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mixas84 [53]
3 years ago
5

Is

Physics
1 answer:
ivann1987 [24]3 years ago
4 0

Answer:

Required energy Q = 231 J

Explanation:

Given:

Specific heat of copper C = 0.385 J/g°C

Mass m = 20 g

ΔT = (50 - 20)°C = 30 °C

Find:

Required energy

Computation:

Q = mCΔT

Q = 20(0.385)(30)

Required energy Q = 231 J

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40 meters divided by 5
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8 meters

Explanation:

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Compare the current in the 8-ohm resistors to the current in the 4-ohm resistors.
Gemiola [76]

Answer:

a)   i₈ = 0.5 i₄,  b)   i₁₀ = 0.3 i₃,    i₁₀ = 0.8 i₈

Explanation:

For this exercise we use ohm's law

       V = i R

        i = V / R

we assume that the applied voltage is the same in all cases

let's find the current for each resistance

         

R = 4 Ω

         i₄ = V / 4

R = 8 Ω

         i₈ = V / 8

we look for the relationship between these two currents

         i₈ /i₄ = 4/8 = ½

         i₈ = 0.5 i₄

R = 3 Ω

        i₃ = V3

R = 10 Ω

         

        i₁₀ = V / 10

   

we look for relationships

       i₁₀ / 1₃ = 3/10

       i₁₀ = 0.3 i₃

       i₁₀ / 1₈ = 8/10

       i₁₀ = 0.8 i₈

7 0
2 years ago
On a frictionless horizontal air table, puck A (with mass 0.254 kg ) is moving toward puck B (with mass 0.367 kg ), which is ini
irinina [24]

Answer:

v_a=0.8176 m/s

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

Explanation:

According to the law of conservation of linear momentum, the total momentum of both pucks won't be changed regardless of their interaction if no external forces are acting on the system.

Being m_a and m_b the masses of pucks a and b respectively, the initial momentum of the system is

M_1=m_av_a+m_bv_b

Since b is initially at rest

M_1=m_av_a

After the collision and being v'_a and v'_b the respective velocities, the total momentum is

M_2=m_av'_a+m_bv'_b

Both momentums are equal, thus

m_av_a=m_av'_a+m_bv'_b

Solving for v_a

v_a=\frac{m_av'_a+m_bv'_b}{m_a}

v_a=\frac{0.254Kg\times (-0.123 m/s)+0.367Kg (0.651m/s)}{0.254Kg}

v_a=0.8176 m/s

The initial kinetic energy can be found as (provided puck b is at rest)

K_1=\frac{1}{2}m_av_a^2

K_1=\frac{1}{2}(0.254Kg) (0.8176m/s)^2=0.0849 J

The final kinetic energy is

K_2=\frac{1}{2}m_av_a'^2+\frac{1}{2}m_bv_b'^2

K_2=\frac{1}{2}0.254Kg (-0.123m/s)^2+\frac{1}{2}0.367Kg (0.651m/s)^2=0.07969 J

The change of kinetic energy is

\Delta K=0.07969 J - 0.0849 J = -0.00521 J

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