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schepotkina [342]
3 years ago
6

Abridgeis36.5mlongandweighs2.56x105N.Atruckweighing 5.25x104Nis10.2 m from one end of the bridge. Calculate the upward force tha

t must be exerted by each pier to support thisweight
Physics
1 answer:
enot [183]3 years ago
3 0

Answer:

Explanation:

Bridge length A B = 36.5 m

weight W = 2.56 x 10⁵ N acting at middle point that is at 18.25 m from either end.

weight of truck W₁ = 5.25 x 10⁴ N  standing at 10.2 m from A .

Let force at A be F₁ and at B be F₂ .

Balancing total upward and downward force

F₁ + F₂ = ( 2.56 + .525 ) x 10⁵ = 3.085 x 10⁵ N

Taking torque of all the forces , for rotational balancing

2.56 x 10⁵ x 18.25 + .525 x 10⁵ x 10.2 = F₂ x 36.5

(46.72 + 5.355 )x 10⁵ = F₂ x 36.5

F₂ = 1.42 x 10⁵ N

F₁ = 1.665 x 10⁵ N

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Magnetic fields exist
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In order to sail through the frozen Arctic Ocean, the most powerful icebreaker ever built was constructed in the former Soviet U
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955.36 seconds ≈ 16 minutes

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6 0
2 years ago
If a hot steel tool of 1200°C was put in a bucket to cool and the bucket contained 15L of water of 15°C, and the water temperatu
Mashcka [7]

3.6 kg.

<h3>Explanation</h3>

How much heat does the hot steel tool release?

This value is the same as the amount of heat that the 15 liters of water has absorbed.

Temperature change of water:

\Delta T = T_2 - T_1= 48\; \textdegree{\text{C}}- 15\; \textdegree{\text{C}} = 33 \; \textdegree{\text{C}}.

Volume of water:

V = 15 \; \text{L} = 15 \; \text{dm}^{3} = 15 \times 10^{3} \; \text{cm}^{3}.

Mass of water:

m = \rho \cdot V = 1.00 \; \text{g} \cdot \text{cm}^{-3} \times 15 \times 10^{3} \; \text{cm}^{3} = 15 \times 10^{3} \; \text{g}.

Amount of heat that the 15 L water absorbed:

Q = c\cdot m \cdot \Delta T = 4.18 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 15 \times 10^{3} \; \text{g} \times 33 \; \textdegree{\text{C}} = 2.06910 \times 10^{6}\; \text{J}.

What's the mass of the hot steel tool?

The specific heat of carbon steel is 0.49 \; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1}.

The amount of heat that the tool has lost is the same as the amount of heat the 15 L of water absorbed. In other words,

Q(\text{absorbed}) = Q(\text{released}) =2.06910 \times 10^{6}\; \text{J}.

\Delta T = T_2 - T_1 = 1200\; \textdegree{\text{C}} -{\bf 48}\; \textdegree{\text{C}} = 1152\; \textdegree{\text{C}}.

m = \dfrac{Q}{c\cdot \Delta T} = \dfrac{2.06910 \times 10^{6} \; \text{J}}{0.49\; \text{J} \cdot \text{g}^{-1} \cdot \textdegree{\text{C}}^{-1} \times 1152\; \textdegree{\text{C}}} = 3.6 \times 10^{3} \; \text{g} = 3.6 \; \text{kg}.

4 0
3 years ago
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