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Sedaia [141]
3 years ago
15

Describe two ways minerals are used by people. Then explain how the increasing human population affects mineral use, and why thi

s could be a problem.
Physics
1 answer:
lianna [129]3 years ago
5 0

1. Just like vitamins, minerals help your body grow, develop, and stay healthy. The body uses minerals to perform many different functions, from building strong bones to transmitting nerve impulses. Some minerals are even used to make hormones or maintain a normal heartbeat.

2. Population growth is the increase in the number of people living in a particular area. Since populations can grow exponentially, resource depletion can occur rapidly, leading to specific environmental concerns such as global warming, deforestation and decreasing biodiversity.

Hope this helps.

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You suspect that a power supply is faulty, but you use a power supply tester to measure its voltage output and find it to be acc
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Answer:

Load

Explanation:

A normal power supply can deliver up to certain amount of power to a load. The output power can be calculated multiplying Voltage (V) x Current (A). It happens that after a certain period of time, the power source's main components begin to wear, thus losing its ability to deliver its nominal power. Normally, when no load its connected to the source, you will get the operating Voltage, but when the load demands power, the ability to deliver power to it may fail to reach nominal levels. When connected, there may be voltage drops (thus, less power output) causing malfunctions turning it into a non-operative power supply.

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Example
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Calculate the accleration of a 25-Kg object that moved with force of 300 N
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F=ma
a=F/m=300/25=12 m/s^2
4 0
2 years ago
A particular heat engine has a mechanical power output of 4.00 kW and an efficiency of 26.0%. The engine expels 8.55 103 J of ex
Ivahew [28]

To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,

\text{Efficiency of the heat engine} = \eta = 26\% = 0.26

\text{Energy taken in by the heat engine during each cycle} = Q_h

\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J

\eta = 1 - \frac{Q_{c}}{Q_{h}}

0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}

\frac{8.55* 10^{3}}{Q_{h}} = 0.74

Q_h = \frac{8.55*10^3}{0.74}

Q_h = 11.554*10^3J

PART A)

Work done by the heat engine in each cycle = W

W = Q_h-Q_c

W = 11.554*10^3J-8.55*10^3J

W = 3004J

According to the value given we have that,

P = 4.0kW

P = 4000W

Power is defined as the variation of energy as a function of time therefore,

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Therefore the interval for each cycle is 0.75s

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3 years ago
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