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larisa86 [58]
3 years ago
12

Please help!!!!!!!!!

Physics
1 answer:
Inessa [10]3 years ago
6 0

A, D is the correct answers


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 A) Muscle Endurance
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Which of the following would NOT affect the level at which a canoe floats in a pond
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A jet liner must reach a speed of 82 m/s for takeoff. If the
SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial (v_{o}) and final speeds (v_{f}), measured in meters per second, of the aircraft and likewise the runway length (d), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed (a), measured in meters per square second:

a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}

If we know that v_{o} = 0\,\frac{m}{s}, v_{f} = 82\,\frac{m}{s} and d = 1500\,m, then the acceleration that the jet must have is:

a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}

a = 2.241\,\frac{m}{s^{2}}

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0
2 years ago
Determine the mass the mass in Kilograms of a Object that has a weight of
34kurt
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Read 2 more answers
A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
sammy [17]

Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

The pressure due to the force F₁ = 60 kN/(50 mm × 10 mm) = 120 MPa

The pressure due to the force F₂ = 110 kN/(70 mm × 10 mm) = 157.142857 MPa

The total pressure applied to the block, P = 120 MPa + 157.142857 MPa = 277.142857 MPa

The required force, F₃ = 277.142857 MPa × (70 mm × 50 mm) = 970 kN

7 0
2 years ago
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