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Tom [10]
3 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]3 years ago
8 0

Answer:

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

Explanation:

Volgvan3 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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Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
I need the solution to this
posledela

Answer:

He could jump 2.6 meters high.

Explanation:

Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

v_0 = \sqrt{2gh}=\sqrt{2\cdot 9.8\frac{m}{s^2}\cdot 1.3m}=5.0\frac{m}{s}

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.

Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

v_0^2 = 2g_{1/2}h\implies \\h = \frac{v_0^2}{2g_{1/2}}=\frac{25\frac{m^2}{s^2}}{2\cdot 4.9\frac{m}{s^2}}=2.6m

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.

6 0
3 years ago
(a) what will an object weigh on the moon's surface if it weighs 100 n on earth's surface
juin [17]
We know the equation

weight = mass × gravity

To work out the weight on the moon, we will need its mass, and the gravitational field strength of the moon.
Remember that your weight can change, but mass stays constant.

So using the information given about the earth weight, we can find the mass by substituting 100N for weight, and we know the gravity on earth is 10Nm*2 (Use the gravitational field strength provided by your school, I am assuming yours in 10Nm*2)

Therefore,

100N = mass × 10
mass= 100N/10
mass= 10 kg

Now, all we need are the moon's gravitational field strength and to apply this to the equation

weight = 10kg × (gravity on moon)
4 0
3 years ago
Calculate the kinetic energy in joules of a 1200 kg automobile moving at 18 m/s .
vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

Ke=\frac{1}{2}Mass*Velocity^2

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.

7 0
3 years ago
Read 2 more answers
What is the acceleration of an object with a constant velocity
valina [46]

Zero.

Acceleration is defined as the change in velocity over time.

Since in your case there is no change, there is no acceleration, so it is zero:

Or in formula: <span>a=<span><span>Δv</span>t</span></span>

Where a=acceleration, <span>Δv</span>=change in velocity and t=time

6 0
3 years ago
Read 2 more answers
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