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1 year ago

A car is traveling at a speed of 37 m/s.

Physics

1 answer:

Vika [28.1K]1 year ago

3 0

1. The speed in kilometers per hour (Km/h) is 133.2 Km/h

2. Yes, the speed is exceeding the 125 Km/h limit

<h3>How to convert 37 m/s to Km/h</h3>

From the question given above, the following data were obtained:

Speed (in m/s) = 37 m/s
Speed (in Km/h) =?

We can convert 37 m/s to kilometers per hour (Km/h) by doing the following:

1 m/s = 3.6 Km/h

Therefore,

37 m/s = 37 × 3.6

37 m/s = 133.2 Km/h

Thus, 37 m/s is equivalent to 133.2 Km/h

<h3>2. How to determine if the speed is exceeding the limit</h3>

Speed of car = 133.2 Km/h
Speed limit = 125 Km/h

From the above, we can see that the speed of the car is greater than the speed limit.

Thus, we can conclude that the speed of the car is exceeding the speed limit.

Learn more about conversion:

brainly.com/question/10893215

#SPJ1

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Find the resultant force of the following forces :

bezimeni [28]

The resultant of the given forces is; 6√2 N

<h3>How to find the resultant of forces</h3>

We are given the forces as;

10 N along the x-axis which is +10 N in the x-direction

6 N along the y-axis which is +6N in the y-direction

4 N along the negative x-axis which is -4N

Thus;

Resultant force in the x-direction is; 10 - 4 = 6N

Resultant force in the y-direction is; 6N

Thus;

Total resultant force = √(6² + 6²)

Total resultant force = 6√2 N

Read more about finding resultant of a force at; brainly.com/question/14626208

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When you look at yourself in the mirror, what is the approximate angle of incidence of the light rays?

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2 years ago

Two ice skaters, each with a mass of 72.0 kg, are skating at 5.45 m/s when they collide and stick together. If the angle between

Rudiy27

Answer:

The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

Explanation:

To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.

By definition it is known that the conservation of the moment is given by:

$m_1v_1+m_2v_2=(m_1+m_2)v_f$

Our values are given by,

$m_1=m_2=72Kg$

As the skater 1 run in x direction, there is not component in Y direction. Then,

Skate 1:

$v_{x1}=5.45m/s$

$v_{y1}=0$

Skate 2:

$v_{x2} = 5.45*cos105= -1.41m/s$

$v_{y2} = 5.45*sin105 = 5.26m/s$

Then, if we applying the formula in X direction:

m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}

75*5.45-75*1.41=(75+75)v_{fx}

Re-arrange and solving for v_{fx}

v_{fx}=\frac{4.04}{2}

v_{fx}=2.02m/s

Now applying the formula in Y direction:

$m_1v_{y1}+m_2v_{y2}=(m_1+m_2)v_{fy}$

$0+75*5.25=(75+75)v_{fy}$

$v_{fy}=\frac{5.25}{2}$

$v_{fy}=2.63m/s$

Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.

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