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docker41 [41]
3 years ago
12

what is the closest to the order of magnitude of the energy rquired to split a nucleus in to its individual protons and neutrons

Physics
1 answer:
skad [1K]3 years ago
5 0

Answer:

More than two million electron volts.

Explanation:

More than two million electron volts energy are needed to break or split a deuteron into a proton and a neutron. Nuclear binding energy is the type of energy that is required to split an atom's nucleus into protons and neutrons. The deuteron is an isotope of hydrogen that is composed of a proton and a neutron and it is a stable particle. Very huge amount of energy is needed for the splitting of nucleus due to the presence of heavy particles i.e. proton and neutron.

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A space shuttle sits on the launch pad for 2.0 minutes, and then goes from rest to 4600 m/s in 8.0 minutes. Treat its motion as
SpyIntel [72]

Answer:

a.) a = 0 ms⁻²

b.) a = 9.58 ms⁻²

c.) a = 7.67 ms⁻²

Explanation:

a.)

    Acceleration (a) is defined as the time rate of change of velocity

                       a = \frac{v_{2} - v_{1} } {t}  

Given data

 Final velocity = v₂ = 0 m/s

 Initial velocity = v ₁ = 0 m/s

  As the space shuttle remain at rest for the first 2 minutes i.e there is no change in velocity so,

                 a = 0 ms⁻²

b.)

     Given data

As the space shuttle start from rest, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 8 min = 480 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {480}

                     a = 9.58 ms⁻²

c.)

    Given data

As the space shuttle is at rest for first 2 min then start moving, So initial velocity is zero

    Initial velocity = v₁ = 0 ms⁻¹

    Final velocity  = v₂ = 4600 ms⁻¹

     Time = t = 10 min = 600 s

By the definition of Acceleration (a)

             a = \frac{v_{2} - v_{1} } {t}  

             a = \frac{4600 - 0 } {600}

                     a = 7.67 ms⁻²

8 0
4 years ago
A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w
trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

KE = PE

\frac{1}{2} mv ^2 = \frac{1}{2} k A^2

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

A = \sqrt{\frac{mv^2}{k}}

Replacing,

A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}

A = 0.0744m

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

T = 2\pi \sqrt{\frac{m}{k}}

Replacing,

T = 2\pi \sqrt{\frac{0.750}{13}}

T= 1.509s

Now the velocity is described as,

v = \frac{2\pi}{T} * \sqrt{A^2-x^2}

v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}

We have all the values, then replacing,

v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}

v = 0.2049m/s

7 0
3 years ago
Is 250 000 miles from the earth to the moon" is a qualitative<br> Observation<br> TRUE<br> Or false
lys-0071 [83]

Answer:

True

Explanation:

3 0
3 years ago
What is the KE if a 10kg mass traveling at 5 m/s
11111nata11111 [884]

Answer: KE = 25 J

Explanation: You must use the formula

KE = 1/2 m v²

to solve this problem.

KE = 1/2 (10 Kg) (5 m/s)

KE = 1/2 (50 kgm/s)

KE = 25 J

7 0
3 years ago
If the caffeine concentration in a particular brand of soda is 2.97 mg/oz, 2.97 mg/oz, drinking how many cans of soda would be l
Ray Of Light [21]

Explanation:

The given data is as follows.

     Concentration of caffeine = 2.97 mg/oz

     Number of oz in a can = 12 oz

Therefore, the concentration of caffeine in one can is calculated as follows.

                 = (12 \times 2.97) mg

                 = 35.64 mg

                 = 35.64 \times 10^{-3} g

Since, it is given that lethal dose is 10.0 g. Hence, number of cans are calculated as follows.

     No. of cans = \frac{\text{Lethal dose}}{\text{concentration in one can}}

                         = \frac{10 g}{35.64 \times 10^{-3} g}

                         = 280.58

                         = 281 (approx)

Thus, we can conclude that 281 cans of soda would be lethal.

5 0
3 years ago
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