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4 years ago

A 5 kg rock is raised 28 m above the ground level. What is the change in its potential energy?

1 answer:

3 0

Let's assume that ground level is the height 0 meters. The change in potential energy is going to be gravitational potential energy, which is given by PE=mgh.
ΔPE=mgh-mgy
=mg(h-y)
=50(28-0)
=1400 J

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Which answer shows a balanced nuclear equation for the alpha decay of plutonium-239

Setler [38]

Explanation:

When plutonium - 239 undergoes an alpha decay, it produces uranium - 235 in the process.

A radioactive occurs when a nuclide spontaneously disintegrates due to instability to produce lighter atoms which the release of radiation and energy.

²³⁹₉₄Pu ⇒ ²³⁵₉₂U + ⁴₂He + energy

In a nuclear reaction, the mass and atomic numbers must always remain conserved. Nuclear reactions do not obey the law of conservation of matter and mass.

The mass number are the superscript before the symbol of the nuclide

The atomic number is the subscript.

The sum of the atomic number on the left hand side equals that on the right hand side. This goes for the mass number too.

learn more:

Balanced nuclear equation brainly.com/question/10094982

#learnwithBrainly

4 0

3 years ago

A cart moves with negligible friction or air resistance along a roller coaster track. The cart starts from rest at the top of a

lina2011 [118]

Answer:

hinit = 17.5 m

Explanation:

Assuming no friction present, the mechanical energy must be conserved, which means that at any point of the trajectory, the sum of the gravitational potential energy and the kinetic energy must keep the same.
At the top of the hill, since it starts from rest, all the energy must be potential, and we can express it as follows:

$E_{o} = U_{o} = m*g*h_{init} (1)$

When the car arrives to the top of the second hill, as we know that it is lower than the first one, the energy of the car, must be part gravitational potential energy, and part kinetic energy.
We can express this final energy as follows:

$E_{f} = U_{f} + K_{f} = m*g* h_{2} + \frac{1}{2} *m*v_{f} ^{2} (2)$

In order to find hinit, we need to make (1) equal to (2), and solve for it.
In (2) we have the value of h₂ (10 m), but we still need the value of the speed at the top of the second hill, vf.
Now, when the car is at the top of the hill, there are two forces acting on it, in opposite directions: the normal force (upward) and the weight (downward).
We know also that there is a force that keeps the car along the circular track, which is the centripetal force.
This force is just the net downward force acting on the car (it's vertical at the top), and is just the difference between the weight and the normal force.
If the cart just barely loses contact with the track at the top of the second hill, this means that at that point the normal force becomes zero.
So, the centripetal force must be equal to the weight.
The centripetal force can be expressed as follows:

$F_{c} = m*\frac{v_{f} ^{2}}{R} (3)$

We have just said that (3) must be equal to the weight:

$F_{c} = m*\frac{v_{f} ^{2}}{R} = m*g (4)$

Simplifying, and rearranging, we can solve for vf², as follows:

$v_{f}^{2} = R*g (5)$

Replacing (5) in (2), simplifying and rearranging in (1) and (2) we finally have:

$h_{init} = h_{2} + \frac{1}{2} R = 10m + 7.5 m = 17.5 m (6)$

7 0

3 years ago

You used a telescope and other mathematics to discover that Jupiter is 5.20 au from the sun. Use the equation to find its orbita

meriva

Answer:

11.9 years

Explanation:

We can find the orbital period by using Kepler's third law, which states that the ratio between the square of the orbital period and the cube of the average distance of a planet from the Sun is constant for every planet orbiting aroudn the Sun:

$\frac{T^2}{r^3}=const.$

Using the Earth as reference, we can re-write the law as

$\frac{T_e^2}{r_e^2}=\frac{T_j^2}{r_j^3}$

where

Te = 1 year is the orbital period of the Earth

re = 1 AU is the average distance of the Earth from the Sun

Tj = ? is the orbital period of Jupiter

rj = 5.20 AU is the average distance of Jupiter from the Sun

Substituting the numbers and re-arranging the equation, we find:

$T_j=\sqrt{\frac{T_e^2 r_j^3}{T_j^2}}=\sqrt{\frac{(1 y)^2 (5.2 AU)^3}{(1 AU)^3}}=11.9 y$

4 0

3 years ago

Read 2 more answers

The planet Uranus has a radius of 25,360 km and a surface acceleration due to gravity of 9.0 m/s^2 at its poles. Its moon Mirand

AlexFokin [52]

Answer:

$8.67791\times 10^{25}\ kg$

$0.34589\ m/s^2$

$0.07903\ m/s^2$

Explanation:

M = Mass of Uranus

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Uranus = 25360 km

h = Altitude = 104000 km

$r_m$ = Radius of Miranda = 236 km

m = Mass of Miranda = $6.6\times 10^{19}\ kg$

Acceleration due to gravity is given by

$g=\dfrac{GM}{r^2}\\\Rightarrow M=\dfrac{gr^2}{G}\\\Rightarrow M=\dfrac{9\times 25360000^2}{6.67\times 10^{-11}}\\\Rightarrow M=8.67791\times 10^{25}\ kg$

The mass of Uranus is $8.67791\times 10^{25}\ kg$

Acceleration is given by

$a_m=\dfrac{GM}{(r+h)^2}\\\Rightarrow a_m=\dfrac{6.67\times 10^{-11}\times 8.67791\times 10^{25}}{(25360000+104000000)^2}\\\Rightarrow a_m=0.34589\ m/s^2$

Miranda's acceleration due to its orbital motion about Uranus is $0.34589\ m/s^2$

On Miranda

$g_m=\dfrac{Gm}{r_m^2}\\\Rightarrow g_m=\dfrac{6.67\times 10^{-11}\times 6.6\times 10^{19}}{236000^2}\\\Rightarrow g_m=0.07903\ m/s^2$

Acceleration due to Miranda's gravity at the surface of Miranda is $0.07903\ m/s^2$

No, both the objects will fall towards Uranus. Also, they are not stationary.

6 0

3 years ago

What is the power generation in a machine that produces 76 j in 3.7s

ICE Princess25 [194]

Power is the ratio between energy and time:
$P= \frac{E}{t}$
In our problem we have E=76 J and t=3.7 s. Therefore, the power is
$P= \frac{76 J}{3.7 s} =20.5 W$

3 0

3 years ago