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2 years ago

How does Newtons second law of motion relate to Track and field (running sport)?

1 answer:

3 0

Newton's second law also helps to explain what happens every time an athlete lands during running. When the foot hits the track, it will decelerate to a stop before leaving the track again. The faster the deceleration, the greater the force of impact on the foot.

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A bus that was moving at a high speed stopped suddenly. One of

fiasKO [112]

Answer:

this doesn't make sinces

7 0

2 years ago

Your friend from France came to visit you when she was packing she went on weather.com and found that the average temperature in

Marina CMI [18]

Answer:

Because there is not as much cold as it was in France.

Explanation:

The average temperature in France during January ranges from 2.7° to 7.2° celsius which makes it the coldest month of the year. But since she comes to know that average temperature in Annville ranges 31° celsius which implies that the temperature is normal there and therefore, she packs sleeveless tops and shorts. Coats would not be required in a hot weather and hence, she does not pack it.

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3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and

garri49 [273]

Answer:

(a) t = 1.14 s

(b) h = 0.82 m

Explanation:

(b)

Considering the upward motion, we apply the third equation of motion:

$2gh = v_f^2 - v_i^2$

where,

g = - 9.8 m/s² (-ve sign for upward motion)

h = max height reached = ?

vf = final speed = 0 m/s

vi = initial speed = 4 m/s

Therefore,

$(2)(9.8\ m/s^2)h = (0\ m/s)^2-(4\ m/s)^2\\$

h = 0.82 m

Now, for the time in air during upward motion we use first equation of motion:

$v_f = v_i + gt_1\\0\ m/s = 4\ m/s + (-9.8\ m/s^2)t_1\\t_1 = 0.41\ s$

(c)

Now we will consider the downward motion and use the third equation of motion:

$2gh = v_f^2-v_i^2$

where,

h = total height = 0.82 m + 1.8 m = 2.62 m

vi = initial speed = 0 m/s

g = 9.8 m/s²

vf = final speed = ?

Therefore,

$2(9.8\ m/s^2)(2.62\ m) = v_f^2 - (0\ m/s)^2\\$

vf = 7.17 m/s

Now, for the time in air during downward motion we use the first equation of motion:

$v_f = v_i + gt_1\\7.17\ m/s = 0\ m/s + (9.8\ m/s^2)t_2\\t_2 = 0.73\ s$

(a)

Total Time of Flight = t = t₁ + t₂

t = 0.41 s + 0.73 s

t = 1.14 s

7 0

2 years ago

-Plz Help Meh- Calculate the acceleration of a car if the force is 450N and the mass is 1300kg.

Allisa [31]

Answer:

Explanation:

F=Ma

450=1300×a

Ans=0.346m/s²

4 0

2 years ago

A baseball 0.145kg is thrown vertically upwards with an initial velocity of 20m/s. Use the law of conservation of energy to find

kvasek [131]

Answer: 20.4m

Explanation:

Mass = 0.145kg

Initial velocity, Vi =20m/s

Initial kinetic energy K =1/2mv^2

Initial potential energy Ui = mgx = 0joules

: From conservation of energy,

Uf + Kf = Ui + Ki ( where f represent (final) )

Thus

mgXf + 0 = 0+1/2 mv^2

Xf = Vi^2/ 2g

= (20m/s) ^2/ 2(9.81m/s)^2

=20.4m

5 0

3 years ago