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tensa zangetsu [6.8K]
3 years ago
14

Emily holds a banana of mass m over the edge of a bridge of height h. She drops the banana and it falls to the river below. Use

conservation of energy to show that the speed of the banana just before hitting the water is v
Physics
1 answer:
bearhunter [10]3 years ago
5 0

Answer:

The mass of the banana is m and it is at height h.

Applying the Law of Conservation of Energy

              Total Energy before fall = Total Energy after fall

                                E_{i}  = E_{f}

Here, total energy is the sum of kinetic energy and potential energy

K.E_{i} + P.E_{i} = K.E_{f} + P.E_{f}       (a)

When banana is at height h, it has

                 K.E_{i} = 0    and    P.E_{i} = mgh          

and when it reaches the river, it has

       K.E_{f}  = 1/2mv^{2}    and   P.E_{f}  = 0

Putting the values in equation (a)

                              0 + mgh = 1/2mv^{2} + 0

                                      mgh = 1/2mv^{2}

<em>cutting 'm' from both sides</em>

<em>                                           </em>gh = 1/2v^{2}

                                          v = \sqrt{2gh}

Hence, the velocity of banana before hitting the water is

                                          v = \sqrt{2gh}

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Answer:

125 kilometers

Explanation:

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Distance= Speed x Time

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1.How far is Object Z from the origin at t = 3 seconds?
GenaCL600 [577]

Answer:

Please find the answer in the explanation

Explanation:

1.) How far is Object Z from the origin at t = 3 seconds

The distance of the object Z from the origin will be the slope of the graph.

Slope = 4/2 = 2m

2.) Which object takes the least time to reach a position 4 meters from the origin ?

According to the graph given to the question above, object Z has the list time which is 2 seconds since object X does not start from the origin.

3.) Which object is farthest from the origin at t = 2 seconds?

The correct answer is still object Z because it has the highest slope.

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3 years ago
How does Newton's second law apply to an "Egg Drop" experiment?
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In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

Therefore, compared to the medium in this question with \theta_{c} = 27^{\circ}, the medium with the larger critical angle \theta_{c} = 32^{\circ} would have a larger \sin(\theta_{c}). such that light would travel faster in that medium.

4 0
3 years ago
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
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Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

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In order to calculate the cutoff wavelength we have to consider that Ek=0

in this case  h*ν=W

(h*c)/λ=4.52 eV

λ= (h*c)/4.52 eV

λ= (1240 eV*nm)/(4.52 eV)=274.34 nm

From this h*ν = Ek+W;  we can calculate the kinetic energy for a radiation wavelength of 198 nm

then we have

(h*c)/(λ)-W= Ek

Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV

Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this  acts to slow down the ejected electrons from the catode.

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