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2 years ago

Which material would provide the

Physics

1 answer:

Bogdan [553]2 years ago

8 0

Answer:

C) Plastic

Explanation:

Copper, silver, and gold all conduct electricity. Therefore, the material that would provide the most resistance for electricity would be plastic.

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What are some possible materials you could use to make your battery? DON’T FORGET TO include information about why it is importa

Dima020 [189]

Explanation:

Commercially available batteries use a variety of metals and electrolytes. Anodes can be made of zinc, aluminum, lithium, cadmium, iron, metallic lead, lanthanide, or graphite. Cathodes can be made of manganese dioxide, mercuric oxide, nickel oxyhydroxide, lead dioxide or lithium oxide. Potassium hydroxide is the electrolyte used in most battery types, but some batteries use ammonium or zinc chloride, thionyl chloride, sulfuric acid or lithiated metal oxides. The exact combination varies by battery type. For example, common single-use alkaline batteries use a zinc anode, a manganese dioxide cathode, and potassium hydroxide as the electrolyt

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2 years ago

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lyudmila [28]

Answer: Answer to the I'm Tall when I'm Young, and I'm Short when I'm Old. What am I? Riddle is a candle

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2 years ago

Read 2 more answers

An unstable particle at rest breaks up into two fragments of unequal mass. The mass of the lighter fragment is equal to 2.90 ✕ 1

motikmotik

Answer:

The speed of the heavier fragment is 0.335c.

Explanation:

Given that,

Mass of the lighter fragment $M_{l}=2.90\times10^{-28}\ kg$

Mass of the heavier fragment $M_{h}=1.62\times10^{-27}\ Kg$

Speed of lighter fragment = 0.893c

We need to calculate the speed of the heavier fragment

Let v is the speed of the second fragment after decay

Using conservation of relativistic momentum

$0=\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}-\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\drac{m_{1}v_{1}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}=\drac{m_{2}v_{2}}{\sqrt{1-\dfrac{v_{1}^2}{c^2}}}$

$\dfrac{2.90\times10^{-28}\times0.893c}{\sqrt{1-(0.893)^2}}=\dfrac{1.62\times10^{-27}v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=\dfrac{2.90\times10^{-28}\times0.893c}{1.62\times10^{-27}\times0.45}$

$\dfrac{v_{2}}{\sqrt{1-\dfrac{v_{2}^2}{c^2}}}=0.355c$

$\dfrac{v_{2}}{1-\dfrac{v_{2}^{2}}{c^2}}=(0.355c)^2$

$\dfrac{1-\dfrac{v_{2}^2}{c^2}}{v_{2}^2}=\dfrac{1}{(0.355c)}$

$\dfrac{1}{v_{2}^2}-\dfrac{1}{c^2}=\dfrac{1}{(0.355c)^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}+\dfrac{1}{0.126c^2}$

$\dfrac{1}{v_{2}^2}=\dfrac{1}{c^2}(1+\dfrac{1}{0.126})$

$\dfrac{1}{v_{2}^2}=\dfrac{8.93}{c^2}$

$v_{2}^2=\dfrac{c^2}{8.93}$

$v_{2}=0.335c$

Hence, The speed of the heavier fragment is 0.335c.

7 0

3 years ago

Two metals of equal mass with different heat capacities are subjected to the same amount of heat. which undergoes the smallest c

liraira [26]

<span>So when two metals of equal mass but different heat capabilities are subjected to same heat quantity, the metal with higher heat capacity have the small temperature change. Heat supplied is determined as heat capacity of the metal times the change in temperature.</span>

8 0

3 years ago

RP 1 - Specific Heat Capacity GCSE Exam Questions (B.S.G)

Stels [109]

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

A sample of 0.50 kg of water boils (reaches 100 °C). After a while, its temperature decreases by 22 °C.

We can calculate the energy transferred to the surroundings from the water in the form of heat (Q) using the following expression.

$Q = c \times m \times \Delta T = \frac{4200J}{kg.\° C} \times 0.50kg \times (-22\° C) \times \frac{1kJ}{1000J} = -46.2 kJ$

where,

c: specific heat capacity of water
m: mass of water
ΔT: change in the temperature

When the temperature of 0.50 kg of water decreases by 22 °C, the energy transferred to the surroundings from the water is -46.2 kJ.

Learn more: brainly.com/question/16104165

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2 years ago