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White raven [17]
3 years ago
5

Roseanne heated a solution in a beaker as part of a laboratory experiment on energy transfer. After a while, she noticed the liq

uid solution began to circulate within the beaker. What process involving heat transfer did Roseanne observe inside the beaker?
Physics
1 answer:
Anna35 [415]3 years ago
5 0
Water boilingis the answer
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What are organs made of?
natita [175]
Organs are made of tissues

Tissues are made of cells though so it’s kinda a weird question lol
7 0
2 years ago
Read 2 more answers
Suppose a rectangular piece of aluminum has a length D, and its square cross section has the dimensions W XW, where D (W x W) to
Ludmilka [50]

Answer:

R₂ / R₁ = D / L

Explanation:

The resistance of a metal is

        R = ρ L / A

Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section

We apply this formal to both configurations

Small face measurements (W W)

The length is

         L = W

Area  

         A = W W = W²

        R₁ = ρ W / W² = ρ / W

Large face measurements (D L)

       Length L = D= 2W

       Area     A = W L

     R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L

The relationship is

    R₂ / R₁ = 2W²/L

6 0
3 years ago
Can i please get some help with this.
Ivenika [448]

the answer is D cuz electricity is a conductive

5 0
3 years ago
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Ne
Furkat [3]

Answer:

the frequency of this mode of vibration is 138.87 Hz

Explanation:

Given;

length of the copper wire, L = 1 m

mass per unit length of the copper wire, μ = 0.0014 kg/m

tension on the wire, T = 27 N

number of segments, n = 2

The frequency of this mode of vibration is calculated as;

F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz

Therefore, the frequency of this mode of vibration is 138.87 Hz

7 0
3 years ago
what is the magnitude of the electric force between charges of 0.25 C and 0.11 C at a separation of 0.88 m? if the separation be
leonid [27]
F = k \cdot \frac{q_1 q_2}{d} = 9 \cdot 10^{9} \cdot \frac{0.25 \cdot 0.11}{0.88} =  144 \cdot 5^{9} \ N.

If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.
6 0
2 years ago
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