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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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2. A force of 156 N acts on a 7.3-kg bowling ball for 0.40 s.
atroni [7]

Answer:

62.4N\cdot s

Explanation:

The change in momentum of the ball is equal to the impulse exerted on it, therefore:

\Delta p = I = F \Delta t

where

\Delta p is the change in momentum

F is the average force exerted on the ball

\Delta t is the time during which the force is applied

In this problem,

F = 156 N

\Delta t = 0.40 s

So, the change in momentum of the ball is

\Delta p =(156)(0.40)=62.4N\cdot s

6 0
4 years ago
In reaching her destination, a backpacker walks with an average velocity of 1.32 m/s, due west. This average velocity results, b
Svetach [21]

Answer:

distance in east is 1273.78 m

Explanation:

given data

average velocity = 1.32 m/s west =

hike = 5.21 km = 5.21 × 10³ m

average velocity = 3.49 m/s west

average velocity = 0.687 m/s east

to find out

distance in east

solution

we consider here distance in east  is = x

so distance from starting point = 5.21 × 10³ - x        ...................1

and we can say time required to reach end

time required = distance / speed

time required = \frac{5.21 *10^3 - x}{1.32}    ................2

and

time required for 6.44 km west

time required = \frac{5.21 *10^3 - x}{3.49}    ................3

and time required for distance x

time required = \frac{x}{0.687}    ................4

so from equation 2 , 3 and 4

\frac{5.21 *10^3 - x}{1.32} = \frac{5.21 *10^3 - x}{3.49}  + \frac{x}{0.687}

x = 1273.78 m

so distance in east is 1273.78 m

8 0
3 years ago
You push a box across the floor with a force of 20 N. You push 10 meters in 5 seconds. How much work did you do? How much power
Veseljchak [2.6K]
Work = 200 i hope this helped
4 0
4 years ago
Read 2 more answers
A 1.7-kg firework is fired from the ground straight up on a planet with whose acceleration due to gravity is 4.8 m/s/s. You want
Sunny_sXe [5.5K]

Answer:

41.0 m/s, 10.6 s

Explanation:

Given:

a = -4.8 m/s²

Δy = 165 m

v = -10.0 m/s

Find: v₀ and t

v² = v₀² + 2aΔy

(-10.0 m/s)² = v₀² + 2(-4.8 m/s²) (165 m)

v₀ = 41.0 m/s

Δy = vt − ½ at²

165 m = (-10.0 m/s) t − ½ (-4.8 m/s²) t²

165 = -10t + 2.4t²

0 = 2.4t² − 10t − 165

Solve with quadratic formula:

t = [ -(-10) ± √((-10)² − 4(2.4)(-165)) ] / 2(2.4)

t = (10 ± √1684) / 4.8

t = 10.6 s

4 0
3 years ago
What is the potential energy of a 2 kg rock sitting at the top of a 10 meter high cliff?
prisoha [69]

Answer:200J

Explanation:E=MGH

= 2*10(assuming you round earth’s gravity to 10)*10

=200

8 0
3 years ago
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