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Afina-wow [57]
3 years ago
8

A plane flying at 70.0 m/s suddenly stalls. If the acceleration during the stall is 9.8 m/s2 directly downward, the stall lasts

5.0 s, and the plane was originally climbing at 25° to the horizontal, what is the velocity after the stall?
Physics
1 answer:
tino4ka555 [31]3 years ago
4 0

Answer:

v = 66.4 m/s

Explanation:

As we know that plane is moving initially at speed of

v = 70 m/s

now we have

v_x = 70 cos25

v_x = 63.44 m/s

v_y = 70 sin25

v_y = 29.6 m/s

now in Y direction we can use kinematics

v_y = v_i + at

v_y = 29.6 - (9.81 \times 5)

v_y = -19.5 m/s

since there is no acceleration in x direction so here in x direction velocity remains the same

so we will have

v = \sqrt{v_x^2 + v_y^2}

v = \sqrt{63.44^2 + 19.5^2}

v = 66.4 m/s

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From the calculations, the force required is 0.437 N.

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C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase ( ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

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we have D = 4 + (4x(1.1 x 10⁻⁵)T = 4 + (4.4x10⁻⁵)T             EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

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we have d = 3.994 + (3.994x(2.4 x 10⁻⁵)T = 3.994 + (9.58x10⁻⁵)T       EQN 2

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