Transverse waves is what its called
Laccolith is a mushroom-shaped pluton that forms by injecting magma between sedimentary strata, forcing the upper layers to arch upward.
Answer:
64 kPa
Explanation:
The pressure exerted by a force on a surface is given by
where
p is the pressure
F is the force
A is the area on which the force is exerted
In this problem, let's call:
F = the weight of the performer, which is the force
A = the area of 1 stilt
At the beginning, the performer is standing on both stilts, so the area on which he exerts pressure is 2A. So the pressure is
(1)
Later, he stands on one stilt only. The force exerted is still the same (his weight), however, the area is now reduced to A; therefore, the new pressure is
which is twice the value calculated in (1); so, the new pressure is
A. The angle at which the arrow must be released to hit the bull's-eye is 20.7 °
B. The arrow will go over the branch.
<h3>A. How to determine the angle</h3>
- Range (R) = 74 m
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = ?
R = u²Sine(2θ) / g
74 = 33² × Sine (2θ) / 9.8
Cross multiply
74 × 9.8 = 33² × Sine (2θ)
725.2 = 1098 × Sine (2θ)
Divide both sides by 1098
Sine (2θ) = 725.2 / 1098
Sine (2θ) = 0.6605
Take the inverse of sine
2θ = Sine⁻¹ 0.6605
2θ = 41.3
Divide both sides by 2
θ = 41.3 / 2
θ = 20.7 °
<h3>B. How to determine if the arrow will go over or under the branch</h3>
To determine if the arrow will go over or under the branch situated mid way, we shall determine the maximum height attained by the arrow. This can be obtained as follow:
- Initial velocity (u) = 33 m/s
- Acceleration due to gravity (g) = 9.8 m/s²
- Angle (θ) = 20.7 °
- Maximum height (H) = ?
H = u²Sine²θ / 2g
H = [33² × (Sine 20.7)²] / (2 ×9.8)
H = 6.94 m
Thus, the maximum height attained by the arrow is 6.94 m which is greater than the height of the branch (i.e 3.50 m).
Therefore, we can conclude that the arrow will go over the branch
Learn more about projectile motion:
brainly.com/question/20326485
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Answer:
Teh velocity of ball with respect to the ground is 29.2 m/s.
Explanation:
Velocity of car, vc = 25 due m/s east
velocity of ball with respect to car, v(b,c) = 15 m/s due Z axis
Write the velocities in the vector form
The velocity of ball with respect to ground is
