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3 years ago

Feeling slightly uncomfortable during exercise

1 answer:

6 0

Feeling slightly uncomfortable during exercise is completely normal and should be expected. If there is unbearable pain however, the exerciser should stop and consult a doctor.

I hope this helped!

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In metal wires the charged particles that move are A. protons B. electrons C. neutrons D. none of the above

VashaNatasha [74]

Answer:

the answer is B: electrons

Explanation:

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4 years ago

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Explain what was ultimately responsible for London’s “killer fog” in 1952.

stellarik [79]

Answer:

possible answer from edge

Explanation:

After World War II the only coal available to burn residentially, either for cooking or warmth, was of poor quality and contained large amounts of contaminants that made its fumes toxic. So when millions of homes in London burned this low quality coal and the local atmosphere over and around London stagnated, there was nowhere for the fumes to go. This stagnation allowed the toxic coal fumes to build up until they became physically harmful to humans and animals.

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3 years ago

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A crate feels a 277 n normal force as it sits on the ground. what is its mass?

Umnica [9.8K]

Answer:

28.3 kg

Explanation:

Assuming the ground is level, the normal force equals the weight.

N = mg

277 N = m × 9.8 m/s²

m = 28.3 kg

5 0

4 years ago

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Sir Dudley the Dextrous is loading cannons on top of the castle's tallest towers in preparation for an expected attack. He slips

jeka94

Answer:

786 J

Explanation:

Let g = 9.81 m/s2. The vertical distance that the canon ball must have traveled within 1.65 s is (ignoring air resistance):

$h = gt^2/2 = 9.81*1.65^2/2 = 13.35 m$

So the work done by gravity would be the product of gravity force and the distance it traveled

[tex]E_p = Ph = mgh = 6*9.81*13.35 = 786 J

7 0

3 years ago

A 10.0 cm object is 5.0 cm from a concave mirror that has a focal length of 12 cm. What is the distance between the image and th

fiasKO [112]

Let's use the mirror equation to solve the problem:
$\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i}$
where f is the focal length of the mirror, $d_o$ the distance of the object from the mirror, and $d_i$ the distance of the image from the mirror.
For a concave mirror, for the sign convention f is considered to be positive. So we can solve the equation for $d_i$ by using the numbers given in the text of the problem:
$\frac{1}{12 cm}= \frac{1}{5 cm}+ \frac{1}{d_i}$
$\frac{1}{d_i}= -\frac{7}{60 cm}$
$d_i = -8.6 cm$
Where the negative sign means that the image is virtual, so it is located behind the mirror, at 8.6 cm from the center of the mirror.

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3 years ago

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