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Yanka [14]
3 years ago
12

To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp

ecified tile surface. A typical athletic shoe has a coefficient of 0.830. In an emergency, what is the minimum time interval in which a person starting from rest can move 3.20 m on a tile surface if she is wearing the following footwear?
Physics
1 answer:
motikmotik3 years ago
8 0

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

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With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by
lara31 [8.8K]

According to the net force, the acceleration of the book is 16.47 m/s².

We need to know about force to solve this problem. According to second Newton's Law, the force applied to an object will be proportional to mass and acceleration. Hence, it can be written as

∑F = m . a

where F is force, m is mass and a is acceleration

From the question above, we know that

m = 3 kg

g = 9.8 m/s²

F1 = 20 N

Find the net force

∑F = F1 + W

∑F = 20 + m . g

∑F = 20 + 3 . 9.8

∑F = 20 + 29.4

∑F = 49.4 N

Find the acceleration

∑F = m . a

49.4 = 3 . a

a = 16.47 m/s²

Find more on force at: brainly.com/question/25239010

#SPJ4

7 0
8 months ago
A man and his dog are “walking” on flat Street. He is pulling on his stubborn dog with a force of 70 N Directed at a 30° angle f
Delvig [45]

Answer:

X component of force is 60.62 N.

Y component of force is 35 N.

Force of gravity on the dog is 245 N.

Magnitude of normal force is 210 N.

Explanation:

Given:

Force of pull is, F=70\ N

Mass of the dog is, m=25\ kg

Angle of inclination is, \theta =30°

Acceleration due to gravity is, g=9.8\ m/s^2

The free body diagram of the dog is shown below.

The X and Y components of force of pull is given as:

F_X=F\cos \theta=(70)\cos (30)=60.62\ N\\F_Y=F\sin \theta=(70)\sin(30)=35\ N

Therefore, the X and Y components of the force are 60.62 N and 35 N respectively.

Force of gravity on the dog is the product of its mass and acceleration due to gravity. Thus,

F_g=mg=25\times 9.8=245\ N

Therefore, the force of gravity on the dog is 245 N.

Now, consider the motion in the vertical direction of the dog. As there is no motion in the vertical direction, the net force along the Y direction is 0. In other words, the total Upward force is equal to the total downward force.

From the free body diagram,

N+F_Y=mg\\N=mg-F_Y\\N=245-35=210\ N

Therefore, the normal force acting on the dog is 210 N.

6 0
3 years ago
A ball is projected vertically upward from the surface of the Earth with an initial speed of +49 meters per second. The ball rea
fenix001 [56]
Answer: 245 meters

Explanation: 49 ? Meters
over over over
1 5 seconds

Cross multiply- 49 times 5 divided by 1 which equals 245
8 0
2 years ago
Which waves are longitudinal waves? Check all that apply.
alekssr [168]
Sound waves.
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Please help homework due tomorrow,,,
Ad libitum [116K]
Cody ...

Everything on this page is solved with the SAME formula !

             Distance = (speed) x (time) .


Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.

'Velocity' is speed AND THE DIRECTION OF THE  MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.

Your teacher may not be happy if you talk about this on your homework,
but that's too bad.  Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.
 

1).  Speed = (distance covered) / (time to cover the distance)

2).  Speed = (distance covered) / (time to cover the distance)

3).  Distance  =  (average speed of travel) x (time traveling at that speed)

4).  Time to cover the distance = (distance) / (speed)

5).  Car's     speed = (distance the car covered)        / (time the car took)
      Sprinter speed = (distance the sprinter covered) / (time the sprinter took)

      Calculate the car's speed.
      Calculate the sprinter's speed.
     
      ... Look at the two speeds.
          Decide which one is faster.
     
      ... Subtract the slower one from the faster one. 
          The difference is the answer to "by how much?" .

6).  Distance  =  (speed) x (time spent moving at that speed)

7).  Average speed  =  (TOTAL distance covered)
                                      divided by
                                    (time to cover the TOTAL distance).
   

8 0
3 years ago
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