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3 years ago

True or false? objects that are sitting still have kinetic energy

1 answer:

5 0

Kinetic energy is the energy of an object due to its motion. An object sitting still isn't moving, therefore it has no kinetic energy. The statement in the question is false.

You might be interested in

What is the speed of terminal velocity?

ivann1987 [24]

The so-called "terminal velocity" is the fastest that something can fall
through a fluid. Even though there's a constant force pulling it through,
the friction or resistance of plowing through the surrounding substance
gets bigger as the speed grows, so there's some speed where the resistance
is equal to the pulling force, and then the falling object can't go any faster.

A few examples:
-- the terminal velocity of a sky-diver falling through air,
-- the terminal velocity of a pecan falling through honey,
-- the terminal velocity of a stone falling through water.

It's not possible to say that "the terminal velocity is ----- miles per hour".
If any of these things changes, then the terminal velocity changes too:

-- weight of the falling object
-- shape of the object
-- surface texture (smoothness) of the object
-- density of the surrounding fluid
-- viscosity of the surrounding fluid .

4 0

4 years ago

n deep space, sphere A of mass 47 kg is located at the origin of an x axis and sphere B of mass 110 kg is located on the axis at

Volgvan

Answer:

a)-1.014x $10^{-7$ J

b)3.296 x $10^{-7$ J

Explanation:

For Sphere A:

mass 'Ma'= 47kg

xa= 0

For sphere B:

mass 'Mb'= 110kg

xb=3.4m

a)the gravitational potential energy is given by

$U_{i$ = -GMaMb/ d

$U_{i$ = - 6.67 x $10^{-11}$ x 47 x 110/ 3.4 => -1.014x $10^{-7$ J

b) at d= 0.8m (3.4-2.6) and $U_{i$ =-1.014x $10^{-7$ J

The sum of potential and kinetic energies must be conserved as the energy is conserved.

$K_{i$ + $U_{i$ = $K_{f$ + $U_{f$

As sphere starts from rest and sphere A is fixed at its place, therefore $K_{i$ is zero

$U_{i$ = $K_{f$ + $U_{f$

The final potential energy is

$U_{f$ = - GMaMb/d

Solving for ' $K_{f$ '

$K_{f$ = $U_{i$ + GMaMb/d => -1.014x $10^{-7$ + 6.67 x $10^{-11}$ x 47 x 110/ 0.8

$K_{f$ = 3.296 x $10^{-7$ J

6 0

3 years ago

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the sys

Archy [21]

Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.

<h3>Conservation of energy</h3>

The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B$

$v_B = \frac{2v_0}{3} \ m/s \ to \ the \ left$

At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.

The complete question is below:

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?

Learn more about conservation of energy here: brainly.com/question/166559

5 0

2 years ago

Read 2 more answers

A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient

gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0

3 years ago

Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a

malfutka [58]

Answer: