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Allisa [31]
3 years ago
9

WILL GIVE BRAINLIEST AND A THANKS TO FIRST CORRECT ANSWER!!!!

Physics
1 answer:
tekilochka [14]3 years ago
5 0

Answer:2 and 4th one

Explanation:

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In a nuclear reactor 1 g of mass is converted into energy. How much energy in Joules is produced?
KengaRu [80]

Answer:

3 x 10^5  J

Explanation:

mass of substance, m = 1 g = 0.001 kg

Velocity of light, c = 3 x 10^8 m/s

According to the Einstein mass energy equivalence, the energy associated with the mass is given by

E = m c^2

E = 0.001 x 3 x 10^8

E = 3 x 10^5  J

5 0
4 years ago
Which applied force will allow a 7.65 kg block of ice to begin sliding on a sheet of ice? The block of ice has a kinetic coeffic
Anni [7]

Answer:

force for start moving is 7.49 N

force for moving constant velocity 2.25 N

Explanation:

given data

mass = 7.65 kg

kinetic coefficient of friction = 0.030

static coefficient of friction = 0.10

solution

we get here first weight of block of ice that is

weight of block of ice = mass  ×  g

weight of block of ice = 7.65 × 9.8 = 74.97 N

so here Ff = Fa

so for force for start moving is

Fa = weight × static coefficient of friction  

Fa = 74.97 × 0.10

Fa =  7.49 N

and

force for moving constant velocity is

Fa =  weight × kinetic coefficient of friction

Fa = 74.97 × 0.030

Fa = 2.25 N

7 0
4 years ago
Read 2 more answers
Three bulbs are connected in series to a 4.5 V battery. The second bulb burns out. The
Oksi-84 [34.3K]

Answer:

0v

Explanation:

3 0
3 years ago
The center of a moon of mass m is a distance D from the center of a planet of mass M. At some distance x from the center of the
Eddi Din [679]

Answer:

x = D (M/M-m) 2.41

Explanation:

a) Let's apply Newton's second law to find the summation of force, where each force is given by the law of universal gravitation

        F = g m₁m₂ / r²

        Σ F = 0

       F1- F2 = 0

       F1 = F2

We set the reference system in the body of greatest mass (M) the planet

       F1 = g m₁ M / x²

       F2 = G m1 m / (D-x)²

      G m₁ M / x² = G m₁ m / (D-x)²

      M (D-x)² = m x²

      MD² -2MD x + M x² = m x²

     x² (M-m) -2MD x + MD² = 0

We solve the second degree equation

     x = [2MD  ±√ (4M²D² - 4 (M-m) MD²)] / 2 (M-m)

     x = {2MD ± 2D √ (M² + (M-m) M)} / 2 (M-m)

     x = D {M  ±  Ra (2M²-mM)} / (M-m)

    x = D (M ± M √ (2-m/M)) / (M-m)

    x = D (M / (M-m)) (1 ±√ (2-m/M)

Let's analyze this result, the value of M-m >> 1, so if we take the negative root, the value of x would be negative, it is out of the point between the two bodies, so the correct result must be taken with the positive root

 

    x = D (M / (M-m)) (1 + √2)

     x = D (M/M-m) 2.41

b) X = 2/3 D

     x = D (M/M-m) 2.41

     2/3 D = D (M/(M-m)) 2.41

     2/3 (M-m) = M 2.41

     2/3 M - 2/3 m = 2.41 M

     1.743 M = 0.667 m

     M/m = 0.667/1.743

     M/m =  0.38

3 0
3 years ago
1. What is the kinetic energy of a frog that has mass of 13kg and can hop<br> 9m/s? *
emmasim [6.3K]

Answer:526.6J

Explanation:I did a test

8 0
3 years ago
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