An umbrella policy primarily provides higher liability coverage that kicks in when the liability on primary insurance policies (like auto, home) are exhausted. It is mainly designed for major personal liability suits rather than covering properties and their components.
The correct option is D.
An umbrella policy is a type of insurance policy, primarily designed to provide extra liability coverage above the limits of homeowners', auto, and boat insurance policies. It kicks in when the liability on these other policies has been exhausted. So, it refers to option D: major personal liability suits.
For example, if you have a car accident and the damages exceed the limit on your car insurance policy, an umbrella policy would help cover the additional costs.
It's not primarily about covering expensive personal property, additional buildings, or properties while traveling. However, remember it is a supplemental form of insurance; therefore, it spans a variety of situations, which may potentially include some cases of property damage or bodily injury claims.
Hence The correct option is D.
Learn more about umbrella policy here:
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Explanation:
To find the takeoff speed of the long jumper, we can utilize the physics principles of projectile motion. Given that the long jumper leaves the ground at a 30-degree angle and travels a distance of 8.50 m, we need to find the initial velocity (takeoff speed) of the jumper.
In projectile motion, we can break down the motion into horizontal and vertical components. The horizontal component remains constant, while the vertical component is affected by gravity.
To solve for the takeoff speed, we can focus on the vertical component of motion. The equation that relates the vertical displacement, initial velocity, launch angle, and acceleration due to gravity is as follows:
Δy = v₀y t + (1/2) g * t²,
where:
- Δy is the vertical displacement (8.50 m),
- v₀y is the vertical component of initial velocity (takeoff speed),
- t is the total time of flight, and
- g is the acceleration due to gravity (approximately 9.8 m/s²).
Since the vertical displacement at the peak of the jump is zero (the jumper is at the highest point), we can rewrite the equation as:
0 = v₀y * t + (1/2) g t².
However, we can derive a relation between the time of flight t and the initial velocity v₀y by using the launch angle θ. The time of flight is given by:
t = (2 v₀y sin(θ)) / g.
Substituting this expression for t in the above equation, we have:
0 = v₀y [(2 v₀y sin(θ)) / g] + (1/2) g [(2 v₀y sin(θ)) / g]².
Now, we can solve for v₀y:
0 = v₀y² (2 sin(θ) + sin²(θ)) / g.
Rearranging and isolating v₀y, we get:
v₀y = √[(g Δy) / (2 * sin(θ) + sin²(θ))].
With the given values:
Δy = 8.50 m,
θ = 30 degrees,
g ≈ 9.8 m/s²,
we can substitute these values into the formula:
v₀y = √[(9
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7. e
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B.proton
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D. compound
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B. lux.
C. lumen.
D. candela.