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spin [16.1K]
3 years ago
8

an object with a mass of 70kilograms is supported at a height 8meters above the ground. what's the potential energy of the objec

t with respect to the ground ?(a)560J(b)2,058J(c)5,488J(d)3,430J
Physics
1 answer:
ra1l [238]3 years ago
3 0
Gravitational Potential Energy = mgh (m=mass; g=gravitational force(9.8N/kg); h = height)

Ug = (70kg)(9.8N/kg)(8m) = 5488J which is C.)
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A 160.-kilogram space vehicle is traveling along a
mario62 [17]

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4 0
3 years ago
g Estimate the number of photons emitted by the Sun in a second. The power output from the Sun is 4 × 1026 W and assume that th
vagabundo [1.1K]

Answer:

The value is N  =  1.107 *10^{45 }  \ photons    

Explanation:

From the question we are told that

   The  power output from the sun is  P_o =  4 * 10^{26} \  W

   The average wavelength of each photon is  \lambda  = 550 \  nm  =  550 *10^{-9} \  m

Generally the energy of each photon emitted is mathematically represented as

        E_c =  \frac{h * c  }{ \lambda }

Here  h is the Plank's constant with value  h  =  6.62607015 * 10^{-34} J \cdot s

          c is the speed of light with value  c =  3.0 *10^{8} \  m/s

So

       E_c =  \frac{6.62607015 * 10^{-34}  * 3.0 *10^{8}  }{ 550 *10^{-9} }          

=>   E_c =  3.614 *10^{-19} \  J          

Generally the  number of photons emitted by the Sun in a second is mathematically represented as

         N  =  \frac{P }{E_c}

=>      N  =  \frac{4 * 10^{26} }{3.614 *10^{-19}}

=>      N  =  1.107 *10^{45 }  \ photons    

5 0
2 years ago
(a) Two ions with masses of 4.39×10^−27 kg move out of the slit of a mass spectrometer and into a region where the magnetic fiel
sammy [17]

Answer:

Part a)

R_1 = 0.072 m

Part b)

R_2 = 0.036 m

Part c)

d = 0.072 m

Explanation:

Part a)

As we know that the radius of the charge particle in constant magnetic field is given by

R = \frac{mv}{qB}

now for single ionized we have

R_1 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(1.6 \times 10^{-19})(0.301)}

R_1 = 0.072 m

Part b)

Similarly for doubly ionized ion we will have the same equation

R = \frac{mv}{qB}

R_2 = \frac{(4.39\times 10^{-27})(7.92 \times 10^5)}{(3.2 \times 10^{-19})(0.301)}

R_2 = 0.036 m

Part c)

The distance between the two particles are half of the loop will be given as

d = 2(R_1 - R_2)

d = 2(0.072 - 0.036)

d = 0.072 m

6 0
3 years ago
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