Do you think the benifits of nuclear power outweigh the potential drawbacks

1 answer:

8 0

More often than not, the pros outweigh the cons when it comes to nuclear energy. It has very low operating costs, produces sufficient energy to meet demand, and has a variety of other economic benefits that can't be ignored

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In a ballistics test, a 24 g bullet traveling horizontally at 1300 m/s goes through a 25-cm-thick 360 kg stationary target and e

MaRussiya [10]

Answer:

A)0.00022s b)40363.6N c) 0.025m/s

Explanation:

Mass = 24g = 0.024kg, distance though the target = thickness of the target = 25cm = 0.25m

Initial speed of the bullet = 1300m/s, final speed = 930m/s

Using equation of motion

Distance = 1/2(vf+vi)*t (time in seconds)

t = 0.25*2/(1300+930) = 0.00022s

B) force exerted on the body

F = ma = m* (vf-vi)/t = 0.024*(930-1300)/0.00022

F = -40363N, it is negative because the body decelerated during this motion

C) using law of conservation of momentum,

M1*U1+ M2*U2(M2and U1 are the mass and initial speed of the body) = M1V1+ M2V2

The target was at rest so initial speed U2 = 0

0.024*1300 + 360*0 = 0.024*930 + 360*V2

31.2 = 22.32+360*V2

31.2-22.33 = 360*V2

V2 = 8.88/360 = 0.025m/s

8 0

2 years ago

An proton-antiproton pair is produced by a 2.20 × 10 3 MeV photon. What is the kinetic energy of the antiproton if the kinetic e

timama [110]

Answer:

K = 80.75 MeV

Explanation:

To calculate the kinetic energy of the antiproton we need to use conservation of energy:

$E_{ph} = E_{p} + E_{ap} = E_{0p} + K_{p} + E_{0ap} + K_{ap} = m_{0p}c^{2} + K_{p} + m_{0ap}c^{2} + K_{ap}$

where $E_{ph}$ : is the photon energy, $E_{0p} and E_{0ap}$ : are the rest energies of the proton and the antiproton, respectively, equals to m₀c², $K_{p} and K_{ap}$ : are the kinetic energies of the proton and the antiproton, respectively, c: speed of light, and m₀: rest mass.

Therefore the kinetic energy of the antiproton is:

$K_{ap} = E_{ph} - m_{0p}c^{2} - K_{p} - m_{0ap}c^{2}$

The proton mass is equal to the antiproton mass, so:

$K_{ap} = E_{ph} - 2m_{0p}c^{2} - K_{p}$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2} - 242.85MeV$

$K_{ap} = 2.20 \cdot 10^{3}MeV - 2(1.67 \cdot 10^{-27}kg)(3\cdot 10^{8} \frac {m}{s})^{2}(\frac{1eV}{1.602 \cdot 10^{-19}J})(\frac{1 MeV}{10^{6}eV}) - 242.85MeV$