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Oliga [24]
3 years ago
7

Help me please answer this

Physics
2 answers:
antiseptic1488 [7]3 years ago
8 0

Answer:

that's fusion

Explanation:

Vinil7 [7]3 years ago
6 0

Answer:

the answer is fusion for this

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you push a book a distance of 5 meters with a force of 10 newtons for 2 seconds how much work did you do on the book
GrogVix [38]

Answer:

50 J

Explanation:

Work, W=Fd where F is the applied force and d is the distance

Substituting 10 N for F and 5 m for d then work done on the book can be expressed as

W=10 N* 5 m=50 Nm= 50 J

Therefore, the work done is equivalent to 50 J

3 0
3 years ago
The mass number of a nucleus (except Hydrogen) is:
netineya [11]
The mass number of a nucleus (except Hydrogen) is: number of protons + number of neutrons.

A=Z+N

A=mass number=protons + neutrons.
Z=atomic number=number of protons.
N=number of neutrons. 

In the case of Hydrogen it depends of isotope of Hydrogen . 
the hydrogen has three isotopes, 
protium : A=Z, because N=0
deuterium:  A=Z+N;   N=1
 tritium: A=Z+N;    N=2 
7 0
3 years ago
Read 2 more answers
Gamma ray technology can be used to do which of the following?
Rudiy27

Answer:

b

Explanation:

radiation can treat tumors.

5 0
3 years ago
Two protons are released from rest when they are 0.750 {\rm nm} apart.
Alex787 [66]

Explanation:

Given:

m = 1.673 × 10^-27 kg

Q = q = 1.602 × 10^-19 C

r = 0.75 nm

= 0.75 × 10^-9 m

A.

Energy, U = (kQq)/r

Ut = 1/2 mv^2 + 1/2 mv^2

1.673 × 10^-27 × v^2 = (8.99 × 10^9 × (1.602 × 10^-19)^2)/0.75 × 10^-9

v = 1.356 × 10^4 m/s

B.

F = (kQq)/r^2

F = m × a

1.673 × 10^-27 × a = ((8.99 × 10^9 × (1.602 × 10-19)^2)/(0.075 × 10^-9)^2

a = 2.45 × 10^17 m/s^2.

4 0
3 years ago
the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity
NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0
4 years ago
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