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True [87]
3 years ago
6

16

Physics
1 answer:
seropon [69]3 years ago
8 0

Answer:  Kinetic energy is proportional to the square of the velocity. If the velocity of an object doubles, the kinetic energy increases by a factor of four. Kinetic energy is proportional to the square

of the velocity. If the velocity of an object

doubles, the kinetic energy increases by a

factor of four.

• Kinetic energy is proportional to the mass. If

a bowling ball and a ping pong ball have the

same velocity, the bowling ball has much

larger kinetic energy.

• Kinetic energy is always positive.

• unit : Joule (J) = kg m

2

/s

2 Example:

If we drop a 3-kg ball from a height of h = 10 m,

the velocity when the ball hits the ground is

given by: v 2 = v0 2 +2a(y− y0 )= 0−2g(0−h)v= 2gh= 2(9.8 m/s 2 )(10 m)=14 m /s Initial:   k = 1 2 mv 2 = 0 Final:    k = 1 2 mv 2= 1 2 (3 kg)(14 m/s) 2= 294 J So as the ball falls, its kinetic energy increases. It is the gravitational force that accelerates the ball, causing the speed to increase. The increase in speed also increases the kinetic energy. The process of a force changing the kinetic energy of an object is called work. Work: Work is the energy transferred to or from an object by mean of a force acting on the object.• energy transferred to an object is positive work, e.g. gravity performs positive work on a

falling ball by transferring energy to the ball, causing the ball to speed up.• energy transferred from an object is negative work, e.g. gravity performs negative work on a ball tossed up by transferring energy from the ball, causing the ball to slow down.• both kinetic energy and work are scalars.• unit: J Work Energy Theorem: The work done is equal to the change in the kinetic energy: ∆K = K f − K i = W In the above example with the ball falling from a height of h = 10 m, the work done by gravity: W = ∆k = k f −k i = 294 J− 0J = 294 J. If a ball rises to a height of h =10 m, the work done by gravity: W = ∆k = k f −k i = 0J−294 J = −294 J. Work Done by a Force: Consider a box being dragged a distance d across a frictionless floor:

d F y x θ v 2 = v0 2 + 2ax (x − x0 ) v 2 = v0 2 +2ax d 1 2 mv 2 = 1 2 mv0 2 +max d 1 2 mv 2 − 1 2 mv0 2 = max d k f −k i = (Fcosθ)d ∴W = (Fcosθ)d• θ is the angle between the force vector and the direction of motion.• If the force is perpendicular to the direction of motion, then the work done: W =(Fcosθ)d = Fdcos90°= Fd×0= 0.• The work energy theorem and the relationship between work and force are valid only if the force does not cause any other form of energy to change, e. g. we can not apply the theorem when friction is

involved because it causes a change in the thermal energy (temperature). Work Done by Multiple Forces: The total work done by many forces acting on an object:Wtot = F1 cosθ 1 d+F2 cosθ 2 d+ F3 cosθ 3 d+L where the angles are the angle between each force and the direction of motion.  The total work is just the sum of individual work from each force:Wtot =W1 +W2 +W3 +L The work energy theorem relates the changes  in the kinetic energy to the total work performed on the object: ∆K =Wtot Example: A 3-kg box initially at rest slides 3 m down a frictionless 30° incline.  What is the work done on the object?  What is the kinetic energy and speed at the bottom?

x y N φ φ mg• The work done is performed by the force in the x direction since there is no motion in the y direction: W = F x d =(mgsinφ)d =(3 kg)(9.8 m/s 2 )(sin30°)(3 m) = 44 J Alternatively, W =(Fcosθ)d = Fcos(90°−φ)d = FsinφdH The first method of using the component of the force in the direction of motion for the calculation is easier.

You might be interested in
What is the unit for work is called
emmasim [6.3K]

Answer:

The unit you should use for work done and energy is the joule (J) which is indeed the same as the newton metre (N m).

There is another physical quantity which is the product of force and distance and that is torque or moment of a force.

The unit you should use for torque is the newton metre (Nm) and not the joule.

Naming the units of work done and torque differently helps to emphasis the fact that work done and torque refer to two different physical quantities although the definitions of both quantities have the product of force and distance in them.

work done=force→⋅displacement→ and torque→=force→×displacement→

Hope I helped

4 0
3 years ago
Wel.
Law Incorporation [45]

Answer:

1 is c 2 is a and 3 is b hope that helped!

4 0
3 years ago
3. Your friend says your body is made up of more than 99.9999% empty space. What do you think?
Nesterboy [21]

Answer:

I would agree with the statement. it's not just the body, but everything that we see is almost 99.9999% empty space

8 0
3 years ago
You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin
Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

          F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

            W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

          r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

         r = Re + 250

We substitute

     W = G m M / r²

      W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

      W = G m M / Re² (1+ 250 / Re)²

      (1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

      (1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

      W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

     W = Wo (1 - 7.85  10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0
3 years ago
Read 2 more answers
The atmospheric pressure on the top of the Engineering Sciences Building (ESB) is 97.6 kPa, while that in Room G39-ESB (ground f
Stells [14]

Answer:

Δ h = 52.78 m

Explanation:

given,

Atmospheric pressure at the top of building = 97.6 kPa

Atmospheric pressure at the bottom of building = 98.2 kPa

Density of air = 1.16 kg/m³

acceleration due to gravity, g = 9.8 m/s²

height of the building = ?

We know,

Δ P = ρ g Δ h

(98.2-97.6) x 10³ = 1.16 x 9.8 x Δ h

11.368 Δ h = 600

Δ h = 52.78 m

Hence, the height of the building is equal to 52.78 m.

6 0
3 years ago
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