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Sonja [21]
3 years ago
6

If a car with mass 750 kg is lifted to a height of 2.3 m how much gravitational potential energy is added to the car

Physics
1 answer:
Jlenok [28]3 years ago
6 0

I found: 16,905J

Explanation:

The Gravitational Potential Energy will depend on the work done against gravity (weight, W=mg) to lift it at height h or:

U = mgh =750 * 9.8 *2.3 =16,905J

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Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det
chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

       The mass of the fly is  m_f =  11 mg =  11*10^{-3} g =  1.1*10^{-5} \ kg

        The extension of the web is  e=  4.00 \ mm = 0.004 \ m

       

The spring constant is mathematically evaluated as

          k = \frac{mg}{e}

substituting values

        k = \frac{1.1 *10^{-5} *9.8}{0.004}

         k = 0.027 \ N/m

The frequency of vibration is

         f =  \frac{1}{2 \pi} \sqrt{\frac{k}{m} }

substituting values

       f =  \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }

      f = 7.9 Hz

         

8 0
3 years ago
The block in the drawing has dimensions L0×2L0×3L0, where L0 =0.5 m. The block has a thermal conductivity of 250 J/(s·m·C˚). In
stira [4]

Answer:

I am sorry but what drawing

3 0
3 years ago
Which of the following is NOT a characteristic of noble gases?
Rudik [331]
Solid at room temperature
8 0
3 years ago
Read 2 more answers
The work function for tungsten metal is 4.52eV a. What is the cutoff (threshold) wavelength for tungsten? b. What is the maximum
Tanya [424]

Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V

Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:

h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.

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.

5 0
3 years ago
Given a die, would it be more likely to get a single 6 in six rolls, at least two 6s in twelve rolls, or at least one-hundred 6s
vichka [17]

Answer:

Explanation:

In first case we are interested in one time 6 in six rolls

Thus probability = number of chances required/Total chances

= 1/6

Similarly in the second case probability = 2/12 = 1/6

In the same way in last case probability = 100/600 = 1/6

The probability is the same . Thus all the cases has equal chances  

4 0
3 years ago
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