do work by virtue of its motion is called the kinetic energy.

A brother and sister are standing next to each other at rest on a surface of frictionless ice. The brother’s mass is exactly twice that of his sister’s.

If the sister's mass is m, the brother's mass is 2m.

The kinetic energy of sister is k =1/2 mv²,

The velocity will also get halved for brother, So, the kinetic energy of brother will be

k' = 1/2 (2m)(v/2)²

On comparing, the relation between the kinetic energy of brother and sister is k' = k/2

Thus, the resulting kinetic energy of the brother is k/2.

Learn more about kinetic energy.

brainly.com/question/12669551

#SPJ4

7 0

1 year ago

A ball is dropped off the side of a bridge. after 1.55 s, how far has it fallen

givi [52]

Answer:

38.64 feet

Explanation:

x=x0 + vx0t + 1/2axt2

x= 0 + 0 + 1/2 X 32.17 ft/sec2 X 1.55 sec2

x = 38.64 feet

7 0

3 years ago

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,

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

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

2 years ago