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polet [3.4K]
3 years ago
9

What is binding energy?

Physics
2 answers:
Bogdan [553]3 years ago
7 0

Answer:

the answer is B i hope it helps :)

Ann [662]3 years ago
3 0
<h3>What is binding energy?</h3>

\huge\color{purple}\boxed{\colorbox{black}{♡Answer}}

B. The energy required to force two nuclei to undergo nuclear fusion. ✅

  • They are usually expressed in terms of \sf\purple{kJ/mole} of nuclei or \sf\pink{MeV's/nucleon}.

\large\mathfrak{{\pmb{\underline{\orange{Happy\:learning }}{\orange{!}}}}}

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You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 12-m-hi
Vlada [557]

Answer:

Vmax=11.53 m/s

Explanation:

from conservation of energy

      E_A} =E_{B}

     Spring potential energy =potential energy due to elevation

   0.5*k*x²= mg(h_{B}-h_{A} )=mgh

   0.5*k*2.3²= 430*9.81*6

         k=9568.92 N/m

For safety reason

                                 k"=1.13 *k= 1.13*9568.92

                                    k"=10812.88 N/m

agsin from conservation of energy

      E_A} =E_{C}

    spring potential energy=change in kinetic energy

   0.5*k"*x²=0.5*m*V_{max}^{2}

      10812.88 *2.3²=430*V_{max}^{2}

           V_{max}=11.53 m/s

5 0
4 years ago
On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s
dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

                                               = 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

                                               = 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2                

                                                = 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0
3 years ago
1.- Se desea elevar un cuerpo de 1500kg utilizando una elevadora hidráulica de plato grande
kvv77 [185]

Answer:

181.48 N

Explanation:

Calculate the area :

Area = pi * r² ;

pi = 3.14 ; r1 = 90cm /100 = 0.9m ; r2 = 10/100 = 0.1m

Area 1, A1 = 3.14 * 0.1² = 0.0314 m²

Area 2, A2 = 3.14 * 0.9² = 2.5434 m²

Force, F = mass * acceleration due to gravity

F2 = 1500 * 9.8 = 14700 N

Force 1 / Area 1 = Force 2 / Area 2

Force 1 = (Force 2 / Area 2), * Area 1

Force 1 = (14700 / 2.5434) * 0.0314

Force = 5779.6650 * 0.0314

= 181.48 N

5 0
3 years ago
Will give brainliest
NNADVOKAT [17]

I love science! if you need any more help let me know i cant guarantee i can help but i will try!

The correct answer is  "Some substances must be dissolved in water before they can be used".  

5 0
3 years ago
A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?
rosijanka [135]

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

5 0
3 years ago
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