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NeTakaya
3 years ago
6

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

nergy?
Physics
1 answer:
frutty [35]3 years ago
5 0

Answer:

The amount of mass that needs to be converted to release that amount of energy is 1.122 X 10^{-7}  kg

Explanation:

From Albert Einstein's Energy equation, we can understand that mass can get converted to energy, using the formula

E= \Delta mc^{2}

where \Delta m = change in mass

c = speed of light = 3 \times 10 ^{8}m/s

Making m the subject of the formula, we can find the change in mass to be

\Delta m = \frac{E}{c^{2}}= \frac{1.01 \times 10^{3} \times 10^{7}}{(3 \times 10^{8})^{2}}= 1.122 \times 10 ^{-7}kg

There fore, the amount of mass that needs to be converted to release that amount of energy is 1.122 X 10 ^-7 kg

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I have an algorithm that runs in O(N ), where N is the size of the problem. For N = 100, the time for the algorithm to run is 1
hodyreva [135]

Answer:

B) 10 minutes

Explanation:

When the size of the problem is 100 then it takes 1 minute to solve

100\ s=1\ min

So, when the size of the problem is 1 then it takes the following minutes

1\ s=\frac{1}{100}\ min

For when the size of the problem is 1000 we have

1000\ s=\frac{1000}{100}\ min=10\ min

So, it will take 10 minutes to solve the algorithm of size 1000

4 0
3 years ago
calculate speed and velocity of the following. the race car was moving for 3.7 hours and during that time it traveled a distance
Ksju [112]

Answer:

v = 135.13 mph

Explanation:

Given that,

The race car was moving for 3.7 hours and during that time it traveled a distance of 500 miles south.

We need to find the speed of the car.

We know that,

Speed = distance/time

So,

v=\dfrac{500\ miles}{3.7\ h}\\\\v=135.13\ mph

So, the speed of the car is equal to 135.13 mph.

7 0
2 years ago
Extend the life of your __________ by avoiding fast starts, stops and sharp turns.
skelet666 [1.2K]

Answer:

Tires.

Explanation:

There are the few steps which are discussed below should be taken to increase or extend the life of tires.

(1) Avoid fast starts: Fast start of the vehicle will increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.

(2)  Avoid fast stop: Fast stop of the vehicle will also increase the pressure on the tires due to the friction between the tires and the road will decrease the life of tires.

(3) Avoid sharp turns: The alignment of the wheels and tires are in such a way that they work properly when vehicle is drive in a straight path but sharp turn will increase the uneven pressure on the tires will lead to decrease the life of tires.

Therefore, the life of tires can be extend by avoiding all the above mention actions such as fast stop, start and sharp turns.

5 0
3 years ago
What voltage would produce a current of 100amps through an aluminum wire assuming that the re sistance of the wire is3.44×10-4 o
agasfer [191]
V = IR

I = current
R = resistance

Voltage = 100 * (3.44x 10^-4) = do the calculation

Hope this helps
8 0
2 years ago
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
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