The type of energy that depends on position is called

If your chunk of gold weighed 1 N in which case would you have the largest mass of gold?

kotykmax [81]

Ah ha ! Very interesting question.
Thought-provoking, even.

You have something that weighs 1 Newton, and you want to know
the situation in which the object would have the greatest mass.

Weight = (mass) x (local gravity)

Mass = (weight) / (local gravity)

Mass = (1 Newton) / (local gravity)

"Local gravity" is the denominator of the fraction, so the fraction
has its greatest value when 'local gravity' is smallest. This is the
clue that gives it away.

If somebody offers you 1 chunk of gold that weighs 1 Newton,
you say to him:

"Fine ! Great ! Golly gee, that's sure generous of you.
But before you start weighing the chunk to give me, I want you
to take your gold and your scale to Pluto, and weigh my chunk
there. And if you don't mind, be quick about it."

The local acceleration of gravity on Pluto is 0.62 m/s² ,
but on Earth, it's 9.81 m/s.

So if he weighs 1 Newton of gold for you on Pluto, its mass will be
1.613 kilograms, and it'll weigh 15.82 Newtons here on Earth.

That's almost 3.6 pounds of gold, worth over $57,000 !

It would be even better if you could convince him to weigh it on
Halley's Comet, or on any asteroid. Wherever he's willing to go
that has the smallest gravity. That's the place where the largest
mass weighs 1 Newton.

3 0

3 years ago

A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b

Softa [21]

Answer:

The net force on the block F(net) = mgsinθ).

Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

F(net) = mgsinθ

The net force on the block F(net) = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

n=mgcosθ

The horizontal component of normal force on the block is equal to force

Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0

3 years ago

If the frequency of a periodic wave is doubled, the period of the wave will be

Lunna [17]

The period of the wave would be halved

5 0

2 years ago

Calculate the wavelength of each frequency of electromagnetic radiation: a. 100.2 MHz (typical frequency for FM radio broadcasti

Natalka [10]

Answer:

a). 100.2 MHz (typical frequency for FM radio broadcasting)

The wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

The wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

The wavelength of a frequency of 835.6 Mhz is 0.35m.

Explanation:

The wavelength can be determined by the following equation:

$c = \lambda \cdot \nu$ (1)

Where c is the speed of light, $\lambda$ is the wavelength and $\nu$ is the frequency.

Notice that since it is electromagnetic radiation, equation 1 can be used. Remember that light propagates in the form of an electromagnetic wave.

a). 100.2 MHz (typical frequency for FM radio broadcasting)

Then, $\lambda$ can be isolated from equation 1:

$\lambda = \frac{c}{\nu}$ (2)

since the value of c is $3x10^{8}m/s$ . It is necessary to express the frequency in units of hertz.

$\nu = 100.2 MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $100200000Hz$

But $1Hz = s^{-1}$

$\nu = 100200000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{100200000s^{-1}}$

$\lambda = 2.99 m$

Hence, the wavelength of a frequency of 100.2 Mhz is 2.99m.

b. 1070 kHz (typical frequency for AM radio broadcasting) (assume four significant figures)

$\nu = 1070kHz . \frac{1000Hz}{1kHz}$ ⇒ $1070000Hz$

But $1Hz = s^{-1}$

$\nu = 1070000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{1070000s^{-1}}$

$\lambda = 280.3 m$

Hence, the wavelength of a frequency of 1070 khz is 280.3 m.

c. 835.6 MHz (common frequency used for cell phone communication)

$\nu = 835.6MHz . \frac{1x10^{6}Hz}{1MHz}$ ⇒ $835600000Hz$

But $1Hz = s^{-1}$

$\nu = 835600000s^{-1}$

Finally, equation 2 can be used:

$\lambda = \frac{3x10^{8}m/s}{835600000s^{-1}}$

$\lambda = 0.35 m$

Hence, the wavelength of a frequency of 835.6 Mhz is 0.35m.

6 0

3 years ago

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experi

melomori [17]

Complete Question:

The following three hot samples have the same temperature. The same amount of heat is removed from each sample. Which one experiences the smallest drop in temperature, and which one experiences the largest drop? Sample A: 4.0 kg of water [c = 4186 J/(kg·C°)] Sample B: 2.0 kg of oil [c = 2700 J/(kg·C°)] Sample C: 9.0 kg of dirt [c = 1050 J/(kg·C°)]

Answer:

A. Smallest B. Largest.

Explanation:

Assuming no heat exchange except for the heat removed from any sample (which we know is the same for the three ones), and that the process is done using only conduction, we can use the equation that relates the heat lost or gained by one object, with the mass of the object and the consequent change in temperature, as follows:

Q = c*m*ΔT, where c, is a proportionality constant called specific heat, which is different for each material.

As we know that the heat removed is the same for the three samples, we can equate the right sides of the equation for each sample, as follows:

cw*mw*ΔTw = co*mo*ΔTo = cd*md*ΔTd

Replacing by the givens, we have:

4.0 kg. 4,186 J/kgºC*ΔT(ºC) = 2.0 kg*2,700 J/kgºC*ΔT(ºC) =9.0kg*1,050J/kgºC*ΔT(ºC)

As the three expressions must be equal each other, it's clear that the unknown term (the drop in temperature) must compensate the product of the mass times the specific heat.

This product is the following for the three samples:

Water: 4.0 kg*4,186 J/kgºC = 16,744 J/ºC

Oil : 2.0 kg*2,700 J/kgºC = 5,400 J/ºC

Dirt: 9.0 * 1,050 J/kgºC = 9,450 J/ºC

Clearly, we see that in order to keep the heat exchange equations equal each other, the water must suffer the smallest drop in temperature, and the oil must experience the largest one.

So, the sample A experiencies the smallest drop in temperature, and sample B does the largest one.

5 0

2 years ago