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Karolina [17]
3 years ago
13

Which of the following is NOT a function of the skeletal system?

Physics
2 answers:
puteri [66]3 years ago
4 0
D
Hope this helps
And
HACTEHA [7]3 years ago
4 0
I think the answer is D
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Question - Does gum affect a student's performance in school?
Mars2501 [29]

Independent variable: the interdependent variable in this example, could be a number of things but the main one is what type of brand is the gum and what kind is it.

Dependent variable: the dependent variable is what you are trying to find so in this case, you are trying to find does gum affect a student's performance.This is the Dependent variable.

Hope I helped!

Have a nice day,

scollier1607

3 0
3 years ago
30. A box with mass 20kg is on a cement floor. The coefficient of static friction between the box and floor is 0.25. A man is pu
Mnenie [13.5K]

Answer:

84.05

Explanation:

  • F=mg×0.25
  • F=20x9.81×0.25
  • f=49.05N
  • F=35N

F=f+F

F=49.05+35

=84.05

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2 years ago
If there is an attractive force between all objects, why do we not feel ourselves gravitating toward massive buildings in our vi
mars1129 [50]
The mass of an object affects how powerful the attractive force is. To feel the pull / gravitation the mass of the object would have to be huge, bigger than that of massive buildings
8 0
3 years ago
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QUESTION 1
lina2011 [118]

Solution :

Magnetic field at the centre due to $I_P$ :

$B_1 = \frac{\mu_0 I_P}{2 \pi d}$

$B_1 = \frac{4 \pi \times 10^{-7} \times 0.2}{2 \pi d}$

$B_1 =10^{-7} \ T$

Its direction will be downwards in the plane of the paper.

Magnetic field at the centre due to $I_Q$ :

$B_2 = \frac{\mu_0 I_Q}{2 \pi d}$

$B_2 = \frac{4 \pi \times 10^{-7} \times 0.6}{2 \pi \times 0.8}$

$B_2 =1.5 \times 10^{-7} \ T$

Its direction will be upwards in the plane of the paper

Magnetic field due to the coil:

$B_3 = \frac{\mu_0 I_{coil}}{2 r}$

$B_3 = \frac{4 \pi \times 10^{-7} \times 0.5}{2 \times 0.2}$

$B_3 =3.14 \times 10^{-7} \ T$

Its direction will be rightwards.

Now the resultant of the magnetic field at the centre.

$B_{net} = \sqrt{(B_2 -B_1)^2+B^2_3}$

$B_{net} = \sqrt{(0.5 \times 10^{-7})^2+(3.14 \times 10^{-7})^2}$

$B_{net} = 3.18 \times 10^{-7} \ T$

Now the direction ,

$\tan \theta = \frac{B_2-B_1}{B_3}$

$\tan \theta = \frac{0.5 \times 10^{-7}}{3.14 \times 10^{-7}}$

Therefore $\theta = 9^\circ$ (from right to upward)

3 0
3 years ago
A weightlifter lifts a set of weights a distance of 2.00 m. If the weights have a mass of 60.0 kg and are pushed vertically upwa
Sergio [31]
The amount of work done on the weights would be approximately 15.0 Kg of work
6 0
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