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3 years ago

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A baseball with a mass of 142 grams is thrown across a field. It accelerates at a rate of 8 m/s^2. What is the force acting on t

he ball?

1 answer:

Bogdan [553]3 years ago

5 0

Answer:

F = 1.14 N

Explanation:

Let F be the force acting on the baseball and let m = 142 g = 142×10^-3 kg and a = 8 m/s^2.

Then according to Newton's second Law of motion, the force is given by:

F = m×a

= (142×10^-3)×(8)

= 1.136 N

Therefore, the force acting on the baseball is 1.14 N.

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Pedro e Maria saíram para passear de carro. Eles partiram de São Paulo às 10 h em direção à Braúna, localizada a 500 km da capit

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Answer:

This can be translated to:

"P edro and Maria went for a drive. They left São Paulo at 10 am towards Braúna, located 500 km from the capital. As the journey was long, they made two 15-minute stops for gas and also spent 45 minutes for lunch. When arriving at the final destination, Maria looked at the clock and saw that it was 6 pm. What is the average speed of the trip?"

Ok, the first thing we know is, that for the average speed we can write:

Speed = Distance/time.

First, we know that Distance = 500km.

And for the time we have two possiblities:

The total average speed will decrease because they were stopped a total of:

15min + 15min + 45min = 75min = 1.25 hours

Then if the travel starts at 10am, and ends at 6 pm, the total time that has passed is:

6pm = 18hs

6pm - 10am = 18 - 10 = 8hs

Then the average speed will be:

Speed = 500km/8h = 62.5 km/h.

Now, if we considerate the average speed only when they are moving, the total time that they are moving is:

Total time in travel - time that they where stopped

8h - 1.25h = 6.75h

Then the average speed would be:

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3 years ago

The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to

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(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration $a_1$ for a total time $t_1$ During this part of the motion, he covers a distance equal to $s_1 = 45 m$ , until he finally reaches a velocity of $v_1 = u + a_1t_1$ . We can use the following suvat equation:

$s_1 = u t_1 + \frac{1}{2}a_1t_1^2$

which reduces to

$s_1 = \frac{1}{2}a_1 t_1^2$ (1)

since u = 0.

- In the second part, he continues with constant speed $v_1 = a_1 t_1$ , covering a distance of $d_2 = 55 m$ in a time $t_2$ . This part of the motion is a uniform motion, so we can use the equation

$s_2 = v_1 t_2 = a_1 t_1 t_2$ (2)

We also know that the total time is 10.0 s, so

$t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)$

Therefore substituting into the 2nd equation

$s_2 = a_1 t_1 (10-t_1)$

From eq.(1) we find

$a_1 = \frac{2s_1}{t_1^2}$ (3)

And substituting into (2)

$s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1$

Solving for t,

$s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s$

So from (3) we find the acceleration in the first phase:

$a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2$

And so the average force exerted on the sprinter is

$F=ma=(66 kg)(2.34 m/s^2)=154.5 N$

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

$v_1 = u +a_1 t_1$

where we have

u = 0

$a_1 =2.34 m/s^2$ is the acceleration

$t_1 = 6.2 s$ is the time of the first part

Solving the equation,

$v_1 = 0 +(2.34)(6.2)=14.5 m/s$

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