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Fiesta28 [93]
3 years ago
6

25 POINTS please help!

Physics
2 answers:
salantis [7]3 years ago
7 0

Answer: b

Explanation:

kow [346]3 years ago
3 0

Answer:

b

Explanation:

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Two small plastic spheres each have a mass of 1.1 g and a charge of -50.0 nC . They are placed 2.1 cm apart (center to center).
Solnce55 [7]

Answer:

Part a)

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Explanation:

Part a)

Electrostatic force between two charged spherical balls is given as

F = \frac{kq_1q_2}{r^2}

here we will have

q_1 = q_2 = 50 nC

here the distance between the center of two balls is given as

r = 2.1 cm = 0.021 m

now we will have

F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

3 0
3 years ago
After a water had boiled, the temperature of the water decreases by 22degrees. The mass of the water in the kettle is 0.50kg.The
kifflom [539]

<u>Answer</u>

46,200 J

The energy of given out when a substance is losing heat is given by:

H = mcΔθ

Where m is the mass of the substance,

            c is the specific heat capacity of the substance and

           Δθ is the temperature change.

H = 0.50 × 22 4200

  = 46,200 Joules.

4 0
4 years ago
Question 4
zysi [14]

Answer:

one of the object is at rest after the collision

3 0
2 years ago
You drop an ice cube into an insulated bottle full of water and wait for the ice cube to completely melt. The ice cube initially
AVprozaik [17]

Answer:

T_{f}  = 17º C

Explanation:

This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)  

Let's reduce the magnitudes to the SI system  

      Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg  

            L = 3.33 105 J / kg  

Water  M = 860 g = 0.860 kg  

           c_{e} = 4186 J / kg ºC

    Q₁ = m L  

     Q₁ = 0.080 3.33 10⁵

     Q₁ = 2,664 10⁴ J

Now let's see what this liquid water temperature is when this heat is released  

      Q = M c_{e} ΔT = M c_{e} (T₀₁ -T_{f1})  

      Q₁ = Q  

     T_{f1} = T₀₁ - Q / M ce  

     T_{f1} = 26.0 - 2,664 10⁴ / (0.860 4186)  

     T_{f1} = 26.0 - 7.40  

     T_{f1} = 18.6 ° C  

The initial temperature of water that has just melted is T₀₂ = 0ª  

The initial temperature of the liquid water is T₀₁= 18.6  

     m c_{e} T_{f} + M c_{e} T_{1} = M c_{e} T₀₁ - m c_{e} T₀₂o2  

         T_{f} = (M To1 - m To2) / (m + M)  

         T_{f} = (0.860 18.6 - 0.080 0) / (0.080 + 0.860)  

T_{f} = 17º C

 

gg

4 0
3 years ago
William drew a diagram of a box containing a gas for his science project. His drawing is shown.
Olin [163]
Question? U need a drawing
7 0
3 years ago
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