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Juli2301 [7.4K]

2 years ago

15

As an object is raised to a higher position,Which type of energy increases?

2 answers:

sertanlavr [38]2 years ago

8 0

As the object is raised higher, its gravitational potential energy increases which is denoted in a formulae as gpe = mgh where m is mass, g is gravitational accelaration and h is height.
Therefore height increases proportionally to gpe if mass and acceleration are kept constant!

If you need further help over this topic, I will be more than happy to help you out, just message me!

Hope this helps!

Marizza181 [45]2 years ago

7 0

By raising a position of an object, Its potential energy increases.

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I don't understand this question at all, can I please get some help?

beks73 [17]

V^2/R=180W
v=root 180R

4 0

3 years ago

Help me answer this question and get yourself some points!

Marrrta [24]

Answer:

A×B=C×D

500×0.5=250×X

250=250×X

X=250/250=1

X=1 m

Explanation:

note: if the force plus two, the distance will be half.

4 0

2 years ago

madreJ [45]

Answer:

$f=6.97\times 10^{14}\ Hz$

Explanation:

Given that,

Wavelength, $\lambda=430\ nm=430\times 10^{-9}\ m$

We need to find the frequency of the violet light.

We know that the relation between frequency and wavelength is given by :

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{430\times 10^{-9}}\\\\f=6.97\times 10^{14}\ Hz$

So, the frequency of violet light is $6.97\times 10^{14}\ Hz$ .

7 0

3 years ago

How much work is accomplished when a force of 250 N pushes a box across the floor for a distance of 50 meters?

IrinaK [193]

Work = force * distance
and newton*meters = Joule

In this case, work = 250N*50m = 12500 J

So the answer is D) 12,500 J

3 0

3 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

<em>Part B </em>

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago