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34kurt
2 years ago
9

Is a personal trainer considered a doctor?

Physics
1 answer:
Temka [501]2 years ago
4 0
No, personal trainers and doctors have very different careers. Though they both do well for the body and health.
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A constant magnetic field can be used to produce an electric current. True or False?
Umnica [9.8K]

this question is true

6 0
3 years ago
Using complete sentences and your own words describe some of the ways humans use water.
Lera25 [3.4K]

Answer:

Humans use water for many different things. We use water to stay hydrated. Our bodies need water to live. Back in the day people used water for transportation and trading. This was a way to become wealthy and exchange goods and ideas from one place to another. We also use water to clean ourselves off. If we don't we can become sick with illnesses that can harm our bodies.

Explanation:

6 0
2 years ago
If V has a magnitude of 14 units and the same direction qs a vector 3i+6j+2k find v​
melomori [17]

Answer:

v = 6i + 12j + 4k

Explanation:

Find the magnitude of the direction vector.

√(3² + 6² + 2²) = 7

Normalize the direction vector.

3/7 i + 6/7 j + 2/7 k

Multiply by the magnitude of v.

v = 14 (3/7 i + 6/7 j + 2/7 k)

v = 6i + 12j + 4k

7 0
3 years ago
Initially a car accelerates at 2 m/s2 for x seconds. The car then travels at a velocity of -6 m/s for x seconds. If the car disp
Luda [366]

Answer:

The time travel is

t=8 s

Explanation:

a= 2 \frac{m}{s^{2} } \\v=-6 \frac{m}{s} \\x=16m

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*t+\frac{1}{2} *2*t^{2}

t^{2}-6*t-16=0\\ using :\\\frac{-b+/-\sqrt{b^{2}-4*c*a } }{2} \\\frac{-(-6)+/-\sqrt{(-6)^{2}-4*(-16)*(1) } }{2}=\frac{3}{2} +/- \frac{10}{2} \\t_{1} = 2s \\t_{2} = 8s

Check

t_{2}=8s

x_{f}=x_{o}+v_{o}*t+\frac{1}{2} *a*t^{2}

x_{f}=0-6*+\frac{1}{2} *2*8^{2}

x_{f}=-48+64\\x_{f}=16

5 0
3 years ago
A house that was heated by electric resistance heaters consumed 1200 kWh of electric energy in a winter month. If this house wer
Marysya12 [62]

Answer:

$84

Explanation:

The coefficient of performance (COP) show the relationship between the power (kW) output of the heat pump and the power (kW) input to the compressor.

The heater consumed by the heater is 1200 kWh.

For a heat pump with a COP of 2.4, the electric input needed to produce an output of 1200 kWh is:

Electric input to heat pump = 1200 / 2.4 = 500 kWh

That means that supplying a heat pump with 500 kWh produces an output of 1200 kWh

The amount of power saved = power consumed by heater - power consumed by heat pump = 1200 - 500 = 700 kWh

Money saved = $0.12/kWh * 700 kWh = $84

6 0
2 years ago
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