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arsen [322]
4 years ago
12

Two soccer players kick a soccer ball back and forth along a straight line. The first player kicks the ball 16 m to the right to

the second player. The second player kicks the ball to the left weakly; it only moves 3.1 m before stopping. (Consider the right to be the positive direction. Where applicable, indicate the direction with the sign of your answer. All distances are in meters.)
Physics
1 answer:
Oliga [24]4 years ago
5 0

Answer:

a

The total distance is d_t =19.1 m

b

The displacement is

  D_t = 12.9m

Explanation:

From the question we are told that

        Distance traveled by the ball for first player d_f = 16m to the right

        Distance  traveled by the ball for second player d_s = 3.1 m to the left

       

The total distance traveled by the ball is mathematically represented as

                   d_t = d_f + d_s

Substituting values

                  d_t = 16 + 3.1

                  d_t =19.1 m

The displacement is mathematically represented as

              D_t  = d_f -d_s

This is because displacement deal with direction and from the question we are told that right is positive and left is negative

          Substituting values  

                D_t = 16 -3.1

                 D_t = 12.9m

     

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Katen [24]

Answer:

<h2>537.6 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 84 × 6.4

We have the final answer as

<h3>537.6 N</h3>

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3 years ago
A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a
3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec            

Explanation:

We have given acceleration of the car a_1=5.9m/sec^2

Acceleration of the stock car a_2=3.6m/sec^2

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So s_1=s_2

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So \frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6

After solving t = 3.56 sec

(b) From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m

(c) From first equation pf motion v = u+at

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0
3 years ago
A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.
galina1969 [7]

Answer:

44.64 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 4.2\times 1180+80.6^2}\\\Rightarrow v=128.01\ m/s

v=u+at\\\Rightarrow 128.01=80.6+4.2t\\\Rightarrow t=\frac{128.01-80.6}{4.2}=11.29\ s

<u>Time taken to reach 1180 m is 11.29 seconds</u>

v=u+at\\\Rightarrow 0=128.01-9.8t\\\Rightarrow t=\frac{128.01}{9.8}=13.06\ s

<u>Time the rocket will keep going up after the engines shut off is 13.06 seconds.</u>

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128.01^2}{2\times -9.8}\\\Rightarrow s=836.05\ m

The distance the rocket will keep going up after the engines shut off is 836.05 m

Total distance traveled by the rocket in the upward direction is 1180+836.05 = 2016.05 m

The rocket will fall from this height

s=ut+\frac{1}{2}at^2\\\Rightarrow 2016.05=0t+\frac{1}{2}\times 9.8\times t^2\\\Rightarrow t=\sqrt{\frac{2016.05\times 2}{9.8}}\\\Rightarrow t=20.29\ s

<u>Time taken by the rocket to fall from maximum height is 20.29 seconds</u>

Time the rocket will stay in the air is 11.29+13.06+20.29 = 44.64 seconds

5 0
3 years ago
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