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kotegsom [21]
3 years ago
13

Why would an insurance company want people to avoid risk, like Nick Cannon having to avoid skydiving?

Physics
2 answers:
Lina20 [59]3 years ago
5 0
An insurance company wants people to avoid risks because if the person meets with an accsident the expense of the medicines and treatment should be covered by the insurance company.
notsponge [240]3 years ago
5 0

Answer:

Researching a little bit, Nick Cannon is a rapper, comedian and record producer, none of his career choices include skydiving as a regular job-hazard activity.

In the eyes of the insurance company, this activity is categorized as a Risk, not only a life threatening risk, but a monetary risk.

When the company covered Nick, they probably didn't care whether or not he would perform a set, or sing a new song, etc., but in skydiving, Nick would be putting himself, and the insurance company, at risk.

If he happened to get injured, the company would be the one covering the hospital's bills, and potentially losing income, that when compared to Nick's income, would not be worth it, he would become an expensive asset.

Yet this is exactly the reason why Nick has an insurance, so he could transfer risk, or the chance of loss, to another party. The Company is thinking on managing and balancing responsibility and risk, Nick in the transfer of risk. The company would like to avoid it to prevent having to lose income.

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A car with an initial velocity of O m/s and a mass of 1500 kg reaches a velocity of 15 m/s in 5
Flura [38]

Answer:the answer can be find out by the following steps;

Explanation:

3 0
3 years ago
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All simple machines are types of
taurus [48]

Answer: Metal. There are six of them.  inclined plane, the wedge, the screw, the lever, the wheel and axle, and the pulley.

Hope this helps!

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Explanation:

6 0
3 years ago
What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Yo
insens350 [35]

Full Question

Consider two point charges located on the x axis: one charge, Q1 = -12.0 nC , is located at x1 = -1.705m ; the second charge, Q2 = 36.5 nC, is at the origin (x=0.0000).

What is the net force exerted by these two charges on a third charge q3 = 49.5 nC placed between q1 and q2 at x3 = -1.170 m ? Your answer may be positive or negative, depending on the direction of the force.

Answer:

6.79E6 N

Explanation:

Given

Q1 is negative and to the left of Q3 the force will be to the left

Q2 is positive and to the right of Q3 the the force will also be to the left

Net Force is calculated as:

Using Coulomb's law

Coulomb's law: F = kqQ / r²

the constant k = 8.99 x 10^9 N m2 / C2

F = -kQ1*Q3/(r1)² -kQ2*Q3/(r2)²)

F = -kQ3(Q1/(r1)² + Q2/(r2)²)

Where

Q1 = -12nC = -12 * 10^-9C

Q2 = 36.5nC = 36.5 * 10^-9C

Q3 = 49.5nC = 49.5 * 10^-9C

x1 = -1.705m

x2 = x = 0

x3 = -1.170m

r1 = x3 - x1

r1 = -1.170 - -1.705

r1 = -1.170 + 1.705

r1 = 0.535

r1² = 0.286225

r2 = x3

r2 = -1.170

r2² = -1.170²

r2² = 1.3689

So,

F = (-8.99 * 10^9)(49.5 *10^-9) [-12 * 10^-9/0.286225 + 36.5 * 10^-9/1.3689]

F = -445.005 (−4.192505895711E−8 + 2.6663744612462E−8)

F = -445.005 * −1.5261314344648E−8

F = -(8.99 * 10^9) * (49.5 * 10^-9) * [ (-12 * 10^-9) /(-1.770 - -1.705)² + (36.5 * 10^-9)/(-1.170)²]

F = -445.005( −0.000002813572941777538)

F = 0.00000679136118994008324

F = 6.79E6 N

3 0
3 years ago
If an object is thrown in an upward direction from the top of a building 160 ft. High at an initial speed of 21.82 mi/h what is
viktelen [127]
To solve this problem we are going to use tow kinematic equations for falling objects.
1. Kinematic equation for final velocity: V_{f}=V_{i}+gt
where
V_{f} is the final velocity 
V_{i} is the initial velocity 
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 
2. Kinematic equation for distance: d=V_{i}t+ \frac{1}{2} gt^2
where
d is the distance 
V_{i} is the initial velocity 
V_{f} is the final velocity
g is the acceleration due to gravity 32 \frac{ft}{s^2}
t is the time 

First, we are going to convert 21.82 mi/h to ft/s:
21.82 \frac{mi}{h} =31.21 \frac{ft}{s}

Next, we are going to use the first equation to find how long it takes for the rock to reach its maximum height.
We know for our problem that the object is thrown in upward direction, so its velocity at its maximum height (before falling again) will be zero; therefore: V_{f}=0. We also know that it initial speed is 31.21 ft/s, so V_{i}=31.21. Lets replace those values in our formula to find t:
V_{f}=V_{i}+gt
0=31.21+(-32)t
-32t=-31.21
t= \frac{-31.21}{-32}
t=0.98seconds

Next, we are going to use that time in our second kinematic equation to find the distance the object reach at its maximum height:
d=V_{i}t+ \frac{1}{2} gt^2
d=31.21(0.98)+ \frac{1}{2} (-32)(0.98)^2
d=15.22ft 

Now we can add the height of the building and the maximum height of the object:
d=160+15.22=175.22ft

Next, we are going to use that height (distance) in our second kinematic equation one more time to fin how long it takes for the object to fall from its maximum height to the ground:
d=V_{i}t+ \frac{1}{2} gt^2
175.22=31.21t+ \frac{1}{2} (32)t^2
16t^2+31.21t-175.22=0
t=2.47 or t=-4.43
Since time cannot be negative, t=2.47 is the time it takes the object to fall to the ground. 

Finally, we can use that time in our first kinematic equation to find the final speed of the object when it hits the ground:
V_{f}=V_{i}+gt
V_{f}=31.21+(32)(2.47)
V_{f}=110.25 ft/s

We can conclude that the speed of the object when it hits the ground is 110.25 ft/s


5 0
3 years ago
If a crow flies west for 60 km and then south for 45 km, what is the direction of its displacement?
son4ous [18]
That's 105 km that he flew, or 65.2 miles !  I'm absolutely positive
that the crow must have landed and gotten some rest when you
weren't looking.  But that had no effect on his displacement when
he got where he was going, so we can continue to solve the problem:


The displacement is the distance and direction from the place
where the crow took off to the place where he landed.

-- It's distance is the hypotenuse of the right triangle whose legs
are 60 km and 45 km.

        D²  =  (60 km)²  +  (45 km)²

              =    3,600 km²  +  2,025 km²  =  5,625 km²

         D  =  √(5625 km²)  =  75 km .    
 
-- It's direction is the angle whose tangent is  (45 S / 60 W).

         tan⁻¹ (45/60)  =  tan⁻¹ (0.75)  =  36.9° south of west

                                                         =  53.1° west of south.

                                                         =  not exactly southwest but close.
7 0
3 years ago
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