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kolezko [41]
3 years ago
15

A 3 kg penguin is pushed by his penguin friends who give him an initial speed vo at the top of a 30 m hill. The penguin is hopin

g this will be enough initial speed that he can make it all the way down this hill and up the next hill which is 45 m tall. Neglecting friction and air resistance, what does the initial speed need to be for him to make it up this second hill?
Physics
1 answer:
Strike441 [17]3 years ago
7 0

Answer:

This question can be answered by using conversation of energy.

K_1 + U_1 = K_2 + U_2

\frac{1}{2}mv_{0}^2 + mgh_1 = 0 + mgh_2

\frac{1}{2}(3)v_0^2 + (3)(9.8)(30) = (3)(9.8)(45)\\\frac{1}{2}(3)v_0^2 = 441\\v_0^2 = 294\\v_0 = 17.14 m/s

Explanation:

Note that we take K_2 = 0 because we are looking for the minimum initial speed for the penguin to reach the top of the second hill. Any other speed more than this will already be enough for him.

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A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v
Pavel [41]

Answer:

65

Explanation:

The resonant frequencies for a fixed string is given by the formula  nv/(2L).  

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e  fn+1 – fn = 390 -325

                      =  65

3 0
3 years ago
What is the longest wavelength of light that will emit electrons from a metal whose work function is 3.70 eV
SSSSS [86.1K]

Answer:

h f = Wf + K

where the total energy available is h f, Wf is the work function or the work needed to remove the electron and K is the kinetic energy of the removed electron

If K = zero then hf = Wf

Wf = h f = h c / λ    or

λ = h c / Wf = 6.63E-34 * 3.0E8 / (3.7 * 1.6E-19)

λ = 6.63 * 3 / (3.7 * 1.6) E-7 = 3.36E-7

This would be 3360 angstroms or 336 millimicrons

Visible light = 400-700 millimicrons

6 0
2 years ago
If the area of an iron rod is 10 cm by 0.5 cm and length is 35 cm. Find the value of resistance, if 11x10^-8 ohm.m be the resist
malfutka [58]

Answer:

Resistance of the iron rod, R = 0.000077 ohms    

Explanation:

It is given that,

Area of iron rod, A=10\ cm\times 0.5\ cm=5\ cm^2 = 0.0005\ m^2

Length of the rod, L = 35 cm = 0.35 m

Resistivity of Iron, \rho=11\times 10^{-8}\ \Omega-m

We need to find the resistance of the iron rod. It is given by :

R=\rho\dfrac{L}{A}

R=11\times 10^{-8}\times \dfrac{0.35\ m}{0.0005\ m^2}

R=0.000077 \Omega

So, the resistance of the rod is 0.000077 ohms. Hence, this is the required solution.

6 0
3 years ago
The drag force that resists the motion of a car traveling at 80 km h^- 1 is 300 N.
kobusy [5.1K]

The power require to keep the car traveling is 6,666 W.

The power of the engine at the given efficiency is 3,999.6 W.

<h3>What is Instantaneous power?</h3>

This the product of force and velocity of the given object.

The power require to keep the car traveling is calculated as follows;

P = Fv

P = 300\ N \ \times  \ \frac{80 \ kmh^{-1}}{3.6 \ km h^{-1}/m/s} \\\\&#10;P = 300 \ N \times 22.22 \ m/s\\\\&#10;P = 6,666 \ W

The power of the engine at the given efficiency is calculated as follows;

E = \frac{P_{out}}{P _{in}} \times 100\%\\\\&#10;60\% = \frac{P_{out}}{6,666} \times 100\%\\\\&#10;0.6 = \frac{P_{out}}{6,666} \\\\&#10;P_{out} = 3,999.6 \ W

Learn more about efficiency here: brainly.com/question/15418098

8 0
2 years ago
If an astronaut can throw a certain wrench 15.0 m vertically upward on earth, how high could he throw it on our moon if he gives
Sever21 [200]

Answer:

hm= 18.2 m

Explanation:

Solution is attached

5 0
3 years ago
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