HELP!!! FAST!!

Which of the following describe circumstances under which an exculpatory clause is generally unenforceable? Select all that appl

Sveta_85 [38]

Answer:

a. When the parties have unequal bargaining power

c. When it covers public transportation

Explanation:

Exculpatory clauses are often found in agreements between a company and a consumer when the activity is in danger, such as at a fitness center or ski resort. The company wants the consumer to understand the risk involved and to avoid lawsuits, so it includes a disclaimer in its contract. These clauses are used to limit liability, so they are not enforced when the parties have unequal bargaining power and when covering public transport.

6 0

3 years ago

in a tall building, a certain window is 500 meters above the ground.A tennis ball is catapulted horizontally out of the window w

expeople1 [14]

The tennis ball lands at a point 40.4 m from the base of the building.

The tennis ball is projected with a horizontal velocity u from a window, which is at a height y from the ground. The ball lands at a distance x from the base of the building. Let the ball take a time t to reach the ground. In the time t ,the ball falls a vertical distance y and also travel a horizontal distance x.

The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction. The ball falls down under the action of gravitational force.

Thus, use the equation of motion,

$y=\frac{1}{2} gt^2$

rewrite the expression for t and calculate the value of t using 9.81 m/s²for g and 500 m for y.

$t=\sqrt{\frac{2y}{g} } \\ =\sqrt\frac{(2)(500m)}{9.81m/s^2} \\ =10.096 s$

The horizontal distance x is traveled using the constant velocity u since no force acts on the ball in the horizontal direction.

Therefore,

$x=ut$

Substitute 4 m/s for u and 10.096 s for t

$x=ut\\ =(4m/s)(10.096s)\\ =40.384m=40.4 m$

Thus, the ball lands at a point 40.4 m from the base of the building.

6 0

3 years ago

The index of refraction of Sophia's cornea is 1.387 and that of the aqueous fluid behind the cornea is 1.36. Light is incident f

shtirl [24]

Answer:

17.85°

Explanation:

To find the angle to the normal in which the light travels in the aqueous fluid you use the Snell's law:

$n_1sin\theta_1=n_2sin\theta_2$

n1: index of refraction of Sophia's cornea = 1.387

n2: index of refraction of aqueous fluid = 1.36

θ1: angle to normal in the first medium = 17.5°

θ2: angle to normal in the second medium

You solve the equation (1) for θ2, next, you replace the values of the rest of the variables:

$\theta_2=sin^{-1}(\frac{n_1sin\theta_1}{n_2})\\\\\theta_2=sin^{-1}(\frac{(1.387)(sin17.5\°)}{1.36})=17.85\°$

hence, the angle to normal in the aqueous medium is 17.85°

7 0

3 years ago

What explains the negative charge on the oxygen atom within the water molecule?

SashulF [63]

To form an ion and bond with hydrogen, the oxygen atom gains electrons. Electrons are negative, therefore the oxygen within the water molecule will be negative.

5 0

3 years ago

Read 2 more answers

The cross-section of a pipe has an area of 2.0 cm2. Calculate the flow rate (in grams per second) of a liquid with p=1.0 g/cm3 t

Maurinko [17]

flow rate = velocity of fluid x area of cross section x density
= 42 x 2 x 1 = 84 g/s

4 0

3 years ago