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Naily [24]
3 years ago
6

Modern rechargeable batteries require rare earth elements to build. Why are modern rechargeable batteries expensive and a concer

n as electric vehicles become more popular?
The batteries are made from a common nonrenewable resource.
The batteries are made from a common nonrenewable resource.

The batteries are made from a common renewable resource.
The batteries are made from a common renewable resource.
Physics
1 answer:
My name is Ann [436]3 years ago
8 0

The batteries being made from a common nonrenewable resource is the

reason why modern rechargeable batteries are expensive.

<h3>What is a Non-renewable resource?</h3>

Non-renewable resource are those in which it is very difficult to replace

them easily through natural methods. This makes it hard to meet up with

customer's use.

This thereby makes it very expensive and not able to meet up with

customer's needs. This is why renewable sources are more preferable.

Read more about Non-renewable resource here brainly.com/question/858812

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A dart is loaded into a spring-loaded toy dart gun by pushing the spring in by a distance d. For the next loading, the spring is
Aliun [14]

Answer:

the work required for the loading of second dart is 64 times greater as work required for loading the first dart.

Explanation:

k = spring constant of the spring loaded toy dart gun

x₁ = compression of spring to load the first dart = d

x₂ = compression of spring to load the second dart = 8 d

E₁ = Work required to load the first dart

E₂ = Work required to load the second dart

Work required to load the first dart is given as

E₁ = (0.5) k x₁² = (0.5) k d²

Work required to load the second dart is given as

E₂ = (0.5) k x₂² = (0.5) k (8d)² = (64) (0.5) k d²

E₂ = 64 E₁

So the work required for the loading of second dart is 64 times greater as work required for loading the first dart

8 0
4 years ago
How does convection current helps cooling the system of engines
Ede4ka [16]
As the engine heats up, a natural circulation starts, as coolant rises through the engine block by convection. It passes through the top hose, and into the radiator. Inside the radiator, heat is removed from the coolant as it falls from the top to the bottom.
Hope this helps!
6 0
3 years ago
How do you change time to kilograms?
enot [183]
You can't change time to a measurement of distance
5 0
3 years ago
Use 3rd law to explain how a fish swims through water. Identify the force pairs and draw a diagram.
vovikov84 [41]
Well as the fish swims he pushes the water behind him which in return push him forward
5 0
3 years ago
You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

3 0
3 years ago
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