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makkiz [27]
1 year ago
13

A ball of mass 5.0kg is lifted off the floor a distance of 1.7m. 1. What is the change in the gravitational potential energy of

the ball? 2. Now you release the ball from rest, and it falls to the floor. What is the kinetic energy of the ball just before it hits the floor? 3. What is the velocity of ball just before it hits the floor?
Physics
1 answer:
tangare [24]1 year ago
4 0

Answer:

Explanation:

Change in gravitational energy of the ball = mgh

5 mutiply 10 multiply 1.7 = 85J

Potential energy at height = Kinetic energy at bottom

KE= 85J

Velocity

v=5.83m/s

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The gold foil experiment led to the conclusion that each atom in the foil was composed mostly of empty space because most alpha
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Answer:

(1) passed through the foil

Explanation:

Ernest Rutherford conducted an experiment using an alpha particle emitter projected towards a gold foil and the gold foil was surrounded by a fluorescent screen which glows upon being struck by an alpha particle.

  • When the experiment was conducted he found that most of the alpha particles went away without any deflection (due to the empty space) glowing the fluorescent screen right at the point of from where they were emitted.
  • While a few were deflected at reflex angle because they were directed towards the center of the nucleus having the net effective charge as positive.
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Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance
scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

x_{1}=\dfrac{\lambda D}{2d}

Second minima from center

x_{2}=\dfrac{3\lambda D}{2d}

The distance between the places where the intensity is zero due to the double slit effect

\Delta x_{d}=x_{2}-x_{1}

\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}

\Delta x_{d}=\dfrac{\lambda D}{d}

Put the value into the formula

\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}

\Delta x_{d}=0.015 =15\times10^{-3}\ m

\Delta x_{d}=15\ mm

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

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2 years ago
To persuade my audience that the use of mobile communication devices by drivers—even when they are hands-free—is contributing to
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Answer:

specific purpose statement      

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It is a specific purpose statement made for a persuasive speech on the question of fact.

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A specific purpose statement is given to audience to persuade on specific information.

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Anna needs to move a box of paperback books across the room. If she applies a force of 20 newtons to the box, what is the magnit
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Newton's third law tells us that for every force there is an equal and opposite force.  This means that if Anna exerts a force of 20 Newtons on the box, the box exerts a force of 20 Newtons on Anna.
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A 1000 kg roller coaster begins on a 10 m tall hill with an initial velocity of 6m/s and travels down before traveling up a seco
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Answer:

10.6 meters.

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In words this means that the initial kinetic energy of the roller coaster plus its gravitational potential energy minus the energy lost due to friction (1700j) must equal to the final kinetic energy at top of the second hill.

Now let us put in the numerical values in the above equation.

m=100kg

h_i=10m

v_i= 6m/s

v_f=4,6m/s

and solve for h_f

h_f= \frac{\frac{1}{2}mv_i^2+mgh_i-1700j-\frac{1}{2}mv_f^2}{mg} =\boxed{ 10.6\:meters}

Notice that this height is greater than the initial height the roller coaster started with because the initial kinetic energy it had.

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