PHYSICS COLLEGE

A Hooke's law spring is mounted horizontally over a frictionless surface. The spring is then compressed a distance d and is used to launch a mass m along the frictionless surface. What compression of the spring would result in the mass attaining double the kinetic energy received in the above situation?

Answers

Answer 1

Answer:

Answer:

The compression is $√(2) \ d$ .

Explanation:

A Hooke's law spring compressed has a potential energy

$E_(potential) = (1)/(2) k (\Delta x)^2$

where k is the spring constant and $\Delta x$ the distance to the equilibrium position.

A mass m moving at speed v has a kinetic energy

$E_(kinetic) = (1)/(2) m v^2$ .

So, in the first part of the problem, the spring is compressed a distance d, and then launch the mass at velocity $v_1$ . Knowing that the energy is constant.

$(1)/(2) m v_1^2 = (1)/(2) k d^2$

If we want to double the kinetic energy, then, the knew kinetic energy for a obtained by compressing the spring a distance D, implies:

$2 * ((1)/(2) m v_1^2) = (1)/(2) k D^2$

But, in the left side we can use the previous equation to obtain:

$2 * ((1)/(2) k d^2) = (1)/(2) k D^2$

$D^2 = (2 \ ((1)/(2) k d^2))/((1)/(2) k)$

$D^2 = 2 \ d^2$

$D = √(2 \ d^2)$

$D = √(2) \ d$

And this is the compression we are looking for

Answer 2

Answer:

Answer:

$d'=√(2) d$

Explanation:

By hooke's law we have that the potential energy can be defined as:

$U=(kd^(2) )/(2)$

Where k is the spring constant and d is the compression distance, the kinetic energy can be written as

$K=(mv^(2) )/(2)$

By conservation of energy we have:

$(mv^(2) )/(2)=(kd^(2) )/(2)$ (1)

If we double the kinetic energy

$2((mv^(2) )/(2))=(kd'^(2) )/(2)$ (2)

where d' is the new compression, now if we input (1) in (2) we have

$2((kd^(2) )/(2))=(kd'^(2) )/(2)$

$2((d^(2) )/(2))=(d'^(2) )/(2)$

$d'=√(2) d$

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Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009

The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
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A loop of wire lies flat on the horizontal surface in an area with uniform magnetic field directed vertically up. The loop of wire suddenly contracts to half of its initial diameter. As viewed from above induced electric current in the loop isa. counterclockwiseb. clockwisec. there is no current in the loop because magnetic field is uniformd. there is no current in the loop because magnetic field does not change

Answers

Complete Question

a. counterclockwise

b. clockwise

c. there is no current in the loop because magnetic field is uniform

d. there is no current in the loop because magnetic field does not change

Answer:

Option A is the correct answer

Explanation:

According to the question the loop of wire contracts to half it initial diameter and will mean that less number of electric field line will pass through the loop and this change in magnetic flux will cause current to flow in the loop of wire and from Lenz's law this current will in the opposite direction of what produced it which is the change in magnetic flux so the current will flow in a counterclockwise direction

How would a spinning disk's kinetic energy change if its moment of inertia was five times larger but its angular speed was five times smaller

Answers

A spinning disk's kinetic energy will change to one-tenth if its moment of inertia was five times larger but its angular speed was five times smaller.

Relation between Kinetic energy and Moment of Inertia:

Rotational kinetic energy is directly proportional to the rotational inertia and the square of the magnitude of the angular velocity.

Now, let's consider moment of inertia = I and angular speed = ω

It is asked that what would be change in Kinetic energy if

moment of inertia = (five times larger)

angular speed = ω/5 (five times smaller)

The kinetic energy of a spinning body is given as:

$K.E.=(1)/(2) I. w^2$

On substituting the values, we will get:

$K.E.= (1)/(2) (5I) ((w)/(5) )^2 \n\nK.E. =(1)/(10) I. w^2$

Kinetic energy will be one-tenth to the kinetic energy before its spinning characteristics were changed.

Learn more:

brainly.com/question/12337396

What is the pH of a solution with a hydrogen ion concentration of 2.0x10^3.(Use 3 digits)

Answers

Answer:

2.70

Explanation:

pH = -log[H+]

pH = -log[2.0x10^-3]

pH = 2.70

Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

Answers

Answer:

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

$Electric FIeld = (Change\: in\: Potential)/(Distance)$

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance= $0.8 * 10^(-2) m$

So, Electric field is

$Electric\: field=(3.8V)/(0.8 * 10^(-2)m )$

$Electric\: field=475 V/m$

So, electric field is 475 Volts per meter.

Note : Also we can say 475 Newtons per coulomb

You know that you sound better when you sing in the shower. This has to do with the amplification of frequencies that correspond to the standing-wave resonances of the shower enclosure. A shower enclosure is created by adding glass doors and tile walls to a standard bathtub, so the enclosure has the dimensions of a standard tub, 0.75 m wide and 1.5 m long. Standing sound waves can be set up along either axis of the enclosure. What are the lowest two frequencies that correspond to resonances on each axis of the shower? These frequencies will be especially amplified. Assume a sound speed of 343 m/s.