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2 years ago

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

2 answers:

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.

You might be interested in

‏Explain how an electron emitted by the photoelectric effect can have kinetic energy less than threshold energy ?

xenn [34]

Answer:

the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

K = E - Ф

Explanation:

The photoelectric effect is the emission of electrons from the surface of a metal.

This was correctly explained by Einstein, in his explanation the energy of the photons (quanta of light) collide with the electrons, these electrons have to overcome the threshold energy that is the energy of union with the metal, and the energy that remains is converted to kinetic energy.

E = hf

E = K + Ф

K = E - Ф

The energy of the photons is given by the Planck relation E = hf and according to Einstein the number of joints must be added

E = n hf

Therefore, depending on the value of this energy, the emitted electrons can have energy from zero onwards.

4 0

2 years ago

La velocidad de un tren se reduce uniformemente de 12m/s a 5m/s. Sabiendo que durante ese tiempo recorre una distancia de 100 m.

MA_775_DIABLO [31]

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

Responder:

Explicación:

Dados los siguientes datos

Valor inicial u = 12 m / s

velocidad final v = 5 m / s

Distancia S = 100 m

Necesario

aceleración del tren.

Usando la ecuación de movimiento

v² = u² + 2as

2as = v²-u²

a = v²-u² / 2s

Sustituyendo los valores dados para obtener la aceleración que tenemos;

a = 5²-12² / 2 (100)

a = 25-144 / 200

a = -119/200

a = -0,595 m / s²

Por tanto, la aceleración del tren durante este período es de -0,595 m / s²

b) Si el tren viaja a una parada desde 5 m / s, su velocidad final será cero y su velocidad inicial u será 5 m / s

Para obtener la distancia durante este período, sustituiremos u = 5 m / s, v = 0 m / sy a = -1,19 m / s² en la ecuación de movimiento anterior;

v² = u² + 2as

0² = 5²-2 (0.595) s

0 = 25-1,19 s

1,19 s = 25

s = 25 / 1,19

s = 21,0 m

Por lo tanto, la distancia que recorre hasta detenerse asumiendo la misma aceleración es 21.0 m

6 0

2 years ago

If you use the same force to push a motorcycle as you would push a bike which one would have more acceleration and why explain u

zaharov [31]

Answer:

The bike would have more acceleration

Explanation:

Accourding to newtons first law a force is equal to its mass multiplied by its acceleration (f=ma) therefore an object with a higher mass compared to an object with a lower mass would experience less acceleration.

Eg.

F=50N

Motorbike M=200kg

F=ma

50=200 x a

50/200=a

0.25m/s/s =a

Bike M=35kg

F=ma

50=35 x a

50/35= a

1.43m/s/s=a

6 0

3 years ago

Read 2 more answers

A man is running on the straight road with the uniform velocity of 3 metre per second calculate the acceleration for produced hi

sp2606 [1]

Theoritically

the body moving with uniform velocity has acceleration zero.

Mathmatically,

u=3m/s

v=3m/s (since body is moving with uniform velocity)

a= v-u/t

3-3/t

0/t

0m/s.s

6 0

2 years ago

A ball is rolled of a 2.14m table at 8m/s. Find the time of flight for the ball and the horizontal range

Mama L [17]

Answer:

0.661 s, 5.29 m

Explanation:

In the y direction:

Δy = 2.14 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²

t = 0.661 s

In the x direction:

v₀ = 8 m/s

a = 0 m/s²

t = 0.661 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²

Δx = 5.29 m

Round as needed.

3 0

2 years ago