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Pavel [41]
3 years ago
6

In human relations, imagined risks are?

Physics
1 answer:
inessss [21]3 years ago
6 0
<span>In human relation, imagined risks are defined as an exageration of a risk which can cause a person to develop a behaviour that negatively affect his or her self confidence. Imagined risks are imaginary and they are created out of fear and doubt. For instance, worrying about the possibility of a nuclear attack is an example of an imagined risk.</span>
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2. In a tug of war contest, one person pulls with a force of 5 Newtons. Another person on the same team pulls with a force of 3
yuradex [85]

Answer:

8 newtons

Explanation:

Balanced forces have an equal amount of newtons. 5+3=8, so the other team will need 8 newtons.

5 0
3 years ago
Read 2 more answers
Give an example of how the relationship between force and acceleration might affect YOU.
IrinaK [193]
Take a car collision as an example, the more you speed up as means of acceleration, the more force will be on impact.
5 0
3 years ago
12. A cross-country race car driver sets out on a 1.00 hour, 100.0 km race. At the halfway marker
Akimi4 [234]

Answer:

A.) Time = 0.625 hrs = 37.5 minutes

B.) Speed required = 133.33km/hr

Explanation:

Given the following :

Total race distance = 100km

Total time required = 1 hour = 60 minutes

Average speed after traveling 50km = 80km/hr

80km/hr : This means it will take one hour to cover a distance of 80km

Therefore, time taken to cove first 50km at that average speed equals :

1 hour = 80 km

t hours = 50km

80t = 50

t = 50/80

t = 5/8 hours

t = 0.625hours

t = 0.625 * 60 = 37.5minutes

B)

Average speed required to complete the race in 1 hour = 100km/hr

Time used to complete first 50km = 0.625 hour

Time remaining: (1 - 0.625) hour = 0.375

Speed required = Distance left / time left

Speed required = 50 / 0.375

Speed required = 133.33km/hr

8 0
3 years ago
g (15 points) Picture taken from problem 13-35. A 2000 kg truck is traveling at 25 meters/second. The driver suddenly hits the b
otez555 [7]

Answer:

Δx = 3.99 m

Explanation:

To determine distance, use kinetic energy

will make it short and easy.  

KE=1/2mv2 and KE=Δxmgμ

 

Set the equations equal to each other

1/2mv2=Δxmgμ            (Note: The masses cancel )

 1/2v2=Δxgμ                  Solve for Δx

where g=9.8  

Δx=v2/(2gμ)                   Δx = 25 / (2 * 9.8 * 0.32)     Δx = 3.99 m

Please let me know if its correct, if not report it so we can correct it.

6 0
4 years ago
A train traveling at 27.5 m/s accelerates to 42.4 m/s over 75.0 s. What is the displacement of the train in this time period
Sergio [31]

Answer:

2621.25 meters

Explanation:

First, write down what we are given.

Initial velocity = 27.5 m/s

Final velocity = 42.4 m/s

Time = 75 seconds

We need to look at the kinematic equations and determine which one will be best.  In this case, we need an equation with distance.  I am going to use v_{f}^{2} = v_{i}^{2} +2ad, but you can also use the other equation, x = v_{o}t+\frac{1}{2}at^{2}

We need to find acceleration.  To find it, we need to use the formula for acceleration: a = \frac{v_{f}-v_{i}}{t}.  Plugging in values, a = \frac{42.4-27.5}{75} = .199\ m/s^{2}

Next, plug in what we know into the kinematics equation and solve for distance.  42.4^{2} = 27.5^{2} + 2(.199)(d)\\d = 2621.25\ meters

7 0
3 years ago
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