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frez [133]
3 years ago
5

Covert 60 mph to SI mks units

Physics
1 answer:
Flauer [41]3 years ago
7 0

26.82m/s

Explanation:

Given:

   speed = 60mph

 problem: convert to m/s

To solve this problem, we have to find the right and appropriate conversion factor which equals to 1 to multiply this unit with:

   we are converting:

      miles to meters

      hours to seconds

    1.609km = 1mile

      1000m = 1km

  60s = 1 min

   60 min = 1hr

Now to convert from mph to m/s

  60 x    \frac{mi}{hr}  x  \frac{1.609km}{1mi}   x   \frac{1000m}{1km}  x \frac{1hr}{60min}  x \frac{1m}{60s}

  =  26.82m/s

   

learn more:

Conversion brainly.com/question/1548911

#learnwithBrainly

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When calculating the net electrostatic force, a negative value indicates that the two charged objects involved __________.
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Answer:

B

Explanation:

The correct answer is B) have unlike charges. Since they are attracted to each other they have to be unlike

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4 years ago
Name two examples where the cohesive force dominates over the adhesive force and vice versa​
lidiya [134]
Attractive forces between molecules of the same type are called cohesive forces. ... Attractive forces between molecules of different types are called adhesive forces. Such forces cause liquid drops to cling to window panes, for example.
5 0
3 years ago
A 12 kg block is accelerating at the rate of 9.0 m/s? while being acted on by two forces.
Aliun [14]

Answer:

See below

Explanation:

F = ma

F = 12 * 9 = 108 N

  108 N  needed      <u> add  30 N more east </u>

8 0
1 year ago
What is the acceleration of a softball if it has a mass of 0.5 kg and hits the cathers glove with a force of 25n
ioda
Acceleration is found if we have the force and mass. 

With the following equation: F = ma, we can find the missing values. 

F = 25n
M = 0.5 kg
a = ?

a = f/m
a = 25/0.5
a = 50

a = 50 m/s

So, the acceleration is 50 m/s^2 
3 0
3 years ago
A block of ice with mass 2.00 kg slides 0.750 m down an inclined plane that slopes downward at an angle of 36.9 degrees below th
zhannawk [14.2K]

Answer: V_{f}=2.96m/s    

Firstly we have to draw the Free Body Diagram (FBD) as shown in the figure attached.

Where the weight w of the block has an x-component and y-component:

w_{x}=wsin(\theta)    (1)

w_{y}=wcos(\theta)    (2)

As well as the Normal Force N:

N_{x}=Nsin(\theta)    (3)

N_{y}=Ncos(\theta)    (4)

In addition, we know N=w, then \sum F_{y}=0

In the X-component:

\sum F_{x}=m.a

m.a=w_{x}    (5)

Substituting (1) in (5):

wsin(\theta)=m.a    (6)

In addition, we know w=m.g, where m is the mass of the block and g the gravity acceleration, which is equal to 9.8m/{s}^{2}  

So:

m.g.sin(\theta)=m.a   (7)

a=g.sin(\theta)    (8)

a=5.88m/{s}^{2}    (9)   >>>>This is the acceleration of the block

On the other hand, we have the following equation that expresses a <u>relation between</u> the distance d with the acceleration a and time t:

d=\frac{1}{2}a{t}^{2}   (10)

We already know the value of  d and calculated a, we have to find t:

t=\sqrt{\frac{2d}{a}}   (11)

t=\sqrt{\frac{2(0.75m)}{5.88m/{s}^{2}}}   (12)

t=0.50s   (13) >>>This is the time it takes to the block to go from the initial velocity V_{o} to its final velocity V_{f}

If the acceleration is the variation of the velocity in time, we can use the following equation to find V_{f}:

V_{f}-V_{o}=a.t   (13)

If V_{o}=0

V_{f}=a.t   (14)

V_{f}=(5.88m/{s}^{2})(0.50s)   (15)

Finally we get the value of the Final Velocity of the block:

V_{f}=2.96m/s    

6 0
3 years ago
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