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Tom [10]
4 years ago
5

Suppose you throw a baseball downward from a roof so that it initially has 120 J of gravitational potential energy, and 10 J of

kinetic energy. What will be true of the kinetic energy at the ground?
A.
It will be 0 J.
B.
It will be greater than 10 J.
C.
It will decrease from 10 J.
D.
It will equal 10 J.
Physics
2 answers:
castortr0y [4]4 years ago
8 0

Answer:

<h2>the correct answer is <u><em>B</em></u></h2>

Explanation:

Volgvan4 years ago
7 0

Answer:

B.

It will be greater than 10 J.

Explanation:

The total mechanical energy of an object is the sum of its potential energy (PE) and its kinetic energy (KE):

E = PE + KE

According to the law of conservation of energy, when there are no frictional forces on an object, its mechanical energy is conserved.

The potential energy PE is the energy due to the position of the object: the highest the object above the ground, the highest its PE.

The kinetic energy KE is the energy due to the motion of the object: the highest its speed, the largest its KE.

Here at the beginning, when it is at the top of the roof, the baseball has:

PE = 120 J

KE = 10 J

So the total energy is

E = 120 + 10 = 130 J

As the ball falls down, its potential energy decreases, since its height decreases; as a result, since the total energy must remain constant, its kinetic energy increases (as its speed increases).

Therefore, when the ball reaches the ground, its kinetic energy must be greater than 10 J.

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Plz i need help fast im timed!!!!!!
professor190 [17]

Answer:

sorry i throght i had the answer

Explanation:

4 0
2 years ago
Read 2 more answers
Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible co
Illusion [34]

Answer:

 q = 8.61 10⁻¹¹ m

charge does not depend on the distance between the two ships.

it is a very small charge value so it should be easy to create in each one

Explanation:

In this exercise we have two forces in balance: the electric force and the gravitational force

          F_e -F_g = 0

          F_e = F_g

Since the gravitational force is always attractive, the electric force must be repulsive, which implies that the electric charge in the two ships must be of the same sign.

Let's write Coulomb's law and gravitational attraction

         k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{r^2}

In the exercise, indicate that the two ships are identical, therefore the masses of the ships are the same and we will place the same charge on each one.

          k q² = G m²

          q = \sqrt{ \frac{G}{k} }    m

we substitute

           q = \sqrt{ \frac{ 6.67 \ 10^{-11}}{8.99 \ 10^{9}} }   m

            q = \sqrt{0.7419 \ 10^{-20}}   m

           q = 0.861 10⁻¹⁰ m

           q = 8.61 10⁻¹¹ m

This amount of charge does not depend on the distance between the two ships.

It is also proportional to the mass of the ships with the proportionality factor found.

Suppose the ships have a mass of m = 1000 kg, let's find the cargo

            q = 8.61 10⁻¹¹ 10³

            q = 8.61 10⁻⁸ C

             

this is a very small charge value so it should be easy to create in each one

6 0
3 years ago
A. Describe the photoelectric effect and explain why it could not be explained by Newtonian physics.
Leno4ka [110]

Explanation:

(a)

The photoelectric effect is the phenomenon in which the light of the particular frequency incidents on the material. Then the emission of the electrons from the surface of the material occurs.

This phenomenon could not be explained by Newtonian physics.

In Newtonian physics, the energy is not discrete. In quantum mechanics, the energy is discrete.  This is the main why the photoelectric effect could not be explained by Newtonian physics.

(b)

Light consists of  photons. The photon is a packet of energy. It is also called quanta. The energies of the photons are quantized.  

When a photon strikes on the surface of metal then the energy of photon is absorbed by an electron in the metal so that it may eject from the surface. This phenomenon is called the photoelectric effect.

(c)

In quantum mechanics, wave-particle duality concept is used to explain the wave-particle nature of the light. Light behaves as particle as well as wave. It shows both nature. The photoelectric phenomenon shows the particle nature of the light. It acts as a particle when it hits the surface of the metal.

In line spectra, the electron is excited to an energy level. In this case energy is transferred from photon to electron. There is a collision between photon and electron. The change in momentum will occur. It shows the particle nature of the light.  

3 0
4 years ago
Read 2 more answers
Conservation of Momentum<br> No one likes you little trolls please send an actual answer
Olin [163]

Hello!

This is an example of an inelastic collision, where the two objects "stick" to each other after their collision. (The Goalkeeper CATCHES the puck).

We can write out the conservation of momentum formula:

m1vi + m2vi = m1vf + m2vf

Let:

m1 = mass of puck

m2 = mass of the goalkeeper

We know that the initial velocity of the goalkeeper is 0, so:

m1vi + m2(0) = m1vf + m2vf

m1vi = m1vf + m2vf

The final velocities will be the same, so:

m1vi = (m1 + m2)vf

Plug in the given values:

(0.16)(40)/ (0.16 + 120) = vf ≈ 0.0533 m/s

Using the equation for momentum:

p = mv

The object with the LARGER mass will have the greater momentum. Thus, the Goalkeeper has the largest momentum as p = mv; a greater mass correlates to a greater momentum since the velocity is the same between the two objects. The puck would have a momentum of p = (.16)(0.0533) = 0.008528 kgm/s, whereas the goalkeeper would have a momentum of

p =  (120)(0.0533) = 6.396 kgm/s.

3 0
2 years ago
Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

8 0
4 years ago
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