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Tresset [83]
4 years ago
7

a 50kg person is standing still in ice skates throws their 2.0kg helmet to the right at 25m/s while on a friction less surface.

what is the final speed of the person
Physics
1 answer:
Roman55 [17]4 years ago
8 0

Answer:

Vp = 1 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the amount of movement. In the same way, analyze the before and after of the actions.

<u>The moment before</u>

The 50kg person is still hold (no movement) with the 2kg helmet

<u>The moment after</u>

The helmet moves at 25[m/s] in one direction, the person moves in the opposite direction, due to the launch of the helmet.

In this way we can apply the principle of conservation of movement, expressing the before and after. To the left we have the before and to the right of the equal sign we have the after.

Σm*V1 = Σm*V

where:

m = total mass = (2 + 50) = 52[kg]

V1 = velocity before the lunch = 0

(50 + 2)*V1 = (25*2) - (Vp*50)

0 = 50 - 50*Vp

50 = 50*Vp

Vp = 1 [m/s]

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Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the required temperature increase to close the gap at C.
Leni [432]

Answer:

T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}

Explanation:

The given data :-

  • Diameter of rod AB ( d₁ ) = 200 mm.
  • Diameter of rod BC ( d₂ ) = 150 mm.
  • The linear co-efficient of thermal expansion of copper ( ∝ ) = 1.6 × 10⁻⁶ /°C
  • The young's modulus of elasticity of copper ( E ) = 120 GPa = 120 × 10³ MPa.
  • Consider the required temperature increase to close the gap at C = T °C
  • Consider the change in length of the rod = бL

Solution :-

\begin{aligned}\sum H& =0\\-R_A+R_C&=0\\R_A&=R_C\\R_A&=R\\R_C&=R\\R_{A}&=\text{reaction\:force\:at\:A}\\R_{C}&=\text{reaction\:force\:at\:C}\\\sigma_{AB}&=\text{axial\:stress\:at\:A}\\\sigma_{BC}&=\text{axial\:stress\:at\:B}\\\sigma_{AB}&=\frac{R}{A_{A}}\\&=\frac{R_{A}}{A_{A}}\\\sigma_{BC}&=\frac{R_{B}}{A_{B}}\\&=\frac{R}{A_{B}}\\\frac{\sigma_{AB}}{\sigma_{BC}}&=\frac{A_{B}}{A_{B}}\\&=\frac{\frac{\pi}{4}\cdot 150^{2}}{\frac{\pi}{4}\cdot 200^{2}}\\&=\frac{9}{16}\end{aligned}

\begin{aligned}\delta L&= (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\0& = (\delta L _{thermal} +\delta L_{axial})_{AB} + ( \delta L _{thermal} +\delta L_{axial})_{BC}\\&=\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{AB}+\left[\alpha\:T\:L+\left(\frac{-RL}{AE}\right)\right]_{BC}\\&=2\:\alpha\:T\:L-\frac{L}{E}(\sigma_{AB}+\sigma_{BC})\\T&=\frac{\sigma_{AB}+\sigma_{BC}}{2\alpha E}\end{aligned}

5 0
4 years ago
What is an example of a non contact force
olchik [2.2K]

Answer: Gravitational force

Explanation:

A non contact force can be described as a force applied to an object by another body that is not in direct contact with it.

For example, an object thrown upwards will return back due to the force of gravity acting on it. So, it means Gravitational force is acting on the body without necessarily being in contact with that body.

8 0
4 years ago
old stars obtain part of their energy by the fusion of three alpha particles to form a 12/6c nucleus, whose mass is 12.0000 u. H
marta [7]

Answer:

11.6532 x 10⁻¹¹ J or 7.3 MeV is given off

Explanation:

Mass of an alpha particle = 4.0026u,   ∴ mass of three = 12.0078u

Find the difference in mass.

Mass of three alpha - Mass of Carbon nucleus

12.0078u - 12u = 0.0078u

Since 1u = 1.66 x 10⁻²⁷ kg

Therefore, 0.0078u = 1.2948 x 10⁻²⁷

Now that we know Mass(m) = 1.2948 x 10⁻²⁷ and Speed (c) 3 x 10⁸ m²s⁻²

Formular for Energy ==> E₀ = mc²

E = (1.2948 x 10⁻²⁷) (3 x 10⁸ m²s⁻²)²

E = (1.2948 x 10⁻²⁷) (9 x 10¹⁶) J

E = 11.6532 x 10⁻¹¹ J

Or, if you need your energy in MeV

1 MeV = 1.60x10⁻¹³ J

Just do the conversion by dividing 11.6532 x 10⁻¹¹ J by 1.60x10⁻¹³ J

It will give you 7.3 MeV

7 0
3 years ago
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