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luda_lava [24]
3 years ago
7

2) Why can oversharing be a danger to yourself?

Physics
1 answer:
lidiya [134]3 years ago
6 0

Sharing is hard to get just right-- share to little, and you're considered stingy. Share too much, and it can just be a danger to yourself.

Oversharing can give the impression to people that you're reckless with your items, or that you'll just give it away. People could also take advantage of you-- if you're too generous, or too nice, they might use that to take lots of favors/money and never return it, or to use it against you. Oversharing, especially to the wrong people, could also gain you enemies, as people might think you're trying to manipulate them by being "too nice". We were raised thinking that everything has a price, and oversharing could make people think that you're trying to harm them<em>. </em>

If you overshare, you could also give away too much of your own things, or own work. If you're too nice and "overshare" your test answers, it could lead you to being expelled for cheating.

In conclusion, oversharing could be much more dangerous than being stingy, and you must always be careful with what you have, and how people might use you to their advantage.

Hope this helps!

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A train travels 59 kilometers in 3 hours, and then 74 kilometers in 5 hours. What is its average speed?
dybincka [34]
Average speed = (total distance) / (total time)

Total distance = 59 km + 74 km = 133 km

Total time = 3 hrs + 5 hrs  =  8 hrs

Average speed  =  (133 km) / (8 hrs)  =  16.625 km/hr
3 0
3 years ago
A 1500 kg car enters a section of curved road in the horizontal plane and slows down at a uniform rate from a speed of 100 km/h
Mandarinka [93]

Answer:

Incomplete question check attachment for diagram

Explanation:

Given that,

Mass of car

M = 1500kg

Enter curve at Point A with speed of

Va = 100km/hr = 100× 1000/3600

Va = 27.78m/s

The car was slow down at a constant rate till it gets to point C at  speed of

Vc = 50km/r = 50×1000/3600

Vc = 13.89m/s

Radius of curvature at point A

p = 400m

Radius of curvature at point B

p = 80m

The distance from point A to point B as given in the attachment is

S=200m

We want to find the total horizontal  forces at point A, B and C exerted by the road on the tire

The constant tangential acceleration can be calculated using equation of motion

Vc² = Va² + 2as

13.89² = 27.78² + 2 × a × 200

192.9 = 771.6 + 400a

400a = 192.9—771.6

400a = -578.7

a = -578.7 / 400

a = —1.45 m/s²

at = —1.45m/s²

The tangential acceleration is -1.45m/s² and it is negative because the car was decelerating

Since the car is slowing down at a constant rate, the tangential acceleration is equal at every point

At point A

at = -1.45m/s²

At point B

at = -1.45m/s²

At point C

at = -1.45m/s²

Now,

We can calculate the normal component of acceleration(centripetal acceleration) at each point since we know the radius of curvature

The centripetal acceleration is calculated using

ac = v²/ p

At point A ( p = 400)

an = Va²/p = 27.78² / 400

an = 1.93 m/s²

At point B (p = ∞), since point B is point of inflection

Then,

an = Vb²/p =  Vb/∞ = 0

an = 0

At point C ( p = 80m)

an = Vc²/p = 13.89² / 80 = 2.41m/s²

an = 2.41 m/s²

Then,

The tangential force is

Ft = M•at

Ft = 1500 × 1.45

Ft = 2175 N.

Since tangential acceleration is constant, then, this is the tangential force at each point A, B and C

Now, normal force

Point A ( an = 1.93m/s²)

Fn = M•an

Fn = 1500 × 1.93

Fn = 2895 N

At point B (an=0)

Fn = M•an

Fn = 0 N

At point C (an= 2.41m/s²)

Fn = M•an

Fn = 1500 × 2.41

Fn = 3615 N.

Then, the horizontal force acting at each point is

Using Vector of right angle triangle

F = √(Fn² + Ft²)

At point A

Fa = √(2895² + 2175²)

Fa = √13,111,650

Fa = 3621 N

At point B

Fb = √(0² + 2175²)

Fb = √2175²

Fb = 2175 N

At point C

Fc = √(3615² + 2175²)

Fc = √17,798,850

Fc = 4218.88 N

Fc ≈ 4219N

6 0
3 years ago
In which of these cases is NO work done on the football? A) You lift a football off the ground. B) You raise a football over you
inn [45]

I believe the answer is C.

8 0
3 years ago
What is the magnitude of the force that must be applied for the system below to be balanced?
Vesnalui [34]
I believe that it is 20 N
8 0
2 years ago
A change
Tamiku [17]

The work done if the forklift is operating at full capacity is 11,900 J.

We have to recall that power is defined as the rate of doing work. The rate of doing work is defined as;

Power = Work done/time taken

When;

Power = 4950 W

Time taken = 2.40 s

Work done = Power × time taken

Work done = 4950 W × 2.40 s

Work done = 11,900 J

Learn more about work done: brainly.com/question/3902440

5 0
3 years ago
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