The upward force exerted on the board by the support is 530.8 N.
<h3>Upward force exerted on the board by the support</h3>
The sum of the upward forces is equal to sum of downward forces;
total downward forces = 52.8 N + 206 N + 272 N = 530.8 N
downward force = upward force = 530.8 N
Thus, the upward force exerted on the board by the support is 530.8 N.
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Answer:
a) P = 44850 N
b) 
Explanation:
Given:
Cross-section area of the specimen, A = 130 mm² = 0.00013 m²
stress, σ = 345 MPa = 345 × 10⁶ Pa
Modulus of elasticity, E = 103 GPa = 103 × 10⁹ Pa
Initial length, L = 76 mm = 0.076 m
a) The stress is given as:

on substituting the values, we get

or
Load, P = 44850 N
Hence<u> the maximum load that can be applied is 44850 N = 44.85 KN</u>
b)The deformation (
) due to an axial load is given as:

on substituting the values, we get

or

Answer:
t = 13.43 s
Explanation:
In order to find the minimum time required by the plane to stop, we will use the first equation of motion. The first equation of motion is written as follows:
Vf = Vi + at
where,
Vf = Final Velocity of the Plane = 0 m/s (Since, the plane finally stops)
Vi = Initial Velocity of the Plane = 95 m/s
a = deceleration of the plane = - 7.07 m/s²
t = minimum time interval needed to stop the plane = ?
Therefore,
0 m/s = 95 m/s + (- 7.07 m/s²)t
t = (95 m/s)/(7.07 m/s²)
<u>t = 13.43 s</u>
Answer:
As the temperature increases, the average kinetic energy increases as does the velocity of the gas particles hitting the walls of the container
Explanation:
5 kg basketball. Inertia is related to the mass of the object.