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s2008m [1.1K]
3 years ago
13

How are the magnetic field lines near the south pole of a magnet affected when a second south pole is brought near it?

Physics
2 answers:
sergejj [24]3 years ago
5 0

Answer:

c

Explanation:

stepladder [879]3 years ago
4 0
The correct answer is c :)
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3 years ago
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7. Police would observe with a helicopter the time it would take for a car to travel between two wide lines placed
masya89 [10]

Answer:

See below

Explanation:

There are 3600 seconds / hr

7.2 s / 3600 s/hr = .002 hr

Car traveled 1/10 mile in .002 hr

    1/10 mi   /  .002 hr   =   50 mi/hr

5 0
2 years ago
1. What different types of shots are taken on the basketball court?
Lynna [10]

Answer:

Here are a few commonly used types of shooting in basketball.

  • Jump Shot. A jump shot is most frequently used for a mid to long-range shots, including shooting beyond the arc. ...
  • Hook Shot. A hook shot is when the shot is made while your body is not directly facing the basket. ...
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8 0
3 years ago
When we breathe, we inhale oxygen and exhale carbon dioxide plus water vapor. Which likely has more mass, the air that we inhale
Solnce55 [7]

Answer:

If the same volume of air is inhaled and exhaled, the air we breathe out normally weighs more than the air we breathe in.

Since the output from the body normally exceeds the input, breathing leads to weight loss.

Explanation:

If equal volumes of gas is inhaled and exhaled, the exhaled gas is heavier.

The inhaled gas contains Oxygen and majorly Nitrogen.

The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.

So, basically, the body exchanges O₂ with CO₂ and H₂O (and some other unwanted gases in the body) in a composition that CO₂, the heavier gas of the ones mentioned here, is prominent.

So, because the mass leaving the body is more than the mass entering, breathing leads to a loss of weight. This is one of the reasons why we need food for sustenance. Breathing alone will wear one out.

5 0
3 years ago
You have a light spring which obeys Hooke's law. This spring stretches 2.92 cm vertically when a 2.70 kg object is suspended fro
ehidna [41]

(a) 907.5 N/m

The force applied to the spring is equal to the weight of the object suspended on it, so:

F=mg=(2.70 kg)(9.8 m/s^2)=26.5 N

The spring obeys Hook's law:

F=k\Delta x

where k is the spring constant and \Delta x is the stretching of the spring. Since we know \Delta x=2.92 cm=0.0292 m, we can re-arrange the equation to find the spring constant:

k=\frac{F}{\Delta x}=\frac{26.5 N}{0.0292 m}=907.5 N/m

(b) 1.45 cm

In this second case, the force applied to the spring will be different, since the weight of the new object is different:

F=mg=(1.35 kg)(9.8 m/s^2)=13.2 N

So, by applying Hook's law again, we can find the new stretching of the spring (using the value of the spring constant that we found in the previous part):

\Delta x=\frac{F}{k}=\frac{13.2 N}{907.5 N/m}=0.0145 m=1.45 cm

(c) 3.5 J

The amount of work that must be done to stretch the string by a distance \Delta x is equal to the elastic potential energy stored by the spring, given by:

W=U=\frac{1}{2}k\Delta x^2

Substituting k=907.5 N/m and \Delta x=8.80 cm=0.088 m, we find the amount of work that must be done:

W=\frac{1}{2}(907.5 N/m)(0.088 m)^2=3.5 J

5 0
3 years ago
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