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Vlad1618 [11]
3 years ago
12

Why is friction important for gymnast working on parallel bars

Physics
1 answer:
igomit [66]3 years ago
8 0
Hey there,
Friction is important to increase the friction between the bars and the gymnast to prevent the gymnast from falling and hurting themselves.

Hope this helps :))

<em>~Top♥</em>
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I.Name two commonly used thermometric liquids.
PtichkaEL [24]

Answer:

mercury and alcohol

ii) used to test temperatures

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3 years ago
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The stone, which weighs 400 g, is thrown upwards at a speed of 20 m / s. Climbed to a height of 12 m. Determine: what is equal t
maxonik [38]

Given that,

Mass of the stone, m = 400 g = 0.4 kg

Initial speed, u = 20 m/s

It is climbed to a height of 12 m.

To find,

The work done by the resistance force.

Solution,

Let v is the final speed. It can be calculated by using the conservation of energy.

v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 12} \\\\v=15.33\ m/s

Work done is equal to the change in kinetic energy. It can be given as follows :

W=\dfrac{1}{2}m(v^2-u^2)\\\\=\dfrac{1}{2}\times 0.4\times (15.33^2-20^2)\\\\=-32.99\ J

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3 0
3 years ago
A) Charge q1 = +5.60 nC is on the x-axis at x = 0 and an unknown charge q2 is on the x-axis at x = -4.00 cm. The total electric
jeka94

Answer:

a) F₃₁ = 63.0 μN  

b) F₃₂ = - 14.0 μN

c) q₂ = - 5.0 nC

Explanation:

a)

  • Assuming that the three charges can be taken as point charges, the forces between them must obey Coulomb's Law, and can be found independent each other, applying the superposition principle.
  • So, we can find the force that q₁ exerts along the x-axis on q₃, as follows:

       F_{31} =\frac{k*q_{1}*q_{3} }{r_{13}^{2}} = \frac{9e9Nm2/C2*5.6e-9C*2.0e-9C}{(0.04m)^{2}}  = 63.0 \mu N   (1)

b)

  • Since total force exerted by q₁ and q₂ on q₃ is 49.0 μN, we can find the force exerted only by q₂ (which is along the x-axis only too) just by difference, as follows:

      F_{32} = F_{3} - F_{31}  = 49.0\mu N  - 63.0\mu N = -14.0 \mu N  (2)

c)

  • Finally, in order to find the value of q₂, as we know the value and sign of F₃₂, we can apply again the Coulomb's Law, solving for q₂, as follows:

      q_{2}  = \frac{F_{32} * r_{23}^{2} }{k*q_{3}} = \frac{(-14\mu N)*(0.08m)^{2}}{9e9Nm2/C2* 2 nC} = - 5 nC  (3)

6 0
3 years ago
How tilted is the axis of the earth's rotation as compared to the plane of the earth's orbit?
Phantasy [73]
C. 23.5 degrees
This tilt in the axis causes the seasons. 
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3 years ago
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He instruction booklet for your pressure cooker indicates that its highest setting is 11.1 psi . you know that standard atmosphe
blagie [28]

Given that,

Atmospheric Pressure = 14.7 psi

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T2 = 25*25.8/14.7

T2 = 43.87 °C

The cooking pressure will be 43.87 °C.

5 0
3 years ago
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