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irakobra [83]
2 years ago
11

In a PE class, the teacher requires his students to monitor periodically their progress towards the fitness goals. Which of thes

e should be done by the students before and after an activity? *
a. weight monitoring
b. monitor pulse rate
c. warm up and cool down activities
d. physical check-up
Physics
1 answer:
pav-90 [236]2 years ago
8 0
The answer would be c warm up and cool down because before you do an activity stretching is very important
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A 100 kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is 10.0 m/s2. A 20 kg ma
DIA [1.3K]

Answer

given,

mass = 100 kg

acceleration = 10 m/s²

A mass 20 kg slides over 100 kg block

acceleration = 3 m/s²

horizontal friction exerted by the 100 kg block on 20 kg

using newton's second law

F - f = 0

F = f

f = ma

f = 20 × 3

f = 60 N

now net force acting on the 100 kg block

F_net = m a

F_net = 100 x 10

F_net = 1000 N

after 20 kg block falls the acceleration of the bock

F = 1000 +60

F = 1060 N

acceleartion on the block

a = \dfrac{F}{m}

a = \dfrac{1060}{100}

a = 10.60 m/s²

3 0
3 years ago
1. Study the map. What do you think divergent boundary means ? What about convergent boundary ?
olasank [31]

Answer:

Convergence generally means coming together, while divergence generally means moving apart. In the world of finance and trading, convergence and divergence are terms used to describe the directional relationship of two trends, prices, or indicators.

Hope this helps! :)

5 0
3 years ago
A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra
lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

m_{1}=5.8 kg is the mass of the bowling ball

V_{1}=4.34 m/s is the velocity of the bowling ball

m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg is the mass of the ping-pong ball

V_{2} is the velocity of the ping-pong ball

Now, the momentum p_{1} of the bowling ball is:

p_{1}=m_{1}V_{1} (1)

p_{1}=(5.8 kg)(4.34 m/s)  

p_{1}=25.172 kg m/s (2)

And the momentum p_{2} of the ping-pong ball is:

p_{2}=m_{2}V_{2} (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

p_{1}=p_{2} (4)

m_{1}V_{1}=m_{2}V_{2} (5)

Isolating V_{2}:

V_{2}=\frac{m_{1}V_{1}}{m_{2}} (6)

V_{2}=\frac{25.172 kg m/s}{0.002214 kg} (7)

Finally:

V_{2}=11369.46 m/s

6 0
2 years ago
If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is A. 600 m/s2 B. 60 m/s2 C. 3 m/s2 D. 6 m/s
enot [183]

Answer:

a

Explanation:

7 0
3 years ago
Read 2 more answers
A particle is moving with velocity v(t) = t2 _ 9t + 18 with distance, s measured in meters, left or right of zero, and t measure
Alex787 [66]
V = t^2 - 9t + 18

position, s
s = t^3 /3 - 4.5t^2 +18t + C

       t = 0, s = 1 => 1=C => s = t^3/3 -4.5t^2 + 18t + 1

Average velocity: distance / time

   distance: t = 8 => s = 8^3 / 3 - 4.5 (8)^2 + 18(8) + 1 = 27.67 m
   Average velocity = 27.67 / 8 = 3.46 m/s

t = 5 s

     v = t^2 - 9t + 18 = 5^2 - 9(5) + 18 = -2 m/s
     speed = |-2| m/s = 2 m/s
 
Moving right
     V > 0 => t^2 - 9t + 18 > 0
     (t - 6)(t - 3) > 0

     => t > 6 and t > 3 => t > 6 s => Interval (6,8)

    => t < 6 and t <3 => t <3 s => interval (0,3)

    

Going faster and slowing dowm

acceleration, a = v' = 2t - 9
     a > 0 => 2t - 9 > 0 => 2t > 9 => t > 4.5 s
     Then, going faster in the interval (4.5 , 8) and slowing down in (0, 4.5)
     


4 0
2 years ago
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